ID: 868475 RSA Lucky #13 July 4, 2018, 11:03 p.m. No.6694   πŸ—„οΈ.is πŸ”—kun

The virtual quantum computer (VQC) is a grid made of infinite yet constructable sets that follow a known pattern. Like a quantum spreadsheet.

The grid is the superposition. The collapse of that superposition will be two input parameters, d and e which can be calculated easily for all integers, c, where c is the difference of two squares. Its purpose and our goal is to learn and show the TRUTH, one of them being P=NP. Cracking RSA will be a consequence.

 

When the integers that are the difference of two squares are arranged into the grid and their corresponding properties are shown, a pattern emerges that shows calculation instead of searching is possible.

 

Glossary

Column

All cells for a given e.

 

Row

All cells for a given n

 

The grid is indexed using e, n, and t, where e is the rows, n is the columns, and t is the specific element in the cell.

 

Entry, record, element

one set of variables that represents one factorization for a number.

an entry = {e:n:d:x:a:b} (e, n, t)

{1:5:12:7:5:29} (1, 5, 4) is a record AKA an element AKA an entry.

 

ab record, nontrivial factorization, prime record

the element that contains the factorization of c that is not 1*c, hence, nontrivial.

 

1c record, trivial factorization

the element generated from setting a=1 and b=c

 

Cell

All entries for a given e,n (not to be confused with an entry itself.)

 

Genesis cell

e,1

 

Remainder Tree

The remainder tree is the result of treating d and e as c's recursively until 1 is reached, creating a tree with several to many branches.

 

Functions

na transform

a movement from a record in (e, n) into (e,1) where n becomes 1 and a becomes a times the n of the (e,n) record. It has also been used to refer to moving n*a records down in a cell.

 

T

T of number or T(input) is the triangle number function. If our input is 7, T(7) returns the 7th triangle number

 

T-1, inverse T

the inverse function of the triangle number function that returns the index of a given triangle number. If our input is the 7th triangle number, the function returns 7.

 

Variables

The map's legend is {e:n:d:x:a:b}, where c is any number that is the difference of two squares, so odd numbers are included. It is the number you want to factor. It is the number that the a and b in an entry multiply to make.

a and b are, to reiterate, the factors of c. a is the smaller factor of c, and b is the larger one.

d is the integer square root of c

e is the remainder of taking the integer square root of c. Unless c is a perfect square, a remainder will be left over.

i is the root of the large square. it is the same thing as (d+n)

j is the root of the small square. it is the same thing as (x+n). i^2 - j^2, difference of squares.

n is what you add to d to be exactly halfway between a and b, and it is the root of the large square. So it takes you from d to the large square.

x is what you add to a to make d. When added to n it makes the root of the small square.

f is what you add to c to make a square. (e is what you subtract from c to make the square below it, f adds to make the square above c.)

g is the square root of c with decimals, opposed to d, which discards decimals.

t is the third coordinate in the VQC, it is a function of x.

u is the base of a triangle that helps us calculate (x+n) for certain c values. simply put, it is a representation of (x+n). 8 times the triangle number of u plus one is x+n.

s was a variable used to demonstrate patterns in (e, 1). See "(e, 1)."

 

When capitalized versions of the variables are used in comparison to lowercase versions, the capitalized versions refer to the variable's value for the trivial record, and the lowercase variables refer to the values for the nontrivial record.

{e:N:d:X:A:B} (e, N, T) is the trivial element.

{e:n:d:x:a:b} (e, n, t) in this context is the nontrivial element, the prime factorization of c.

ID: 868475 July 4, 2018, 11:14 p.m. No.6695   πŸ—„οΈ.is πŸ”—kun   >>6769 >>7025

Rules

Each cell of the grid (e,n) has infinite elements or ZERO elements.

Each cell with one value has infinite elements, since every element can make a new one.

By induction, a cell only needs one value to make infinite values, that's part of the power of this and is why it is a virtual quantum computer as a whole.

The t variable is what will allow you to walk across these infinite elements.

If a grid cell has elements, all elements are constructable from a finite set of root elements.

Thus, only three variables are required to identify an element: e, n and t.

All products of odd numbers and all products of pairs of even numbers are the difference of two squares.

 

(1, 1) - the key

The values of a and b at 1,1 are related to the length of the longest side in right angled triangles.

The values here can be used to create the entire grid.

The values here determine the values of the rows to the left and right, which determine the values of the whole column.

 

Columns

Each cell at n=1 contains the roots of products in the column.

If c is a prime number, it will appear in one column exactly once.

If c is the product of two prime numbers that do not equal eachother, c will appear in two cells of one column.

All products (integers) c that are the sum of two squares appear (only) in columns where e=0,1,4,9,16,25…

All factors in a column are factors of the elements of the first cell in their column.

All Fermat primes (except) 3 appear in column one.

 

(e, 1)

If a number at position t has a factor s, then s is a factor at (t+s), (t+2s) and so on for a at (e,1).

Also, if a number at position t has a factor s at (e+1), then s is a factor at (s+1-t), (2s+1-t), etc for a at (e,1).

na and nb for any c can be found n places apart in the cell at (e,1).

 

(1, n)

The cells in row one where n=1 have a relationship with the cells 2n to the right and 2n to the left.

Each "a" from the first row equals na because xx+e = 2na and na is half of that. That's BIG part of the KEY

Each element in a cell can be generated by moving up (t-1 = x-2) or down (t+1 = x+2). Other variables can be generated from x.

 

For more of these rules, see the grid patterns thread.

 

Useful Equations and Notation

ab = c

dd + e = c

(d + n)(d + n)-(x + n)(x + n) = c

a + 2x + 2n = b

a = d - x

d = a + x

d = floor_sqrt(c)

e = c - (dd)

b = c / a

n = ((a + b) / 2) - d

d + n = i

x = d - a

x = (floor_sqrt(( (d+n)*(d+n) - c))) - n

x + n = j

j^2 = 8*T(u) + 1

f = e - 2d + 1

u = (x+n) / 2

 

if (e is even) t = (x + 2) / 2

if (e is odd) t = (x + 1) / 2

ID: 868475 July 4, 2018, 11:17 p.m. No.6696   πŸ—„οΈ.is πŸ”—kun

Code

 

C#

BigInteger Square Root β€”β€” https://pastebin.com/rz1SdACZ

VQC code w/ Bitmap β€”β€” https://pastebin.com/hMTtJF6E

PMA's tree generator β€”β€” https://pastebin.com/ZH9fSWu2

Original VQC code β€”β€” https://pastebin.com/XFtcAcrz

Unity Script β€”β€” https://pastebin.com/QgAXLQj3

Unity Script 2 β€”β€” https://pastebin.com/Y38nVWgT

 

Java

Traverse the VQC cells in real-time β€”β€” https://anonfile.com/TeH6q3d8bd/VQCGUI_v2.7z

Tree Generator β€”β€” https://pastebin.com/VZnQQR2i

VQCGenerator β€”β€” https://pastebin.com/Dgu9aP1h

VQC Triangle Number Methods β€”β€” https://pastebin.com/NCQ3HK2K

 

NodeJS

BigInteger Library and Sqrt β€”β€” https://pastebin.com/y8AXtFFr

 

Python

3D VQC [V2] β€”β€” https://pastebin.com/wZM5Thzu

Useful methods from CollegeAnon β€”β€” https://pastebin.com/d8xZZnm0

Create the VQC β€”β€” https://pastebin.com/NZkjtnZL

Fractal cryptography β€”β€” https://pastebin.com/XuN4U7Dv

GAnon's Viewable Grid code β€”β€” https://pastebin.com/czpK8A4j

Generate any cell in (0,1) and (0,2) β€”β€” https://pastebin.com/gRTYpdMU

Generate cells for a (and more) β€”β€” https://pastebin.com/iAizgLFF

Generate genesis cell β€”β€” https://pastebin.com/GKzcCpMF

Generate positive AND negative genesis cells β€”β€” https://pastebin.com/9ixjRyxt

VQC + t β€”β€” https://pastebin.com/Lgufk0db

RSA & PGP key wrapper β€”β€” https://pastebin.com/vNqnPRJR

 

Rust

Additional VQC code β€”β€” https://play.rust-lang.org/?gist=50def916ad48400bc5d638fbf119ae85&version=stable

Check if a number is prime β€”β€” https://huonw.github.io/primal/primal/fn.is_prime.html

Create Bitmap using the VQC Generator [V2] β€”β€” https://pastebin.com/zGSusyz5

Generate the VQC β€”β€” https://play.rust-lang.org/?gist=6b6beb372b6b931f1abd30642a35a80c&version=stable

 

Static Java/C# class with all RSA numbers β€”β€” https://pastebin.com/XYFpsDWE

 

Factorization methods (Java)

Binary search for i β€”β€” https://pastebin.com/TAt5bDsR

GCDFactor β€”β€” https://pastebin.com/70GJSMrv

Calculate factors using -x jumps β€”β€” https://pastebin.com/gKX9GW9r

Count down from t of 1c element β€”β€” https://pastebin.com/xxYa946V

Mirrors 1c until e=(-x+n^2) β€”β€” https://pastebin.com/WJBqPM4P

Shor's Algorithm (enter a random number < c as m) β€”β€” https://pastebin.com/RD83RTNc

 

Other Threads

Fermat's Last Theorem β€”β€” https://archive.fo/iTneU

Grid Patterns β€”β€” https://archive.fo/isamV

 

RSA #0 β€”β€” https://archive.fo/XmD7P

RSA #1 β€”β€” https://archive.fo/RgVko

RSA #2 β€”β€” https://archive.fo/fyzAu

RSA #3 β€”β€” https://archive.fo/uEgOb

RSA #4 β€”β€” https://archive.fo/eihrQ

RSA #5 β€”β€” https://archive.fo/Lr9fP

RSA #6 β€”β€” https://archive.fo/ykKYN

RSA #7 β€”β€” https://archive.fo/v3aKD

RSA #8 β€”β€” https://archive.fo/geYFp

RSA #9 β€”β€” https://archive.fo/jog81

RSA #10 β€”β€” https://archive.fo/xYpoQ

RSA #11 β€”β€” https://archive.fo/ccZXU

RSA #12 β€”β€” https://archive.fo/VqFge

AA !dTGY7OMD/g ID: b1f68c July 7, 2018, 1:15 a.m. No.6747   πŸ—„οΈ.is πŸ”—kun

I figured we could go with this one since it has the pastebin link dump and all the other introductory information, and because >>6746. Hopefully you all found your way here.

Anonymous ID: fae3ab July 7, 2018, 1:26 a.m. No.6750   πŸ—„οΈ.is πŸ”—kun

ISP here, but no trip for now. I'm stocked with this stuff, but I'm also traveling for a few days so I have no means of participation. For safe keeping if someone archives or as new stuff and shares it (bunker? discord?) In the event that we get shut down I'll be able to review in a couple of days. Either way godspeed anons and as always Hello VQC! Always a pleasure to see you.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 7, 2018, 1:28 a.m. No.6751   πŸ—„οΈ.is πŸ”—kun

Now to bless this Bread.

 

Baruch The Scribe

Was A Guy

Who at one time

May have baked a bread.

-mumbles off-

Aaaaaand that's as far as I know.

Can't match syllables to a prayer I was never actually taught.

I'm like… the worst Juden ever.

 

Mazelbrot!

AA !dTGY7OMD/g ID: b1f68c July 7, 2018, 1:46 a.m. No.6753   πŸ—„οΈ.is πŸ”—kun   >>6754 >>6759 >>6766 >>6794 >>6916

>>6746

I'm happy for you to wait for someone else to confirm as well (not to mention I'm currently in my bedroom while the other people who live here are having a party I'll have to show my face at eventually), but this seems far more important than not drinking around a bunch of drunk people.

 

>Next, we show how to calculate BigN of an odd number.

>All odd numbers are the difference of two squares.

>Every odd number is the difference of two consecutive squares.

1 = 1-0, 3 = 4-1, 5 = 9-4, etc.

>The product of two primes, is the difference two sets of squares.

Semiprimes are all odd since odd*odd, so they fit that pattern.

>The value of n for the product of 1 and c is defined now as BigN for odd numbers.

>Since d-a=x and x+n is the smaller square in the difference of two squares, then for odd numbers, BigN is ((c-1)/2)-x

>x is d-1, since a=1

This is fairly straight forward (for me, anyway). Algebra.

 

>Anyone want to show:

>RSA 2048

>c=RSA 2048

25195908475657893494027183240048398571429282126204032027777137836043662020707595556264018525880784406918290641249515082189298559149176184502808489120072844992687392807287776735971418347270261896375014971824691165077613379859095700097330459748808428401797429100642458691817195118746121515172654632282216869987549182422433637259085141865462043576798423387184774447920739934236584823824281198163815010674810451660377306056201619676256133844143603833904414952634432190114657544454178424020924616515723350778707749817125772467962926386356373289912154831438167899885040445364023527381951378636564391212010397122822120720357

 

>d=

I copied the C# code to Java so hopefully I did it all accurately. Obviously you'll be able to confirm that this is all correct.

floor(sqrt(above big number)) = 158732191050391204174482508661063007579358463444809715795726627753579970080749948404278643259568101132671402056190021464753419480472816840646168575222628934671405739213477439533870489791038973166834068736234020361664820266987726919453356824138007381985796493621233035112849373047484148339095287142097834807844

 

>e=

c - dd = 149730827186590819409161975355863102499674499386960841184873248110419525705142414412560340457818822465695973580080183089801154577153055685206470161899420076208154943348665238647007714095089342669905666884517354400122833485897064098153958317993833609635063079613328500485188782423277886515290863800949716792021

 

>BigN=

x is d-1 since a=1, so x = 158732191050391204174482508661063007579358463444809715795726627753579970080749948404278643259568101132671402056190021464753419480472816840646168575222628934671405739213477439533870489791038973166834068736234020361664820266987726919453356824138007381985796493621233035112849373047484148339095287142097834807843

 

n = ((c-1)/2)-x = 12597954237828946747013591620024199285714641063102016013888568918021831010353797778132009262940392203459145320624757541094649279574588092251404244560036422496343696403643888367985709173635130948187507485912345582538806689929547850048665229874404214200898714550321229345908597559373060757586327316141108434993615859020166427425368088424069958780819853230147577508164643339364712441831390650677628862077837124697517250971910788373374647441598985076306038901094587160385923033013611772476591818466822702222519806172328865872316642926190459725502720591581076567956723729060778728578126316270798047266909911419313225552335

 

>Remember BigN is the row in column e, where we always find (e,n,d,x,1,c)

So (1,c) for RSA2048 is at cell (149730827186590819409161975355863102499674499386960841184873248110419525705142414412560340457818822465695973580080183089801154577153055685206470161899420076208154943348665238647007714095089342669905666884517354400122833485897064098153958317993833609635063079613328500485188782423277886515290863800949716792021, 12597954237828946747013591620024199285714641063102016013888568918021831010353797778132009262940392203459145320624757541094649279574588092251404244560036422496343696403643888367985709173635130948187507485912345582538806689929547850048665229874404214200898714550321229345908597559373060757586327316141108434993615859020166427425368088424069958780819853230147577508164643339364712441831390650677628862077837124697517250971910788373374647441598985076306038901094587160385923033013611772476591818466822702222519806172328865872316642926190459725502720591581076567956723729060778728578126316270798047266909911419313225552335).

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 7, 2018, 1:52 a.m. No.6754   πŸ—„οΈ.is πŸ”—kun   >>6755

>>6753

"WHERE YOU BEEN BROSEF?!"

 

"Oh, y'know. Savin' the world by exposing a paradigm shifting bit of information and breaking the Qabal."

 

I've got a programmer knocked out on my couch at the moment, personally.

 

Priorities, my nigga.

AA !dTGY7OMD/g ID: b1f68c July 7, 2018, 1:56 a.m. No.6755   πŸ—„οΈ.is πŸ”—kun   >>6757

>>6754

Oh for sure. Although I'm not sure that I'd tell a bunch of normalfags that an anonymous stranger on the internet has been trying to teach me and some others integer factorization for the last 7 months. I think I recognize that picture. Is it from a painting of a UFO? Or is it something completely different?

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 7, 2018, 2:12 a.m. No.6756   πŸ—„οΈ.is πŸ”—kun

It's an egg on a moped on a tightrope with an audience and a safety net.

Seemed like a fitting analogy for what's going on here.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 7, 2018, 2:19 a.m. No.6757   πŸ—„οΈ.is πŸ”—kun   >>6768

>>6755

Prolly shoulda just included this in the first place.

I'm the photographer.

I've also put a vinyl horse head on Sophie Dee, the porn star.

A picture of her signing the mask is fine.

But I can't rightly post the fun ones.

Gotta see me in person for those.

 

Cuuuuuuz I could.

And I'm ridiculous.

Anonymous ID: caab4e July 7, 2018, 3:08 a.m. No.6759   πŸ—„οΈ.is πŸ”—kun   >>6767 >>6769

>>6753

Thanks AA.

I can verify once I'm at a desktop.

Now, quick reminder on triangular numbers and the construction of the grid.

Column 0.

The values of a[t] at 0,0 are twice the square numbers.

The values of d[t] at 0,0 are 4 multiples by the triangular numbers. All other cells at row 1 can be constructed from these values by adding to them or subtracting from them.

GAnon !OTLXnHjvTY ID: 28ac3a July 7, 2018, 5:04 a.m. No.6763   πŸ—„οΈ.is πŸ”—kun   >>6764

>>6760

I understand the last post you had. It's the same as PMA's calculation for the next record but worded a bit differently. Also if you go into negative x then back into positive you can get the record in between c and c'. Could be irrelevant though

3DAnon !!!YmM2NGExNzVlMDQ2 ID: b00103 July 7, 2018, 7:20 a.m. No.6766   πŸ—„οΈ.is πŸ”—kun   >>6768

>>6753

Wrote some python for this before noticing the full bread or reading your post. Got the same results as you:

c =

25195908475657893494027183240048398571429282126204032027777137836043662020707595556264018525880784406918290641249515082189298559149176184502808489120072844992687392807287776735971418347270261896375014971824691165077613379859095700097330459748808428401797429100642458691817195118746121515172654632282216869987549182422433637259085141865462043576798423387184774447920739934236584823824281198163815010674810451660377306056201619676256133844143603833904414952634432190114657544454178424020924616515723350778707749817125772467962926386356373289912154831438167899885040445364023527381951378636564391212010397122822120720357

d =

158732191050391204174482508661063007579358463444809715795726627753579970080749948404278643259568101132671402056190021464753419480472816840646168575222628934671405739213477439533870489791038973166834068736234020361664820266987726919453356824138007381985796493621233035112849373047484148339095287142097834807844

e =

149730827186590819409161975355863102499674499386960841184873248110419525705142414412560340457818822465695973580080183089801154577153055685206470161899420076208154943348665238647007714095089342669905666884517354400122833485897064098153958317993833609635063079613328500485188782423277886515290863800949716792021

BigN =

12597954237828946747013591620024199285714641063102016013888568918021831010353797778132009262940392203459145320624757541094649279574588092251404244560036422496343696403643888367985709173635130948187507485912345582538806689929547850048665229874404214200898714550321229345908597559373060757586327316141108434993615859020166427425368088424069958780819853230147577508164643339364712441831390650677628862077837124697517250971910788373374647441598985076306038901094587160385923033013611772476591818466822702222519806172328865872316642926190459725502720591581076567956723729060778728578126316270798047266909911419313225552335

 

Now I need to go get some coffee…

AA !dTGY7OMD/g ID: d6da98 July 7, 2018, 7:42 a.m. No.6767   πŸ—„οΈ.is πŸ”—kun   >>6769

>>6759

Guess I'll keep doing the deconstruction thing if nobody else is.

 

>Column 0.

>The values of a[t] at 0,0 are twice the square numbers.

Not sure I understand this one, actually. So here's the first 7 records of (0,0).

{0:0:1:0:1:1}, t=1, f=-3

{0:0:2:0:2:2}, t=1, f=-5

{0:0:3:0:3:3}, t=1, f=-7

{0:0:4:0:4:4}, t=1, f=-9

{0:0:5:0:5:5}, t=1, f=-11

{0:0:6:0:6:6}, t=1, f=-13

{0:0:7:0:7:7}, t=1, f=-15

>The values of a[t] at 0,0

t=1 for all of these records. That means there aren't multiple t values, so a[t] is always a[1], even though there are multiple values of it.

>are twice the square numbers

a and b are equal for all of these, but they just increment: 1, 2, 3, 4, 5, 6, 7. So, I mean, if you're referring to the fact that every square multiplied by 2 will come up in this infinite set of a values, then sure, but I'm not sure if that's what you're actually talking about.

>The values of d[t] at 0,0 are 4 multiples by the triangular numbers.

d also increments by 1 upwards from 1 like a and b. It would also contain every instance of a number being equal to a triangle number multiplied by 4. Is that what you mean? I can't tell.

>All other cells at row 1 can be constructed from these values by adding to them or subtracting from them.

I do remember that being the case but I don't remember the specific rules. Does this have to do with the a[t] and d[t] values?

 

You might have to verify this one before moving on maybe.

AA !dTGY7OMD/g ID: 52eb76 July 7, 2018, 7:53 a.m. No.6768   πŸ—„οΈ.is πŸ”—kun

>>6766

Thanks for double checking. I did it in a bit of a hurry so I did wonder if it was completely accurate.

 

>>6757

Man, that is not what I expected. Not just the picture too; I mean, I know you all as text on a screen so seeing actual pictures with one of you on the other end of the camera and a weird story related to it is always strange.

 

See you all in about 8 hours.

PMA !!y5/EVb5KZI ID: 47b0a9 July 7, 2018, 8:05 a.m. No.6769   πŸ—„οΈ.is πŸ”—kun   >>6778 >>6787

>>6695

Thanks baker!

 

>>6736

Morning VQC.

 

The formulas posted work to construct infinite elements in a chain and where n=1 will generate all records.

 

For n>1, there will be gaps.

 

This is equivalent to creating elements by ent where t=t+n.

 

>>6759

>>6767

These d[t] and a[t] rules are for (0,1).

Teach !!UgZAPoSXEk ID: 2259c7 July 7, 2018, 2:07 p.m. No.6771   πŸ—„οΈ.is πŸ”—kun   >>6772 >>6776

>>6726

Thanks! Its great to see you posting more often, and I really appreciate the guidance.

I'm personally very interested in your path VQC. I have so many questions - may I ask a few?

 

Based on the date of the youtube post, you obviously discovered this years ago - what happened when you first found this solution?

Were you working in university research? NZ, I assume, at the time? How did your peers and superiors respond?

I imagine many people would go very far to get their hands on this. Were you safe, or at risk?

I'm really very curious about your story.

 

I understand that many of these questions you may not be able to answer. But thank you!

 

>>6728

I've read your post GA, still trying to process it into what it means. I'll definitely continue to think on it.

Sorry that I have nothing to add on this at the moment, great post though :)

 

>>6723

I couldn't imagine this community without you Topol.

Thanks for defending all of us.

 

>>6730

Thank you for your thoughts. I've been here in spirit! But I too missed being part of this math family, I missed you all!

Teach !!UgZAPoSXEk ID: 2259c7 July 7, 2018, 2:49 p.m. No.6773   πŸ—„οΈ.is πŸ”—kun   >>6775

>>6772

Thats where I went with it too.

I actually share a homeland with someone who has been in exactly that situation…

I guess my next thought is, of all the people in the world, who knows this algorithm? Just 1 person? Any governments? Or NSA/military?

What do you think Topol?

VQC !!/aJpLe9Pdk ID: caab4e July 7, 2018, 3:06 p.m. No.6774   πŸ—„οΈ.is πŸ”—kun   >>6777 >>6779 >>6783 >>6784 >>6787 >>6794 >>6813 >>6942 >>6967 >>7319 >>7341

Tomorrow we will go through where the product of BigN and c are found in (e,1) and where the other value of c is in (e,1) and what information that gives us.

Then we will look at the key that is made by column -f with the locations of c in (-f,1) and how to find x for an and a(n-1).

Then we build the algorithm.

Then one of you anons can edit the Wikipedia page with the remaining RSA numbers and their new factors.

VQC !!/aJpLe9Pdk ID: caab4e July 7, 2018, 3:14 p.m. No.6776   πŸ—„οΈ.is πŸ”—kun   >>6787

>>6771

This was all done alone.

From scratch.

I realised that if I learned to build my own software for multiplying large numbers together and trying new views of number properties, I could see something where information was lost.

The first truly new idea came out of reading Revelation one afternoon in 2011.

The red text in the first chapter looked like description of odd (prime) numbers in binary, since "I" is the first and last number of all binary odd numbers.

I think binary is not new.

It looked like a coded algorithm using allegories to keep it secret.

The first version of the grid, looks very different, was in binary.

ISP !kIkD/SqZ4s ID: 34fb33 July 7, 2018, 3:30 p.m. No.6785   πŸ—„οΈ.is πŸ”—kun   >>6786

>>6783

I agree, but I have a feeling that we're just getting started. Integer factorization doesn't prove P = NP. It will prove P = BQP and thus prove that quantum computers can be simulated on a classical machine (self verifying the VQC). So we still have work to do.

 

I mean, we got Mandelbrot set as phase 2, right?

Teach !!UgZAPoSXEk ID: 2259c7 July 7, 2018, 3:57 p.m. No.6787   πŸ—„οΈ.is πŸ”—kun   >>6790 >>6791 >>6792

>>6769

Hey PMA! The d[t] pattern is found in (0,1), but the a[t] pattern repeats!

Each n = a perfect square ((0,1), (0,4), (0,9)…), a[t] is 2 * a square.

 

>>6774

Looking forward to the process. Thank you.

 

>>6775

I've thought about this so much. If other people know, how many and how is it kept secret? Military organizations are structured in a way that embeds secrecy, so I can understand those groups, as I can secret organizations. But in the business sector, I really have no idea.

But if John knew (>>6776), then it's likely others have known over time.

 

>>6777

I wonder if we can anonymously update the wikipedia page via vpn's or tor or a public wifi access point. A starbucks in a major city would be pretty anonymous.

 

>>6776

Thanks for sharing this.

I'm really curious about the religious connections, but I'm very ignorant about religion unfortunately. Your quotes from Revelations took me down a research path, one that I intend to continue. One of the things I found interesting was the etymology of apocalypse - "to uncover".

Do you have any info you could share on following your interpretation of binary in Revelations?

Teach !!UgZAPoSXEk ID: 2259c7 July 7, 2018, 4:58 p.m. No.6793   πŸ—„οΈ.is πŸ”—kun   >>6795 >>6801

>>6789

I love it.

 

>>6791

I don't at the moment, should I? Am I missing much over there?

 

>>6790

Yep, good find.

I think this has to do with another link, I'll try my best to describe it:

  • in (0, 1) we know a[t] = tt2.

  • we also know how to "move" up and down (0,1) whereby t=p+t (I think thats the equation… please correct me if i'm off-by-one)

  • This means that if we try to find all the a[t] that are divisible by 3, we can simply list:

t = 3, 6, 9, 12, etc.

  • This means that at (0,9) for example, we'll see the x[t] values are equal to the x[t] values in (0,1) at t = 3, 6, 9, etc.

 

>>6792

I'm trying to follow this one, but i'm missing something, can you give an example or something PMA?

PrimeAnon !!!ZmMwNmUyMmUyMzY5 ID: a2c3ea July 7, 2018, 4:59 p.m. No.6794   πŸ—„οΈ.is πŸ”—kun

>>6753

Here's some Java code for testing values using your method:

https://pastebin.com/PxKGHme8

 

I'm using a different Sqrt function–the link from the source is in the code; I modified it a bit. So far it's working as intended.

 

The css file is included in case anyone wants to skin it–just comment out the relevant line in the code if not (located in the "start" method).

 

I've got to go over things and see how everything relates. Had a big revelation yesterday after a prayer in frustration (see "old bread rehash"). I think I'm almost caught up :)

 

>>6774

Thanks again VQCβ€”the more I look at the grid, the more I find. So many relationships and patterns; it's magnificent :)

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 7, 2018, 5:17 p.m. No.6795   πŸ—„οΈ.is πŸ”—kun   >>6796 >>7110

>>6793

It's more of a speed thing.

We're able to spitball over there and that's how we started making progress again after the lull.

 

Also, there's less pressure to come up with a post that goes far enough to be "worth adding to the bread yet".

 

Super useful for getting on the same page and I make sure everyone remembers to post findings on the board.

 

Simple lock and key mechanism to get into the server:

Post something (UNIQUE) in two parts.

Post one half with your trip here, and then send the other half to Ta.io in a private message (server owner):

https://discord.gg/pQDaMkU

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 7, 2018, 5:35 p.m. No.6798   πŸ—„οΈ.is πŸ”—kun   >>6799 >>6800

>>6796

Could be a obscure ass picture, an output style that only you do, specific something where the reference doesn't make sense until you send it to me… is whatever.

 

I'm just making sure you hit the correct points along a curve.

PMA !!y5/EVb5KZI ID: 47b0a9 July 7, 2018, 5:47 p.m. No.6801   πŸ—„οΈ.is πŸ”—kun   >>6802

>>6793

Teach.

 

First pic is the pattern you pointed out in (0,n) where n is a perfect square. Examples show the first few records for n=1,4,9,16,81,169. The a values repeat. 0,2,8,18,32,etc.

 

Second pic, is the pattern mentioned above >>6792 where the a[t] value from (0,1) becomes the n value in (0,n), and the perfect squares move to a[t].

 

Think this is justing show how the n and a values are interchangeable in certain circumstances.

Teach !!UgZAPoSXEk ID: 2259c7 July 7, 2018, 9:25 p.m. No.6807   πŸ—„οΈ.is πŸ”—kun

Ok, so I was investigating d[t] at (0,n) and its relationship to triangle numbers…

so, lets say we're at n = 15, what are the set of d[t]?

 

which rows are in (0,1) have 15 as a factor?

 

start with t = 16:

 

t=16: a = 450 = 21515 x = 30 = 215 d = ( 15 +1)( 15)2

t=31: a = 1800 = 421515 x = 60 = 2215 d = (215 +1)(215)2

t=46: a = 4050 = 921515 x = 90 = 3215 d = (315 +1)(315)2

t=51: a = 7200 = 1621515 x = 120 = 4215 d = (415 +1)(415)2

 

move to (0,15):

t= 1: a = 30 = 215 x = 30 = 215 d = 2215 *1

t= 2: a = 120 = 4215 x = 60 = 2215 d = 2215 *3

t= 3: a = 270 = 9215 x = 90 = 3215 d = 2215 *6

t= 4: a = 480 = 16215 x = 120 = 4215 d = 2215 *10

 

What this is all saying is, for odd n, d[t] = n4T(t)

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 8, 2018, 12:47 a.m. No.6809   πŸ—„οΈ.is πŸ”—kun   >>6811

1:8 - "I am Alpha and Omega", says the Lord God, "who is and who was and who is coming, the Almighty."

 

Imagination is the Beginning and the End.

Er… let me rephrase that.

The Beginning, the Infinite Now, and the Continuance.

Imagine that.

<3

~The Phither

Anonymous ID: a2c3ea July 8, 2018, 3:31 a.m. No.6812   πŸ—„οΈ.is πŸ”—kun   >>6813 >>6814

I'm thinking this may be where the recursive algorithm starts to come in? With an 'e' over 300 digits in length, it seems like a few more rounds of squaring and storing results might be called for. It's so satisfying to watch that number get chopped down to something comprehendable, even if it's just play-acting.

PrimeAnon !!!MDg4Y2Y1MDc5ZWMy ID: a2c3ea July 8, 2018, 3:53 a.m. No.6814   πŸ—„οΈ.is πŸ”—kun   >>6815 >>6816 >>6817

>>6812

Forgot trip.

 

VQC, I don't think I can edit the article as I'm basically barebacking. But don't you want credit for this? And if so, how should you be given it? Twitter handle?

 

I'm sure it will all come in time. As magnanimous as you are, though, do you think it might be a good idea to claim credit?

 

Consider how many brilliant inventors there are that most people have never heard of–maybe you're comfortable knowing that God sees your good acts, and J know this to be true and reward enough. But there is something to be said about the hijacking of the message; we have no idea how Doug Engelbart saw things, but we know exactly what the men who pirated his ideas think–they're still venerated to this day.

 

Then again, maybe that has more to fo with corruption than anything. I can't imagine there will be much joy from the bad actors in Europe once we fly their dirty laundry.

 

I hope this is not what you mean when you speak of sacrificing yourself to this. I really hope you're just referencing the money you stand to lose by sharing the idea rathing than capitalizing on it…

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 8, 2018, 4:28 a.m. No.6815   πŸ—„οΈ.is πŸ”—kun

>>6814

Dearest Fagwit:

 

STFU.

I'm putting this where it needs to be.

I'll deal with the media.

I'm not looking to famefag but I got this becuase…

 

Torture me all they like, I don't know aaaanything of value.

 

I just say Trangles funny.

 

;)

I still want Tucker, but that's quite the request.

 

Especially since his Hair could do the whole interview.

VQC !!/aJpLe9Pdk ID: d99457 July 8, 2018, 8:27 a.m. No.6817   πŸ—„οΈ.is πŸ”—kun   >>6818 >>6844

>>6814

Hi, thanks for the question. No I don't want any personal credit for this. I have a family, a career and mild Aspergers. I'm also a recovering alcoholic/addict. Fame, infamy or money would only end in a quick death pursuing women, drugs and alcohol into the gates of death. It wouldnt be my intention, but that is exactly where it would lead after just one drink. That's all it would take. But it is valuable to demonstrate that there is a whole world of maths and engineering that is waiting and using mathematical objects to solve problems in the virtual and real world in this way could open up a new dimension, so to speak.

Other anons can make the billions but I don't have any needs except the thrill of finding no problems to solve and helping the group effort to make the world better where I can.

The only way forward for me is to be helpful and share ideas.

If it sounds altruistic, it's closer to self preservation.

Maybe with a few years of recovery under my belt?

Prolly not.

 

Now, my computer is frozen (the new one) as soon as it's up and running, I'll write the e column definitions with RSA 100 and 2048, to show how easily everything scales up and to give examples for knowns and unknowns.

VQC !!/aJpLe9Pdk ID: d99457 July 8, 2018, 8:31 a.m. No.6819   πŸ—„οΈ.is πŸ”—kun   >>6821 >>6822

Please also anons, choose caution as the temptation to use this maths for hacking is huge.

Bitcoin has been talked about before.

Bear in mind that this maths can lead to shortcutting the mining process before it can be used even to do anything else.

VQC !!/aJpLe9Pdk ID: d99457 July 8, 2018, 9:39 a.m. No.6820   πŸ—„οΈ.is πŸ”—kun   >>6822

Second laptop is not functioning.

Before I attribute it to malice, I will let it run to apply background updates.

Hasn't been switched on in a while.

Do not worry.

I have this week off work too.

I will keep you updated.

Hobo !!1yNgQ3NlCs ID: ca2d33 July 8, 2018, 10:03 a.m. No.6821   πŸ—„οΈ.is πŸ”—kun   >>6822 >>6823 >>6829 >>6844

>>6819

Some thoughts on all this. Of course everyone is thinking about crypto-currencies with this process. They are a HUGE target and will be destroyed but, IMO they are a small portion of where this will go. Of primary importance is the idea that this creation cannot be lost. I have been backing up the breads here to the best of my ability. So many technologies have been suppressed and buried over the years. Much of the troubles we have in the world are related to that fact. It is frustrating and difficult to accept but I am pretty much convinced this is true. I fear that going after BTC as an early example of this will cause a colossal shit-storm. I haven’t mentioned much about cryptos since coming here but I have been following them since around 2010. I managed to mine a tiny bit in the early days. Certainly not enough to be called wealth but it was something. Most importantly I learned a LOT and found an accepting community. Everyone will want to go for the Satoshi wallets… or simply mine the rest of the block chain in a single day or something like that. I caution against this action for the following reason.

 

Those will be high-profile events. Especially the mining. If anyone here managed to crank out the code to pre-mine BTC someone will notice. Instead, simply identify and crack the private keys on β€œlost” wallets. What I mean about lost wallets is this. There are tons of early BTC wallets whose private keys have been lost. Many of these wallets are vast by today’s standards. Not a lot of people are looking at them. We could all do this for YEARS with no one noticing. There are block chain forensic firms that WILL notice but it will be nothing but a curiosity to them probably for YEARS because of the un-believability of the cracking of this encryption. They will write it off as a super-hodler finally opening that wallet. The reason I say this is because I hope a lot of good will come from this as opposed to it getting buried like all those other technologies. Topol, I think you would be a good face to speak to the media about this but, honestly, they won’t care. They certainly won’t want to hear a REAL story like the one coming from this board and report it as such. It won’t happen. Instead they will try HARD to bury this and bury it DEEP (probably all of you along with it) and probably in the worst way.

 

I do not have the skills to actually use this tool in its current form but if I did or (if one of you helped me to do it) I would use much of what I gathered cracking lost wallets to try to release buried technologies and to push for Full Disclosure. I would support a man named Steven Grier and do so in a huge way. I want to live in a world with advanced energy technologies at the very least. This is a noble goal in my opinion and one that is worth fighting for. It is the one I will fight for if I have the means to do so. I encourage you to all find something similar to actually help build a new world with this tool. I, again, caution you all deeply to consider carefully before approaching ANY mainstream news or government organization with this in any form at all. Always remember that their goal will be to use the vast (and terrible) resources at their disposal to destroy and hide this VQC tool. If it is given to them FIRST the THEY WILL CONTROL IT. Instead I propose first, each of us set ourselves up so none of us have to waste time at menial jobs and such and can live in some degree of comfort then BUILD STUFF OF VALUE with this thing and give that away for free, essentially under a GNU license or some such thing. Let the world figure it out that way as opposed to pursuing an easily suppressible fame-fest where this tool will be captured, hidden, and then used for nefarious purposes. The media and the government are not our friends, they are the greatest enemy. They WILL use this tool to do harm as opposed to good.

Teach !!UgZAPoSXEk ID: 2259c7 July 8, 2018, 10:21 a.m. No.6822   πŸ—„οΈ.is πŸ”—kun

>>6820

Thanks VQC.

 

>>6819

>>6821

Well said Hobo. I agree with you.

I'm glad we're talking about this, I'd love to continue the conversation further about how to use this for good, and not cause total destruction in the process.

 

I've been following the NIST post-quantum crypto RFP: https://csrc.nist.gov/Projects/Post-Quantum-Cryptography/Round-1-Submissions

and: https://en.wikipedia.org/wiki/Post-Quantum_Cryptography_Standardization

 

My fear is that all the world's data will be hackable until we have better crypto, and I'm not sure we have anything that is proven to be secure yet.

AA !dTGY7OMD/g ID: add144 July 8, 2018, 11:46 a.m. No.6823   πŸ—„οΈ.is πŸ”—kun   >>6828

>>6821

>I caution against this action for the following reason.

>Those will be high-profile events.

>their goal will be to use the vast (and terrible) resources at their disposal to destroy and hide this VQC tool

Now, I actually think these are a little contradictory. It depends on our goals (more importantly, in my opinion, VQC's goals), but if we aim to make the VQC public knowledge, the perfect way would be through a massive shitstorm. If it's the goal to make the VQC public knowledge, cracking into the Satoshi wallets and mining the rest of the blockchain in one day would do exactly that, and there would be no way to stop it from becoming public. That would be exactly the kind of event that would guarantee the public would learn of the existence of the VQC without the possibility of diversion by the media.

 

Of course, I don't think it's a good idea for other reasons. Firstly, I would think the only way to do it would be to take all of this huge mass of bitcoins and then not do anything with them. If anyone tried to cash them in for money, obviously they would be mobbed with publicity. Not to mention, that kind of money very easily corrupts people. Just leave this ridiculous scenario of a bitcoin wallet overflowing with bitcoins without actually doing anything with them as a public testament to the ridiculousness of it all. Secondly, and also in regards to your lost wallet idea, while that actually sounds like a brilliant way to fund the rest of this, keep in mind, there are anons lurking. While you can go through the threads and get a list of regular contributors together, and then distribute large sums of cash around to the people who still contribute, there are going to be people who feel entitled to this money. This is where fighting, manipulation and sabotage come into play. It's a dangerous thing. If we try to use this with lost wallets for a long period of time, like you said, and someone isn't getting a cut, what's stopping them from ruining our funding plans and getting the lost Satoshi wallets just to fuck it all up? Not to mention, it's going to be public by then, so anyone else could do the same thing.

 

Personally, while I think that idea could be very useful, and I can't predict the future, I'm holding out for this thing VQC briefly mentioned a couple months ago. He said he wanted to start a company making futuristic tech. I don't know how he plans on funding it if he isn't going after this money, but if it'll provide a viable and stable source of income, I would be part of that. I couldn't agree more about your motivations for the lost wallet idea, and if there's one main thing keeping me here it's the fact that we've all been given an extremely unique opportunity to change the world for the better, so I feel like it's a responsibility at this point to stick with it since there are so few of us. Not to mention, it would give me the funding and the free time to pursue music full time. That's the only selfish reason I have for being here, as far as I can tell.

 

>Steven Grier

This is just speculation, but if you've seen the end of Undisclosed, he had an interview with John Podesta, talking as if voting for Hillary would have led to disclosure (yeah fucking right). I definitely don't think he's a bullshit artist, but I don't know how helpful sending him a bunch of money would actually be if he has the ability to get someone as heavily cabal as John Podesta to be in his movie talking about UFO disclosure, you know?

VQC !!/aJpLe9Pdk ID: d99457 July 8, 2018, 12:03 p.m. No.6824   πŸ—„οΈ.is πŸ”—kun   >>6825 >>6827

Computer that can be connected to the internet is now functioning.

Wife is binge watching The Handmaid's Tale right now with it, which rewards her patience this very hot afternoon and allows me to check it's not spiking on CPU or network traffic as this will pause the show.

Once she is done (still recovering from jetlag) I'll install VS C# and it will be a number storm for a couple of post.

We'll be looking at values in a[t] a lot with examples.

As an exercise, can you find BigN-1 in column -f in terms of t? This is unfair as I said I would walkthrough but need to balance a wife who has gone the extra mile and patient anons.

Meanwhile, Q has been very quiet.

Something big happening?

Teach !!UgZAPoSXEk ID: 2259c7 July 8, 2018, 12:08 p.m. No.6825   πŸ—„οΈ.is πŸ”—kun   >>6826

>>6824

>As an exercise, can you find BigN-1 in column -f in terms of t? This is unfair as I said I would walkthrough but need to balance a wife who has gone the extra mile and patient anons.

Family first.

 

even d:

t = d/2 + 1

 

odd d:

t = (d+1)/2

 

I think this is right?

AA !dTGY7OMD/g ID: add144 July 8, 2018, 12:15 p.m. No.6827   πŸ—„οΈ.is πŸ”—kun

>>6824

If we're waiting around then, could you possibly elaborate on that idea you posted a while ago about starting a company making futuristic tech?

>BigN-1 in column -f in terms of t

Did you mean like Teach did above, or do you mean the cell in which we would find BigN-1 in column -f for RSA2048?

Hobo !!1yNgQ3NlCs ID: ca2d33 July 8, 2018, 12:17 p.m. No.6828   πŸ—„οΈ.is πŸ”—kun

>>6823

"there would be no way to stop it from becoming public"

Well, AA I hadn't thought of it that way. It would be Sooo easy for them to say something like "Russian Hackers… blah blah" though then just shoot us all and use the tool to end what Trump is doing. It will have to be done in a way that the tool is released to TONS of people with a really easy way to use it. It will simply destroy the cryptos, then be reworked by smart people and will wipe all cryptography that is not air-gapped or pen and paper.

 

Yeah I see your point on the wallet thing. Boy, what a fascinating situation this is. How would we do mass distribution. We will need a normie-tier tool to get this done. I mean a 1 click crypto cracker that is easy to use by anyone. Something like "input public wallet address " Click button. Output private key for that wallet. Then distribute it to people who do crypto for a living or something like that, They will spread it automatically. The tool will have to have a back-end description of what is happening behind the scenes so it is spread.

 

As for Grier, Yeah, I have no idea if he really is legit. What I DO know is he is far more legit than other personalities out there (Gaia/etc). Maybe Catherine Austin Fitts would be able to help? She has a pretty negative view of BTC and has been attacked relentlessly by the big guys. I could easily see her having some ideas of where to go from here? Yeah, I just urge caution. I do like the idea of leaking it as far and as wide as possible as fast as possible but the opportunity for manipulation of WHAT it is is sooooo easy. Soooo many normies. It has to be super-easy to use/understand. Perhaps a new thread for this conversation since this is not really research-related?

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 8, 2018, 12:22 p.m. No.6829   πŸ—„οΈ.is πŸ”—kun   >>6832

>>6821

Considering the effects and situations on the Q Phenomenon that I keep having and finding myself in…

I shoulda been buried a long time ago if that's what's going to happen.

 

Cuz you have to figure…

General Nakasone Satoshi-senpai has our backs, like Adm. Rogers before him.

 

And, again, the main reason I volunteer as Tribute, because somebody WILL ask about this, is because I'm already doxxed anyway, and y'all would prefer to maintain your privacy.

 

(Artists have to doxx themselves to get work on the interwebs, so I came in doxxed to begin with.)

VQC !!/aJpLe9Pdk ID: d99457 July 8, 2018, 12:25 p.m. No.6830   πŸ—„οΈ.is πŸ”—kun   >>6831

While we wait, just to go over the solution.

The Grid called The End is a superposition of all products where those products are the difference of two squares. This includes all odd numbers and even numbers that are zero mod 4.

Row 1, n=1 is special, since all factors of all products are present in it and by knowing d and e, and calculating f, this superposition collapses into a method to use cells e and -f in the first row, which are unique to d and e.

If there are more than two prime factors, d and e are the same, you can just find the largest two first, one will just be the product of further primes, divide and go again.

The problem we are trying to solve appears to be one where we don't have enough variables known in the equations we have.

The different factors (n-1) in -f, and n in e, coupled with BigN in e AND BigN-1 in -f as well as their 'shadow' partners where they exist at a[c+1-t] give us enough information to determine 1. Is the number prime 2. The value of x.

Remember, x is related to t.

So, for this problem, the 'superposition' collapses on two variables, d and e.

That was my assumption back in 2010 as it happens, that these two properties of c would be enough to factorise.

Integers are not a number line.

They are a family that grows like a fractal tree.

This work can probably be optimised.

There are more ways to do this, as implied, one just uses column zero. The other uses the diagonals but all are related. Like turning a higher dimensional object in 3D. It can look different.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 8, 2018, 12:40 p.m. No.6831   πŸ—„οΈ.is πŸ”—kun   >>6832

>>6830

Speaking of which… something that was brought up but never really touched on…

 

I have a vision in my head that RSA is akin to the shadow of ECC in 3-space.

 

Is that anywhere close to what you were saying when you said they were related, if not the same thing from different "perspectives"?

VQC !!/aJpLe9Pdk ID: d99457 July 8, 2018, 12:52 p.m. No.6832   πŸ—„οΈ.is πŸ”—kun   >>6833

>>6829

>>6831

I would love to buy physical copies of your work Tops.

Elliptic curves are DEFINITELY related.

The universality of the extended Riemann zeta function, the Mandelbrot Set. Related.

Whether you like to get lost in the beauty of colourful representations or found in the ruthless logic of it all, there is something for everyone.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 8, 2018, 1:02 p.m. No.6833   πŸ—„οΈ.is πŸ”—kun   >>6834

>>6832

I have a variant I made of Lady Killer Miller, pic related.

 

Second post, I didn't make any of that, but if I ever make a coffee table book; you're totally getting a signed copy.

 

I'll sign it at the Top instead of the bottom right like I usually do.

 

I wouldn't accept payment, and, when we're done, your money's no good, anyway. :D

 

Right now I'm trying to comfort Jordan Sathers because David Wilcock left GAIA cuz luciferianism, and because 5G went active in some cities.

 

Let's see them pull of their 5G bullshit when we're done here.

Better yet, if it can effect people as is described…

Let's take it over and improve everyones situation instead of using 5G as a slow/quick kill like we're told it's supposed to be.

 

And just for fun… rando images, but if I feel the need for more rando I'll head to the EZ Bake.

VQC !!/aJpLe9Pdk ID: d99457 July 8, 2018, 2:32 p.m. No.6834   πŸ—„οΈ.is πŸ”—kun   >>6835 >>6836 >>6840 >>6841

>>6833

Nice.

Thanks for waiting just a bit longer.

Looking at morning UK time to get started as I am still limited to a phone.

A few extra reminders…

t is the order of members in a cell

In the first row where n=1, t gives a values of x that alternates between odd and even.

The value of x in e and -f will never both be even of odd since the gap between the column is always odd or 2d+1

a[t] is half the square of x plus e for the column

When e is larger than x by more than 2x+1, the value at a[t] will be found in a previous column in the first row.

For odd e, x=2t-1; even e, x=2(t-1)

a[t]=(((2t-1)(2t-1)+e)/2, (((2(t-1)2(t-1))+e)/2 for odd, even e

d[t]=a[t]+x[t]

 

In the rows below the first, where n>1, have you noticed that there is sometimes more than one set of products in the cell?

If there are two, both are growing by 2n added to x in that cell. The first set you can spot the same starting value of x from the value of x in the first cell at n=1. The second begins at -x until adding 2n makes it positive. I'll show this in the other thread after the walkthrough, if anyone is still interested at that point!

GAnon !OTLXnHjvTY ID: 28ac3a July 8, 2018, 8:15 p.m. No.6837   πŸ—„οΈ.is πŸ”—kun   >>6838

So I've been looking into the frequency of repeating A[t] values and B[t] values in (e,1), (-f,1) and other cells with same c but different d. Look at the following records images. Pay attention to the block on the right. The center dot is (e,1,t) for whatever record we have at the moment with the same x or t. Column to the left would be -f, then the next would be for d=d+2, etc. Then if you go to the right each column is d=d-1. Going down you have higher t values, and going vertical up you have lower t values. If there is a red dot that means it shares a as factor of A[t], green dot means it shares b as factor of A[t] and cyan is c as factor. Notice that for any entry in with the t=t+n shift, the location of that c row (which is origianlly the cyan one going down to the right) turns into a B line for the next record at t+n, then it is always a B line for the rest of those records. This is pretty neato I think.

PMA !!y5/EVb5KZI ID: 47b0a9 July 8, 2018, 9:29 p.m. No.6840   πŸ—„οΈ.is πŸ”—kun   >>6843

>>6834

>>6836

>are there only 2 sequences at n > 1

Answering my own question. No, there can be more than 2 sequences.

 

Example for c287 at (31,128) attached shows 4 sequences incrementing by t+n starting from t=8, 57, 72, and 121.

 

Not sure of the pattern yet to determine the number of sequences within each n.

ID: caca72 July 8, 2018, 9:29 p.m. No.6841   πŸ—„οΈ.is πŸ”—kun

>>6834

Could you elaborate more on Revelation? Where do you see roots and remainders in it, and the other variables? I've already gotten the idea that it has variables corresponding to the churches and found 5461 in there, which is 7 0's and 1's

PrimeAnon !!!ZmMwNmUyMmUyMzY5 ID: a2c3ea July 8, 2018, 10:47 p.m. No.6844   πŸ—„οΈ.is πŸ”—kun

>>6816

That was not my point. My point was this: when a message is received, if the messenger isn't there to interpret, pretenders will certainly be more than willing to make the interpretation. This is trivially easy to prove: the New Testament is rife with Jesus condemning the Pharisees and Sadducees for twisting the Word of God to their benefit; similarly, we have the Catholic Church's storied history, Buddhist monks in designer fashions, hundreds of nuclear bombs for every nuclear reactor, etc.

 

There is a reason why the membership of the Council on Foreign Relations is structured as it is: those in media, social media and higher ed twist our perception of right and wrong, while bankers, international corporations and government officials do wrong. By placing their members at the critical distribution points of our shared reality, they've managed to put rings in our collective noses and bring us to heel.

 

Rest assuredβ€”in the absence of an intervention of Biblical proportions, this knowledge will be used (in large part) to deliver porn before it is used to save lives. Maybe it really wouldn't matter if Chris revealed himself–Doug Engelbart wasn't hiding; it's just that nobody cared to solicit his views.

 

>>6817

I understand, in a very personal way. I went down that path and ended up just short of homelessness. I got your "three rivers"(?) reference awhile back, but didn't want to bring it up–shitty way to introduce oneself.

 

It's your message to give–God gave you the understanding for a reason, and this student will respect the teacher's wishes. At this point, I have no idea what I will (or even can) do with the knowledge, but I pledge to you that, so long as I make money off of it, and I have a way of contacting you (or vice-versa), neither you or any member of your family will want for their share.

 

>>6821

Thank you for sharing your insight, Hobo–these are wise considerations. I know nothing about cryptocurrencies; I would be lying if I didn't hope to get a few bitcoins out of this, but I never expected there to be any left by the time I caught up. It sucks, given that I've probably invested over 2k hours toward the Q movement and this by now; that could've translated into enough scholarships that I wouldn't have to work while going to school. But money can't buy this (unless Chris is giving private lessons somewhere).

 

I honestly think it is inevitable that cryptocurrencies go down, once this knowledge is out–if not from one of us, then from the next ones who draw the connection. We have no idea who is monitoring this thread; my guess is that some are prepared just as you have described. Once the news spreads that RSA has been broken, there will be a rush to sell. It is Chris' wish to let this news out, none here will deny it; it would be best to cash out asap, imho.

 

Also, as you have mentionedβ€”I think there's a lot to be said for doing the right thing. From the outset, I felt that the real test here was not whether we could be one of the first to understand the language of Creation, but what we did with that knowledge. This point in time is precarious–it feels, for lack of a better word, "pregnant." If we are reaching a pivotal point in our existence–one in which the Veil is lifted and we come face-to-face with our Creator, I wouldn't want to have to explain why $2 Billion worth of bitcoin was spent developing dildos that could read one's mind and be several places at once.

VQC !!/aJpLe9Pdk ID: d99457 July 9, 2018, 12:06 a.m. No.6846   πŸ—„οΈ.is πŸ”—kun   >>6847 >>6848 >>6849 >>6850 >>6853 >>6854 >>6870

Morning anons, have woken in the night with the kind of sore throat I haven't had since I was little. Horrible timing. This should only push things to tomorrow or hopefully this afternoon. Feels like swallowing broken glass.

Thanks for the questions since last night, I will get to them once I can.

Anyone have recommendations besides Manuka honey?

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 9, 2018, 12:52 a.m. No.6853   πŸ—„οΈ.is πŸ”—kun

>>6846

Oooooooh…

You know what would reeeeeally make you feel better?

VQCrumbs!

 

Once you're smiling because we're even further ahead, you'll be feeling better in no time, ol' chum!

Yes sir! Posting crumbs and answering questions while you lie on the couch feeling icky is just what the doctor ordered!

 

Cuz a lot of teas have caffeine, so if you were hoping to sleep it off, you'll wanna pick a sleepytime tea.

But if it's too late for that and you've got energy but you feel boo…

 

OH!

And a hot, steamy shower.

That should also help.

 

Do it for Albino Morpheus; cuz you're totally not him.

Then we'll take a peek at all his cables and surely something in there will make your throat feel better.

VQC !!jhli0.Qcik ID: d99457 July 9, 2018, 3:43 a.m. No.6855   πŸ—„οΈ.is πŸ”—kun   >>6857

Brilliant ideas.

My wife made a manuka honey drink with a berocca, plus coffee, plus paracetamol, plus this throat numbing thing.

Just doing the install of VS so can use some proper numbers.

I cannot quite believe the amount of the Spyware that comes these days on a cheap laptop. They don't even try and hide it any more.

Customer experience is second to full on data gibs.

VQC !!/aJpLe9Pdk ID: d99457 July 9, 2018, 3:53 a.m. No.6858   πŸ—„οΈ.is πŸ”—kun   >>6860 >>6929 >>6967

When we start, we'll build some functions.

 

We want to get a[t] (some na) for a column. We supply the column e and the t. The return is a[t].

 

Then we want to show that if p is a factor of a[t], p is also a factor of a[p+1-t] for some column e.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 9, 2018, 3:53 a.m. No.6859   πŸ—„οΈ.is πŸ”—kun   >>6862

>>6856

Guy figures out how to devastate all of encryption…

Can't secure a reliable machine, let alone with viable opsec.

 

That's like…

Why is God always asking for tithes and donations?

Screw the idea that the Creator of All is somehow bad with money…

JEWS ARE THE CHOSEN PEOPLE!

FFS, THERE'S GOTTA BE ONE ACCOUNTANT OR WEALTH MANAGER OUT THERE WHO COULD HANDLE THIS!

VQC !!/aJpLe9Pdk ID: d99457 July 9, 2018, 4:02 a.m. No.6862   πŸ—„οΈ.is πŸ”—kun   >>6863 >>6864 >>6865

>>6857

Good question.

>>6859

Another good question.

Should be ok until the moment the correct factorisation of RSA 2048 appears or just before.

This isn't about hacking or breaking encryption, that's a consequence. This is solving a math problem.

Nothing illegal is occuring here.

Also, nothing attracts attention like having linux pop up from a USB drive all over the place. That's got hacker written all over it, right?

Hackers don't use Β£200 laptops with Micro$oft on them.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 9, 2018, 4:11 a.m. No.6863   πŸ—„οΈ.is πŸ”—kun

>>6862

Meh.

That screams "burner phone", to me, but I see the logic. No sense buying a shit ton of Macs unless you have a serious hard on for ditching/destroying them.

 

Still… surely there's a way you could preload a wipe that quickly boils it down to its barebones onto a something. With as often as you do this and with as blatant as they are about it, there has to be a faster way.

 

Granted, you just kinda killed your "muh compy!" schtick, but thank you for essplaining.

GAnon !jPVzzZOz2c ID: 28ac3a July 9, 2018, 6:46 p.m. No.6867   πŸ—„οΈ.is πŸ”—kun

Guys I looked at the frequencies with which a, b, c and their squares appear as factors in A[t] in the (e,1) cell. The gist of the pictures is each column represents a column (e,1). The column to the left is NOT (e-1,1), but it is (-f,1). The next to the left of that is the cell with the same a,b but different d, so each column represents a different e for a d value for the same c value. The row/column that are white are the row that has the same x value as the goal cell. For (1,5,12,7,5,29) this would be (1,1) with x=7 (or t=4). In the top half, red means that A[t] is divisible by a, green means A[t] is divisible by b, and cyan means its divisible by c. Then in the bottom half, this is the same but the divisibility of each square. There are very cool patterns in this square part.

 

Looking at this, we can see that the factors of A and B are fairly easy to predict, but the factors of AA, BB, and D*D are these weird patterns. That repeat.

 

Also for each shift in a column the e shifts by 2d+1 changes d by one, so this is not a linear relation but the pattern looks linear. Maybe its a derivative or something?

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 9, 2018, 8:21 p.m. No.6869   πŸ—„οΈ.is πŸ”—kun   >>6876

>>6866

Let you back in?

Can you not log into your account or something?

 

I believe I still see your screen name in the list but if yer not, just follow the instructions I gave teach.

ID: 7ab02a July 10, 2018, 12:05 p.m. No.6872   πŸ—„οΈ.is πŸ”—kun   >>6876

>>6866

You have to log back into your account. If you didn't complete making the account you have to make another. The discord URL is on one of the older threads, just DM ta.io and ask for the vqc server.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 10, 2018, 12:43 p.m. No.6873   πŸ—„οΈ.is πŸ”—kun   >>6874

Update from last night's shadow boxing:

Their, the nerds, mental kung fu seems to have leveled up after meditation.

 

Which you'd kinda think they'd be better at considering sci-fi, neckbeards, and tabletop/MasterRacePC gaming…

 

Maybe I'll help show them how to larPLAY during a lull as a continuance of the visualization walkthrough I started yesterday.

 

Basic gist was (Kindergarten Cop Arnie) "Which are your Trangulata's and what do zey dooo?"

 

Now… The Alpha∞End Grid…

From the side it'd look like a line with a point on it.

Maybe it's that thing we desire. Buuuut…

The Grid. Let's call that our base"line" even though we just jumped to 2D, because puns.

There's a Trangle on it and a mid point but you don't know derp about it cuz you ain't got no references.

 

So let's make the grid an array of 4 grids by defining (0,0) {ooooor (0,1)? moving on.}

 

Seems like we have an idea of what's going on a lil' bit better now.

though… where did we place the common "midpoint" of the trangle?

At (0,0) or (0,1)? I guess if it's a mid"point" it's have to have a value or something, like binary, maybe. It's me, moving on.

 

So let's say we put it at (0,0)…

Wait… did you just assume my midpoint?

What if we go all Neo and push the grid back into space… (0,0) should turn into (0,0,0), with a newer, "bigger yet same" 0 in this new… Matrix.

Since we can imagine The Grid as a Square, we get a Cube, so the midpoint of that.

 

Now… that's fine and dandy, but if our Trangulated Common-Midpoint is (0,1) and we bust into the Third dimension, something happens to the Trangle.

The first time would have had it going dimensionally batshit trying to maintain the common 0 and its relationship to the three other points in this "new" third dimension.

With the TCM at (0,1), and assume this just plays out like you're opening a translucent accordion with a pop-up book style Trangle within it that locks upon full expansion of the Mathcordion.

And while we're here, you pick 3 notes/chords/combinations, or however that works, and upon expansion you can see where they reside in 3space with the revealed position of (0, 1, z-value).

Maybe it's that the chords have (x,y,z) values (cuz chords) and we're decrypting the third note for (0, 1, z) out of that. MOVING ON!

 

Toooo waveforms and frequencies…

Take our 3D The Griddles and turn it into a hypercube. (0, 1, z, βˆ†)

Which makes me wonder if locking our (Trangle) gives us the ability to see in 4space (or wherever I am)… because without it we'd lose (0,0,0,0) in the permutations (however they "operate) of the hypercube.

 

From the Chris-fu lessons last night:

"in a static equilibrium yes but the center is moving".

 

That way, no matter how the Trangle mutates with the permutations of the cube, we aways know that our TCM is always (0,1,,) away from it.

 

Aaaaaand that might kinda be how you move through "time/4D". Maybe.

 

-wanders off-

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 10, 2018, 7:30 p.m. No.6878   πŸ—„οΈ.is πŸ”—kun   >>7237

I don't remember where we were putting Mandelbrot and Pi stuff, but here's a bunch of mandelbrot stuff that someone out in the open waters was aksin' aboot.

 

Had to do with mutating/permutating/transforming a Flower of Life into a Mandelbrot or vice a verso.

Something like that.

 

3d Mandelbrot

https://www.youtube.com/watch?v=kaGcXUIg2qQ

 

Hidden areas in Julia and Mandelbrot sets (3d)

https://www.youtube.com/watch?v=vfteiiTfE0c

 

Times Table Modulo 200 (vid related)

https://www.youtube.com/watch?v=qhbuKbxJsk8

PMA !!y5/EVb5KZI ID: a44f6d July 10, 2018, 11:13 p.m. No.6881   πŸ—„οΈ.is πŸ”—kun   >>6887 >>6900

>>6654

>When does c first appear at a[t]?

>When is the second time it appears?

 

Not sure about the first time yet, but the second time is always in the same (e,n) at t = t + n.

 

Stumbled upon the third time c appears at a[t].

 

Borrowing from QVC's variable style >>6736, the following formulas represent the move to the third occurrence for odd e only, starting from the 1,c entry record:

 

e' = e

n' = 2x + n + 2

x' = 3x + 2n + 2

a' = b

d' = a' + x'

b' = a' + 2x' + 2n'

 

t' = (c+1)/2 + t

 

Examples:

 

entry

(1,61,6) = {1:61:12:11:1:145} = 145

 

2nd a[t]=c

(1,61,67) = {1:61:278:133:145:533}

 

3rd a[t]=c

(1,85,79) = {1:85:302:157:145:629}

 

n' = 2*11 + 61 + 2 = 85

 

entry

(31,128,8) = {31:128:16:15:1:287} = 287

 

2nd a[t]=c

(31,128,136) = {31:128:558:271:287:1085} = 311395 (second occurrence)

 

3rd a[t]=c

(31,160,152) = {31:160:590:303:287:1213} = 348131 (third occurrence)

 

n' = 2*15 + 128 + 2 = 160

 

Not sure if this is useful other than to show an example of how n can be calculated in terms of x and n.

PrimeAnon !!!ZmMwNmUyMmUyMzY5 ID: a2c3ea July 11, 2018, 12:52 a.m. No.6883   πŸ—„οΈ.is πŸ”—kun

I was looking over that wave equation originally posted in general #2β€”without knowing all of the functions, you can still get some idea of how everything is working. The key is what followed the equation:

>as |fi| approaches ∞, acos(((1-(fi/(fi-1))).5))5)/3) approaches Ο€, and, of course, acos(((1-(fi/(fi-1))).5))5)/3)-Ο€ then approaches 0"

 

Once you substitute Ο€ for the 'acos(…)" equation, it renders down into a pretty basic trig function which is in turn modified by other functions whose effects (if not methods) are pretty familiar. For instance, the 'l' function will affect the amplitude of the wave, and the 'k*' part will affect the phase. The 'j' function seems to affect the period, but I wouldn't know without more research. The 'j' and 'l' functions are time varying as well.

 

It's worth noting that if the 'acos(…)' function tends toward Ο€, with its ratio of the i'th iteration of the Fibonacci sequence over the i'th iteration of the Fibonacci sequence - 1, there is a similar function just to the right of it that is getting closer to Ο€ as well–just at a much slower rate. But the difference between them, even after only 8 iterations, is relatively small.

 

I made a graph on Desmos with some dummy variables for j, k, and l; one of them is set to the 8th iteration, the other one tending toward a large number of iterations at a point where the left-side equation is about 500x closer than the right toward infinity.

 

So it's four series/summations running in tandem, some of them functions dealing with time and other factors, and as those numbers grow larger it gets closer and closer to what I would expect is as good of an approximation as anyone might ever need–kind of like going from an 8-bit image, to 16, 32, 64, etc. I'm pretty sure anybody familiar with quantum mechanics at that point would have decided it was time to listen closely.

 

Here's the link:

https://www.desmos.com/calculator/3wkhcwllmx

Anonymous ID: a332a2 July 11, 2018, 2:28 a.m. No.6887   πŸ—„οΈ.is πŸ”—kun   >>6888

>>6881

I'm looking a bit into this as well, and while my results aren't great. I did find something that appears as a minor pattern, maybe?

 

I've only tested this on a few cells, but from what I've seen it is interesting.

 

Take a=7, b=37, c=259. The first occurrence of 259 in (e, 1) is at t=101. a = 20202, but note a % 4 = 2. We need to divide 20202 by 2, which gives us 10101.

 

So why is this interesting?

 

10101/259 = 39. And the cell for a=39, b=259 is:

{101:49:100:61:39:259}

 

Let's try this again with another cell like a=19, b=61, c=1159.

The first occurrence in (3, 1) appears at t = 502 =(3, 1, 504009, 1003, 503006, 505014)

503006 % 4 = 2 =We need to divide by two => 251503 which mod 4 = 3.

 

Then we do 251503 / 1159 = 217. We get the cell for a=217, b=1159 and lo and behold, we get:

(502, 187, 501, 284, 217, 1159).

 

What I've seen so far is that the e will always be equal to the t where c first occurs. The d will always be equal to e - 1 (or e is equal to d + 1?).

 

I've only run this through a few of the (3, 6) numbers, but so far it checks out.

 

The first occurrence of c in (e, 1) will have a corresponding cell with b=c at (t, k) for some k. I haven't locked down any patterns relating the cells, except for the e and b. So far I'm unable to "predict" where it would be given just a c.

 

And it also appears that the a at a[t] where a is the first occurrence of c, a % 4 == 2. Again, I've only tested cells from (3, 6), but so far it looks promising.

Anonymous ID: a332a2 July 11, 2018, 2:43 a.m. No.6888   πŸ—„οΈ.is πŸ”—kun

>>6887

I might have jumped the gun a bit fast. I is an interesting pattern, but does not hold for (1, 5) or (27, 6). That is, the e=t does not hold. There might be some differences between odd n and even n? Maybe the parity of e also affects such a pattern, I'm not entirely sure yet.

 

I'm still looking into it though. I'll post any new observations.

Anonymous ID: a332a2 July 11, 2018, 3:33 a.m. No.6889   πŸ—„οΈ.is πŸ”—kun

Also >>6735

 

>Because the RoD approach has Big Oh the same as the square root function, it doesn't matter how big they are, they would be insecure and too impractical to use.

 

RoD? Recursion on D?

AA !dTGY7OMD/g ID: 7ddfda July 11, 2018, 7:25 a.m. No.6890   πŸ—„οΈ.is πŸ”—kun

So obviously he's not doing what he said he would again. I've been having a look at the RoD thing. Over the next few posts, I'm going to post every cell containing a and b values that multiply together to produce a given c value for the sequence of c values starting from RSA2048 replacing c with d recursively until reaching the lowest possible cell (obviously excluding RSA2048, since we don't know its a and b). It's a lot of numbers, so give me a bit. He said look at big semiprimes to see the pattern, so why not use the biggest semiprime we’ve dealt with so far?

AA !dTGY7OMD/g ID: 7ddfda July 11, 2018, 7:27 a.m. No.6891   πŸ—„οΈ.is πŸ”—kun   >>6898

This is the sequence of c values where each time we replace c with d (excluding RSA2048 itself):

 

158732191050391204174482508661063007579358463444809715795726627753579970080749948404278643259568101132671402056190021464753419480472816840646168575222628934671405739213477439533870489791038973166834068736234020361664820266987726919453356824138007381985796493621233035112849373047484148339095287142097834807844, 12598896421924866518220704506279920262503827873668266660431014021495031828185862096976477460760574959532030199266536614252274764167453116999041838483272614, 112244805768128382102848738141332806509123753930790945631219843869937264197591, 335029559543823509028205969685172744250, 18303812705112110303, 4278295537, 65408, 255, 15, 3, 1, 1

AA !dTGY7OMD/g ID: 7ddfda July 11, 2018, 7:52 a.m. No.6892   πŸ—„οΈ.is πŸ”—kun   >>6898

Let's start at the bottom because I'm waiting for this webtool to factor the second d (I somehow doubt I'll get to the first d, but who knows). To begin with, I accidentally added an extra 1. The lowest cell we find is c = 1. Above that, c = 3, then above that c = 15, then c = 255, then c = 65408, then c = 4278295537, then c = 18303812705112110303, then c = 335029559543823509028205969685172744250, then c = 112244805768128382102848738141332806509123753930790945631219843869937264197591, then c = 12598896421924866518220704506279920262503827873668266660431014021495031828185862096976477460760574959532030199266536614252274764167453116999041838483272614 (and I'm probably not going to do the original d since it's quite large, and then the original c we want to factor is obviously too large itself). I'll do 65408 etc in the next post because that one is even so it has a lot of factors.

 

c = 1, factors are 1

(0, 0, 1) = {0:0:1:0:1:1} f = -3

 

c = 3, factors are 1 and 3

(3, 2, 0) = {3:2:0:-1:1:3}, f = 4 (which has a negative x value and t = 0, so it doesn't exist)

 

c = 15, factors are 1, 3, 5 and 15

(6, 5, 2) = {6:5:3:2:1:15}, f = -1

(6, 1, 1) = {6:1:3:0:3:5}, f = -1

 

c = 255, factors are 1, 3, 5, 15, 17, 51, 85 and 255

(30, 113, 8) = {30:113:15:14:1:255}, f = -1

(30, 29, 7) = {30:29:15:12:3:85}, f = -1

(30, 13, 6) = {30:13:15:10:5:51}, f = -1

(30, 1, 1) = {30:1:15:0:15:17}, f = 1

AA !dTGY7OMD/g ID: 3e05dd July 11, 2018, 8:08 a.m. No.6893   πŸ—„οΈ.is πŸ”—kun   >>6894

c = 65408, factors are 1, 2, 4, 7, 8, 14, 16, 28, 32, 56, 64, 73, 112, 128, 146, 224, 292, 448, 511, 584, 896, 1022, 1168, 2044, 2336, 4088, 4672, 8176, 9344, 16352, 32704 and 65408

(383, 32449, 127) = {383:32449:255:254:1:65408}, f = -126

(383, 16098, 127) = {383:16098:255:253:2:32704}, f = -126

(383, 7923, 126) = {383:7923:255:251:4:16352}, f = -126

(383, 4420, 124) = {383:4420:255:248:7:9344}, f = -126

(383, 3837, 124) = {383:3837:255:247:8:8176}, f = -126

(383, 2088, 121) = {383:2088:255:241:14:4672}, f = -126

(383, 1797, 120) = {383:1797:255:239:16:4088}, f = -126

(383, 927, 114) = {383:927:255:227:28:2336}, f = -126

(383, 783, 112) = {383:783:255:223:32:2044}, f = -126

(383, 357, 100) = {383:357:255:199:56:1168}, f = -126

(383, 288, 96) = {383:288:255:191:64:1022}, f = -126

(383, 229, 91) = {383:229:255:182:73:896}, f = -126

(383, 93, 72) = {383:93:255:143:112:584}, f = -126

(383, 64, 64) = {383:64:255:127:128:511}, f = -126

(383, 42, 55) = {383:42:255:109:146:448}, f = -126

(383, 3, 16) = {383:3:255:31:224:292}, f = -126

 

c = 4278295537, factors are 1, 4561, 938017 and 4278295537

(89073, 2139082361, 32704) = {89073:2139082361:65408:65407:1:4278295537}, f = -41742

(89073, 405881, 30424) = {89073:405881:65408:60847:4561:938017}, f = -41742

 

c = 18303812705112110303, factors are 1, 73, 229, 16717, 1094922097572059, 79929313122760307, 250737160344001511 and 18303812705112110303

(3197991934, 9151906348277759615, 2139147769) = {3197991934:9151906348277759615:4278295537:4278295536:1:18303812705112110303}, f = -5358599139

(3197991934, 125368575893705255, 2139147733) = {3197991934:125368575893705255:4278295537:4278295464:73:250737160344001511}, f = -5358599139

(3197991934, 39964652283084731, 2139147655) = {3197991934:39964652283084731:4278295537:4278295308:229:79929313122760307}, f = -5358599139

(3197991934, 547456770498851, 2139139411) = {3197991934:547456770498851:4278295537:4278278820:16717:1094922097572059}, f = -5358599139

Hobo !!1yNgQ3NlCs ID: 93dfa0 July 11, 2018, 2:15 p.m. No.6896   πŸ—„οΈ.is πŸ”—kun   >>6897 >>6899

>>6894

You are my hero AA. Get some rest man!

So I have been thinking about this distribution problem and I see it like this.

Assumptions:

  1. Chris is not working with/for Q. He is independent but still on the side of the "good guys".

  2. He does not know the best way to distribute this thing without hurting (and helping) a hell of a lot of people.

  3. He is certain that the release will do more good than bad overall. (I deeply agree) and wants us to figure the details out.

The worse-case scenario is that the "Bad guys" (Think Hillary Clinton/Podesta/Pizzagate-tier bad guys) capture this tool and use it to crush what Trump is doing to fix the US and the world. If this thing is in the hands of just a few bad guys this is very possible, likely even.

Best case scenario is that we drop this thing in the right way. In a positive way that helps more than hurts people and the good guys use it to clear many of the bad things out of the system.

My guess is we will be somewhere in-between. See the movie β€œSneakers” if you want a fun example of this exact problem. My voice is my passport, Verify Me!

The obvious target here is BTC. When I mentioned a one-click tool the reason I did was because this is easy to distribute to many people and would he hard hitting and attention getting. Most importantly though it could easily include the raw source code as well as the source files from this place compiled in text format so someone with a large enough brain could reproduce the work all in one package. It would be OUT. Think about the old keygen’s that were common in the software piracy days of the late 90s and early 0’s except you input a public key, hit generate and a private key pops out.

This would protect all of us from the men with the dead eyes, silenced pistols and blue gloves, which is very real risk going forward. Also it would make sure that no one could lock it up, which if you look into so many of these amazing inventions that have never come out is a very likely possibility that could happen. In fact, I think that is still the most likely possibility at this point. There will soon come a critical time where we must distribute far and wide or die trying. To simply squash BTC like a bug and post the code around may wipe cryptos from the face of the earth but that is just a tiny fraction of what is possible. A gnat buzzing around the ear of the elephant. What will be built and discovered with this tech is what really matters. THAT is what we are fighting for here weather we know it or not. Cryptos are simply a way to get there.

AA mentioned just making a script that rips all the bitcoin wallets to zero and sends the coins all to one wallet, essentially destroying the chain. Chris mentioned updating Wikipedia. I think the Wikipidia idea is VERY dangerous to alone. Definitely a way to get wacked in a hot-minute. It solves nothing at all related to distribution and puts up a gigantic bullseye on us all. I think that is the right move AFTER WE HAVE DISTRIBUTED THE CODE FAR AND WIDE. Wikipedia would be a good step-3 right there with Topol's Media release. Step 1 and step 2 though need to be more subtle and broader in scale. Step 1: This thing needs to be sent to people who UNDERSTAND what the hell we are doing here. This is tough for me because I barely understand what we are doing here but that is what needs to happen first. This means crypto-anarchists and netsecurity guys/math nerds at the academic level. Very shortly thereafter we need to use other broad distribution methods such as the BTC cracker I mentioned. Something that will be shared and shared far and wide. Something everyone would want and everyone would want to share. Bit Torrent as well as stuff like 'chan boards and social media. Lastly we drop the hammer and update Wikipedia/post all over social media.

I think a source of β€œPeople who understand this stuff” would be in the Net Security community as well is the crypto currency community. This could be done with a white paper. Then the BTC cracker, then the academic-tier Wikipedia release and official release to the media as Topol suggested. That is what I think is the best direction to go. What do you guys think?

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 11, 2018, 6:47 p.m. No.6897   πŸ—„οΈ.is πŸ”—kun

>>6896

  1. We already know that there's a mini VQC team. It's at least 2 or 3 people.

Beyond that, timing. Chris doesn't time things willy nilly. Can't get ahead of the plan unless organically.

 

  1. Yes he does.

You're looking at it.

 

  1. Trust the Plan.

This part of the Plan. The OVER ALL Plane.

Cicada -Q -> VQC -> Great Awakening Begins (When we finish, but we reeeeeally don't have much longer according to "the whispers".)

 

Release:

Wikipedia

White Hats do what they were planning to do with this the entire time.

(Chris is just teaching multiple facets of the situation.)

Then if I need to, I talk to people and y'all maintain your privacy.

 

This means no more BTC and Assange's PGP signature gets used to unlock everything that key goes to.

 

But really… everything's fine.

Y'all calms ya tiddies.

AA !dTGY7OMD/g ID: df77d5 July 11, 2018, 7:07 p.m. No.6898   πŸ—„οΈ.is πŸ”—kun

>>6894

There are honestly probably over a hundred factors of c = 335029559543823509028205969685172744250. This is not a useful way for me to spend my time.

 

>>6891

>>6892

My point in all of this was to show that if we are meant to use d as c recursively (which got the thumbs up from VQC, so we probably are meant to), we can create a path from the lowest possible c in this sequence upwards, but there are differing numbers of (e,n,t) cells for each one. I think it would be very useful to find where d[t]-d = b(n-1), where BigN * c pops up in (e,1), and the triangle bases for each of the cells I've posted, and see if there's some way in which we can use one of the cells one lower to generate the next one up. That is the point of recursion, after all.

AA !dTGY7OMD/g ID: df77d5 July 11, 2018, 7:19 p.m. No.6899   πŸ—„οΈ.is πŸ”—kun

>>6896

I have a feeling he's holding it back because of the Wikileaks cables. There was someone on 4chan last year at some point claiming to be from Wikileaks saying that the huge US-wide internet blackout was caused by international powers trying to stop Julian Assange's dead man's switch. This person was on the run trying to get to a place where they could release it themselves, and they said Wikileaks was now compromised. This is around when weird shit started happening to do with Wikileaks. They didn't say this, but putting it together in hindsight with what we've learned from Q, it would have possibly meant releasing information that would have given everyone what we now refer to as the 40,000ft view. It would have crashed economies and started global conflicts, apparently. If DJT and Q really know what they're doing, this giant international campaign to stop JA's dead man's switch could be considered a good thing. Q et al are trying to get this information out gradually, because if it all came out at once without the media changing its tune and without key players having their power taken away, the world would turn to chaos. So think about it: is the stage set? Because once we figure this out, we can read those Wikileaks cables.

 

That's why I think VQC is stalling for a good reason without saying it out loud, rather than just being a super unreliable person (if you take everything he says completely literally). VQC did say he would give us time but keep up with the happenings and release it when the time is right. Once we solve the VQC puzzle, depending on the contents of these cables, the major worldwide events we've been waiting for are unstoppable. So the world needs to be ready. I think that might be fairly soon, personally, considering the huge growth in the Q thing publicly, as well as that stuff with the server.

PMA !!y5/EVb5KZI ID: a44f6d July 11, 2018, 9:24 p.m. No.6900   πŸ—„οΈ.is πŸ”—kun

>>6881

Figured out the indexing problem for even e to find the 3rd occurrence of a[t] = c, starting from the original c entry record.

 

Formula for n stays the same:

 

n' = n + 2x + 2 .

 

t shortcuts are as follows:

 

odd e: t' = n' - t

even e: t' = n' + 1 - t

 

The attached pic shows the starting c record and the 2nd and 3rd occurrences for a number of test cases.

 

The remaining columns show the u values from the 2nd a[t] and 3rd a[t], their difference, and a number of ways to represent that difference.

 

u' = u + 2d'

u' = u + 2(x'+1)

 

or, without going back to the starting record at all:

 

u' = u + 2*(sqrt(a''))

 

Where ' is the 3rd record, is the 2nd record, and ' is the starting record. (Please excuse the notation).

VA !!Nf9AmQNR7I ID: 3a32d9 July 11, 2018, 9:36 p.m. No.6901   πŸ—„οΈ.is πŸ”—kun

Ok. Here’s the idea I'm working on:

At RSA sized numbers, movements between elements are still dictated by the earliest elements of (e,1)

And their ascending factors

So what is the average distance between a and b and c at an RSA sized number ??

It’s relative

But based on observable patterns

Multiples reduce the infinite search to a limited number of possibilities

Which we can create at will using the known rules

We can create any element surrounding (na transform) for any c.

We can search those elements surrounding for Multiples

Such as an, bn, etc.

My theory is that they should still be relatively close together.(edited)

VA !!Nf9AmQNR7I ID: 3a32d9 July 11, 2018, 9:41 p.m. No.6902   πŸ—„οΈ.is πŸ”—kun

Searching the factor tree in (e,1) is the main idea.

Constructing only what we need

Moving up and down the factor tree

οΏΌοΏΌ

At 100000000000 numbers, how can (bn) and b(n-1) still be one t value apart ???????

WTF

Instead, Iteration by Creation.

Creation of each element we need, and nothing more.

Lean program.

Killer.

Start at c, then (1,c), then (na transform) then -f na transform.

 

The reason why the element movements surrounding (na transform) are key is because they hold many factors.

We have a clear blueprint of the pattern of factors we are looking for

Therefore we search the patterns of factors surrounding na transform.

(bn) and b(n-1) are only one x value apart.(edited)

We could base a search just on that.

But we have to start at (1,c)

Then na transform

Then (-f,1)

Then crate the elements surrounding (e,1) and (-f,1)

In an expanding quadrilateral

With bn being in the lower right corner of that shape

If no match, then expand the quadrilateral

 

We don’t even need the Grid. We just need the rules of adjacent element Creation FROM the Grid. That’s gonna be a badass algorithm!!!!!οΏΌοΏΌοΏΌ

You guys agree?? Is it possible??

SHIT!! It’s so beautiful!

Adapting the Grid rules for element Creation into a group of strings, one for each element.

Expanding out from the na transform

Looking for the lock

On an, ab, a(n-1), and b(n-1)(edited)

An expanding square with an offset of 1 in the bottom right corner

Which is bn

Shape is quadrilateral

But not a trapezoid

I love you @everyone ️

This is so FUN!! οΏΌ

Fuck, I love Math(s)

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 11, 2018, 9:51 p.m. No.6904   πŸ—„οΈ.is πŸ”—kun

>>6903

I was honestly kind of surprised as to how many people didn't notice that until it was so overtly pointed out.

Then again, part of my journey was r/T_D, so I rarely miss a Ben Garrison.

AA !dTGY7OMD/g ID: b4c9e6 July 11, 2018, 11:41 p.m. No.6905   πŸ—„οΈ.is πŸ”—kun   >>6906 >>6907

Has anyone found anything to do with this clue?

>where the product of BigN and c are found in (e,1)

 

I'm using a=7, b=29, c=203 in this example. I'll do another example soon.

 

(7,4,4) = {7:4:14:7:7:29} f = -22

(7,88,7) = {7:88:14:13:1:203} f = -22

Product of BigN and c = 88*203 = 17864

Here’s (e,1) for this cell:

(7,1,1) = {7:1:5:1:4:8} f = -4

(7,1,2) = {7:1:11:3:8:16} f = -16

(7,1,3) = {7:1:21:5:16:28} f = -36

(7,1,4) = {7:1:35:7:28:44} f = -64

(7,1,5) = {7:1:53:9:44:64} f = -100

(7,1,6) = {7:1:75:11:64:88} f = -144

(7,1,7) = {7:1:101:13:88:116} f = -196

(7,1,8) = {7:1:131:15:116:148} f = -256

(7,1,9) = {7:1:165:17:148:184} f = -324

(7,1,10) = {7:1:203:19:184:224} f = -400

(7,1,11) = {7:1:245:21:224:268} f = -484

(7,1,12) = {7:1:291:23:268:316} f = -576

88 appears as b in (7,1,6) and as a in (7,1,7). 203 appears as d in (7,1,10). I don't know if 17864 itself appears though.

AA !dTGY7OMD/g ID: b4c9e6 July 11, 2018, 11:56 p.m. No.6906   πŸ—„οΈ.is πŸ”—kun   >>6907

>>6905

a=13, b=43, c=559

(30,5,6) = {30:5:23:10:13:43} f = -17

(30,257,12) = {30:257:23:22:1:559} f = -17

Product of BigN and c = 257*559 = 143663

Here’s (e,1) for this cell:

(30,1,1) = {30:1:15:0:15:17} f = -1

(30,1,2) = {30:1:19:2:17:23} f = -9

(30,1,3) = {30:1:27:4:23:33} f = -25

(30,1,4) = {30:1:39:6:33:47} f = -49

(30,1,5) = {30:1:55:8:47:65} f = -81

(30,1,6) = {30:1:75:10:65:87} f = -121

(30,1,7) = {30:1:99:12:87:113} f = -169

(30,1,8) = {30:1:127:14:113:143} f = -225

(30,1,9) = {30:1:159:16:143:177} f = -289

(30,1,10) = {30:1:195:18:177:215} f = -361

(30,1,11) = {30:1:235:20:215:257} f = -441

(30,1,12) = {30:1:279:22:257:303} f = -529

…

(30,1,17) = {30:1:559:32:527:593} f = -1089

257 appears as b in (30,1,11) and as a in (30,1,12). 559 appears as d in (30,1,17). This does appear to be a pattern. I’m going to quickly put together a spreadsheet to extend these into further ts and see if I can actually find the product of BigN and c as the product, rather than the two separate variables.

AA !dTGY7OMD/g ID: b4c9e6 July 12, 2018, 12:16 a.m. No.6907   πŸ—„οΈ.is πŸ”—kun   >>6909

Success with the product of BigN and c in (e,1)

 

>>6905

Here are some relevant cells for a=7, b=29, c=203

(7,1,94) = {7:1:17675:187:17488:17864} f = βˆ’35344

(7,1,95) = {7:1:18053:189:17864:18244} f = βˆ’36100

So for this one, the product of BigN (88) and c (203), which is 17864, does appear in (e,1). It appears as b at (7,1,94) and as a in (7,1,95). We can also say based on the algebra of x=d-a and the rules in (e,1) that a[t]=b[t-1] that BigN*c = d[94]+x[95] = d[95]-x[95]. So the increase in d from the first of these cells to the next is equal to 2x. These rules also apply to the example below.

 

>>6906

Here are some relevant cells for a=13, b=43, c=559

(30,1,268) = {30:1:143127:534:142593:143663} f = βˆ’286225

(30,1,269) = {30:1:144199:536:143663:144737} f = βˆ’288369

So for this one, the product of BigN (257) and c (559), which is 143663, does appear in (e,1). It appears as b at (30,1,268) and as a in (30,1,269). As I said, same rules with d[t]+x[t+1] and d[t+1]-x[t+1] apply.

AA !dTGY7OMD/g ID: 520224 July 12, 2018, 1:57 a.m. No.6909   πŸ—„οΈ.is πŸ”—kun

>>6907

People have been asking on /qr/ and Discord, so just FYI, this image wasn't anything important. When you paste something from the clipboard, sometimes 8ch will create an image containing the thing you pasted. It was just an unformatted version of one of those cells.

Anonymous ID: a332a2 July 12, 2018, 6:41 a.m. No.6916   πŸ—„οΈ.is πŸ”—kun   >>6918

Okay, I'll bite.

 

>>6742 >>6739

Checks out. It will generate the next cell in the the sequence (or chain as I've referred to them before).

 

>>6746

INFO[0000] Starting [Big]VQC program

 

INFO[0000] E: 149730827186590819409161975355863102499674499386960841184873248110419525705142414412560340457818822465695973580080183089801154577153055685206470161899420076208154943348665238647007714095089342669905666884517354400122833485897064098153958317993833609635063079613328500485188782423277886515290863800949716792021

 

INFO[0000] N: 1259795423782894674701359162002419928571464106310201601388856891802183101035379777813200926294039220345914532062475754109464927957458809225140424456003642249634369640364388836798570917363513094818750748591234558253880668992954785004866522987440421420089871455032122934590859755937306075758632731614110843499361585902016642742536808842406995878081985323014757750816464333936471244183139065067762886207783712

4697517250971910788373374647441598985076306038901094587160385923033013611772476591818466822702222519806172328865872316642926190459725502720591581076567956723729060778728578126316270798047266909911419313225552335

 

INFO[0000] D: 158732191050391204174482508661063007579358463444809715795726627753579970080749948404278643259568101132671402056190021464753419480472816840646168575222628934671405739213477439533870489791038973166834068736234020361664820266987726919453356824138007381985796493621233035112849373047484148339095287142097834807844

 

INFO[0000] X: 1587321910503912041744825086610630075793584634448097157957266277535799700807499484042786432595681011326714020561900214647534194804728168406461685752226289346714057392134774395338704897910389731668

34068736234020361664820266987726919453356824138007381985796493621233035112849373047484148339095287142097834807843

 

INFO[0000] A: 1

 

INFO[0000] B: 2519590847565789349402718324004839857142928212620403202777713783604366202070759555626401852588078440691829064124951508218929855914917618450280848912007284499268739280728777673597141834727026189637501497182469116507761337985909570009733045974880842840179742910064245869181719511874612151517265463228221686998754918242243363725908514186546204357679842338718477444792073993423658482382428119816381501067481045

1660377306056201619676256133844143603833904414952634432190114657544454178424020924616515723350778707749817125772467962926386356373289912154831438167899885040445364023527381951378636564391212010397122822120720357

 

>>6753

Appears to check out.

Anonymous ID: a332a2 July 12, 2018, 9:04 a.m. No.6917   πŸ—„οΈ.is πŸ”—kun

Adding again, but this time I'm adding the following:

 

(1, c)

(1, cN)

(-f, c)

(-f, c(N-1))

 

If anyone can double-check them, that would be appreciated.

 

Added in pastebin due to size of text: https://pastebin.com/6jn9Dmni

VA !!Nf9AmQNR7I ID: c7075d July 12, 2018, 5:39 p.m. No.6918   πŸ—„οΈ.is πŸ”—kun   >>6919

>>6916

Nice work!

Here's my working idea for order of operations.

 

Start at c (of course!)

Create (1,c)

Create na transform records for (e,1) and (-f,1)

Then begin expanding outward from the na transform records.

We already have a starting t value from the na transforms

We know an and a(n-1) have a lower t value

We know bn and b(n-1) have a larger t value

We also know that at the correct t values, we can move from an to bn, and they are [t+n] elements apart

We also know that at the correct t value, a(n-1) and b(n-1) are [t+n-1] elements apart

So we β€œiterate” (not really, because we’re moving by multiples and factors), looking for the 4 elements that match and give us our an, bn, a(n-1), and b(n-1) values.

It’s definitely an o(log t) way to search

Thoughts, Anons?

GAnon !jPVzzZOz2c ID: 28ac3a July 12, 2018, 9:11 p.m. No.6920   πŸ—„οΈ.is πŸ”—kun   >>6921 >>6926

So I'm looking at the A[t] divisibilities for (e,1) and related e columns (namely the e's for various d's). So here is a gif with the divisibilities for (1,c) for c=1,3,5,…,149

 

Its symettrical. Also you'll notice those little 3x2 boxes. They switch axes. It starts they switch after 2 numbers, then they switch after 4 numbers twice, then they switch after 6 numbers twice, then after 8 numbers twice etc. Basically each row is ABD(1,c,d) for various d's. (the original D is in the center of the pic). Also this just looks dank.

 

So the colors are yellow if its divisible by c and black if its divisible by a (a=1 in all these cases).

VA !!Nf9AmQNR7I ID: c7075d July 12, 2018, 9:35 p.m. No.6928   πŸ—„οΈ.is πŸ”—kun

>>6919

and looking for the an, ab, a(n-1), and b(n-1) records. They occurs in a semi predictable pattern surrounding (na transform).

 

We're searching the factor tree now, niggaz.

All that "iterate" stuff pales in comparison.

PMA !!y5/EVb5KZI ID: a44f6d July 12, 2018, 10:11 p.m. No.6929   πŸ—„οΈ.is πŸ”—kun   >>6945

>>6619

>>6643

>>6858

Continuing with the gcd approach from the previous thread, and based a little bit on VQC's function idea, pics attached represent a comparison matrix between a[t] and a[p+1-t] records in both (-f,1) and (e,1).

 

Examples attached are for c145, c297, c6107.

 

Starting from easy to calculate records (starting c, na, nb, nc, 2nd a[t]=c, 3rd a[t]=c) the comparison shows for both -f and e columns:

 

1) The record at t=p+1-t based on each factor (p).

2) the gcd(a[t], a[p+1-t]) result.

3) the different between t values.

4) the gcd(c, t diff) - noticed that sometimes this gives the correct factor.

 

Records are displayed just showing (e,n,t) index, and the relevant a value used in the calculations (just to save space).

 

The gcd(-f,e) column at the end is the result of gcd( (-f, 1) a[t], (e,1) a[t] ), and records are filtered to where any of the gcd results matches p.

 

Given that these comparisons are against offsets of known factor values (p), there are a couple of things that stand out:

 

a) The gcd solution sometimes comes from the (-f,1) column and sometimes from (e,1).

b) For c6107, the factor shows up when comparing (e,n) a[t] instead of (e,1) a[t].

c) For c145, funny enough, the factor shows up in the very first gcd calculation between the na record an it's -x offset.

 

Think the point of this post is to perhaps illustrate that we have almost enough records to solve the problem.

AA !dTGY7OMD/g ID: adf860 July 13, 2018, 5:54 a.m. No.6930   πŸ—„οΈ.is πŸ”—kun   >>6950

Here's a tired thought off the top of my head that I've applied no context to: e and -f in n=1 are 2d+1 apart. This is equivalent to 2d(n-0) + nn from the (x+n)(x+n), since n=1. We're finding (e,1) and (-f,1) based on a set d, e and f value with an assumed n. So what if we instead assumed n=2, and we found (e,2) and (-f,2) and so on upwards? Would this relationship hold (2d(n-0) + nn gap between (e,2) and (-f,2))?

Anonymous ID: 6a2b09 July 13, 2018, 8:36 a.m. No.6932   πŸ—„οΈ.is πŸ”—kun

>>6931

I've been looking more into this, but I can't help but feel a bit stuck.

 

I feel a bit bad. I'm sure VQC is sitting here watching us while ripping his hair out because the solution is right in front of us.

VA !!Nf9AmQNR7I ID: daac05 July 13, 2018, 10:31 a.m. No.6933   πŸ—„οΈ.is πŸ”—kun   >>6934

>>6931

Hello Anon! The asymmetry is that (-f,1) contains factors on n-1, and (e,1) contains factors of n.

 

Check out the a values in the image I'm attaching. The upper highlighted bar has a values that are 5(5-1) = 20 = a(n-1) and 5(5) = 25 = a * n

The lower highlighted bar contains a values that are 29(5-1) = 116 = b(n-1) and 29(5)=145 = bn

Anonymous ID: a332a2 July 14, 2018, 12:17 a.m. No.6935   πŸ—„οΈ.is πŸ”—kun   >>6940

>>6934

> That n-1 is a factor of col -f and n is a factor of e will show you a shortcut to the triangle solution.

 

We're still doing triangular numbers, but now we're looking for a shortcut with -f and e.

 

When e is odd and -f is even, then every a-value in -f is a square multiplied by two and subtracted f/2. Every e is a triangle multiplied by two and added e/2.

 

Equation for even -f: 2tt - f/2

Equation for odd e: 2t(t+1) + (e/2)

 

But how this is leading us to a shortcut I'm still not sure.

 

We know we have n and n-1. We know for example that n(n-1)/2 is a triangle and is also the sum up to n-1. I was thinking that maybe we can find this n(n-1)/2 triangle somewhere either in e or maybe column 0, which should allow us to do some kind of shortcut. Not sure yet though.

Anonymous ID: a332a2 July 14, 2018, 12:25 a.m. No.6939   πŸ—„οΈ.is πŸ”—kun

>>6938

I guess we'll see. Worst case we will just keep at it. By the sounds of VQC we've gotten close (although close doesn't necessarily mean close to a solution, but rather closer to a solution).

 

Either way I'm not go anytime soon.

Anonymous ID: a332a2 July 14, 2018, 1:57 a.m. No.6941   πŸ—„οΈ.is πŸ”—kun   >>6954

>>6940

We have our equation nn + 2d(n-1) + f - 1 which is nice and dandy.

 

For -f when our a, b is in (-f, n-1) the equation still holds, since we are having -30 it turns into +30

 

(n-1)(n-1) + 2d'(n-1) + f - 1 where f is now defined as 2*d + 1 -(-f)) turning it into a + f.

 

For example, take a=7, b=37

 

For (3, 6) the equation will be: 66 + 216*5 + 30 - 1

While for (-30, 5) it will be: 55 + 217*4 + 65 - 1

 

We get 65 by doing 2*17 + 1 - (-30).

 

Essentially by moving to f, we move another part of 2d(n-1) into f.

AA !dTGY7OMD/g ID: 7d2b9b July 14, 2018, 2:01 a.m. No.6942   πŸ—„οΈ.is πŸ”—kun   >>6943 >>6949

Working on what VQC said he was going to explain to us seemed to be enough for him to taunt us with his presence yesterday, so I'll continue.

 

>>6774

>Then we will look at the key that is made by column -f with the locations of c in (-f,1)

I'll start with the same two examples from my last posts, c=203 (729) and c=559 (1343). I'll just do the first in this post and probably do the other in the next post.

 

c=203, d=14, e=7, f=-22

Relevant n=1 records are (7,1) and (-22,1).

559 doesn't seem to come up as a, b or d in (-22,1), nor does it appear to come from some kind of equation between different variables. So I thought the only way I would see it explicitly come up was by extending the records until x=203 or t=203. x is always even for (-22, 1), so c only appears in relation to x as x+n or x-n (since n=1).

 

(-22, 1, 101) = {-22:1:20593:202:20391:20797} f=41163

(-22, 1, 102) = {-22:1:21001:204:20797:21207} f=41979

Blindly looking for c here, the difference between d values here is 408, which is (203+1)2, the difference between a values is 406, which is just 2032, and the difference in f values here is 816, which is (203+1)*4. So I guess you can technically find c here, when x is adjacent to c.

 

The only other way you can explicitly find c in (-f,1) is when t=c.

(-22, 1, 203) = {-22:1:82813:406:82407:83221} f=165603

Here also, we have 406 pop up again, which is 203*2.

 

So, yeah, c does kinda come up in (-f,1), as 2c when you make t=c. I'll have to see about the other one. I also don't see how this could be useful at all, but he mentioned it, so I guess I'll keep looking.

AA !dTGY7OMD/g ID: 7d2b9b July 14, 2018, 2:14 a.m. No.6943   πŸ—„οΈ.is πŸ”—kun   >>6946 >>6949

>>6942

So then for c=559 (13*43), d=23, e=30, f=-17.

Again, 559 doesn't appear as an a, b or d value, nor as an equation of the variables (that I can find). a, b and d stay the same parity, too, so considering 559 is odd, which through addition or subtraction would change the parity of an incrementing number, it isn't going to appear as the gap between successive values either. So, I'll do the same thing: x=c, t=c. x is odd this time, so 559 does appear as an x value.

 

x=c

(-17, 1, 280) = {-17:1:156791:559:156232:157352} f=313564

The gap between a and b here is 1120, which is (559+1)*2. I can't really see anywhere else that I could get 559 from this record.

 

t=c

(-17, 1, 559) = {-17:1:624953:1117:623638:626072} f=1249888

The only thing close to being c here is 1117, which is actually one less than 559*2. But that isn't consistent with the other example in my previous post.

 

In conclusion, I have no idea what VQC was talking about when he said

>Then we will look at the key that is made by column -f with the locations of c in (-f,1)

because the only definite location of c in (-f,1) is when t=c, and the relationship between the variables at that record and our c is inconsistent.

PMA !!y5/EVb5KZI ID: a44f6d July 14, 2018, 9:13 a.m. No.6945   πŸ—„οΈ.is πŸ”—kun   >>6949 >>6972

>>6929

Found and fixed a bug.

 

The negative x offset formula for a[p+1-t] needs to be adjusted depending on odd or even e.

 

odd e: a[p+1-t]

even e: a[p+2-t]

 

Based on this change, attached are revised "matrix" views for c145, c287, c6107 and a new test for c96133.

 

The results are now more consistent across both -f and e columns with gcd solutions being found for various a[t]=c records.

PMA !!y5/EVb5KZI ID: a44f6d July 14, 2018, 9:25 a.m. No.6946   πŸ—„οΈ.is πŸ”—kun   >>6948 >>6949

>>6943

>Then we will look at the key that is made by column -f with the locations of c in (-f,1)

Might be able to add a bit of clarity here.

 

Attached are the a[t]=c relevant records for c145 in both e and -f, and then the na transforms for each in the c145-occurrences-na image.

 

We can currently get to all of these records reliably except the first occurrence of a[t] = c, labeled as "1st_c_at".

 

For c145, those 1st occurrence records are (1,1,9), (-24,6,22), and their corresponding na records of (1,1,9) and (-24,1,22).

 

Notice that the other records between e and -f columns have the same t values +/- 1.

 

What causes the first occurrence records to behave differently in -f vs e?

How do we determine the first occurrence?

 

Don't have these answers yet, but do believe this is the key VQC was talking about.

GAnon !jPVzzZOz2c ID: 28ac3a July 14, 2018, 10:37 p.m. No.6950   πŸ—„οΈ.is πŸ”—kun   >>6951

>>6930

This graph is centered horizontally at d=0 and vertically at t=0. Down on the graph is a higher t. To the right on the graph is a higher d. So these correspond to (e,1) values for e's that are changing depending on the d values.

 

>>6347

 

This is basically the post because each d value is paired with a certain e, and they are both unique to c values. Center dot is (c,1,0) for (e,n,t), to the right is (c-1,1,0), down from that would be (c-1,1,2) etc..

 

The first picture is (1,c) so black means that the A[t] value is divisible by a for that (e,1) cell [note in this first one its black because a=1]. Then yellow means it is divisible by c. Then red means it is divisible by b and white means its not divisible by anything. This first picture is just for(1,c) values, so this is kind of what we're working with for the moment. You'll notice they're in little pods of 6 and then there are lines from the center of the pods to the next one diagonally. These are ways to visualze the factors of any value in (e,1) rows for any d. How it starts is with c=3. If the top left corner of one of the little 6 packs is at (d,t) then there is the top left corner at (d+2ci, t+cj) for any integers i,j. Also if I'm at (d,t) and im' on the top left corner of a little pod, I can always do (d-2i, t-i) for any i and the same goes for the other corners. You're in like a little rectangle web. These create the diamonds and the diamonds are always centered at d=0,t=0. Moreover, these little 6 packs are only on top and bottom of the center cell, or the left and right. If we have c=5 or 7, they start horizontally, on the left and right. Then as the web expands out to c=9, 11, 13, 15, the little 6 packs are above and below the origin. Then for c=17, 19, 21, 23 they are horizontal again. Then for c=25,27,29,31,33,35 they are vertical again, then for c=37,39,41,43,45,47 they are horizontal again. I think it would continue on this pattern out switching above and below. Also for these there are random little dots that I can't figure out the pattern on (I'll look into it soon enough).

 

Now since these are all factors of odd numbers, in the second pic we can see how they fit together, becaues we have the product of 2 odd numbers. So to visualize these we just overlay two other graphs like this. These are multiples of 5 notice how the grid mostly stays the same and the other grid shifts out. The black is the 5 grid and the red is other odd values. yellow is still c. So we can see that the c values are just the juxtaposition of these two regular graphs that we can solve for for any c (mostly).

 

The third pic is the factorization of all odd c up to like 150 I think. If I could factor it I did and the pictures are very cool for them. These patterns are just a superposition of the two regular graphs and the c values come from where they intersect. This one is pretty cool

 

Last pic is squares and seeing that these juxtapositions of these graphs makes cool almost sinusoidal movement.

Anonymous ID: a332a2 July 15, 2018, 9:19 a.m. No.6954   πŸ—„οΈ.is πŸ”—kun   >>6955 >>6956

>>6941

Just to build a bit on this.

 

I was reading over some older crumbs by our beloved mentor, also known as VQC.

 

Attached are two triangles based on the record a=7, b=37.

 

First one is based on (3, 6) the second one is based on (-30, 5). I used the equations in my previous post for this stuff.

 

I opted for the original nn + 2d(n-1) + f - 1 as I'm starting to suspect that it is irrelevant how we structure it. I don't think it's about how we structure it as much as it is about us understanding the triangle. Although I realized now that I did add f before 2d(n-1), but oh well.

 

Legend is:

nn = red (darker shade of red is remainder)

f-2 = yellow (darker shade of yellow is remainder)

2d(n-1) = blue (had no remainder)

 

The center square doesn't exists since I'm removing it (nn + 2d(n-1) + f - 1 = 8Tu + 1, moving the one over and we're left with 8 triangles).

 

I've also balanced the remainders on the opposite sides as a way of balancing out the distribution of the elements.

Anonymous ID: a332a2 July 15, 2018, 9:23 a.m. No.6955   πŸ—„οΈ.is πŸ”—kun

>>6954

Note I've also only divided by 8 as I haven't yet understood what VQC was talking about when he increased the part we divide for. I still think about >>5965 but VQC hasn't acknowledged those and has afterwards stated that we should keep the diagrams we have. Reason for that is that it aligns well with his first posts about triangles from thread #10 >>4342.

VA !!Nf9AmQNR7I ID: 2a785c July 15, 2018, 6:09 p.m. No.6957   πŸ—„οΈ.is πŸ”—kun   >>6958 >>6960

Ok, so for c6107, here are the values I've been yammering on about. Posting PMA's output for key records.

 

an = 31 * 36 = 1116

a(n-1) = 31 * (36-1) = 1085

na transform record: a[39] = 2976.

bn = 197 * 36 = 7092

b(n-1) = 197 * (36-1) = 6895

So can one of you Excellent Programmers help me locate all these in the (-f,1) and (e,1) cells / elements please ?? I'm working by hand. Need help to generate the full elements for each record.

So that we can analyze the t differences, and look for patterns.

VA !!Nf9AmQNR7I ID: 2a785c July 15, 2018, 6:11 p.m. No.6958   πŸ—„οΈ.is πŸ”—kun   >>6959

>>6957

also (prime) d = 78, and SQRT d = 8 remainder 14 = 8^2 +14

I'm gonna predict that at least of one the key values will be [t] +/- 8 from the na transform at [t]=39

so what's going on at [t]=31 ??

Do we find a(n-1) and (an) ??

VA !!Nf9AmQNR7I ID: 2a785c July 15, 2018, 6:46 p.m. No.6959   πŸ—„οΈ.is πŸ”—kun

>>6958

Here's VQC teaching us how to create new cells to handle the an, bn, a(n-1), b(n-1) construction of needed cells and elements.

 

Anyone understanding what I'm talking about?? We are moving into (-f,1) and (e,1) and need to construct the elements surrounding the (na transform) record.

GAnon !jPVzzZOz2c ID: 28ac3a July 15, 2018, 8:02 p.m. No.6963   πŸ—„οΈ.is πŸ”—kun

>>6961

It looks like if e is odd, then bn and b(n-1) have the same t value. If that is the case then t for an is one less than the t for a(n-1).

If e is even, then an and a(n-1) have the same t value and bn has t one greater than b(n-1).

Anonymous ID: a332a2 July 16, 2018, 4:45 a.m. No.6964   πŸ—„οΈ.is πŸ”—kun   >>6965

VQC, hoping for a correction from you here.

 

I was looking into:

> If an integer p is a factor of a[t], then p will be a factor of a[p+1-t] for ALL cells in row n=1.

 

From the thread Grid patterns, you posted that in >>6561

 

I'm thinking that we generate Nc in (e, 1). We then know that N and c are integer factors of Nc.

 

That should mean that N is a factor in a[N + 1 - t] and that c is a factor in a[c + 1 - t]. However I'm struggling to actually make it work.

 

Take the example a=7, b=37. c=259. big N = 114.

 

We generate Nc in (e, 1) which is {3:1:29769:243:29526:30014} which is at t = 121.

 

Then 114 should be a factor in a[114 + 1 - 121] and 259 again in a[259 + 1 - 121]. Neither are.

 

The cell for a[114 + 1 - 121] is {3:1:51:-11:62:42} and for a[259 + 1 - 121] has cell {3:1:39201:279:38922:39482}.

 

This lead me to think it only applies to prime factors of a[t] which lead me to factorize 114 to test. 114 = 2319.

 

So this means a[2 + 1 - 121], a[3 + 1 - 121] and a[19 + 1 - 121] should have 2, 3 and 19 as factors.

 

It holds true for a[2 + 1 - 121], but not for a[3 + 1 - 121] or a[19 + 1 - 121].

 

What I'm wondering is if I've misunderstood it or I'm applying it wrongly. My initial though when using it was that when p is less than t, we enter negative x space, but you've stated earlier that we should be afraid of that.

 

Of course if anyone else knows I'd love a reply. I can't quite grasp how this is supposed to work.

Anonymous ID: a332a2 July 16, 2018, 7:45 a.m. No.6967   πŸ—„οΈ.is πŸ”—kun   >>6968 >>6970

>>6774

 

Product of Big N and c are important, as it combined with the "other" value of c from (e, 1) will give us some extra information. This combined with -f and the locations of c will somehow lead us to x for an, a(n-1). Does this mean we want to find x for either of them, both?

 

>>6858

 

We want to make a function to get a[t] from some column (Presumably a[t] = Nc). Then we want to take advantage of a[p - (t+1)] = kp for some k, p.

 

I'm speculating that we want to get Nc, use a[c - (t+1)] to get another location for c and from here move over to -f and then find the value of x for either/or an or a(n-1). But if we're actually going to find x or t I'm not sure.

 

Hell I don't know the step to go from knowing different locations of c in (e, 1) and (-f, 1) to solve for x, but given that these are the last posts he made I'm willing to speculate that they are somehow related.

Anonymous ID: a332a2 July 16, 2018, 8:06 a.m. No.6968   πŸ—„οΈ.is πŸ”—kun   >>6970

>>6967

One thing I've seen, which catches my fancy:

 

Take c, generate a[t] = Nc in (e, 1). Now generate record for a[c - t - 1]. This a[c - t - 1] = kc for some k. If we divide kc/c we get k. If we then go to -f, and repeat we will end up with (k+1)c.

 

Make record (1, c)

Generate record for a[t] = Nc in (e, 1)

Generate record for a[c - t - 1] in (e, 1)

a[c - t - 1] = kc for some value of k.

kc/c = k

 

Now make the -f record for (1, c).

Find the a[t] = (N - 1)c

Here, in -f, we have a[c - t - 1] = (k+1)c

 

For some reason I would assume it would be (k-1). So do we have another layer of asymmetry?

 

Attached shows the case of k, k+1 for (e, 1) and (-f, 1)

Anonymous ID: a332a2 July 16, 2018, 9:40 a.m. No.6970   πŸ—„οΈ.is πŸ”—kun

>>6966

>>6965

>>6967

>>6968

 

Chalk this one up to Anon's happy mistake and not VQC. I used the wrong calculation on x to t and t to x, indexing the first record in (e, 1) as t=0. As a result, a[p + 1 - t] wouldn't work but a[p - 1 - t] worked. I've corrected my function and a[p + 1 - t] works fine now.

PMA !!y5/EVb5KZI ID: 52c782 July 16, 2018, 2:45 p.m. No.6972   πŸ—„οΈ.is πŸ”—kun   >>7340

>>6945

Some additional details, and perhaps some clarification, of how the different p formulas can be used to navigate the grid.

 

Attached pic shows a range of records in (1,145) for both positive and negative x.

 

The columns p and px represent "pointers" to other "t" values and are calculated as:

 

p column uses the formula "p+t".

px column uses the formula "p+1-t" for odd e, and "p+2-t" for even e.

 

In this case, the "p" value is being set to n. So the p column represents moves of 145 + t, and the px column represents moves of 145 + 1 - t.

 

The two records in the rectangle on the left are the 2nd and 3rd occurrences for a[t] = c, transformed by switching a and n (we can always do this because xx+e = 2na).

 

From there, the underlined p, px, and t values, show how we are currently able to jump between records using these p and px "pointers".

 

Within (1,145), there are 4 sequences or chains of records connected by a and b values.

 

These "pointer" values currently enable moving between 2 of the 4 sequences.

 

The ability to navigate to the first sequence would lead to a solution.

PrimeAnon !!!ZmMwNmUyMmUyMzY5 ID: a2c3ea July 16, 2018, 6:23 p.m. No.6973   πŸ—„οΈ.is πŸ”—kun   >>7002

Guys, pretty sure I got something big.

 

Pics related use the following color code:

 

Red for primes

Dark red for Fermat primes

Blue for semiprimes

Green for squares, cubes, etc, with darker green == higher power

 

In the 2nd image, notice how certain squares seem to follow the curve of the spiral on the left side . If the spiral can largely be described by n >= (e * e / 8), some of the squares on the left look like they could be described by a similar functions. I decided to look for similar patterns inside the spiral, and I found some. Lots of them.

 

"BigTreeShrunk.png" is a smaller version of a huge file I've uploaded here:

https://anonfile.com/W5pdsff8b9/BigTree.png

(~7k x 28k, but only 2.8 MB)

 

In the larger version, you can zoom in on the values for the first 't' entry. If you follow the pathways down, you'll notice that they're almost all continuous & hit sequential prime numbers.

 

I used the following method: starting from a prime, I would add an even number 'g'. From then on, I added g + 2, g + 4, g + 6, g + 8…etc. In that fashion I was able to hit as many as 29 primes in a row–from 41, 43, 47, …, to 971. It might have gone even further if I'd made a larger image.

 

The line in the center is somewhat of an exception–it hit a Fermat number (257), then a square, semiprime, two primes, semiprime, four primes, semiprime, then 6 primes when it hit the bottom; it was like a motorcycle that hit a bump then started righting itself.

 

Several sequences end up on squares–I usually didn't bother taking them further. Also, it may be confirmation bias but it seemed to me like a lot of the sequences converged on Fermat numbers.

 

There's a lot to learn from this–I haven't even looked for patterns in the other direction, and there are tons more than I can see but haven't mapped out yet. Thus far I've only looked at things in terms of 'c'. But I'm pretty blown away.

 

Please tell me you guys didn't figure this out seven months ago.

PrimeAnon !!!ZmMwNmUyMmUyMzY5 ID: a2c3ea July 16, 2018, 7:01 p.m. No.6974   πŸ—„οΈ.is πŸ”—kun

Side note, I came across an interesting article yesterday:

https://www.quantamagazine.org/mathematicians-discover-prime-conspiracy-20160313/

 

>Among the first billion prime numbers, for instance, a prime ending in 9 is almost 65 percent more likely to be followed by a prime ending in 1 than another prime ending in 9. In a paper posted online today, Kannan Soundararajan and Robert Lemke Oliver of Stanford University present both numerical and theoretical evidence that prime numbers repel other would-be primes that end in the same digit, and have varied predilections for being followed by primes ending in the other possible final digits.

 

Paper mentioned in article is attached.

 

Also, the sequence from 41, 43, 47, …, consists of 40 primes, all the way to 1681.

 

Here's a quick code snippet for anyone wanting to test how far a sequence goes:

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

public static void startSequence(){

int n = <your first prime>;

int m = <the first even number to add to your prime>;

int count = 0;

while(n < 4096 && (PRIMES4096.contains(n) || FERMATPRIMES.contains(n))){

count++;

n += m;

m += 2;

}

System.out.println("count: " + count + ", Last n value: " + n);

}

GAnon !jPVzzZOz2c ID: 28ac3a July 17, 2018, 8:19 p.m. No.6975   πŸ—„οΈ.is πŸ”—kun   >>6976

Remember 2na = xx + e. If n=1 then xx + e = 2a and so a = (xx+e)/2. Since we know that we can reverse engineer this to put our c value in the a position, so we can find where c is in A[t] in n=1.

c = (xx + e)/2 - 2c = xx + e

 

So we do the same calculation the square root then the remainder with the first thing. So for 145 we would get 2c = 290 = xx + e = 17*17 + 1.

Also 145 = 1212 + 1. Also we know that (of course) 2c - c = c. This gives us the equation 1717 + 1 - (1212 + 1) = 145, which simplifies to 1717 - 12*12 = 145, the correct answer to our problem, because d+n = 17, x+n = 12, d-x = 5 (I think this is why VQC liked this number). This got me to look into more equations and they aren't always that simple. Here is another:

 

17*41 = 697

697 = 26 * 26 + 21

697*2 = 37 * 37 + 25

 

3737 + 25 - 2626 - 21 = 697

3737 - 2626 + 4 = 697

 

So if anything this may just give us a close approximation of a difference of two squares. But we may be able to notice that when you shift d up one by either of these e gets shifted by 2d+1 or 2d-1 depending on the direction. Maybe we could use that and algebra to match up the columns to whatever makes e values equal then when e is equal we have the solution.

Anonymous ID: a332a2 July 17, 2018, 9:25 p.m. No.6977   πŸ—„οΈ.is πŸ”—kun

I've been looking into a[p + 1 - t] and I wanted to know what was happening under the hood. So I wrote it up using algebra and I got the following:

 

For odd e:

2(p + 1 - t)(p + 1 - t - 1) + ((e +1) / 2)

 

This can be rewritten as:

2p(p - 2t + 1) + 2t(t-1) + (e+1)/2

 

We can immediately see that 2t(t-1) + (e+1)/2 is the value of a[t] (the t we're doing a[p + 1 - t] from). Meaning we're doing 2p(p - 2t + 1) + a[t].

 

Then I was wondering what (p - 2t + 1) and 2p(p - 2t + 1) would look like for a few values and I wrote up a test. Attached is an image showing the test cases with values from (e, 1) and (-f, 1).

 

For these tests I did the na, nb and nc transformations on records from (3, 6) starting in the sequence a=7, b=37 and the 9 following records.

 

Note, aT = t where a[t] = na, bT = t where a[t] = nb and cT = t where a[t] = Nc. (Same for the negative records)

 

What sticks out is that b - 2bT + 1 = d (the d for our record) and c - 2cT + 1 = d (same d as our record d) while a - 2aT + 1 = d'

 

Note I marked it d' because it refers to the previous record in the sequence of (7, 37)s' d. The first record in the list is also the first "valid" record in the sequence which is probably why it shows d = -2, but you can see in the following records that a - 2aT + 1 will match the previous record in the sequences d.

PMA !!y5/EVb5KZI ID: a44f6d July 17, 2018, 10:58 p.m. No.6980   πŸ—„οΈ.is πŸ”—kun   >>6981 >>6983 >>6992

>>6654

>When is the first time a squared appears?

>What is the factor it is multiplied by?

>What is the first time b squared appears?

>The second?

 

Circled back to VQC's Independence Day post regarding the first and second occurrences of a^2 and b^2 in e and -f.

 

Attached pics are test cases for c145, c287, and c551 and show the first 2 records where a[t] mod a^2 or b^2 = 0, and a[t] is multiplied by another factor, in both positive and negative x.

 

The output shows -f details on the left and e on the right, and includes:

 

p column is the [p+t] formula where p is either a^2 or b^2

px column is the [p+1-t] formula, same a^2 or b^2 values.

n=a[t]/p just shows how n is derived for a[t] for known factors.

 

The p and px columns again show valid jumps between sequences of records.

 

The reason for this post, however, is that there is a pattern between the two a^2 and two b^2 records that enables us to predict the correct t values.

 

For example, in c145:

 

between (-24,1,10) and (-24,1,17)

 

10 + 17 - 2 = 25

 

between (-24,1,312) and (-24,1,531)

 

531 + 312 - 2 = 841

 

between (1,1,22) and (1,1,29)

 

((22 + 29) - 1)/2 = 25

 

between (1,1,821) and (1,1,862)

 

((821 + 862) - 1)/2 = 841

 

And the generalized formulas:

 

even e:

t1 + t2 - 2 = p

 

odd e:

(t1 + t2 - 1)/2 = p

 

In the case of odd e especially, it's interesting to note that the genalized t1 and t2 formula gives different results than the p and px pointer formulas.

Anonymous ID: a332a2 July 17, 2018, 11:17 p.m. No.6981   πŸ—„οΈ.is πŸ”—kun   >>6982

>>6980

I was trying to replicate with a=7, b=37.

 

The t-values for the first occurrences of a^2 in (e, 1) was 19 and 31. However 19 + 31 - 1 == 49. Same for b^2: First occurrences 582 and second at 788, 582 + 788 = 37^ + 1.

 

I suspect then that there is more to it than just odd/even e, maybe odd/even n's as well?

 

In your second image where e=odd and n=even you can see what I'm talking about.

Anonymous ID: a332a2 July 17, 2018, 11:50 p.m. No.6983   πŸ—„οΈ.is πŸ”—kun   >>6992

>>6980

If you look at the record a=91, b=169 you'll also see that 91^2 appears first at t=2175, second at t=3020 and the third time at t=5262

 

2175 + 3020 = 5195 which is not correct, while 3020 + 5262 = 8282 = 91^2 + 1

AA !dTGY7OMD/g ID: 56e3af July 18, 2018, 6:47 a.m. No.6984   πŸ—„οΈ.is πŸ”—kun

Thinking this through again:

>>6629

>As stated at the start, the square root of (D)avid and what remains are the key to unsealing the The End, the grid.

I'm thinking it isn't that we treat d as c recursively (even though VQC's reply to my post saying that implied it was), but maybe it's literally as he said, and it the square root of d and the difference between d and its square root.

AA !dTGY7OMD/g ID: 56e3af July 18, 2018, 6:52 a.m. No.6985   πŸ—„οΈ.is πŸ”—kun

I tried seeing if this stuff PMA has been posting examples of with the first three times a[t]=c shows any relationship with these root(d) and remainder values but it's quite hard actually finding examples without just making a brute force algorithm. I do have part of another example if anyone's interested.

 

7*29=203

d=14, root(d)=3, 14-3=11 (before you get any ideas, I realize this is equal to e+n but this is definitely not the case universally)

The true cells for this are {7:4:14:7:7:29:4} -22} and {-22:3:15:8:7:29:4} 7}

-f β€” a[t] = c

First: {-22:63:363:160:203:649} t=80 e=703

Second: {-22:87:391:188:203:753} t=94 e=759

Third: {-22:117:421:218:203:873} t=109 e=819

Fourth: {-22:149:449:246:203:993} t=123 e=875

e β€” a[t] = c

First: {7:88:392:189:203:757} t=95 f=-778

Second: {7:116:420:217:203:869} t=109 f= -834

Third: I couldn't find

 

I have no idea what any of this has to do with 3 or 11. Possibly coincidentally, x[t] = c at {7:64:525:203:322:856} t=102 f=-1044, which is the cell in the sequence that exists in the negative space but was missed in the positive space. I also did this with 13*43=559 and found x[t]=c at (30,1) when the first cell where a[t]=c in the negative space was (-17,4), so I don't know if this is actually relevant or a pattern.

GAnon !jPVzzZOz2c ID: 28ac3a July 18, 2018, 5:59 p.m. No.6990   πŸ—„οΈ.is πŸ”—kun   >>6991

>>6987

 

Okay so this is a snippet of the code and it isn't necessarily the beginning, the opening bracket indicates that it is the start of a function or a loop or an if statement or something, it seems like there have been a few things defined already, namely e and minus_f and they have been defined before the '{'. Again this could be after an 'if' in a loop or something where e and minus_f change in the loop. Then you take the sqrt of 2d which could mean to recursively do the function on 2d [145 = 1212 + 1, 24 = 44 + 8, 8 = 22 + 4, 4 = 22, 4 = 2*2, etc]. under some circumstance. I'm trying to read between the lines here

PMA !!y5/EVb5KZI ID: a44f6d July 18, 2018, 9:22 p.m. No.6992   πŸ—„οΈ.is πŸ”—kun

>>6980

>>6983

Thanks for double checking.

 

In reviewing further, I think I may have just restated how the p+t formulas work in a different way.

 

Was looking for a way to jump sequences, but don't believe these formulas enable that kind of movement.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 18, 2018, 10:53 p.m. No.6994   πŸ—„οΈ.is πŸ”—kun   >>6995

>>6987

Here's a question…

How warped is this image?

 

Are they supposed to be rectangles or squares?

Cuz adjusting for squares makes things screwballs.

 

Example… the Red area in the Big "Square"…

Is that supposed to be the same Area as in the red or is it not a 1:1?

 

I'm trying to channel Teh Maffs and send that vibe to the nerds so they can find whatever it is they're looking for, or think they should be looking for, or maybe recognize what they're finding as what they're actually looking for.

 

I'm not sure what they need right now.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 18, 2018, 10:58 p.m. No.6995   πŸ—„οΈ.is πŸ”—kun

>>6994

Ignore the 11^2, that came from when I had 12 blue instead of 11.

 

I feel like this was skewed on purpose to obfuscate something.

 

Also, the floating blue block in the top left of the small square on the original is offset by half a block, for some reason.

 

Something feels "on purpose, to obscure a thing" here.

Anonymous ID: a332a2 July 19, 2018, 8:20 a.m. No.6997   πŸ—„οΈ.is πŸ”—kun   >>6998

>>6996

I typed this out too fast. Let me rephrase it.

 

If we multiply c by 4, which was a crumb by VQC in >>6445 then the d off that record will equal the 2*d from c.

 

Example:

{595:170:4698:1105:3593:6143}

 

Here d = 4698, 2*d = 9396. The record for 4 * 3595 * 6143 = {2380:11026505:9396:9392:4:22071799}

 

So what he is doing is taking the square root of the d for 4c. Why I don't know yet, but since this is related to the 4c crumb I suspect there is more to that crumb than we figured out.

AA !dTGY7OMD/g ID: 21dd9c July 20, 2018, 1:28 a.m. No.6998   πŸ—„οΈ.is πŸ”—kun   >>6999

>>6997

>>6996

I'm sure you're already aware, but since we're using floors of square roots, 2d doesn't equal d of 4c in every case. Here's an example:

 

c=145, d=12, e=1, 2d=24

4c=580, d=24, e=4, 2d=48

4c=2320, d=48, e=16, 2d=96

4c=9280, d=96, e=64, 2d=192

4c=37120, d=192, e=256, 2d=384

 

This is where it stops following the pattern.

 

4c=148480, d=385, e=255, 2d=770

4c=593920, d=770, e=1020, 2d=1540

 

It then also resets here too.

 

4c=2375680, d=1541, e=999, 2d=3082

4c=9502720, d=3082, e=3996, 2d=6164

 

And so on, every time e becomes greater than d.

AA !dTGY7OMD/g ID: 21dd9c July 20, 2018, 2:23 a.m. No.6999   πŸ—„οΈ.is πŸ”—kun   >>7000

>>6998

I thought this sequence could potentially eventually end up with c being a square (there's probably a really obvious reason why it wouldn't do that but whatever), then creating some kind of sequence, but it just seems to go on forever. Pic related.

Anonymous ID: a332a2 July 20, 2018, 5:57 a.m. No.7002   πŸ—„οΈ.is πŸ”—kun

>>6973

Yeah, I think we were doing those back in thread #3-#4? I remember a lot of talks about Chirstmas trees (as a joke since it looks like a christmas tree).

 

I think you'd be better off if you started skimming all the threads.

Anonymous ID: a332a2 July 21, 2018, 12:15 a.m. No.7003   πŸ—„οΈ.is πŸ”—kun   >>7004 >>7005 >>7008

I think I realized a crumb from way back in thread #11 that deals with this 4c.

 

So some thoughts regarding c and 4c. Right now we're focusing on odd squares (x+n)^2. By multiplying c with 4 we're creating a new record and distributing the 4 (2x2) to both a and b, giving us 2a, 2b. This also gives us an even square to work with 2(x+n)^2.

 

The crumb from post >>5491

>"The 'heart' of the problem is that breaking the problem down involves two sets of triangle solving (otherwise it is prime), with one set of triangles (say half of the eight) working on a triangular problem that is one unit longer than the other set. This is one reason why no current approach I have seen works."

 

Using 4c we now have another 2(x+n)^2 square However, since this is even we can't divide it into 8 perfect triangles like we can with our odd (x+n)^2. Luckily for us, PMA has done some great work on even squares at >>6272 and >>6273 and showed that an even (x+n)^2 can be expressed as 4*(Tu + T(u-1)).

 

This gives us two equations for our x+n squares:

Odd (x+n)^2 = 8(Tu) + 1

Even (x+n)^2 = 4(Tu + T(u-1))

 

We now have two triangle problems to solve. For an even square, half of the triangles is one unit longer than the other half.

 

Now for some insight into our equations (This is just for insight, not sure if it can be used for solving):

8Tu = 8(u*(u+1)/2) =4u(u+1) => 2(2u*(u+1)).

 

2u(u+1) should be something we all recognize as it's the same method we calculate a's or d's depending on parity of e (Missing + (e + 1)/2 or e/2). This should mean (I might be wrong) that we are using values from (0, 1) as the 4T(u), as in 4T(u) should appear in (0, 1) as d. Just to re-iterate T(u) is the triangle with a base of u, has the equation u(u+1)/2. But by multiplying 4 we get 4(u(u+1)/2) =2u(u+1). This should mean that our odd (x+n)^2 - 1 = 2(d[u + 1]) from (0,1).

 

For even squares we have the equation:

4(Tu + T(u-1)) which is also equal to 4Tu + 4T(u-1) =2u(u+1)) + 2(u)(u-1). Here we have two d's from (0, 1) with a difference of t by 1. Essentially d[u+1] + d[u].

 

I think this might be why VQC has said that column (0, 1) is so special. Our x+n squares, regardless of parity, can be expressed using d's from (0, 1).

Anonymous ID: a332a2 July 21, 2018, 12:41 a.m. No.7004   πŸ—„οΈ.is πŸ”—kun

>>7003

 

Assuming we're trying to figure out the triangles using a similar method Get_N_From_… we need two new functions to work with even triangles.

 

We have GetNFromOddTriangleBase and we now need GetNFromEvenTriangleBase and a way to get the remainder needed for even squares:

def GetNFromEvenTriangleBase(bs, c, d): big_triangle = triangleN(bs) small_triangle = triangleN(bs-1) eight_base = big_triangle * 4 + small_triangle * 4 XPN = eight_base # (x + n)(x + n) DPN = XPN + c # (d + n)(d + n) return math.floor(math.sqrt(DPN)) - ddef GetRemainderEven2dnm1(bs, d, n, f): big_triangle = triangleN(bs) small_triangle = triangleN(bs - 1) eight_base = big_triangle * 4 + small_triangle * 4 XPN = eight_base # (x + n)(x + n) two_d = d + d nm1 = n - 1 two_d_m1 = two_d * nm1 XPN_mfp1 = XPN - f - 1 resultpN = XPN_mfp1 - two_d_m1 result = resultpN - (n * n) return result

Anonymous ID: a332a2 July 21, 2018, 1:18 a.m. No.7005   πŸ—„οΈ.is πŸ”—kun   >>7006 >>7009

>>7003

 

Just a quick unverified thing:

 

When we multiply our c with 4 we do 2a, 2b. This gives us a new square (x+n)^2 that is even.

 

If we divide this by 4 we end up with our original x+n square, however that means that our original x+n square, which is defined to by (x+n)^2 = 8Tu + 1, can also be expressed as the sum of two triangle numbers, with a difference of a unit of one. Specifically the triangle numbers used in the even square.

 

The one we're after

(x+n)^2 = 8Tu + 1

 

The one we get after multiplying 4

(x+n)^2 = 4(Tu + T(u-1))

 

Dividing the new square after multiplying 4 gives us the (x+n)^2 from our a, b. This also means that our (x+n)^2 = Tu + T(u-1).

 

So we have two equations for out x+n.

 

(x+n)^2 = 8(Tu) + 1

(x+n)^2 = Tu' + T(u' - 1).

 

Note the two u's are different, hence the u' vs u.

 

I haven't had time to properly verify it, but I think it holds.

VA !!Nf9AmQNR7I ID: e07318 July 21, 2018, 6:09 p.m. No.7006   πŸ—„οΈ.is πŸ”—kun

>>7005

Hey Isee! Nice work.

 

I'm still plugging on the an, bn, a(n-1), and b(n-1) records. Here's my WIP for c145 and c287. Lots of cool patterns, but no breakthrough as of yet.

 

Interesting find: for c145, the c value for b(n-1)=17400, and the c val for (bn) = 26245. When you subtract them, you get 8845, which is BigN * c = 61 * 145 = 8845

 

I tried it on c287 but the pattern didn't hold. Just wanted to share a cool find tho! :)

 

I'm gonna keep plugging and thinking. Anyone else got some work in progress??

AA !dTGY7OMD/g ID: 6c488c July 21, 2018, 10:46 p.m. No.7007   πŸ—„οΈ.is πŸ”—kun

Dissecting these code snippets a little bit.

 

>>6910

>//Function. RoD key pairs

>returns one pair for primes two for product of different primes, etc

So this function is only returning the n values. It doesn't seem to be doing anything with t of the correct (e,n). That means there's something inherent to the root of d that is also inherent to n, right? Because that seems to be the point of this function (given e, -f and d, return valid n values).

>given two columns, which values of n are separated by 1, where e(n)>-f(n)

It also means that something about e, -f and d tells us the number of factors that a number has. Every n value in the entire positive e space has a corresponding n-1 somewhere in the negative f space, but obviously all the ns in one e column aren't going to all be in the same -f column. So it returns every instance of an n in a given e having an n-1 in the given -f of our c. That means it returns every possible n value for our c. That means it knows how many n values there are for our c (and thus how many factors it has).

 

>>6987

I don't think these are the same function. This starts with an open parenthesis, but the last piece of code he posted was comments followed by a variable. That variable from the last one was the list of n and n-1 values for a given e and -f. That's definitely a local variable rather than a global variable, so those comments were inside its function. That means this has to be the beginning of a separate function. So we have one function to find all of the n values for our c, and another to do whatever this one's doing (which seems to be creating a list of BigIntegers (as opposed to a list of BigInteger arrays with 2 values each) and doing something to them that has something to do with sqrt(2d)).

Anonymous ID: a332a2 July 22, 2018, 12:26 a.m. No.7008   πŸ—„οΈ.is πŸ”—kun

>>7003

Just to add some more insight into even (x+n)^2.

 

Even (x+n)^2 = 4(Tu + T(u-1))

4(Tu + T(u-1)) = 4(u(u+1)/2 + u(u-1)/2)

4(u(u+1)/2 + u(u-1)/2) = 2(u(u+1) + u(u-1))

2(u(u+1) + u(u-1)) = 2(uu + u + uu - u) = 2(2uu)

 

This means our even (x+n)^2 is 2(a[u]) from (0, 1) and 2(d[u]) from (1, 1).

 

In this case, based on some testing u = x+n.

Anonymous ID: a332a2 July 22, 2018, 12:37 a.m. No.7009   πŸ—„οΈ.is πŸ”—kun

>>7005

Also, just to point out (x+n)^2 = Tu' + T(u' - 1) makes sense. I think I was a bit tired or fatigued. Two triangles (one being a unit longer than the other) added together makes a square, in this case the sum of Tu' and T(u' - 1) is equal to our x+n.

Anonymous ID: a332a2 July 22, 2018, 3:03 a.m. No.7011   πŸ—„οΈ.is πŸ”—kun

I'm probably re-iterating something we already know, but the nn + 2d(n - 1) + f - 1 = (x+n)^2 comes from (d+n)^2 - d*d.

 

(d+n)^2 - d^2 = (d+n)(d+n) - dd

(d+n)(d+n) = dd + dn + nd + nn

dd + 2dn +nn - dd = (x+n) + e

2dn + nn = (x+n) + e

nn + 2dn - e = (x+n)

 

Then we "steal" from 2dn to create f, since f = 2d + 1 - e, we steal a 2d

 

nn + 2d(n-1) + 2d - e = (x+n)^2

 

Then we add 1 to both sides

 

nn + 2d(n-1) + 2d - e + 1 = (x+n)^2 + 1

nn + 2d(n-1) + f = (x+n)^2 + 1

nn + 2d(n-1) + f -1 = (x+n)^2

AA !dTGY7OMD/g ID: 430395 July 22, 2018, 4:07 a.m. No.7012   πŸ—„οΈ.is πŸ”—kun

I've made a brute-force version of this algorithm

>>6910

It doesn't use the root of d (since obviously we don't know how you're meant to use that to do this), but it's useful for considering the context in which this function fits within the entire algorithm. It doesn't do (1,c) for even cs since n would have .5 at the end with an odd a+b (so those cells aren't in the grid).

 

import java.util.;public class brute{ public static void main(String[] args){ int e = Integer.parseInt(args[0]); int f = Integer.parseInt(args[1]); int d = (e-f-1)/2; int c = (dd)+e; int twod = d2; int roottwod = (int)Math.floor((int)Math.sqrt(twod)); System.out.println("c="+c+", d="+d+", e="+e+", f="+f+", 2d="+twod+", sqrt(2d)="+roottwod); for(int i=0; i<((((1+c)/2)-d)+1); i++){ int aplusb = (i+d)2; int x = ((int)Math.floor((int)Math.sqrt(((d+i)*(d+i))-c)))-i; int testa = d-x; int testb = aplusb-testa; if(testa*testb==c){ System.out.println(i + " is a valid n, with a=" + (int)testa + " and b=" + (int)testb); } } }}

So you input e and f values (f has to be negative) and it outputs every valid n value for that pairing

AA !dTGY7OMD/g ID: d2681d July 23, 2018, 3:18 a.m. No.7019   πŸ—„οΈ.is πŸ”—kun   >>7020 >>7023

Okay guys, I think found something interesting. I was just messing around with that brute force algorithm, comparing variables. I’m not sure what this means, but it’s a thing.

 

The t where a[t]=a(BigN-1) in (-f,1) and the t where a[t]=c*BigN in (e,1) are BigN or BigN-1 apart, depending on the parity of e.

If e is even, BigN + (the t where a[t]=a(BigN-1) in (-f,1)) - 1 = the t where a[t] = c*BigN in (e,1)

If e is odd, BigN + the t where a[t]=a(BigN-1) in (-f,1) = the t where a[t] = c*BigN in (e,1)

 

Here are some examples. First with odd e:

>c=209, e=13, BigN=91

>a[t]=a(BigN-1) at (-16,1,7) = {-16:1:104:14:90:120}, t=7

>a[t]=c*BigN at (13,1,98) = {13:1:19214:195:19019:19411}, t=98

>91+7=98

 

>c=403, e=3, BigN=182

>a[t]=a(BigN-1) at (-38,1,10) = {-38:1:201:20:181:223}, t=10

>a[t]=c*BigN at (3,1,192) = {3:1:73729:383:73346:74114}, t=192

>182+10=192

 

Now with even e:

>c=559, e=30, BigN=257

>a[t]=a(BigN-1) at (-17,1,12) = {-17:1:279:23:256:304}, t=12

>a[t]=c*BigN at (30,1,268) = {30:1:144199:536:143663:144737}, t=268

>257+12-1=268

 

>c=551, e=22, BigN=253

>a[t]=a(BigN-1) at (-25,1,12) = {-25:1:275:23:252:300}, t=12

>a[t]=c*BigN at (22,1,264) = {22:1:139931:528:139403:140461}, t=264

>253+12-1=264

 

I’ve tested this on over 20 semiprimes at it holds true for all of them. I would have to rewrite my entire algorithm in order to accommodate for non-semiprimes, so I’ve only tested it on semiprimes.

Anonymous ID: a332a2 July 23, 2018, 9:21 a.m. No.7020   πŸ—„οΈ.is πŸ”—kun   >>7021

>>7019

Isn't this this just the asymmetry we've talked about before with regards to a(n-1), an, b(n-1), bn combined with a[p + t]?

 

Or are you referring to the parity effecting it?

 

I also see a difference in how you estimate t for even numbers. This is something we've gone over a few times, it seems like we're all uneven on how to estimate t from x. PMA and I had a conversation about this a few days ago as I used a different equation (indexing t=0 instead of t=1).

 

I believe the "correct" calculations are:

even x: x/2 + 1 = t

odd x: (x+1)/2 = t

 

So for your examples with even e the t should be 269 and 265 respectively. Assuming I'm not messing up with the t-calculations again.

VA !!Nf9AmQNR7I ID: 395389 July 23, 2018, 6:22 p.m. No.7023   πŸ—„οΈ.is πŸ”—kun   >>7024

>>7019

Hello AA! I'm going over this again, and working to understand. You definitely found something cool. You seem to have proved that VQC's rules for an, a(n-1), bn, b(n-1) also hold true for BigN-1 and BigN where they appear as a[t] values. Any further insight into how we can use this in (e,1) ??

AA !dTGY7OMD/g ID: c59906 July 24, 2018, 12:10 a.m. No.7024   πŸ—„οΈ.is πŸ”—kun   >>7026

>>7022

It's good that none of us knows what we're doing, so we all feel inadequate together, huh?

 

>>7023

I suppose all it really means is that the a(n-1) thing works for more than just the n value we're looking for. So if you have a c with more than two n values (i.e. it has more than two prime factors plus one and itself), all of them will do that thing. But we could have figured that out by just sort of thinking about it outside of the context of semiprimes and testing it. Most of these rules do apply to more than just semiprimes. This is what happens when I do math when I should be asleep. I forget that the things I find are things I already know to be true.

AA !dTGY7OMD/g ID: c2fe35 July 24, 2018, 1:59 a.m. No.7025   πŸ—„οΈ.is πŸ”—kun   >>7026

I've been going over this VQC post VA brought up on Discord as well as the fourth image in >>6695.

 

From the attached image:

>The final steps involve finding a base using (f-1)/8 for each of the eight triangles (some a bit bigger than others by a unit in length) which is larger than (n-1)/8 BUT smaller than (x+n)/8.

>The choices reduce β€˜exponentially’ and with less complexity than find the root of C, so the last steps do not push the overall complexity Big Oh above Big Oh for finding the root.

 

From the image at the beginning of the thread:

>Since f can be divided by 5, we can make a base for each triangle which is made of (f/40), still with 4 left over.

 

Are we meant to be factoring f recursively? We're meant to "use" (f-1)/8 to find the base of the triangles in nn. f can be smaller or larger than n, so we can't just "use" f to find it, since it doesn't seem to scale with any of the other related variables. But if we multiply a factor of f by 8, we can find a base with a remainder less than 8 that works. And I remember when he was talking about that people were asking where this arbitrary 5 came from. Maybe we're meant to find the factors of f. f is made of 2d and e, which also have square roots in the latest code snippet.

PMA !!y5/EVb5KZI ID: a44f6d July 25, 2018, 10:55 p.m. No.7031   πŸ—„οΈ.is πŸ”—kun   >>7383

Not quite sure where to go with the sqrt(2d) hints, so in the meantime have integrated the iterative search solution for both odd/even squares into (-f,1) and (e,1) records.

 

Attached pics are test cases for c85, c145, and c15120 and show each factor record next to the (-f,1) or (e,1) record with a matching u.

 

For even e:

 

  • the XPN is calculated as 1+8T(u).

  • the corresponding n=1 record can be found at t = u + 1.

 

For odd e:

 

  • the XPN is calculated as 4T(u) + 4T(u-1).

  • the corresponding n=1 record can be found at t = u.

 

u is simply iterated from 0 to a max value from the entry record.

 

Rather than being a pathway to a solution, this was just an exercise to integrate the even square triangle formulas and visualize perhaps how factor records "bounce" between -f and e (c15120).

AA !dTGY7OMD/g ID: 0cec15 July 27, 2018, 3:36 a.m. No.7034   πŸ—„οΈ.is πŸ”—kun

At risk of pointing out things we already know again, I'll begin posting any diagonal patterns I notice while the rest of you are still asleep (or until I go to sleep).

 

First one is the relationship between d and f when moving diagonally. When you increase e and n by 1, at the same t, the increase in f is equal to 2*(the increase in d)+1.

Examples:

(19,14,41) = {19:14:316:81:235:425} f=-614

(20,15,41) = {20:15:294:80:214:404} f=-569

d decreases by 22, f increases by 45 (22*2 + 1).

 

(19,14,6) = {19:14:16:11:5:55} f = -14

(20,15,6) = {20:15:14:10:4:54} f = -9

d decreases by 2, f increases by 5 (2*2 + 1).

 

(60,40,56) = {60:40:262:110:152:452} f = -465

(61,41,56) = {61:41:262:111:151:455} f = -464

d stays the same, f increases by 1 (0*2 + 1).

AA !dTGY7OMD/g ID: 0cec15 July 27, 2018, 3:46 a.m. No.7035   πŸ—„οΈ.is πŸ”—kun

When you decrease e by 1 but increase n by 1, the increase in f is equal to 2d-1.

Examples:

(83,42,43) = {83:42:172:85:87:341} f = -262

(82,43,43) = {82:43:167:84:83:337} f = -253

d decreases by 5, f increases by 9 (2*5 – 1).

 

(75,34,93) = {76:34:683:184:499:935} f = -1291

(75,35,93) = {75:35:675:185:490:930} f = -1276

d decreases by 8, f increases by 15 (2*8 – 1).

 

(61,23,40) = {61:23:216:79:137:341} f = -372

(60,24,40) = {60:24:206:78:128:332} f = -353

d decreases by 10, f increases by 19 (2*10 – 1).

VA !!Nf9AmQNR7I ID: b0abce July 27, 2018, 6:41 p.m. No.7036   πŸ—„οΈ.is πŸ”—kun

Hello Anons! :)

Our Definite Purpose is that WE are going to solve this, no matter how long it takes!

And Infinite Intelligence will come to our Aid as we keep our purpose Clear and Strong.

Ok, so we start with C

We create BigN and BigN-1

Those two are the upper limit of our search for n and n-1

They are also the upper limit for x and t.

(for our given C)

We know that the na transform into e and -f puts us right between two important groups in e and -f

Those two groups are:

  1. a(n-1) and (an)

  2. b(n-1) and (bn)

Here's c287( attached). Dark green is na transform in -f and e.

As you all can see, the a(n-1) and (an) records are 3 elements above both the -f and e na transform elements.

Also, you can all see the b(n-1) and (bn) records are 4 and 5 elements down from the na transform elements.

3,4,5 ???

Lol, Pythagorean triangle, at least for this solution for c287

3^2 + 4^2 = 5^2

VQC is also hinting at the ability to Scan/Search all c287 columns for -2 and 31, looking in those offset columns for a pair that matches our C.

Starting with Big N and Big N-1, then moving to row 1.

The offset between b(n-1) and (bn) is the ONLY similar pattern we currently know in row 1 that compares to the n and (n-1) offset.

Easy way to search is to move down (higher x and t) from the na transform in -2 and 31 looking for the b(n-1) and (bn) records.

Which are offset by one.

All we have to do is subtract each offset pair and do c / b.

We are looking for a remainder of 0,

Where the answer is a.

Thoughts, Anons ??

AA !dTGY7OMD/g ID: 1bb682 July 29, 2018, 11:48 p.m. No.7037   πŸ—„οΈ.is πŸ”—kun   >>7038 >>7041

No posts in two days. I guess I'm not the only one who has absolutely no fucking idea what diagonals have anything to do with. I had a few ideas that I tested but they didn't lead to anything useful.

Anonymous ID: a332a2 July 30, 2018, 1:37 a.m. No.7038   πŸ—„οΈ.is πŸ”—kun   >>7039 >>7133

>>7037

I've been looking into x and negative x, but I haven't found much of use yet.

 

What I've seen is that the number of chains / sequences in (e, n) appears to depend on the number of factors in n, but it doesn't always seem to hold. For some cases, we'll have k^2 number of chains / sequences where k is the number of factors in n, but again it doesn't always hold (For a prime number it appears they usually have only 2 chains, but c=15 which has two factors, 3 and 5, also only has two chains).

 

It appears that these sequences / chains come in pairs, and despite not appearing to be connected, they are at -x. Take for example c=3367 (a=37, b=91). It has 8 chains (and 3 factors):

 

e:

0 (3, 3367, 207, 201, 6, 7142)

1 (3, 3367, 847, 761, 86, 8342)

2 (3, 3367, 3165, 2347, 818, 12246)

3 (3, 3367, 4935, 3309, 1626, 14978)

4 (3, 3367, 5167, 3425, 1742, 15326)

5 (3, 3367, 7245, 4387, 2858, 18366)

6 (3, 3367, 11271, 5973, 5298, 23978)

7 (3, 3367, 12871, 6533, 6338, 26138)

 

-f:

0 (-114, 3367, 333, 318, 15, 7385)

1 (-114, 3367, 1837, 1502, 335, 10073)

2 (-114, 3367, 1989, 1606, 383, 10329)

3 (-114, 3367, 4933, 3308, 1625, 14975)

4 (-114, 3367, 5169, 3426, 1743, 15329)

5 (-114, 3367, 9033, 5128, 3905, 20895)

6 (-114, 3367, 9297, 5232, 4065, 21263)

7 (-114, 3367, 12529, 6416, 6113, 25679)

 

The indentation shows the connection between them. Move into negative x for any of these pairs and you'll find the chain / sequence with the same indentation level.

 

For example, take e and chain #3. (3, 3367, 4935, 3309, 1626, 14978)

 

Create the cell for -3309 and you will get: (3, 3367, -1683, -3309, 1626, 1742). Then we'll use the x + 2n and create the cell for e=3, n=3367, x=-3309 + 2*3367 and you will get the cell: (3, 3367, 5167, 3425, 1742, 15326) which is equal to chain #4 (meaning they are a connected pair).

 

The connected pairs here is (0, 7), (1, 6), (2, 5), (3, 4) each have the sum of 7 (indexing with 0 and 9 indexing with 1) and the sum of x in these connected pairs appear to always be 2*n (n=3367 =2*3367 = 6754).

 

The difference of x between the connected chains also maps to a d-value, meaning each chain is the result of a*b where 3367 is a factor of these. The chains in the middle (#3, #4) seems to always represent (1, c).

 

To show an example of this:

The difference between x in chain #0 and it's connected pair chain #7 is 6533 - 201 = 6332. 6332/2 = 3166.

 

3166^2 + e = 10023559 and 10023559/3367 =2977.

 

Another way of getting to the connected pairs is through a[p + 1 - t]. Take c and create the cell for Nc in (e, 1). In this case, a[p + 1 - t] = kc where k is the first a in the connected chain. For example: c = 259

It has the cell (3, 114, 16, 15, 1, 259) which gives us the Nc cell: (3, 1, 29769, 243, 29526, 30014) which is at t=122. If we now make the cell at 259 + 1 - 122 =(3, 1, 38089 275, 37814, 38366), we get a = 37814 => 37814/259 = 146.

 

The sequences in (3, 259) is:

1 (3, 259, 279, 201, 78, 998)

2 (3, 259, 357, 243, 114, 1118)

3 (3, 259, 421, 275, 146, 1214)

4 (3, 259, 511, 317, 194, 1346)

 

This means that the a, b we're after is also connected to another sequence, maybe we're going for a little hunt? Like go from our c to the alternate sequence and from that go to our proper sequence?

Anonymous ID: a332a2 July 30, 2018, 2:01 a.m. No.7040   πŸ—„οΈ.is πŸ”—kun

>>7039

Column 0 is special. I haven't quite grasped it yet, but there are several interesting parts going on here.

 

Take c=259, in (0, 259) there is only one chain / sequence.

 

1 (0, 259, 1036, 518, 518, 2072)

 

But by multiplying c with odd multiples of 2, we increase the number of sequences:

 

(0, 2*259):

1 (0, 518, 777, 518, 259, 2331)

2 (0, 518, 2072, 1036, 1036, 4144)

 

(0, 2**3 * 259):

1 (0, 2072, 1295, 1036, 259, 6475)

2 (0, 2072, 3108, 2072, 1036, 9324)

3 (0, 2072, 5439, 3108, 2331, 12691)

4 (0, 2072, 8288, 4144, 4144, 16576)

 

(0, 2**5 * 259):

1 (0, 8288, 2331, 2072, 259, 20979)

2 (0, 8288, 5180, 4144, 1036, 25900)

3 (0, 8288, 8547, 6216, 2331, 31339)

4 (0, 8288, 12432, 8288, 4144, 37296)

5 (0, 8288, 16835, 10360, 6475, 43771)

6 (0, 8288, 21756, 12432, 9324, 50764)

7 (0, 8288, 27195, 14504, 12691, 58275)

8 (0, 8288, 33152, 16576, 16576, 66304)

 

(0, 2**7 * 259):

1 (0, 33152, 4403, 4144, 259, 74851)

2 (0, 33152, 9324, 8288, 1036, 83916)

3 (0, 33152, 14763, 12432, 2331, 93499)

4 (0, 33152, 20720, 16576, 4144, 103600)

5 (0, 33152, 27195, 20720, 6475, 114219)

6 (0, 33152, 34188, 24864, 9324, 125356)

7 (0, 33152, 41699, 29008, 12691, 137011)

8 (0, 33152, 49728, 33152, 16576, 149184)

9 (0, 33152, 58275, 37296, 20979, 161875)

10 (0, 33152, 67340, 41440, 25900, 175084)

11 (0, 33152, 76923, 45584, 31339, 188811)

12 (0, 33152, 87024, 49728, 37296, 203056)

13 (0, 33152, 97643, 53872, 43771, 217819)

14 (0, 33152, 108780, 58016, 50764, 233100)

15 (0, 33152, 120435, 62160, 58275, 248899)

16 (0, 33152, 132608, 66304, 66304, 265216)

 

(0, 2**9 * 259):

1 (0, 132608, 8547, 8288, 259, 282051)

2 (0, 132608, 17612, 16576, 1036, 299404)

3 (0, 132608, 27195, 24864, 2331, 317275)

4 (0, 132608, 37296, 33152, 4144, 335664)

5 (0, 132608, 47915, 41440, 6475, 354571)

6 (0, 132608, 59052, 49728, 9324, 373996)

7 (0, 132608, 70707, 58016, 12691, 393939)

8 (0, 132608, 82880, 66304, 16576, 414400)

9 (0, 132608, 95571, 74592, 20979, 435379)

10 (0, 132608, 108780, 82880, 25900, 456876)

11 (0, 132608, 122507, 91168, 31339, 478891)

12 (0, 132608, 136752, 99456, 37296, 501424)

13 (0, 132608, 151515, 107744, 43771, 524475)

14 (0, 132608, 166796, 116032, 50764, 548044)

15 (0, 132608, 182595, 124320, 58275, 572131)

16 (0, 132608, 198912, 132608, 66304, 596736)

17 (0, 132608, 215747, 140896, 74851, 621859)

18 (0, 132608, 233100, 149184, 83916, 647500)

19 (0, 132608, 250971, 157472, 93499, 673659)

20 (0, 132608, 269360, 165760, 103600, 700336)

21 (0, 132608, 288267, 174048, 114219, 727531)

22 (0, 132608, 307692, 182336, 125356, 755244)

23 (0, 132608, 327635, 190624, 137011, 783475)

24 (0, 132608, 348096, 198912, 149184, 812224)

25 (0, 132608, 369075, 207200, 161875, 841491)

26 (0, 132608, 390572, 215488, 175084, 871276)

27 (0, 132608, 412587, 223776, 188811, 901579)

28 (0, 132608, 435120, 232064, 203056, 932400)

29 (0, 132608, 458171, 240352, 217819, 963739)

30 (0, 132608, 481740, 248640, 233100, 995596)

31 (0, 132608, 505827, 256928, 248899, 1027971)

32 (0, 132608, 530432, 265216, 265216, 1060864)

 

Say we take 2^17, then we should have 2^((17+1)/2) number of chains (which checks out). As for doing it an even number, then it appears to have 2^(n/2) number of chains.

Anonymous ID: 08d201 July 30, 2018, 7:07 a.m. No.7041   πŸ—„οΈ.is πŸ”—kun   >>7042

>>7037

Hmmmmmm….

Curiouser and curiouser….

 

"RSA is all about so-called dissimilarity matrices: square, symmetric matrices with a zero diagonal that encode the (dis)similarity between all pairs of data samples or conditions in a dataset. We compose a little helper function to plot such matrices, including a color-scale and proper labeling of matrix rows and columns."

 

http://www.pymvpa.org/examples/rsa_fmri.html

VQC !!/aJpLe9Pdk ID: 49f8ff July 30, 2018, 1:41 p.m. No.7050   πŸ—„οΈ.is πŸ”—kun   >>7051 >>7056 >>7057 >>7069

Looks like at least two parties have solved this and one has extended to elliptic curves.

Watch crypto currencies. Mining, exchanges and embedding/retrieving messages…

Unsealing(s) coming.

The ReVeal.

The End is near for me, friends.

Look after each other anons.

God wins.

L(o)VE is on the inside.

PMA !!y5/EVb5KZI ID: 47b0a9 July 30, 2018, 9:35 p.m. No.7055   πŸ—„οΈ.is πŸ”—kun   >>7124

Just posting a work in progress understanding of the RoD algorithm.

 

Attached pics are for c145, c287, c551, and c6107, and include the (-f,n-1) and (e,n) records, and factor tree.

 

Aside from the sqrt(2d), sqrt(e), and sqrt(f) values - still not quite sure how these apply - the only new piece is the (e,1) record at x=f or x=f-1. >>7047

 

Initial tests indicate that the x record will always be at f-1. Which essentially means that we are finding a record at (e,1) where x+n = f.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 31, 2018, 10:16 a.m. No.7062   πŸ—„οΈ.is πŸ”—kun

>>7061

Nigga lay the fuck off it.

It's about timing.

VQC also told us his name was "Chris Curtis" after a dead drummer.

Told us that his family is gone forever but they'll be joining him for dinner.

Who goes to the zoo BY THEMSELVES AS A GROWN MAN?

His compy that never works? Even though that's not the case at all?

Wasn't he going to splay out RSA2048 in a manner of days?

 

Misdirection is necessary for the timing to work out.

You want it work out faster? SOLVE THIS SHIT!

And believe, I'm aware of AA's thoughts on the matter.

I run the Discord server, 'member?

 

If you wanna vent, go over to the EZ Bake.

This isn't tone policing, but pissing and moaning has nothing to do with solving the problem.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 July 31, 2018, 10:58 a.m. No.7063   πŸ—„οΈ.is πŸ”—kun

>>7061

On top of that…

I only go where the wind takes me.

My main contributions are that I see patterns, "motivate", and make connections.

I was directed a long time ago to use my tripnamefaggotry to direct anons where they need to go.

 

Shall Capwn wasn't working on anything directly related to this.

When they showed me what they were working on and how they were approaching it,

I put 2 and 2 together because its' the same 2 from different perspectives, it seemed.

And Lord knows that 2 is one of the most mind bending numbers in Maths.

 

If complaining accomplished anything, I'd do more of it.

 

Whatchu workin' on?

AA !dTGY7OMD/g ID: f0b540 July 31, 2018, 4:48 p.m. No.7064   πŸ—„οΈ.is πŸ”—kun   >>7065 >>7066 >>7067

>>7061

I didn't swear at him. He doesn't owe us anything. We still have every right to be fucked off if he bails, but you've got to keep in mind, OP (as a concept) barely ever delivers. We're lucky we got this far. I'm still confident we'll at least get to learn about sonoluminescence. After that, I mean, I hope he sticks around, but who knows what he wants to do.

Anonymous ID: 2a395f July 31, 2018, 7:47 p.m. No.7065   πŸ—„οΈ.is πŸ”—kun   >>7066 >>7067

>>7064

Yeah, you’re right. Hell of a ride either way. I thought he was dropping all the code piece by piece, and when I saw that post I was like YOU MOTHERFUCKER. I will choose to take a broader view as you suggest. Sorry.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 Aug. 1, 2018, 10 a.m. No.7068   πŸ—„οΈ.is πŸ”—kun

>>7044

Doin' muh quantum intuition and collective conscious thing… I've been looking at how to do the "tetrabola" stuff from other perspectives.

 

The goal is to find the Rosetta Stone/common language between that and what we've been doing with the VQC.

I was going with my visualization approach of "how can I construct that…" and it seemed like it had to do with reflections, rotations, and an offset from 0, as well as Limits..

 

Should produce a Eureka Moment, if my figgerin's correct.

 

1st Set:

Parabola -Tetrabola

 

2nd Set:

Hyperbola -Tetrabola-Reciprocal

Hobo !!1yNgQ3NlCs ID: 4c7894 Aug. 2, 2018, 7 p.m. No.7078   πŸ—„οΈ.is πŸ”—kun   >>7079

>>7076

Well If Flat Earth shit is being posted here, there is really no greater proof that this is all a threat. I always saw shills as a sign of success and truth as opposed to a sign of failure. Remember, Putin gave Trump 160tb of data whose encryption was cracked… Just let that sink in a bit…

VA !!Nf9AmQNR7I ID: 61127e Aug. 2, 2018, 7:28 p.m. No.7079   πŸ—„οΈ.is πŸ”—kun   >>7080 >>7081

>>7078

To be fair, Saga reposted one of OUR memes, which was meant to be a joke from the beginning. I think Topol was high as SHIT and Saw Red like a bull in the arena. He fought off the Flat Earth shills in /qr/ so many times, he's like the "Round But Hollow Earth Warrior" now. I learned my lesson about losing my temper. Tryna keep it coo, niggaz.

 

Saga, you'll build a good reputation here when you post ideas about what we're currently on. You are Most Welcome! Contribute. Even if you're a bit lost, just do your best, man! Everyone here has posted MANY ideas that didn't pan out.

 

#WWG1WGA

#WWF1WFA

Saga ID: 82319e Aug. 2, 2018, 8:02 p.m. No.7080   πŸ—„οΈ.is πŸ”—kun

>>7079

Thanks VA, I appreciate your welcome and am having fun just finding relations, I hope to solve this puzzle but I'm satisfied with what I learn here everyday.

 

I'm working in kind of a backwards way going column wise and row wise on the grid, and I've only been getting into the +ve side of it. I'll post what I have as soon as I finish typing it in text format, currently working on paper as I'm more used to that.

 

Any finding that I think would help you guys out in any way I would post, don't worry about that.

Hobo !!1yNgQ3NlCs ID: 4c7894 Aug. 2, 2018, 8:06 p.m. No.7081   πŸ—„οΈ.is πŸ”—kun   >>7082

>>7079

That VQC inspiration. Those are words to live by. Yeah Topol is our Pit-bull. He has really kept this place clean unlike anywhere on the chans ever.

I have thought this from the beginning here… The journey is far more important than the destination. I have never observed a better group of people doing math that I don't understand than I have observed here, in this place…

Saga ID: 85c377 Aug. 2, 2018, 9:07 p.m. No.7083   πŸ—„οΈ.is πŸ”—kun   >>7085 >>7086

In the second pic, the "n" values that have a dash - before them are ones that don't follow the simple Pythagorean triplet numbers listed in A001844. They have a complex pattern of modulus that I couldn't find (yet). I'll go to the next column so I can make sense of any of this.

Saga ID: 85c377 Aug. 2, 2018, 10:33 p.m. No.7088   πŸ—„οΈ.is πŸ”—kun   >>7089

>>7087

I'm trying, but I don't want to hinder your current work so your priority should be on your current level.

 

First pic is quite obvious, the numbers {x,y} denote the occurrence modulus of the "t". Second pic has 3 columns: (n,c) which is "n", triplets associated with said "n" pythagorean number, and the previously defined {x,y}. range and other values aren't very important but just to roughly help me arrange pairs of numbers if the "n" happens to have more than one pattern.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 Aug. 3, 2018, 1:10 a.m. No.7089   πŸ—„οΈ.is πŸ”—kun   >>7090

>>7088

We've met, right?

These nerds are also on the Q boards.

And doing whatever else with their lives.

 

Don't worry about distracting them.

"Back to Basics" never hurts and can be very useful sometimes.

AA !dTGY7OMD/g ID: 6be8a1 Aug. 4, 2018, 6:25 a.m. No.7092   πŸ—„οΈ.is πŸ”—kun   >>7093 >>7094 >>7104

I swear this isn't pessimism (no Topol, I am not grumpy, please stop saying that), but somehow I feel like these latest crumbs have left me with more questions than directions. We still have no idea what diagonals have to do with anything. We can get the cell where x=f-1, sure, and more cells are always good for looking for patterns, but what value is he talking about that we're meant to find and what is it relevant to? There are a bunch of values associated with x=f-1. I haven't found any useful ones in any of my tests. In what case does x=f? In every single case I've seen, it's only valid for x to equal f-1. He wouldn't say that if it wasn't true somewhere, I wouldn't think. Considering this was the latest thing posted (including "build the tree recursively", which also seemingly comes out of nowhere) right after the thing about the diagonals, are they meant to have some relevance to each other? Because I'm not seeing any relevance in my tests. And where was this information 9 months ago? This is what gets me the most. If we've had enough information to figure it out this whole time, and if we've been so close to the eureka moment for so long, how were we meant to get there without any of these ideas about diagonals and x=f-1, and how were we meant to come up with these ideas? Is this disinfo to intentionally confuse and frustrate us for some hidden plot to prove our loyalty or to delay it until the 'stage is set' or something? It could be. I'm personally feeling far more curious than I am frustrated as a result of all of this. It just makes me wonder what's going on behind the scenes.

 

These are genuine questions. This isn't just me going on some meaningless rant. With how few work-related posts there have been here on the board lately, I know I'm not the only one who feels this way. Maybe stepping back and considering these newest points in a broader context is a useful way to wrap our heads around them. And as an aside, considering all the theories that go back and forth about who or what VQC really is and why so much confusing bullshit goes on, there are infinite different secret scenarios that could be going on here. Maybe they're waiting for us to ask the right question. Maybe they actually are intentionally trying to see if any of us quit. That kind of thing.

 

So does anyone have literally any idea what we're meant to be doing with these latest posts? Because none of my ideas panned out, and there are barely any other posts so far. What have you all been working on? Or, you know, VQC, if you're feeling generous, maybe just be open about the fact that we need to wait another month or whatever weird shit you're not telling us.

Anonymous ID: a332a2 Aug. 4, 2018, 8:51 a.m. No.7093   πŸ—„οΈ.is πŸ”—kun   >>7094 >>7104

>>7092

I can relate to this. I feel pretty stuck my self and with the latest "hints", the lines of code and comments I'm kind of stuck. The square root things just popped seamingly (Not the d, though. That we've speculated on for a while, not that we have any clue on how to use it) out of nowhere and got me a bit confused.

 

Maybe it is being postponed, maybe VQC really wants us to figure it out ourselves, maybe something else.

 

I've been dividing my time between trying to figure out what square roots of things are used for, x=f-1 and just looking for more patterns in the grid. Mostly looking for more patterns and writing down my thoughts to cement them.

 

Regarding the square root of d, f and e I was thinking in the terms of pythagorean theorem, that this big thing could just boil down to a few triangles that we need to triangulate and that with those values, we have a start position and need to figure out the rest. But that is just a wild guess, like anything else.

 

The recursive tree is still lost on me, I never understood it when it was first introduced and I still don't see how we're going to use it with regards to x=f-1. All in all, I agree and I'm feeling lost in all of these hints, patterns and triangles.

VA !!Nf9AmQNR7I ID: 779e23 Aug. 4, 2018, 11:56 a.m. No.7094   πŸ—„οΈ.is πŸ”—kun   >>7104

>>7092

>Is this disinfo to intentionally confuse and frustrate us for some hidden plot to prove our loyalty or to delay it until the 'stage is set' or something? It could be. I'm personally feeling far more curious than I am frustrated as a result of all of this. It just makes me wonder what's going on behind the scenes.

 

I feel the same way. More curious than anything else. I've been re-reading all my notes and studying all the past crumbs looking for greater understanding. Just like the Q maps, every time I re-read the crumbs I understand things I missed or didn't get before. I encourage all Anons to go re-read the VQC maps today.

 

>>7093

>Maybe it is being postponed, maybe VQC really wants us to figure it out ourselves, maybe something else.

 

Yeah, we have no clue what's going on behind the scenes, but it sure will be cool when we solve this! MSM meltdown will be hilarious "The Hacker Known As 8Chan has cracked ALL THE THINGS!!"

 

I'm gonna keep working, Lads! We learned so much already, and plus it's fun to be here working on Math with you all. It keeps me out of trouble on Cuckbook, pissing off all the mind slaves stuck in the Matrix. It's a better investment of my time to be here learning about cool math. If I'm honest, I've also invested so much time here It would be pointless to quit now, lol.

 

Onward, Faggots! ;)

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 Aug. 4, 2018, 2:28 p.m. No.7097   πŸ—„οΈ.is πŸ”—kun   >>7104

>>7095

This kind of became a general intro for newcomers…

 

Welcome!

Please enjoy your stay, look around, and feel free to contribute and ask questions.

 

Nothing holds us back, nothing slows us down, and we'd love to bring you up to speed and integrate your knowledge for a fuller perspective.

 

If you see a thing that pops out and your intuition leads to put 2 and 2 together… Post it.

 

We have a progressive RSA Bread Relay, as well as breads for Trangles, Curves, Sonic Rainbooms, Shitzen Posten, and we can happily make others if a point of interest so requires.

 

Things that may not be immediately related, may be deeply related upon staring at it.

 

Think of this as AA for those with a very serious math problem. Or Set of Problems.

Or like… a whole array of them.

 

We're here for you, my bruddah.

 

WHERE WE FACTOR 1!

WE FACTOR ∞!

Saga ID: 2b46e6 Aug. 5, 2018, 2:25 a.m. No.7106   πŸ—„οΈ.is πŸ”—kun

I don't know if you guys know or not, but each column has a "simple" and complex formula for their existence. The complex includes all values of "n" that exist at a given "e" column. Say for the second column, n exists at integers whose prime divisors are congruent to 1 or 3 (mod 8) (A225771). That's the complex formula. The simple is a(x) = 2*x^2+1. (A058331). Why did I name them simple and complex? because when you find the x for a given simple cell, you can find the occurrence modulus for its t. Example shown in pic related.

Anonymous ID: 358135 Aug. 5, 2018, 7:02 a.m. No.7112   πŸ—„οΈ.is πŸ”—kun

>>27

This is basically the Lorentz transformation in graphic form 2D, By changing the angle of what I called the light square from 45 degrees. See https://8ch.net/qresearch/res/2126184.html#2463036

Anonymous ID: 358135 Aug. 5, 2018, 7:38 a.m. No.7114   πŸ—„οΈ.is πŸ”—kun   >>7116

>>197

X^2 is also the square to the circle. The smaller circle radius a^2 to the radius d^2, and the difference is x^2. See ArcTriangles in https://8ch.net/qresearch/res/2126184.html#2463036

Anonymous ID: 358135 Aug. 5, 2018, 1:34 p.m. No.7121   πŸ—„οΈ.is πŸ”—kun

>>4649

You forgot the torque angle, in non perfect squares) created by the tensors rotating the central tensor square. Note the area lost in moving to a rectangle is converted to area in Z axis.

PMA !!y5/EVb5KZI ID: 6a88a2 Aug. 5, 2018, 3:51 p.m. No.7124   πŸ—„οΈ.is πŸ”—kun   >>7125

>>7047

>>7055

Using x=f-1 as the basis to chain records together enables large jumps in both (e,1) and (-f,1).

 

Attached pics are for c145 and c551 and show only the (e,n,t) values in their x=f-1 sequences.

 

For c145, the starting (1,1,12) record represents the x=f-1 move from (1,61,6) where f=24. Records for subsequent moves upwards are created by e,n,x where x=f-1, using f from the current record.

 

For c551, the starting (-25,1,37) record represents the x=f move from (22,253,12) where f=25. Note the use of x=f when moving from the e column to -f column. Subsequent moves up then follow the x=f-1 formula from the current record.

 

When moving in reverse, both columns behave similarly.

 

The previous d value can be calculated as:

 

previous_f = x + 1;

previous_d = ( previous_f + e - 1 ) / 2;

 

And then records can be created using the e,n,d values.

PMA !!y5/EVb5KZI ID: 6a88a2 Aug. 5, 2018, 4:20 p.m. No.7125   πŸ—„οΈ.is πŸ”—kun

>>7124

Attached pics are additional output of the x=f-1 sequences for c145, c259, c287, and c6107.

 

These tests show both (e,1) and (-f,1) columns including negative x. Again, only displaying (e,n,t) values for brevity.

 

The entry into (-f,1) was created via x=f, initial entry in (e,1) created via x=f-1. All other records created using x=f-1 from the previous record in the column.

 

The (-f,1) prev and (e,1) prev columns show the backwards movements, except where (0,1,1) is displayed - haven't yet figured out how to extend into negative x.

 

Finally, the t diff, div and mod columns show a continued relationship between records in -f and e to the starting f value even while the t values grow independently.

 

Also interesting to note the t value patterns. See c259 in (e,1) and c6107 in (-f,1) for examples. Perhaps binary would reveal a closer connection.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 Aug. 5, 2018, 5:47 p.m. No.7126   πŸ—„οΈ.is πŸ”—kun

Aaaaah the classics.

 

http://archive.is/ND7yN

 

Searching "vqc+++" is always a fun trip down memory lane and sometimes I find new writings from back in the day depending on what the algorithm feels like showing me at the moment.

Anonymous ID: 38e508 Aug. 5, 2018, 10:14 p.m. No.7127   πŸ—„οΈ.is πŸ”—kun   >>7128

Some transparency from VQC would go a very long way right about now… hint hint

 

I didn't think it would be super useful to post what I'd been working on since none of it went anywhere, but I guess I might as well. We're meant to calculate the "value" at (e,1) where x=f-1. One specific value. There are several values, so who knows not only which value we're looking for but also why we're looking for it. So I thought I'd just generate some of these cells and see if any values stand out. e doesn't change, so it wouldn't be that. n will always be 1 at this cell, obviously, so it wouldn't be that. We know x is equal to the original f, so it wouldn't be that, since we wouldn't need to go to a new cell to figure that out. t is directly related to x, so it wouldn't be that either. So it's going to be something to do with d, a, b, c or f. I wouldn't think it would be f, since if we continually do this f becomes exponential, and that's the opposite of what we want (logarithmic). I can't see anything else about d, a, b or c that has any relevance to literally anything else, though. I have found that in quite a few test cases, c'/a is a whole number. It isn't in every case, but it has been in most of my test cases. I don't think that's useful but I was blindly looking or something.

 

I was also thinking about diagonals again, while looking at this image >>6952 here. Maybe whatever diagonals we're looking for for whatever obscure reason we're looking for them have a gradient of 1/2, rather than being a direct 1:1 relationship between the change in e and the change in d. There are diagonals in this image but they're all across two down one.

Anonymous ID: a332a2 Aug. 6, 2018, 9:18 a.m. No.7132   πŸ—„οΈ.is πŸ”—kun

I was thinking about the asymmetry regarding -f, e and an, a(n-1), bn and b(n-1).

 

I haven't figured out how to determine which row is off by one, but it does make sense and what I find actually more strange is that an and a(n-1) occurs at the same t.

 

So if you think about it, bn and b(n-1) being off by 1 makes sense. Since we know that when a[t] = an, a[t + n] = bn then for b(n - 1) it would be a[t + n - 1] = b(n-1). If it didn't then -f wouldn't abide by the grid / column rules.

 

Another thing which seems interesting (but I'm not sure if it will hold for all cases) is that if you were to increase n by d (essentially setting d=0, giving you the column=c) then at the correct a[t] where a[t] = a(n + d), x[t] = a.

Anonymous ID: a332a2 Aug. 6, 2018, 11:08 a.m. No.7133   πŸ—„οΈ.is πŸ”—kun

(1/3)

 

I hit a doozy of a fun pattern here. Since it contains a lot of different elements to it, I'm going to split it up into a few posts describing the different patterns in it.

 

I have been looking into jumping up and the consequences of this. As you jump up (increasing n), d and x decreases (I've posted about this before >>6648) and e increases. After d jumps up, e = c, as d = 0. If you keep jumping up, e will start to decrease as d becomes negative. After a total of 2d jumps, you'll be back to the original e, however, our n value will now be different. It will be 2d + n, which also refers to a valid record in e, which is at (e, 2d + n). Here both d and x is negative, but our a and b value will remain normal. Keep jumping from here on out and it will grow in the negative space forever, but the interesting part is when it bounces back to column e after 2d jumps.

 

I made another set of posts a while ago where I described how sequences / chains in (e, n) are connected. If the number of sequences is greater than 1, then it will be paired with another set of sequences which will meet at negative x. After jumping 2d times, we end up in the paired n.

 

A few examples showing how this jumping works:

 

For this example I'll use a=7, b=37, c=259. The record for (1, c) is:

{3:114:16:15:1:259}

 

If we look at the column/row for (e, 259) we will see a total of four sequences, one is 114 and the other is 146. I showed in >>7038 that these are connected in negative x. With the example record above, when we jump 2*16 times, we end up with the record:

(3:146:-16:-17:1:259}

 

So the paired sequences meet, and our c is now in the negative x-space of the connected pair.

Anonymous ID: a332a2 Aug. 6, 2018, 11:09 a.m. No.7134   πŸ—„οΈ.is πŸ”—kun

(2/3)

 

This is a neat pattern, but does it bring us any value? Yes it does. Once I saw this I started playing with other records, so let's use the record for our actual a=7,b=37:

{3:6:16:9:7:37}

 

Now we jump 2*16 times and we end up with:

{3:38:-16:-23:7:37}

 

38 is a connected pair of 6 and also exists at (3, 7) in the record a=6, b=38. Now here comes the fun part.

This means we have actually TWO different an/bn values. We have 76 and we have 738 and we also have 376 and now 3738.

 

What makes this pattern even more fun is the following property. Let's find the record at t where a[t] = 7*38 in (3, 1).

{3:1:289:23:266:314}

And now for bn:

{3:1:1459:53:1406:1514}

 

So you might think, oh cool more numbers to work with, but this isn't something THAT awesome. Except it is. The difference in t, between an and bn is not 38, but it's 15. Why does 15 make it FUN? Because 15 is equal to x + n. That's right, for these records in negative x-space, with the connected n, the difference between a[t] = an and a[t] = bn is x + n.

Anonymous ID: a332a2 Aug. 6, 2018, 11:12 a.m. No.7135   πŸ—„οΈ.is πŸ”—kun   >>7149

(3 / 3)

 

To move on a bit. Let's look at -f, because here is also fun time. Our records in -f usually has n = n - 1, but for these they exists in n = n + 1, and also in negative x-space. There also isn't any asymmetry between them. They exists at a[t] = an, a[t] = a(n + 1) in both e and -f. Same for a[t] = bn and a[t] = b(n - 1).

 

Now all of this made me think more about >>6645

> When a appears as "an" it appears another time

Take the record for a[t] = 7x6:

{3:1:51:9:42:62}

 

After jumping 2d times, it exists again in:

{3:103:-51:-93:42:62}

 

Same holds true for bn:

{3:1:243:21:222:266}

{3:487:-243:-465:222:266}

 

To summarize a bit:

By jumping a record up by 2d you will end up with a new n-value. This means we have 2 an-values, but with different n-values. Connected through negative x. The difference between this a[t] = an' and a[t] = bn' is x + n.

Anonymous ID: 358135 Aug. 6, 2018, 5:08 p.m. No.7136   πŸ—„οΈ.is πŸ”—kun   >>7137

Link Mandelbrot to Euclidean?

 

Tetrahedrons? 4 isosceles triangles to find center of next larger circle and 3 of the smaller circles. Note these 2X circles are shifted 1/2 radius of 1X circles to the right. Mandelbrot has both scale shift and shift of self similar pattern like this diagram.

AA !dTGY7OMD/g ID: 8541f4 Aug. 7, 2018, 7:05 p.m. No.7142   πŸ—„οΈ.is πŸ”—kun   >>7143

I've been making bitmaps again because I have absolutely no fucking idea what else to do at this point until VQC comes back (seems like nobody else does either, but I hope that isn't true).

 

This is interesting. It turns out n=1 isn't the only time f=(x+n). So while f does always equal (x+n) when n=1 like from that thing I did probably months ago at this point (fuck), there are other cases that seem to follow patterns.

VA !!Nf9AmQNR7I ID: 703e75 Aug. 7, 2018, 7:13 p.m. No.7143   πŸ—„οΈ.is πŸ”—kun   >>7144

>>7142

Yeah, me too AA. I'm reviewing my notes and all the crumbs on loop, looking for new insight or things I've missed. I think we have everything we need, but are missing that crucial connection. VQC seemed to think we were VERY close.

 

I'm still mentally engaged with finding the jump from (-f na transform) and (e na transform) to the (an), a(n-1), (bn), and b(n-1) elements.

 

So where is the recursive piece of data that tells us how far to jump/move to arrive at the correct element or t value? That's what I'm looking for.

AA !dTGY7OMD/g ID: 8541f4 Aug. 7, 2018, 7:15 p.m. No.7144   πŸ—„οΈ.is πŸ”—kun

>>7143

In all honesty, right now I'm just looking for patterns again. I really doubt I'll find anything useful but I have things to procrastinate right now. Obviously that missing link thing is way more important but I really don't have any ideas. Good luck.

GAnon !jPVzzZOz2c ID: e906b3 Aug. 7, 2018, 8:09 p.m. No.7147   πŸ—„οΈ.is πŸ”—kun   >>7148 >>7150 >>7152 >>7165 >>7184

So I was trying to think of something and I was visualizing our records as sums of consecutive odd numbers. This is very evident through d+n and x+n if you look at L numbers. Another thing that helps this is if you know that a square is a sum of consecutive odd numbers. Anyway here is the picture I did it in excel. lmk if unclear in any way. Basically I'm proposing a new pathway to study. Maybe these records are linked in another way that we can visualize through these pictures and this idea but we can see in the records in another way.

Anonymous ID: a332a2 Aug. 7, 2018, 10:07 p.m. No.7149   πŸ—„οΈ.is πŸ”—kun   >>7151 >>7152 >>7157

>>7135

I was THINKING about this and I realized I haven't been thinking that much. Looking back I feel like I've been more brute forcing, more rushed and not enough stepping back and thinking about it.

 

So to give some intuition about this, what we're actually doing is to remember that the square root of any given positive integer is NOT just another positive integer. The square root of any positive integer has two results, sqrt(d) = k AND sqrt(d) = -k.

 

The reasoning is simple, the square root of a positive integer is equal to k and -k. Any negative number squared will "lose" it's negativity and become a positive integer.

 

What I'm doing above is looking at the negative part of the square root of c, which does have a different set of values for n, d, and x. This in turn gives us a different set of an and bn to work with that is different from the record where a=a, b=b and d = the positive result from sqrt(c).

Anonymous ID: 358135 Aug. 8, 2018, 5:49 a.m. No.7153   πŸ—„οΈ.is πŸ”—kun   >>7154

>>7152

Then Factors.

The 30,45,60 and 90 are all factors of 15 so break all those out into 15 degree rectangles in a new geometric plane and view. Then 15 has factors of 3 and 5 so you can break 3 and 5 out into a new geometric plane and plot those. Are you seeing my self similar geometric progression here? I place all planes at 0,0,0 for symmetry and so I don't wind up with something like the Mandelbrot.

GAnon !jPVzzZOz2c ID: e906b3 Aug. 8, 2018, 9:37 a.m. No.7154   πŸ—„οΈ.is πŸ”—kun

>>7153

I wasn't thinking about the circle with this. It would be an iterative approach. After work I'm gonna try to find relevant records and map it out more clearly.

VQC !!/aJpLe9Pdk ID: 815a08 Aug. 8, 2018, 12:49 p.m. No.7155   πŸ—„οΈ.is πŸ”—kun   >>7158 >>7161 >>7163 >>7184 >>7216 >>7414

One last hint before the code. Try for larger numbers too.

What do you see when subtract n from BigN? In any e column.

Remember, BigN-1 - (n-1) is the same number in column -f

Biggest hint yet.

Last one.

Reverse of that is the answer to finding the solutions.

Smooth numbers?

Shortcut to the general number field sieve?

Calculation instead of search.

Anonymous ID: 358135 Aug. 8, 2018, 4:01 p.m. No.7159   πŸ—„οΈ.is πŸ”—kun   >>7160

>>7156

Yes we can graph it as I plane or call it z' or what every you want, but we in the normal sphere don't have the programs to create a 4D and 5D graph we have to rotate and overlap it on X or Y.

AA !dTGY7OMD/g ID: 077d1d Aug. 8, 2018, 4:56 p.m. No.7161   πŸ—„οΈ.is πŸ”—kun

>>7155

Hey VQC, I understand what you're trying to do by giving us as much ability to solve this as we can before you potentially post it yourself, but I've got to ask. This stuff about diagonals and x=f-1 seems to have thrown everybody off significantly. Were we meant to have figured that out ourselves? We still don't seem to know what to do with either clue. Considering you've said many times that we have enough information to figure this out, did we have enough information to figure out that diagonals and x=f-1 are significant?

Anonymous ID: 358135 Aug. 8, 2018, 5:20 p.m. No.7162   πŸ—„οΈ.is πŸ”—kun   >>7164

>>7160

Interesting.

The slider thing up to the 3D definitely follows the sin and cos wave functions. After that since the dimensional origin area starts poking back into the the lower dimensions. Thus the Hopf sphere. Sorry second picture is EM but basically shows Cos as X axis and Sin as Y axis.

ID: 96e43e Aug. 8, 2018, 6:58 p.m. No.7166   πŸ—„οΈ.is πŸ”—kun

Some papers on GNFS:

 

http://www.ams.org/notices/199612/pomerance.pdf

 

An explanation of how it works is underway. It uses smooth numbers, though. In the 4th paper there is a part on smooth numbers

Anonymous ID: a332a2 Aug. 9, 2018, 5:29 a.m. No.7176   πŸ—„οΈ.is πŸ”—kun

I was thinking about our numbers in (e, 1), and for odd e's they consist of 4 triangles. What does this really mean?

 

I was playing with it and to build some intuition about it I drew up four of these triangles. Instead of multiplication, let's say we want to simply add them together. In the image above I used t=8 for (3, 1) which is the two N's that belong to c=259.

 

So for an odd e, we have the equation:

2t(t-1) + (e + 1)/2

 

That is 4 triangles plus half ( + 1) of the remainder. So if we want to add the two values of this cell we would get:

 

2t(t-1) + (e + 1)/2 + 2t(t+1) + (e + 1).

 

This is the same as:

2t(t+1) + 2t(t-1) + e + 1. This means adding this rows will give us a total of 8 triangles, half of which are one unit longer than the other half. Image should clear this a bit up.

 

I am contending now that take a number c, if this number has an odd remainder, then adding 1 to it will give you a number who's sum can be expressed as the sum of a and b from a single record in (e, 1).

 

So for 259, add 1 and you get 260. This number is equal to the sum of it's two big N values (Remember turns out each number has 2 n values as one belongs to a positive d and the other to the negative d).

 

Note: In the image I added 114 + 146, both are 4 triangles, one a unit longer than the other (114 = orange, 146 = green), but I omitted e + 1 =3 + 1 = 4. That is, there are 4 left over.

 

You can also see how it consists of 4 squares each of which has the square root of 8, because it belongs on t=8.

 

I don't think this is anything new, but it is just to help people get a better grasp of the values in (e, 1). For even e's every a and b consists of two squares added together, so a + b from any cell[t] in (e, 1) where e is even is the same as 4 squares added together. Quite similar to what we have in the image.

Anonymous ID: 358135 Aug. 9, 2018, 4:53 p.m. No.7179   πŸ—„οΈ.is πŸ”—kun

I noticed my Naked Wave Axis were a little too warped to understand real well so I cleaned them up - probably loosing mathematical precision.

VA !!Nf9AmQNR7I ID: 1b7977 Aug. 9, 2018, 5:59 p.m. No.7181   πŸ—„οΈ.is πŸ”—kun

>>7178

We're the Agartha conspiracy theory recruits for another go round of Cicada 3301. So they haven't solved it yet ?? I like the VQC way, it's fun and enjoyable to work on.

VA !!Nf9AmQNR7I ID: 1b7977 Aug. 9, 2018, 6:29 p.m. No.7182   πŸ—„οΈ.is πŸ”—kun

>>5690

 

>For odd (x+n) where:

>nn + 2d(n-1) + f - 1

>This is eight triangles surrounding a single unit square.

>Using an arbitrary divisor for f, then each of the eight triangles will have one OR one of two (the latter when c is large and the product of two different prime numbers) configurations in each triangle. The difference between those two configurations of a portion of f are that they are both staircase numbers where the base of one is a unit longer than the other. E.g. (3,4,5) and (4,5,6)

>This might help visualise and I hinted at this in an earlier diagram.

>https://en.wikipedia.org/wiki/Polite_number

>This will REALLY help if you visualise the smaller (x+n) squares and throw in a few much larger (x+n) squares. Focus on the patterns in the way f is distributed in each of the 8 triangles. Look at symmetries and how to construct the triangles when you know you need the two different portion sizes of f. This affects the values of 2d(n-1)

>There is a pattern to these that matches the grid which is why later you'll look back at the grid (The End) and understand all the puns and analogies of using that term.

 

for c6107, f = 134, f-1= 133

(f-1)/8 = 16

(f-1) mod 8 = 5

sqrt 16 = 4

T(4) = 10

T(3) = 6

T(4)+T(3) = 16

8 * (T(4)+T(3)) + 5 = 133

So, the polite triangle numbers are 4+3+2+1=10 and 3+2+1=6

I think the mod = 5 needs to be accounted for as well. So for each time we fill each of the 8 triangles with 16, we also add 5. This is the "One unit asymmetry" VQC was referring to.

 

Thoughts, Anons?

GAnon !jPVzzZOz2c ID: e906b3 Aug. 9, 2018, 6:57 p.m. No.7184   πŸ—„οΈ.is πŸ”—kun   >>7185

>>7147

>>7165

Realized these are the same two algorithms but the first was a little optimized (skipped every other). The goal from this perspective is to find a number K which is equal to (d+n)^2 - d^2 BUT this number must also be equal to e + m^2

 

(d+n)^2 - d^2 = e + m^2

dd + 2dn + nn - dd = e + m^2

n(2d+n) = e + m^2

 

for c=145

 

5(212+5) = 1 + m^2

5*(24+5) = 1 + m^2

5*(29) = 1 + m^2 =m = d [convenient because n=a]

 

From start (c=145):

(1,61,12,11,1,145)

61*(24+61) = e + m^2

 

13,31

[3, 2, 20, 7, 13, 31]

 

2(202+2) = 3 + m^2

2*(42) = 3 + m^2

84 = 3 + m^2

m = 9

 

84/4 = 21

 

This might be a path to a recursive solution of sorts but you need the solution to generate the numbers (as far as I know yet)

 

Also I've been thinking about this

>>7155

We need a way to use this info to our advantage. I think that we can use this to look at the different values in (e,1) and (-f,1). So we know that at the correct x in (e,1), A[t] = na. At the same x+1 in (-f,1), A[t]=(n-1)a. Also I've done some looking and I remember (pic related) that if A[t] is divisible by a, then A[t+a] is divisible by a. This also means that if A[t] is divisible by n, then A[t+n] is divisible n. Maybe we can multiply that stuff out. Here is my attempt it might be good.

 

n = 1

 

2na = xx + e

 

e%2==0:

X[t] = 2(t-1)

A[t] = (X[t]*X[t] + e)/2

= (2(t-1)*2(t-1) + e)/2

= (4(t-1)(t-1) + e)/2

= (4(tt-2t+1) + e)/2

= (4tt-8t+4+e)/2

= 2tt - 4t + 2 + e/2

 

e%2==1:

X[t] = 2t-1

A[t] = (X[t]*X[t] + e)/2

= ((2t-1)(2t-1) + e)/2

= (4tt - 4t + 1 + e)/2

= 2tt - 2t + (e+1)/2

 

diff even e:

 

eA[T]a - eA[t] = fA[T]a - fA[t]

 

[2TT - 4T + 2 + e/2]a - [2tt - 4t + 2 + e/2] = [2TT - 2T + (-f+1)/2]a - [2tt - 2t + (-f+1)/2]

 

2TTa - 4Ta + 2a + ea/2 - 2tt + 4t - 2 - e/2 = 2TTa - 2Ta + (-f+1)a/2 - 2tt + 2t - (-f+1)/2

 

  • 4Ta + 2a + ea/2 + 2t - 2 - e/2 = - 2Ta + (-f+1)a/2 - (-f+1)/2

  • 4Ta + 2a + ea/2 + 2t - 2 - e/2 = -2Ta + (-(2d+1-e)+1)a/2 - (-(2d+1-e)+1)/2

  • 4Ta + 2a + ea/2 + 2t - 2 - e/2 = -2Ta + (-2d + e)a/2 - (-2d+e)/2

  • 4Ta + 2a + ea/2 + 2t - 2 - e/2 = -2Ta + -da + ea/2 + d - e/2

  • 4Ta + 2a + 2t - 2 + ea/2 - e/2 = -2Ta + -da + d + ea/2 - e/2

  • 4Ta + 2a + 2t - 2 = -2Ta + -da + d

  • 2Ta + 2a + 2t = -da + d

  • 2a(T+1) + 2t = d - da

 

Since our e is even, our x value is even. This means 2(t-1)=x, so 2t-2=x, so 2t=x+2)

 

  • 2a(T+1) + x + 2 = d - da

 

Now we have T to decode.

eA[T] = N (because we multiply by a outside the function)

N = 2TT - 4T + 2 + e/2

N - e/2 = 2TT - 4T + 2

0 = 2TT - 4T + (2 + e/2 - N)

 

quad

 

4 +/- sqrt(16 - 42(2 + e/2 - N))

/2*2

 

4 +/- sqrt(16 - 16 - 4e + 8N)

/4

1 +/- sqrt(8N - 4e)/4

 

So T = 1 +/- sqrt(8N - 4e)/4

 

So now we can calculate big T whenever we want.

 

  • 2a(T+1) + x + 2 = d - da

  • 2(d-x)(T+1) + x + 2 = d - d(d-x)

x + 2 - d = 2(d-x)(T+1) - d(d-x)

x + 2 - d = (d-x)(2(T+1) - d)

x + 2 - d = (d-x)(2T + 2 - d)

x + 2 - d = 2dT + 2d - dd - 2Tx - 2x + xd

2 - d - 2dT - 2d + dd = - 2Tx - 2x - x + xd

(2 - d - 2dT - 2d + dd)/(- 2T - 2 - 1 + d) = x

 

Someone check this??

AA !dTGY7OMD/g ID: 24e214 Aug. 10, 2018, 3:37 a.m. No.7192   πŸ—„οΈ.is πŸ”—kun   >>7193

>>7191

I've tested this on non-semiprimes too now (in fact every odd number up to 100, and a bunch of bigger ones), and it seems to hold true for every set of a and b where c is odd (not evens since they don't have BigN). Here's some example output.

 

54321BigN cell: (32,26928,117) = {32:26928:233:232:1:54321}, f = -435(32,272,89) = {32:272:233:176:57:953}, f = -435BigN - n = 26656(BigN-n)/(b-1) = 28, (BigN-n)%(b-1) = 0(a-1)/2 = 28(32,1206,108) = {32:1206:233:214:19:2859}, f = -435BigN - n = 25722(BigN-n)/(b-1) = 9, (BigN-n)%(b-1) = 0(a-1)/2 = 9(32,8822,116) = {32:8822:233:230:3:18107}, f = -435BigN - n = 18106(BigN-n)/(b-1) = 1, (BigN-n)%(b-1) = 0(a-1)/2 = 1

 

I don't know if this is actually useful in any way, but it does seem pretty weird to me that the distance between BigN and n even has any relationship with a and b values that aren't actually related to BigN.

AA !dTGY7OMD/g ID: 24e214 Aug. 10, 2018, 4:46 a.m. No.7193   πŸ—„οΈ.is πŸ”—kun   >>7194 >>7201

>>7191

>>7192

Something’s fishy about this. It’s definitely true that (BigN-n)/(b-1) = (a-1)/2 for every n, but it doesn’t make algebraic sense. So why is it the case?

 

(BigN-n)/(b-1) = (a-1)/2

((((c+1)/2)-d) – (((a+b)/2)-d)) / (b-1) = (a-1) / 2

(((c+1)/2)-d) – (((a+b)/2)-d) = ((a-1)/2)(b-1)

((c+1)/2)-d = (((a-1)/2)(b-1)) + ((a+b)/2) – d

(c+1)/2 = ((a-1)/2)(b-1) + (a+b)/2

c+1 = (a-1)2(b-1) + a+b

c+1 = (a-1)(2b-2) + a+b

c+1 = 2c-2a-2b+2 + a+b

c+1 = 2c-a-b+2

c+a+b-1 = 2c

 

The difference between c and 2c is a+b-1, according to the algebra. But it definitely isn't true.

c559

559 =/= 13+43-1

c203

406 =/= 7+29-1

Anonymous ID: 358135 Aug. 10, 2018, 5:03 a.m. No.7196   πŸ—„οΈ.is πŸ”—kun   >>7197

>>7195

Well I noticed in reviewing the original stuff that it looked like diagonals reaching away in space that would make triangles (and squares) or dimension shifts. I notice everyone treating them like points or lines but not connecting the different points.

AA !dTGY7OMD/g ID: 24e214 Aug. 10, 2018, 5:10 a.m. No.7197   πŸ—„οΈ.is πŸ”—kun   >>7198

>>7196

Each of the points in that first picture is a pair of e and n values. e is the x axis, n is the y axis. The thing I'm talking about involves two points with the same e value but two different n values. It would be a vertical line. Also here's a more hi-res version since you seem to have saved a thumbnail by mistake. I don't mean to be patronizing, but I'm asking this because I'm not sure who you are: do you know much about what we're doing here? Do you know about all of the different variables, how they relate to each other, and the other relationships we've been finding? VQC has mentioned diagonals being important, so you're probably going in a pretty useful direction with your thinking, but I don't think it's related to the thing I was posting about.

Anonymous ID: 358135 Aug. 10, 2018, 5:21 a.m. No.7198   πŸ—„οΈ.is πŸ”—kun   >>7199 >>7202

>>7197

You are right I got lost as the algebra got more complex and just put it down to encryption stuff; and that is not my thing. But same e and two different n could be a right triangle of 45,45,90 degrees if the numbers fit.

AA !dTGY7OMD/g ID: 29572d Aug. 10, 2018, 7:37 a.m. No.7202   πŸ—„οΈ.is πŸ”—kun   >>7203 >>7207

>>7201

I wrote it kinda weird.

(c+1)/2 = ((a-1)/2)(b-1) + (a+b)/2

Multiply each thing by 2

c+1 = 2((a-1)/2) * 2(b-1) + (a+b)

Simplify

c+1 = (a-1)*(2(b-1)) + a + b

Which is what I meant by

c+1 = (a-1)2(b-1) + a+b

which is typed out wrong. The logic's still there.

 

>>7198

>>7199

>>7200

If you know what to do, feel free to do it. I don't think I really understand how you'd do that but it could be useful. It's an avenue I don't think any of us have explored, so I'm sure there's a lot to discover. Some examples of points would be (e,n) = (30,5) and (30,257) for a=13, b=43, and then (e,n) = (7,4) and (7,88) for a=7, b=29.

Anonymous ID: 358135 Aug. 10, 2018, 4:24 p.m. No.7207   πŸ—„οΈ.is πŸ”—kun   >>7208

>>7202

I am a bit limited with points. But here are the options with two points and two numbers that can be flipped one or the other way for points. a and b could also be a line segment but I don't know where to put that line segment.

AA !dTGY7OMD/g ID: 92669d Aug. 10, 2018, 10:37 p.m. No.7211   πŸ—„οΈ.is πŸ”—kun   >>7212

Something I’ve been going over with baker/Jan (and briefly VA) on Discord:

 

So (x+n)(x+n) = 2d(n-1) + nn + f – 1. We’ve known this for months now. You can create the smaller square that you’d add to c to make the bigger square with 2d(n-1), nn, f and -1. I was just thinking about why VQC brought up sqrt(2d) and sqrt(e). 2d(n-1)+nn+f-1 is always going to be a square with a valid cell. So if we take the square root of 2d(n-1)+nn+f-1, we’ll get x+n, right?

x+n = sqrt(2d(n-1)+nn+f-1)

This is true, and can be verified with test cases. e.g. c559, (30:5:6) = {30:5:23:10:13:43} f=17

x+n = sqrt(2d(n-1)+nn+f-1)

10+5 = sqrt(46(4)+25+17-1)

15=sqrt(225) (this is true)

 

What’s weird is when you use algebra to try to isolate sqrt(2d) or any of the other elements.

x = sqrt(2d(n-1)) + sqrt(f) - 1

This can also be represented as

x = sqrt(2d(n-1)) + sqrt(2d) – sqrt(e)

which isolates sqrt(2d) and sqrt(e), two newer elements that VQC brought up probably about a month ago now. This seems like it makes algebraic sense, right? I mean, as an example, 16=28, and sqrt(16)=sqrt(2)sqrt(8). So if you plugged in the numbers, surely you’d calculate the correct x, right? Well, this x should be 10. Let’s plug in some numbers. First with the sqrt(f) equation:

x = sqrt(2d(n-1)) + sqrt(f) – 1

x = sqrt(46(4)) + sqrt(17) – 1

x = 16.687765592

Clearly that isn’t correct. Let’s also do it with the 2d and e equation.

x = sqrt(2d(n-1)) + sqrt(2d) – sqrt(e)

x = sqrt(46(4)) + sqrt(46) – sqrt(30)

x = 14.869764374

This not only also isn’t equal to 10 but it also isn’t equal to the sqrt(f) version of the equation.

 

So does anyone know why this would happen? Is it us making a mistake with the algebra, or is there something weird going on with these squares?

Anonymous ID: 7b4f3f Aug. 10, 2018, 11:57 p.m. No.7214   πŸ—„οΈ.is πŸ”—kun   >>7215

>>7213

Well, I mean, it's algebraically possible to start from the (x+n)^2 equation and rearrange it to solve for sqrt(2d), but the result isn't useful since it still depends on x, n, and f.

Both d and e are easily calculated directly from c, so I'm not sure what good isolating sqrt(2d) or sqrt(e) does.

AA !dTGY7OMD/g ID: 15edb9 Aug. 11, 2018, 12:56 a.m. No.7215   πŸ—„οΈ.is πŸ”—kun   >>7219 >>7220

>>7214

The point is that VQC suddenly brought up sqrt(2d), sqrt(e) and sqrt(f) out of the blue when he started posting code comments, and he didn't explain why he brought them up at all. Since 2d(n-1) + nn + f - 1 is a square (since it equals (x+n)(x+n)), taking the square root will give us (x+n). Taking the square root of f means taking the square root of 2d+1-e. That means this could potentially be why he brought up sqrt(2d) and sqrt(e). So if we can isolate sqrt(2d) and/or sqrt(e) from that equation that we use to calculate x+n, we might know what it's used for. So do you think you could do that? Because I can't remember how you're meant to isolate a variable from sqrt(x+y), and neither does the only other person I've gone through this with here. For all I know you might be able to do it with logs but it's been a while since I've done anything with logs.

Anonymous ID: a332a2 Aug. 11, 2018, 3:37 a.m. No.7216   πŸ—„οΈ.is πŸ”—kun   >>7238

>>7155

I'm not too familiar with the General Number Field Sieve, but smooth numbers appear to be a big part of it.

 

Now first simplify our calculations for our two N's (one that is 2d "jumps" away, or negative d)

 

Our normal 'n' is defined as:

n1 = (a + b)/2 - d

 

But our paired n is:

n2 = (a + b)/2 + d

 

These two n's exist in (e, a) as a and b for some t.

 

But, over to the smooth numbers.

 

We know our formula for n and big N, big N is:

 

bigN1 = (1 + a*b)/2 - d

bigN2 = (1 + a*b)/2 + d

 

If we subtract bigN1 - n1 and bigN2 - n2 we have the same number, which appear to always be a smooth number.

 

When we do bigN1 - n1 we actually do:

 

((1 + ab)/2 - d) - ((a + b)/2 - d)

 

Here the d's cancel each other out and we're left with

 

(1 + ab)/2 - (a + b)/2 = k (for some smooth number k)

1 + ab - a - b = 2k

ab - a - b + 1 = 2k

(ab - a - b + 1)/2 = k

 

This can also be rewritten to:

(a - 1)(b - 1)/2 = k

or

(a - 1)(b - 1) = 2k

 

Also note that (a + b)/2 is equal to (d + n). That means if we remove (d + n) from our big (d + n) we get the smooth number we want(?)(Unless we want to remove smooth numbers from our big (d + n) to get our smaller (d+n)).

GAnon !jPVzzZOz2c ID: e906b3 Aug. 11, 2018, 7:55 a.m. No.7217   πŸ—„οΈ.is πŸ”—kun   >>7218 >>7221

>>7151

>>7156

 

BIG post

 

We can at least begin to understand the complex plain.

 

Don't think of it as a + bi, think of it as Ae^(iΞΈ)

 

e^(iΞΈ) = cos(ΞΈ) + i * sin(ΞΈ)

 

[proof]

Probably the coolest proof of all time. Conceptually its out of this world.

 

Taylor Series [look up for e, sin, cos to validate]

 

e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + …

e^(iΞΈ)= 1 + (iΞΈ) + (iΞΈ)^2/2! + (iΞΈ)^3/3! + (iΞΈ)^4/4! + …

e^((iΞΈ) = 1 + (iΞΈ) - ΞΈ^2/2! - iΞΈ^3/3! + ΞΈ^4/4! - …

e^2(iΞΈ) = (1 - ΞΈ^2/2! + ΞΈ^4/4! - ΞΈ^6/6! + …) + i(ΞΈ - ΞΈ^3/3! + ΞΈ^5/3 - …)

 

e^2(iΞΈ) = cos(ΞΈ) + isin(ΞΈ)

 

Then, from this, we can get any angle from ΞΈ because these are all the points on a circle. At that point if we multiply by a constant, we 'shoot' or expand the line out in its current direction. Then numbers are Ae(iΞΈ).

 

If we want to multiply two numbers, the way we can think about it is you multiply the LENGTHS together an that is your A value, then you add the ΞΈs and that is your angle. I made you people some pics to show you how it works. The vertical axis is +i, horizontal is real numbers

 

The first is all different numbers with the same radius and different thetas randomly made. It's made to be viewed each pic seperately. Basically you notice that both first points are on the same inner circle and no matter the angle they always end up on the circle of radius^2.

 

The second is to give you an idea of the angles and I made it like a clock with hours and minutes because why not.

 

The third and fourth are to help you imagine the complex conjugate numbers. These are of the form,

so they are on the same real axis and have opposite imaginary numbers.

This means vertically they differ (in the pics) but horizontally they are the same

(a+bi)(a-bi) = aa +abi - abi -bibi = aa - bb

So this will always land you in the real numbers.

Moreover, if the radius of the number and its conjugate is equal to the radius of others, then they will land on the same point pics related.

These look like the death star.

 

Lastly are the 4 pics for squaring a number.

These look the coolest. They always end up making a cardioid with other pics related, which looks like the rebel alliance

http://demonstrations.wolfram.com/

ModularMultiplicationOnACircle/

 

https://www.youtube.com/watc

h?v=qhbuKbxJsk8

Anonymous ID: 7b4f3f Aug. 11, 2018, 9:34 a.m. No.7219   πŸ—„οΈ.is πŸ”—kun

>>7215

Sure, let me do the algebra, though again I warn you that I don't think the result is particularly elegant or useful. We start with the (x+n)^2 equation you had:

 

(x+n)(x+n) = 2d(n-1) + nn + f – 1

x^2 + 2nx + n^2 = 2d(n-1) + n^2 + f - 1

x^2 + 2nx - f + 1 = 2d(n-1)

(x^2 + 2nx - f + 1) / (n-1) = 2d

sqrt( (x^2 + 2nx - f + 1) / (n-1) ) = sqrt(2d)

 

That's what solving for sqrt(2d) gives you.

 

A quick check with the c=559 record:

 

sqrt( (10^2 + 2510 - 17 + 1) / (5-1) ) = sqrt(2*23)

sqrt( (100 + 100 - 16) / 4 ) = sqrt(46)

sqrt( 184 / 4 ) = sqrt(46)

sqrt(46) = sqrt(46)

Anonymous ID: a332a2 Aug. 11, 2018, 10:09 a.m. No.7220   πŸ—„οΈ.is πŸ”—kun

>>7215

Keep in mind that:

 

d = (a + b)/2 - n

2d = a + b - 2n

sqrt(2d) = sqrt(a + b - 2n)

 

Make the record for (1, c). 2d here could also point to the record for 4c (or a=2, b=2c) as the d there is 2d.

 

dd + e = c

dd = c - e

dd = ((a + b)/2 - n)*((a + b)/2 - n)

dd = ((a + b - 2n)^2)/4

dd = ((2d)^2)/4

((2d)^2)/4 = c - e

Anonymous ID: 358135 Aug. 11, 2018, 11:48 a.m. No.7221   πŸ—„οΈ.is πŸ”—kun

>>7217

I will happily pick your brain on this as sin and cos are the answer (ie wave axis), this is what I am working on right now. The basic gravity deviation of Einstein tensor that would create a symmetrical 100 unit space time deviation and how it might relate to sub 4D dimensions. One anomaly is the 100 degree angle and 50.73 length. Still working on below 4D line measurements and angles.

AA !dTGY7OMD/g ID: 39485d Aug. 11, 2018, 8:15 p.m. No.7223   πŸ—„οΈ.is πŸ”—kun   >>7231

Just speaking personally here but I've exhausted all of my ideas related to BigN-n. That (BigN-n)/(b-1) = (a-1)/2 thing was the only thing I could find, and I don't see any way to use it considering every variable other than BigN is an unknown (and algebra doesn't seem to change that). Anyone else getting anywhere? I think I remember PMA saying his thing didn't go anywhere but I might be wrong.

GAnon !jPVzzZOz2c ID: e906b3 Aug. 11, 2018, 10:52 p.m. No.7224   πŸ—„οΈ.is πŸ”—kun   >>7225 >>7226

>>7222

Okay so here is what I did for that picture I wanted to see what squaring complex numbers would be like. And I know that complex numbers typically deal with circular stuff so I decided to keep them all the same length and just change their angle. Then I multiplied the number by itself. Then I drew a line from 0 to the point and one from the point to the square of the point. As I changed the number (went around counter clockwise from 3:00) I went from red to green.

 

Or do this:

 

iterations = 200 #arbitarary

radius = 4 #arbitrary

 

for i in range(iterations):

theta = 2.0pii/iterations

point = radius * (cos(theta) + i*sin(theta))

square = point * point

draw_line_series(0,point,square)

Anonymous ID: 358135 Aug. 11, 2018, 11:18 p.m. No.7226   πŸ—„οΈ.is πŸ”—kun

>>7224

And when you change angles across dimensions it changes the line length via Lorentz.

Also note when you create a square you also create a grid in a new dimension.

I get your process. It is like when I rotate around the axis a set (form) You are rotating - remember to calculate the angle rotated.

Anonymous ID: 358135 Aug. 11, 2018, 11:47 p.m. No.7229   πŸ—„οΈ.is πŸ”—kun

>>7227

But when you do switch dimensions remember there is two angles the one on the plane, the one to the new dimension and then the angles related on the new dimension. Plane, Plane shift Then new plane

Anonymous ID: a332a2 Aug. 12, 2018, 12:29 a.m. No.7231   πŸ—„οΈ.is πŸ”—kun

>>7223

 

I think the point is that big N - n is a smooth number, and they're used in number field sieving.

 

I've been reading a bit about GNFS and from what I can understand is that it goes through several steps to setup the algorithm before actually running, which is why it isn't used for smaller numbers. One of the steps in the algorithm is that it sieves for smooth numbers. I wonder if we're either going to find the smooth number (Big n - n) through calculations, or if this is just a hint towards finding the speed up algorithm which will lead us to solving the calculation.

 

We know that big N - n (column e) is the same as (big N - 1) - (n - 1) (column f). I wonder if that means we're going to find this number in two calculations using different columns. The question is if this was always about finding u, the triangle base created by x + n. I assumed it was, but now it seems we're stopping by GNFS, so I'm not sure anymore. Maybe we're stopping by GNFS before finally arriving at the calculation of our triangle base?

ID: 96e43e Aug. 12, 2018, 1:01 a.m. No.7232   πŸ—„οΈ.is πŸ”—kun

We were analyzing 2d today. Below are some tasks I think are worth looking at. Probably for rsa100 it should be done, so irrelevant patterns go away.

 

sqrt0 = floor_sqrt

analyze sequence f(x) = sqrt0(x*d), starting from x=1

analyze sequence f(x) = sqrt0(x*e), starting from x=1

analyze sequence f(x) = sqrt0(x*f), starting from x=1

 

hypothesis 1

the first and second sequence always repeat

 

if they don't repeat or repeat in too many terms, you can tack on another sqrt (just like the tree) and they will repeat.

 

my idea for analyzing these sequences comes from looking at sqrt(2d) and asking are 3d and 4d significant as well? combined with the remainder tree and the fact that finding where a sequence repeats solves the number in Shor's algorithm.

ID: 96e43e Aug. 12, 2018, 1:05 a.m. No.7233   πŸ—„οΈ.is πŸ”—kun

hypothesis 2

if 1 in f=2d+1-e is actually A in (e, N, T) then maybe f2=2d+a-e is significant too.

 

hypothesis 3

if the 2nd hypothesis is correct, then maybe 1 in (n-1) is actually (n-A) and therefore (n-a) is significant too

 

f2 - f = a - A

Anonymous ID: 358135 Aug. 12, 2018, 7:45 a.m. No.7235   πŸ—„οΈ.is πŸ”—kun

Finished the calculations below 4D line. Note the triangles below 4D line are spun and define the 10 degree shift noted inside the 4D sphere.

Anonymous ID: a332a2 Aug. 13, 2018, 9:03 a.m. No.7238   πŸ—„οΈ.is πŸ”—kun   >>7239 >>7255 >>7346

>>7216

I haven't had too much time to work on this lately, but I have some more patterns related to our two n-values.

 

We have two n-values for any given record. One is for positive d, the other is for negative d. The difference between the n-values is 2d.

 

For positive d, we have n. For negative d we have n' (n MARK).

 

For n we have the known pattern of:

a[t] = an, a[t + n] = bn

For n' we have:

a[t] = an', a[t + x + n] = bn' (Note t + x + n, as in NOT n')

 

We can create an f for the cell for n', this will be negative (while our normal f is positive).

 

The difference between these two, f and f' is e + 1.

 

Now some fun patterns crossing these two sets:

a[t] = an

a[t + d + n] = bn' (Again, note the [t + d + n], it is NOT n')

a[t] = bn

a[t + d] = bn'

a[t] = an'

a[t + x + n] = bn'

 

The difference in t between a[t] = an and a[t] = an' is b.

The difference in t between a[t] = bn and a[t] = bn' is d.

 

Would love it if someone verified.

Anonymous ID: a332a2 Aug. 13, 2018, 10:33 a.m. No.7239   πŸ—„οΈ.is πŸ”—kun   >>7241

>>7238

Minor correction:

 

a[t] = an

a[t + a] = an'

 

That means the difference in t between a[t] = an and a[t] = an' is not b, it is a.

 

a[t] = bn'

a[t - b] = bn (Might be in the negative part of the grid =-x)

Anonymous ID: a332a2 Aug. 13, 2018, 10:39 p.m. No.7241   πŸ—„οΈ.is πŸ”—kun

>>7239

Just adding some more patterns:

 

a[t] = an'

a[a + 1 - t] = an (Might be negative x)

 

a[t] = bn'

a[b + 1 - t] = bn (Might be negative x)

 

We know we have sequences in (e, n), sometimes we have 1, other times there is more than 1 sequence. But our record is always tied to a sequence within (e, n).

 

a[t] = an

a[a + 1 - t] = ak

 

This time k belongs to a different sequence and points to a record where our 'a' exists as b and k is some value of from that sequence.

 

Example:

Take a=7,b=37

 

a[5] = 42 (7x6 = an for record above)

a[7 + 1 - 5] = 14 (7x2, this record exists in (3, 2) where a=1, b=7)

 

Another example:

Take a=37, b=91

 

a[11] = 222 (37x6 = an for record above)

a[37 + 1 - 22] = 1406 (37x38, this record exists in (3, 38) where a=7, b=37. Note this is also the second n (where d=-16) for the record a=7,b=37).

 

So a[p + 1 - t] allows us to move to a record where our a becomes b, notice also how they swap back again, 37 + 1 - 22 = 16, 37 + 1 - 16 = 22. So we can think of a[p + 1 - t] to move between records, as if they were a linked list.

 

Since a[p + 1 - t] points to ak, where k is some value of (e, k), we should have enough data to make that record. We start with the record: {3:6:124:33:91:169}

 

a[t] = an =a[17] = 546

a[91 + 1 - 17] = a[75] = 11102. 11102 / 91 = 122. We then create a record using ENB for 3, 122, 91.

{3:122:-58:-95:37:91}

a[t] = an =a[48] = 4514

a[37 + 1 - 48] = 222. 222/37 = 6

{3:6:16:9:7:37}

a[t] = an =a[5] = 42

a[7 + 1 - 5] = 14, 14/7 = 2

{3:2:2:1:1:7}

Anonymous ID: a332a2 Aug. 14, 2018, 10:13 p.m. No.7246   πŸ—„οΈ.is πŸ”—kun   >>7247

I've been thinking more about the number of records that we are interested in that exists ONLY in column e, for some c.

 

I've compiled a list:

 

The structure of the list is as follows:

(a, b) denotes the record where a=a, b=b.

(1, c) denotes the record where a=1, b=c

(e, 1) denotes the records that exists in column e at row 1

 

n = The n we know

n' = The shadow n

 

Column e:

(a, b)

  • (e, n)

  • (e, n')

(1, c)

  • (e, n)

  • (e, n')

(e, 1)

  • an

  • bn

  • an'

  • bn'

  • N

  • N'

  • cN

  • cN'

 

This gives us a total of 12 records for any given a, b that are of interest to us. Out of these 12 we can already generate 6 of them. Below is the same list only filtered based on what we can generate from c alone:

 

(1, c)

  • (e, n)

  • (e, n')

(e, 1)

  • N

  • N'

  • cN

  • cN'

 

This is not including records from -f, so we're really dealing with a lot more cells here. We should be dealing with another 12 records for -f, knowing 6 of them. So using e and -f, we should have a total of 12 records that we can generate based on c alone, out of a total of 24 records.

Anonymous ID: a332a2 Aug. 14, 2018, 10:20 p.m. No.7247   πŸ—„οΈ.is πŸ”—kun

>>7246

Since the board is drying up a bit and people are stuck, I'm throwing out a few ideas I've been working on.

 

I've been trying to figure out a way of using the recursive tree by generating x=f or x=f-1 based on the columns we get from recursing on d and e. However, I've been using too small a numbers and I'm getting a lot of false positiives (think gcd on values in these records where x=f, x=f-1).

 

Looking at patterns of t from the 12 records we can create. Use bigger numbers to remove irrelevant patterns (of course when looking for patterns you can generate the other 12 records).

 

Defining a problem statement, I'm as guilty as you guys, but we should definitely do this as it's been a big hint. We could try to formulate one together here and I'm pretty convinced now that we're missing a variable that will help us.

 

Look for more patterns. Since I've felt a bit stuck I decided to go back to the grid and look for more, which is how I found the shadow n and the patterns within it. Those are damn cool and if you're stuck and out of ideas, go back and verify those patterns. Maybe I missed something, maybe you can find more and deeper patterns within them.

 

 

I'm not going to sit around and wait for VQC to reveal it all, I'd rather try and fail until he does.

ID: 96e43e Aug. 15, 2018, 1:07 a.m. No.7248   πŸ—„οΈ.is πŸ”—kun   >>7249

I've been testing triangle numbers on rsa100.

 

If you calculate T-1(N) and take the difference of T(T-1(N)) and T(T-1(N)+1) you get a number that is extremely close to T-t on the order of same amount of digits (the difference in elements between na transform and solution in e,1)

 

Also if you recursively calculate a summation like so

 

T(1)+T(2)+T(4)+T(11)+T(68)+T(2407)+T(2899431)+T(4203353390611)+T(8834089863183560067072041)+T(39020571855401265512289573339484371018905006900193)+T(N)

 

and

 

T(1)+T(2)+T(4)+T(11)+T(68)+T(2407)+T(2899431)+T(4203353390611)+T(8834089863183560067072041)+T(39020571855401265512289573339484371018905006900193)+T(N-1)

 

where each preceding term from N and N-1 are the triangle inverses of the last term, then the difference between those sequences is N.

 

if you remove N and N-1 as terms and take the difference with N like so you also get a number that is extremely close to T-t on the order of same amount of digits

 

T(1)+T(2)+T(4)+T(11)+T(68)+T(2407)+T(2899431)+T(4203353390611)+T(8834089863183560067072041)+T(39020571855401265512289573339484371018905006900193) - N

 

-

 

N - T(1)+T(2)+T(4)+T(11)+T(68)+T(2407)+T(2899431)+T(4203353390611)+T(8834089863183560067072041)+T(39020571855401265512289573339484371018905006900192)

 

the two numbers close to T-t in question being

16822699634989797327123095165092932420211999031884

and

16822699634989797327123104264258038130140741377251

where T-t (if you can calculate this from c you win) =

18987613968471836961404436377722813927282768319099

ID: 96e43e Aug. 15, 2018, 1:14 a.m. No.7250   πŸ—„οΈ.is πŸ”—kun   >>7251 >>7256

I was motivated to check these triangle number sequences because in the case of 6107, N=64 and 64 = T(7)+T(8) and 7+8=15 which is the difference in t between the solution record and na transform in e,1.

Anonymous ID: a332a2 Aug. 15, 2018, 10:40 a.m. No.7254   πŸ—„οΈ.is πŸ”—kun   >>7255

>>6996

For the shadow n, we have the very neat pattern of:

 

a[t] = an' (shadow n)

a[t + x + n] = bn' (x, n from solution record where a=a, b=b and n' is shadow n).

 

The difference here is x + n, which means the difference in x between these two records is 2(x+n), which reminds me a lot about VQC's hint regarding 4c, or the record for a=2, b=2*c.

 

Example:

a=37, b=91

{3:6:58:21:37:91}

shadow n = 122

 

a[48] = 37x122

a[75] = 91x122

 

75 - 48 = 21 + 6 = 27.

 

The records for t=48 and t=75 is:

{3:1:4609:95:4514:4706}

{3:1:11251:149:11102:11402}

 

149 - 95 = 54 = 2(21 + 6).

 

Not sure if these are connected in a clever way, as in a solution-wise clever way.

ID: 96e43e Aug. 15, 2018, 12:46 p.m. No.7256   πŸ—„οΈ.is πŸ”—kun   >>7257

>>7250

This is false. I made a mistake there, N for c6107 is way larger. N = 2976

 

However I have another observation to make up for that mistake in this triangle number analysis. x of (-f, 1, T) = T-1(N)

 

7461

{65:3645:86:85:1:7461} (65, 3645, 43) (e, N, T)

{65:1:3730:85:3645:3817} (65, 1, 43) (e, 1, T)

{-108:1:3558:84:3474:3644} (-108, 1, 43) (-f, 1, T)

 

βˆ†-1(N) = 84

 

6107

{23:2976:78:77:1:6107} (23, 2976, 39) (e, N, T)

{23:1:3053:77:2976:3132} (23, 1, 39) (e, 1, T)

{-134:1:2897:76:2821:2975} (-134, 1, 39) (-f, 1, T)

 

βˆ†-1(N) = 76

 

93801

{165:46595:306:305:1:93801} (165, 46595, 153) (e, N, T)

{165:1:46900:305:46595:47207} (165, 1, 153) (e, 1, T)

{-448:1:46288:304:45984:46594} (-448, 1, 153) (-f, 1, T)

 

βˆ†-1(N) = 304

 

145

{1:61:12:11:1:145} (1, 61, 6) (e, N, T)

{1:1:72:11:61:85} (1, 1, 6) (e, 1, T)

{-24:1:48:10:38:60} (-24, 1, 6) (-f, 1, T)

 

βˆ†-1(N) = 10

 

287

{31:128:16:15:1:287} (31, 128, 8) (e, N, T)

{31:1:143:15:128:160} (31, 1, 8) (e, 1, T)

{-2:1:111:14:97:127} (-2, 1, 8) (-f, 1, T)

 

βˆ†-1(N) = 15

 

rsa100c

 

{61218444075812733697456051513875809617598014768503:761302513961266680267809189066318714859034057480651323757098846024549192056136964455981270339102876:39020571855401265512289573339484371018905006900194:39020571855401265512289573339484371018905006900193:1:1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139} (61218444075812733697456051513875809617598014768503, 761302513961266680267809189066318714859034057480651323757098846024549192056136964455981270339102876, 19510285927700632756144786669742185509452503450097) (e, N, T)

 

{61218444075812733697456051513875809617598014768503:1:761302513961266680267809189066318714859034057480690344328954247290061481629476448827000175346003069:39020571855401265512289573339484371018905006900193:761302513961266680267809189066318714859034057480651323757098846024549192056136964455981270339102876:761302513961266680267809189066318714859034057480729364900809648555573771202815933198019080352903264} (61218444075812733697456051513875809617598014768503, 1, 19510285927700632756144786669742185509452503450097) (e, 1, T)

 

{-16822699634989797327123095165092932420211999031886:1:761302513961266680267809189066318714859034057480612303185243444759036902482797480084962365332202681:39020571855401265512289573339484371018905006900192:761302513961266680267809189066318714859034057480573282613388043493524612909457995713943460325302489:761302513961266680267809189066318714859034057480651323757098846024549192056136964455981270339102875} (-16822699634989797327123095165092932420211999031886, 1, 19510285927700632756144786669742185509452503450097) (-f, 1, T)

 

βˆ†-1(N) = 39020571855401265512289573339484371018905006900193

 

Same observation holds except it is the x of the e,1 record.

ID: 96e43e Aug. 15, 2018, 2:31 p.m. No.7258   πŸ—„οΈ.is πŸ”—kun   >>7259

rsa110 analysis

 

{7251398426599644794623954759043454469676218891789649338:17897117089862934387495903916284227701501889012114113090783625819274031122082020851638527741916458186435175261:5982828275968304004100317854118230313685793843723609073:5982828275968304004100317854118230313685793843723609072:1:35794234179725868774991807832568455403003778024228226193532908190484670252364677411513516111204504060317568667} (7251398426599644794623954759043454469676218891789649338, 17897117089862934387495903916284227701501889012114113090783625819274031122082020851638527741916458186435175261, 2991414137984152002050158927059115156842896921861804537) (e, N, T)

 

{7251398426599644794623954759043454469676218891789649338:1:17897117089862934387495903916284227701501889012114113096766454095242335126182338705756758055602252030158784333:5982828275968304004100317854118230313685793843723609072:17897117089862934387495903916284227701501889012114113090783625819274031122082020851638527741916458186435175261:17897117089862934387495903916284227701501889012114113102749282371210639130282656559874988369288045873882393407} (7251398426599644794623954759043454469676218891789649338, 1, 2991414137984152002050158927059115156842896921861804537) (e, 1, T)

 

{-4714258125336963213576680949193006157695368795657568809:1:17897117089862934387495903916284227701501889012114113096766454095242335126182338705756758055602252030158784333:5982828275968304004100317854118230313685793843723609073:17897117089862934387495903916284227701501889012114113090783625819274031122082020851638527741916458186435175260:17897117089862934387495903916284227701501889012114113102749282371210639130282656559874988369288045873882393408} (-4714258125336963213576680949193006157695368795657568809, 1, 2991414137984152002050158927059115156842896921861804537) (-f, 1, T)

 

βˆ†-1(N) = 5982828275968304004100317854118230313685793843723609072

βˆ†-1(N-1) = 5982828275968304004100317854118230313685793843723609072

βˆ†-1(n) = 56415892823691513151002588

βˆ†-1(n-1) = 56415892823691513151002588

 

{7251398426599644794623954759043454469676218891789649338:1:9303852447694684408344911826978140118950296432398266895204888436423619628647716470410476187019755759685209:136410061562149325263764671139067929487183338122546740:9303852447694684408344911826978140118950296432398266758794826874274294364883045331342546699836417637138469:9303852447694684408344911826978140118950296432398267031614949998572944892412387609478405674203093882231951} (7251398426599644794623954759043454469676218891789649338, 1, 68205030781074662631882335569533964743591669061273371) (e, 1, t)

 

{-4714258125336963213576680949193006157695368795657568809:1:9303852447694684408344911826978140118950296432398261048786674030268940792094533491248091988409250158622877:136410061562149325263764671139067929487183338122546741:9303852447694684408344911826978140118950296432398260912376612468119615528329862352180162501225912036076136:9303852447694684408344911826978140118950296432398261185196735592418266055859204630316021475592588281169620} (-4714258125336963213576680949193006157695368795657568809, 1, 68205030781074662631882335569533964743591669061273371) (-f, 1, t)

 

{7251398426599644794623954759043454469676218891789649338:1591376481547123786477396152060299026390686203475593:5982828275968304004100317854118230313685793843723609073:136410061562149325263764671139067929487183338122546740:5846418214406154678836553182979162384198610505601062333:6122421090493547576937037317561418841225758554253106999} (7251398426599644794623954759043454469676218891789649338, 1591376481547123786477396152060299026390686203475593, 68205030781074662631882335569533964743591669061273371) (e, n, t)

 

T-1(N) = x{e, 1, T} or x{-f, 1, T} observation holds.

 

triangleSummation(T-1(N)) = ts

ts = T(1)+T(2)+T(4)+T(13)+T(100)+T(5078)+T(12897767)+T(83176210704686)+T(3459141013595226072509109016)+T(5982828275968304004100317854118230313685793843723609072)

 

ts = 17897117089862934387495903916284227701501889012114113096132169019926664730920520256395609944257205578415628169

ts N

ts - N = 5348543200652633608838499404757082202340747391980452908

 

triangleSummation(T-1(N) - 1) = ts2

ts2 = T(1)+T(2)+T(4)+T(13)+T(100)+T(5078)+T(12897767)+T(83176210704686)+T(3459141013595226072509109016)+T(5982828275968304004100317854118230313685793843723609071)

 

ts2 = 17897117089862934387495903916284227701501889012114113090149340743958360726820202402277379630571411734692019097

N ts2

N - ts2 = 634285075315670395261818449361148111345046451743156164

 

num1 = ts - N

num2 = N - ts2

 

(num1 - num2)/2 =

2991414137984152002050158927059115156842896921861804536

T-t =

2923209107203077339418276591489581192099305252800531166

 

Approximation of T-t reached from c.

ID: 96e43e Aug. 15, 2018, 4:21 p.m. No.7260   πŸ—„οΈ.is πŸ”—kun

In cases where our approximation is too far away to be an approximation, 2 other approximations can be calculated, which work for rsa129 and rsa120 (the approximations are close, but the search space is so massive that iteration is still computationally unfeasible, so the approximation needs to be improved or learned from to get closer.)

 

rsa120 analysis

{276798643817788533350132260296625874116995171400783089588379:113505240647718681667129980473746834447937668233042390019086152667335248378823675608062518130457849501431800035015938730:476456169332959066192086833057656566733872508443700132335510:476456169332959066192086833057656566733872508443700132335509:1:227010481295437363334259960947493668895875336466084780038173258247009162675779735389791151574049166747880487470296548479} (276798643817788533350132260296625874116995171400783089588379, 113505240647718681667129980473746834447937668233042390019086152667335248378823675608062518130457849501431800035015938730, 238228084666479533096043416528828283366936254221850066167755) (e, N, T)

 

{276798643817788533350132260296625874116995171400783089588379:1:113505240647718681667129980473746834447937668233042390019086629123504581337889867694895575787024583373940243735148274239:476456169332959066192086833057656566733872508443700132335509:113505240647718681667129980473746834447937668233042390019086152667335248378823675608062518130457849501431800035015938730:113505240647718681667129980473746834447937668233042390019087105579673914296956059781728633443591317246448687435280609750} (276798643817788533350132260296625874116995171400783089588379, 1, 238228084666479533096043416528828283366936254221850066167755) (e, 1, T)

 

{-676113694848129599034041405818687259350749845486617175082642:1:113505240647718681667129980473746834447937668233042390019085676211165915419757483521229460473891115628923356334883603219:476456169332959066192086833057656566733872508443700132335508:113505240647718681667129980473746834447937668233042390019085199754996582460691291434396402817324381756414912634751267711:113505240647718681667129980473746834447937668233042390019086152667335248378823675608062518130457849501431800035015938729} (-676113694848129599034041405818687259350749845486617175082642, 1, 238228084666479533096043416528828283366936254221850066167755) (-f, 1, T)

 

e%4 = 3

βˆ†-1(N) = 476456169332959066192086833057656566733872508443700132335508

 

triangleSummation(T-1(N)) =

T(1)+T(2)+T(5)+T(16)+T(142)+T(10281)+T(52863280)+T(1397263246090468)+T(976172289437637198279784066271)+T(476456169332959066192086833057656566733872508443700132335508)

113505240647718681667129980473746834447937668233042390019086252496098005964090096585348899032777024250358969739065683157

 

triangleSummation(T-1(N) - 1) =

T(1)+T(2)+T(5)+T(16)+T(142)+T(10281)+T(52863280)+T(1397263246090468)+T(976172289437637198279784066271)+T(476456169332959066192086833057656566733872508443700132335507)

113505240647718681667129980473746834447937668233042390019085776039928673005023904498515841376210290377850526038933347649

 

T-t approximation = 238228084666479533096043416528828283366936254221850066167754, 60 digits

T-t approximation 2 = 138399321908894266675066130147925964192187327052146016423327, 60 digits

T-t approximation 3 = 476456169332959066192086833057656566733872508443700132335509, 60 digits

T-t = 163707277846749007875573151874570744031821201620085731703441, 60 digits

Closest approximation: approximation 2, 60 digits

Closest approximation difference: 25307955937854741200507021726644779839633874567939715280114, 59 digits

βˆ†-1(n) = 260470505313768807462186889446

 

[1/2]

ID: 96e43e Aug. 15, 2018, 4:23 p.m. No.7261   πŸ—„οΈ.is πŸ”—kun

{276798643817788533350132260296625874116995171400783089588379:1:11106701298127191112847355240730405916226091301428275464843399758107464187140818760660545833177599981455232730223775381:149041613639461050440940529308515078670230105203528668928627:11106701298127191112847355240730405916226091301428275464843250716493824726090377820131237318098929751350029201554846754:11106701298127191112847355240730405916226091301428275464843548799721103648191259701189854348256270211560436258892704010} (276798643817788533350132260296625874116995171400783089588379, 1, 74520806819730525220470264654257539335115052601764334464314) (e, 1, t)

 

{-676113694848129599034041405818687259350749845486617175082642:1:11106701298127191112847355240730405916226091301428275464842774260324491767024185733298179661532195878841585501422511243:149041613639461050440940529308515078670230105203528668928626:11106701298127191112847355240730405916226091301428275464842625218710852305973744792768871146453525648736381972753582617:11106701298127191112847355240730405916226091301428275464842923301938131228074626673827488176610866108946789030091439871} (-676113694848129599034041405818687259350749845486617175082642, 1, 74520806819730525220470264654257539335115052601764334464314) (-f, 1, t)

 

{276798643817788533350132260296625874116995171400783089588379:33922442069205032282149019766764497978931624740053939704438:476456169332959066192086833057656566733872508443700132335510:149041613639461050440940529308515078670230105203528668928627:327414555693498015751146303749141488063642403240171463406883:693342667110830181197325401899700641361965863127336680673013} (276798643817788533350132260296625874116995171400783089588379, 33922442069205032282149019766764497978931624740053939704438, 74520806819730525220470264654257539335115052601764334464314) (e, n, t)

 

To brainstorm that missing piece in the approximation I would look at the closest value and its difference.

 

[2/2]

ID: 96e43e Aug. 15, 2018, 4:27 p.m. No.7262   πŸ—„οΈ.is πŸ”—kun

rsa 129

{10127537895024395905251173100883802246370188433498376141602081316:57190812878944433834617889988073306005109148360621181281280921457158533038780477389984354088670452233171543980194516501823035186:10694934584086471525314207693308900296322993593605128511616736585:10694934584086471525314207693308900296322993593605128511616736584:1:114381625757888867669235779976146612010218296721242362562561842935706935245733897830597123563958705058989075147599290026879543541} (10127537895024395905251173100883802246370188433498376141602081316, 57190812878944433834617889988073306005109148360621181281280921457158533038780477389984354088670452233171543980194516501823035186, 5347467292043235762657103846654450148161496796802564255808368293) (e, N, T)

 

{10127537895024395905251173100883802246370188433498376141602081316:1:57190812878944433834617889988073306005109148360621181281280921467853467622866948915298561781979352529494537573799645013439771770:10694934584086471525314207693308900296322993593605128511616736584:57190812878944433834617889988073306005109148360621181281280921457158533038780477389984354088670452233171543980194516501823035186:57190812878944433834617889988073306005109148360621181281280921478548402206953420440612769475288252825817531167404773525056508356} (10127537895024395905251173100883802246370188433498376141602081316, 1, 5347467292043235762657103846654450148161496796802564255808368293) (e, 1, T)

 

{-11262331273148547145377242285733998346275798753711880881631391855:1:57190812878944433834617889988073306005109148360621181281280921467853467622866948915298561781979352529494537573799645013439771770:10694934584086471525314207693308900296322993593605128511616736585:57190812878944433834617889988073306005109148360621181281280921457158533038780477389984354088670452233171543980194516501823035185:57190812878944433834617889988073306005109148360621181281280921478548402206953420440612769475288252825817531167404773525056508357} (-11262331273148547145377242285733998346275798753711880881631391855, 1, 5347467292043235762657103846654450148161496796802564255808368293) (-f, 1, T)

 

e%4 = 0

βˆ†-1(N) = 10694934584086471525314207693308900296322993593605128511616736583

βˆ†-1(N-1) = 10694934584086471525314207693308900296322993593605128511616736583

 

triangleSummation(T-1(N)) =

T(1)+T(2)+T(5)+T(19)+T(195)+T(19232)+T(184947519)+T(17102792626580353)+T(146252757813905659871375936577926)+T(10694934584086471525314207693308900296322993593605128511616736583)

57190812878944433834617889988073306005109148360621181281280921457442231383311515200015871384883089959790751131906158453282803705

 

triangleSummation(T-1(N) - 1) =

T(1)+T(2)+T(5)+T(19)+T(195)+T(19232)+T(184947519)+T(17102792626580353)+T(146252757813905659871375936577926)+T(10694934584086471525314207693308900296322993593605128511616736582)

57190812878944433834617889988073306005109148360621181281280921446747296799225043674701663691574189663467757538301029941666067122

 

T-t approximation = 5347467292043235762657103846654450148161496796802564255808368291, 64 digits

T-t approximation 2 = 5063768947512197952625586550441812421542289645090922304348599772, 64 digits

T-t approximation 3 = 10694934584086471525314207693308900296322993593605128511616736584, 65 digits

T-t = 1028759532095353253114134196990247228746418031676057044411877127, 64 digits

Closest approximation: 5063768947512197952625586550441812421542289645090922304348599772, 64 digits

Closest approximation difference: 4035009415416844699511452353451565192795871613414865259936722645, 64 digits

βˆ†-1(n) = 190419700934841194543373870443698

βˆ†-1(n-1) = 190419700934841194543373870443698

ID: 96e43e Aug. 15, 2018, 4:30 p.m. No.7263   πŸ—„οΈ.is πŸ”—kun   >>7264

{10127537895024395905251173100883802246370188433498376141602081316:1:37302473431668114358131420783420175468441264382066034437953513370074778641178128081332420460685355563492329655197776241440137438:8637415519895765019085939299328405838830157530253014422792982330:37302473431668114358131420783420175468441264382066034437953513361437363121282363062246481161356949724662172124944761818647155108:37302473431668114358131420783420175468441264382066034437953513378712194161073893100418359760013761402322487185450790664233119770} (10127537895024395905251173100883802246370188433498376141602081316, 1, 4318707759947882509542969649664202919415078765126507211396491166) (e, 1, t)

 

{-11262331273148547145377242285733998346275798753711880881631391855:1:37302473431668114358131420783420175468441264382066034437953513368017259576987421575104152066704861105999493591845662152616383184:8637415519895765019085939299328405838830157530253014422792982331:37302473431668114358131420783420175468441264382066034437953513359379844057091656556018212767376455267169336061592647729823400853:37302473431668114358131420783420175468441264382066034437953513376654675096883186594190091366033266944829651122098676575409365517} (-11262331273148547145377242285733998346275798753711880881631391855, 1, 4318707759947882509542969649664202919415078765126507211396491166) (-f, 1, t)

 

{10127537895024395905251173100883802246370188433498376141602081316:18129831252057180249554918905369072823951862938527912049817621466:10694934584086471525314207693308900296322993593605128511616736582:8637415519895765019085939299328405838830157530253014422792982330:2057519064190706506228268393980494457492836063352114088823754252:55592012608096597043509984803375451783056877000913967034044961844} (10127537895024395905251173100883802246370188433498376141602081316, 18129831252057180249554918905369072823951862938527912049817621466, 4318707759947882509542969649664202919415078765126507211396491166) (e, n, t)

 

[2/2]

 

Approximation didn't work as well for rsa129. Still carving out a methodology here.

Anonymous ID: a332a2 Aug. 15, 2018, 9:58 p.m. No.7267   πŸ—„οΈ.is πŸ”—kun

>>7255

Just looking a bit more into this.

 

The x-values for the record a=2, b=2*c for both the normal big N and shadow big N is equal to the two f's we have in (e, 1).

 

The value of x for normal n is equal to f with regards to a positive d and the value of x for the shadow n is equal to f with regards to a negative d.

ID: 96e43e Aug. 16, 2018, 12:13 a.m. No.7269   πŸ—„οΈ.is πŸ”—kun

>>7268

>>7265

>>7266

Elaborating on this, he gave me polynomial for even and odd e

 

For even e:

nd = 2t^2 + 2nt + (e/2)

For odd e:

nd = 2t^2 + 2(n-1)t - (n - ((e+1)/2))

 

And when graphed the points which are whole integers will be elements, each point being (t, n). So two elements exist for semiprime c in this curve.

 

Pics are the points for c6107.

 

A way to find the preceeding integer point on this curve would be a solution. The problem is known in mathematical circles as non-trivial but not computationally infeasible.

ID: 96e43e Aug. 16, 2018, 12:22 a.m. No.7270   πŸ—„οΈ.is πŸ”—kun   >>7273

I'm still going to be working on the approximations if anyone has thoughts on those let me know. If we can calculate more accurate approximations we can get within the realm of solving these numbers (and maybe a solution like Chris described, since this approximation comes from a recursive triangle some of input T-1(N)).

 

Like Chris said there are different solutions just like a 4d object viewed in 3d space looks like different 3d objects. So do not be discouraged if it seems like things are going in different directions/different paths.

Anonymous ID: a332a2 Aug. 16, 2018, 9:15 a.m. No.7271   πŸ—„οΈ.is πŸ”—kun   >>7272

Thinking more about this one too >>6469

 

I haven't "gotten" anything out of it yet, but I decided to look into it.

 

We know d = (a + x)

We know f = 2d + 1 - e

We know f = 2(a + x) + 1 - e

 

We know how to compute elements in -f.

a[t] in even negative f is calculated:

2tt - e/2

In this case e = f

2tt - f/2

2tt - (2d + 1 - e)/2

2tt - (2(a + x) + 1 - e)/2

2tt - (a + x + (1 - e)/2)

2tt - a - x - (1 - e)/2

 

Not sure where I am going with this, though.

ID: 96e43e Aug. 16, 2018, 9:57 p.m. No.7275   πŸ—„οΈ.is πŸ”—kun   >>7276

c6107 is special because n is a triangle number.

 

6107

{23:2976:78:77:1:6107} (23, 2976, 39) (e, N, T)

 

{23:1:3053:77:2976:3132} (23, 1, 39) (e, 1, T)

 

{-134:1:2897:76:2821:2975} (-134, 1, 39) (-f, 1, T)

 

e%4 = 3

2d = 156

f = 134

βˆ†-1(N) = 76

βˆ†-1(N-1) = 76

 

triangleSummation(T-1(N)) =

T(1)+T(2)+T(4)+T(11)+T(76)

3006

 

triangleSummation(T-1(N) - 1) =

T(1)+T(2)+T(4)+T(11)+T(75)

2930

 

T-t approximation = 38, 2 digits

T-t approximation 2 = 8, 1 digits

T-t approximation 3 = 77, 2 digits

T-t = 15, 2 digits

Closest approximation: 8, 1 digits

Closest approximation difference: 7, 1 digits

βˆ†-1(n) = 8

βˆ†-1(n-1) = 7

 

{23:1:1163:47:1116:1212} (23, 1, 24) (e, 1, t)

 

{-134:1:1037:46:991:1085} (-134, 1, 24) (-f, 1, t)

 

{23:36:78:47:31:197} (23, 36, 24) (e, n, t)

ID: 96e43e Aug. 17, 2018, 12:21 a.m. No.7277   πŸ—„οΈ.is πŸ”—kun

Some experimental data for safekeeping.

 

c = rsa100

N = 761302513961266680267809189066318714859034057480651323757098846024549192056136964455981270339102876

βˆ†-1(N) = 39020571855401265512289573339484371018905006900193

n = 14387588531011964456730684619177102985211280936

βˆ†-1(n) = 169632476436630285119505

triangleSummation(T-1(N)) =

T(1)+T(2)+T(4)+T(11)+T(68)+T(2407)+T(2899431)+T(4203353390611)+T(8834089863183560067072041)+T(39020571855401265512289573339484371018905006900193)

 

triangleSummation(T-1(N) - 1) =

T(1)+T(3)+T(10)+T(67)+T(2406)+T(2899430)+T(4203353390610)+T(8834089863183560067072040)+T(39020571855401265512289573339484371018905006900192)

 

triangleSummation(T-1(N) - sqrt(d)) =

T(1)+T(2)+T(8)+T(63)+T(2372)+T(2898174)+T(4203351809684)+T(8834089863181060738192547)+T(39020571855401265512289567092839523150469841441251)

 

triangleSummation(T-1(N) - sqrt(2d)) =

T(1)+T(2)+T(8)+T(62)+T(2371)+T(2898118)+T(4203351666598)+T(8834089863180587847385815)+T(39020571855401265512289564505394507835344939828152)

 

βˆ†1(βˆ†-1(n)) = 14387588531011964456730676802388333031208282265

βˆ†2(βˆ†-2(n)) = 14387588530975214951661888938891904772729775106

βˆ†3(βˆ†-3(n)) = 14387555083370601538478896472570573635945074246

βˆ†4(βˆ†-4(n)) = 14273631718948863438208830586582694349507198856

βˆ†5(βˆ†-5(n)) = 9488836738160187798300677759237642920019716506

βˆ†6(βˆ†-6(n)) = 53327858858072239299138189326373798809103885

βˆ†7(βˆ†-7(n)) = 2076895351339769460477611370186681

ID: 96e43e Aug. 17, 2018, 12:30 a.m. No.7278   πŸ—„οΈ.is πŸ”—kun

c = rsa110

N = 17897117089862934387495903916284227701501889012114113090783625819274031122082020851638527741916458186435175261

βˆ†-1(N) = 5982828275968304004100317854118230313685793843723609072

n = 1591376481547123786477396152060299026390686203475593

βˆ†-1(n) = 56415892823691513151002588

triangleSummation(T-1(N)) =

T(1)+T(2)+T(4)+T(13)+T(100)+T(5078)+T(12897767)+T(83176210704686)+T(3459141013595226072509109016)+T(5982828275968304004100317854118230313685793843723609072)

 

triangleSummation(T-1(N) - 1) =

T(1)+T(3)+T(12)+T(99)+T(5077)+T(12897766)+T(83176210704685)+T(3459141013595226072509109015)+T(5982828275968304004100317854118230313685793843723609071)

 

triangleSummation(T-1(N) - sqrt(d)) =

T(1)+T(3)+T(11)+T(92)+T(5026)+T(12895116)+T(83176203672128)+T(3459141013595176615638324526)+T(5982828275968304004100317851672248245892102105984803468)

 

triangleSummation(T-1(N) - sqrt(2d)) =

T(1)+T(3)+T(11)+T(92)+T(5025)+T(12894997)+T(83176203035627)+T(3459141013595167258046486330)+T(5982828275968304004100317850659089300090567771214500056)

 

βˆ†1(βˆ†-1(n)) = 1591376481547123786477396143030577099882291366850166

βˆ†2(βˆ†-2(n)) = 1591376481546920421154689382900985625636153296142100

βˆ†3(βˆ†-3(n)) = 1591373857158754367809961369378063355314169683653515

βˆ†4(βˆ†-4(n)) = 1585747893594404099564413340729515765120866343772760

βˆ†5(βˆ†-5(n)) = 1338420359857405075273316351571726222383145912185046

βˆ†6(βˆ†-6(n)) = 9989452752508151923071579773982579506854290142986

βˆ†7(βˆ†-7(n)) = 30700002123226936025189367747945843590228731690

Anonymous ID: 358135 Aug. 17, 2018, 4:43 a.m. No.7280   πŸ—„οΈ.is πŸ”—kun   >>7284

Off Topic But…..

Last night in bed it occurred to me that the 90 degree dimensional shift can be seen as an arc moving from grid plane to new point outside grid plane. The arcangle of the arc will define the perspective from the original plane and the graph area from the original 4D plane. The non graph area of the new triangle/square will be in the imaginary plane. The angle of the arc/curve close to 45 will give a high graphable area; An angle closer to 0 or 90 will give a very small graphable area (or I may have that reversed).

ID: 96e43e Aug. 17, 2018, 4:04 p.m. No.7281   πŸ—„οΈ.is πŸ”—kun

I created a new tree algorithm that uses T-1(input) as the d values and input - T(T-1(input)) as the e values.

 

Triangle Tree {

145 (c)

145 (c)

| 9 (e)

| | 3 (e)

| | | 2 (d)

| | | | 1 (e)

| | | | 1 (d)

| | 3 (d)

| | | 2 (d)

| | | | 1 (e)

| | | | 1 (d)

| 16 (d)

| | 1 (e)

| | 5 (d)

| | | 2 (e)

| | | | 1 (e)

| | | | 1 (d)

| | | 2 (d)

| | | | 1 (e)

| | | | 1 (d)

}

 

Remainder Tree {

145 (c)

| 1 (e)

| 12 (d)

| | 3 (e)

| | | 2 (e)

| | | | 1 (e)

| | | | 1 (d)

| | | 1 (d)

| | 3 (d)

| | | 2 (e)

| | | | 1 (e)

| | | | 1 (d)

| | | 1 (d)

}

ID: 96e43e Aug. 17, 2018, 4:05 p.m. No.7282   πŸ—„οΈ.is πŸ”—kun

N for 145 = 61

 

Triangle Tree(N) {

61 (c)

| 6 (e)

| | 3 (d)

| | | 2 (d)

| | | | 1 (e)

| | | | 1 (d)

| 10 (d)

| | 4 (d)

| | | 1 (e)

| | | 2 (d)

| | | | 1 (e)

| | | | 1 (d)

}

 

Remainder Tree(N) {

61 (c)

| 12 (e)

| | 3 (e)

| | | 2 (e)

| | | | 1 (e)

| | | | 1 (d)

| | | 1 (d)

| | 3 (d)

| | | 2 (e)

| | | | 1 (e)

| | | | 1 (d)

| | | 1 (d)

| 7 (d)

| | 3 (e)

| | | 2 (e)

| | | | 1 (e)

| | | | 1 (d)

| | | 1 (d)

| | 2 (d)

| | | 1 (e)

| | | 1 (d)

}

ID: 96e43e Aug. 17, 2018, 4:23 p.m. No.7283   πŸ—„οΈ.is πŸ”—kun

Take 16^2 + 9 from the 145 triangle tree

 

Triangle Tree(16^2+9) {

265 (c)

| 12 (e)

| | 2 (e)

| | | 1 (e)

| | | 1 (d)

| | 4 (d)

| | | 1 (e)

| | | 2 (d)

| | | | 1 (e)

| | | | 1 (d)

| 22 (d)

| | 1 (e)

| | 6 (d)

| | | 3 (d)

| | | | 2 (d)

| | | | | 1 (e)

| | | | | 1 (d)

}

 

22^2 + 12 = 31st triangle number = 496

 

Triangle Tree(T(31)) {

496 (c)

| 31 (d)

| | 3 (e)

| | | 2 (d)

| | | | 1 (e)

| | | | 1 (d)

| | 7 (d)

| | | 1 (e)

| | | 3 (d)

| | | | 2 (d)

| | | | | 1 (e)

| | | | | 1 (d)

}

AA !dTGY7OMD/g ID: c2fe35 Aug. 18, 2018, 6:59 a.m. No.7288   πŸ—„οΈ.is πŸ”—kun   >>7321

I've spent way too much of today putting together a program that I intend to have do the following:

>use the nested i and j loop from the grid to generate a bunch of cs (only semiprimes in this case, because they only have two n values and I haven't accounted for varying numbers of possible ns)

>brute force every single number related to c that we've been told about

>using this big list of variables, see which ones are equal, and do some arbitrary math (like mods and additions and thing) to pairs of variables to see if there are any relationships we aren't aware of

>do this to every pair of i and j, generating a big list of relationships for each one and storing them in an array

>find the relationships that are common to every single pair of i and j

VQC said enumerate the patterns, so that's what I'm trying to do with this. When I tried to do it with an Excel spreadsheet, he said I was getting very close. That Excel spreadsheet didn't go anywhere because it was far too clunky and hard to use. I've been trying to figure out how to do this in the most efficient way possible since then. I would post my code, but the number of things I've done in really stupid or lazy ways is probably embarrassing, and it's 579 lines of code anyway. Pic related is what it currently outputs. These are all pretty obvious and well known to us now (or they're just stupid redundancies that I haven't bothered to code out yet), but I've only implemented basic stuff so far. I think 55 is the number of i and j semiprime pairs it went through. I'm about to implement the first three cells where a[t]=c in (e,n) and the an, bn, a(n-1) and b(n-1) cells. Then I'll start adding other variables.

 

We keep going through this cycle (I think I've mentioned this before) of working on getting a big list of the patterns together, seeing a new VQC post mentioning something we weren't aware of already, and then spending all of our time on the new VQC thing or something else he didn't mention that we think will be useful and forgetting about our main objective here. If anyone would like to help me to revive the whole "enumerate the patterns" thing, please, I would really love to not be the only one doing that. If not, I'll still do it myself. Another thing that would also be useful, if anyone reads this, would be math ideas. I'm thinking there might be obscure things we aren't thinking of, like d of the na transform cell being equal to the x from the cell where a=1 and b=n or some weird bullshit like that, but I'm also thinking maybe there's a less obvious mathematical relationship like the thing I mentioned about (BigN-n)/(b-1)=(a-1)/2 (which is less obvious just looking at it). So I'm thinking if I make a big list of every variable mod every other variable, or every variable mod (every variable +1), and so on, we might find something. So what else? Maybe (variable+1)/2 and stuff. It will be more useful to first work on getting every known pattern together but I'm still thinking it might bring up things we wouldn't have otherwise discovered.

Anonymous ID: 358135 Aug. 18, 2018, 3:09 p.m. No.7290   πŸ—„οΈ.is πŸ”—kun   >>7291

hi

A quantum wave return is the solid evidence seen. it is the matter or physical evidence that shows something is 'real'. The quantum wave is all possibilities that exist for that query. Which is real as the first is just a probability?

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 Aug. 18, 2018, 5:39 p.m. No.7291   πŸ—„οΈ.is πŸ”—kun   >>7292

>>7290

YO!

We talked about this out in the open waters.

 

Please relate what you're working on to RSA, in a way that someone other than you can understand, or take it to another thread, or start your own.

 

Thank you.

Anonymous ID: 358135 Aug. 18, 2018, 7:41 p.m. No.7292   πŸ—„οΈ.is πŸ”—kun   >>7293 >>7294

>>7291

I guess I cannot prove it. What thread? Basically what I see VQC is that question - the quantum wave and what I have seen shown is basically a multi dimensional grid that had diagonals and rarely some circle components (the early stuff) when graphed. My stuff I am trying to see if anons can figure the pattern of that multidimensional grid as I think it is the answer to the TOE. Why do I think that I think that the graphing in that manner will pass Einstein and show that fractal and Euclidean connect and give us the pattern of how space time works in toto. What thread should that be in?

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 Aug. 18, 2018, 7:53 p.m. No.7293   πŸ—„οΈ.is πŸ”—kun   >>7295 >>7298

>>7292

You're already posting in this one, which is good:

https://8ch.net/vqc/res/5774.html

 

This one is about physics, paradoxes, and physics paradoxes, according to VQC:

https://8ch.net/vqc/res/5913.html

(This may be the one for you, actually… give that one a read.)

 

And you can post whatever you'd like to in here, just keep it… not hardcore lewdness, plz.

https://8ch.net/vqc/res/846.html

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 Aug. 18, 2018, 8:44 p.m. No.7300   πŸ—„οΈ.is πŸ”—kun   >>7301

>>7299

WELCOME TO VQC!

Start your own thread, and bread it down for folks.

Perhaps your explaining in its own thread,

People will be able to understand it better and it might give them the boost they need to correlate it, or vice a verso.

 

I'm not the only whose invited you to do so, please start your own bread.

Your head's in the right spot, just not the posts at the moment.

 

And believe me, things have a of circling back around, here.

(Recursion puns.)

Anonymous ID: a332a2 Aug. 19, 2018, 8:17 a.m. No.7316   πŸ—„οΈ.is πŸ”—kun   >>7318 >>7367

Just an interesting observation, not entirely sure what it means yet or if it is a pattern of use.

 

Take a record with an odd e.

a[t] = bn

 

Then we now know that d[t - 1] - d = b(n - 1). I was thinking about d[t] - d[t - 1] and noticed the pattern. For odd e's the d in (e, 1) is calculated as 2tt + (e + 1)/2. But when we are subtracting d[t] - d[t - 1] the (e + 1)/2 is removed and we're left with 2tt - 2(t-1)(t-1). Essentially, the number of integers between the two squares. What I noticed is that this is actually related to the d of our record and of the next record in the sequence in (e, n) where a and b exists.

 

Say we have a record in (e, n) for a, b where e is odd. Let's say we use the first chain, so we choose the record at t=1 within (e, n). We then have a second record in that chain and we know the method to calculate this (VQC also posted this in >>6736). Let's call the record at (e, n, 1) for r_1 and the next record in the sequence for r_2.

 

There is a t in (e, 1) where a[t] = bn from r_1. We know the difference between d[t] - d[t - 1] = 2 times the number of integers between the squares. If you add r_1.d + (d[t] - d[t - 1]) you will get the d from r_2. This means the difference between the d's in a sequence within (e, n) is two times the number of integers between the squares for t and t-1 where a[t] = bn.

 

This also means that b + x + 2n = 2 times the number of integers between two squares (paraphrasing d' = a' + x') >>6736).

 

To repeat, we have two records r_1 and r_2. They belong to the same sequence in (e, n) (for odd e) with a t-difference of 1. If we were given r_1 we can construct r_2 (>>6736). The difference in d between these two records is equal to 2tt - 2(t-1)(t-1) where t is the t-value belonging to r_2 (That is t = (x + 1)/2 for an odd x, t = x/2 + 1 for even x).

 

Example:

r_1 = {627:6:1310:117:1193:1439}

r_2 = {627:6:1568:129:1439:1709}

 

In (627, 1) we have:

a[65] = 1439*6, which gives the record: {627:1:8763:129:8634:8894}

The previous record is at t 64: {627:1:8505:127:8378:8634}

 

The difference in d[t] - d[t - 1] = 8763 - 8505 = 258.

 

1310 + 258 = 1568.

 

For even e's this holds true for a-values (In an even e, d's are 4 triangles with a base of t).

Saga ID: dfa1af Aug. 19, 2018, 8:55 a.m. No.7317   πŸ—„οΈ.is πŸ”—kun

I don't want to put anyone's hopes down but the two formulas stated here >>7268 are merely the same as

 

x^2 + e = 2n (d-x)

 

which can be derived from this equation:

 

(d+n)^2 - (x+n)^2 = d^2 - e

 

Pic for c20737

Anonymous ID: a332a2 Aug. 19, 2018, 1:44 p.m. No.7318   πŸ—„οΈ.is πŸ”—kun

>>7316

 

I think this is actually quite amazing, but not in the sense that it is going to help us solve our factorization problem, but rather as insight. It essentially allows us to get a better understanding of how the values in (e, n) relate.

 

We know that dd + e = c. What this does is show a relationship between this c and the next one in (e, n). This is important to get a better understanding of the grid, which I think is the actual main goal here. I'm not here anymore to factorize, but rather to understand the grid. This is more about learning how numbers truly are related and less about breaking cryptography.

 

This shows us a big part of the relationship between the records in a sequence. We can take a record and observe how the differences in d is related to the integers between two perfect consecutive squares. It is quite beautiful.

 

What we now know is that given a record, the difference between d^2 + e = ab and (d + k)^2 + e = bi (where i is the next b' in the sequence) is equal to the difference in squares multiplied by two, where the squares are defined to be the t^2 and (t-1)^2 of the next sequence.

AA !dTGY7OMD/g ID: 8e8f2b Aug. 19, 2018, 9:28 p.m. No.7319   πŸ—„οΈ.is πŸ”—kun   >>7320 >>7341

Hey guys, so you remember VQC mentioning the cell in (e,1) where a[t] = c*BigN? He mentioned it >>6774 here. PMA called it nb on Discord just now, if that helps. I just discovered that we can actually calculate this cell straight from c, rather than going back and forth between cells. In every case, the x of that cell = c-d.

 

c=1441, d=37

c*bigN=985644, which first appears in (e,1) as a at (72,1,702) = {72,1,987048,1404,985644,988454}, f=-1974025

1441-37=1404

 

c=4031, d=63

c*bigN=7872543, which first appears in (e,1) as a at (62,1,1984) = {62,1,7876511,3968,7872543,7880481}, f=-15752961

4031-63=3968

 

c=4601, d=67

c*bigN=10278634, which first appears in (e,1) as a at (112,1,2267) = {112,1,10283168,4534,10278634,10287704}, f=-20566225

4601-67=4534

Anonymous ID: a332a2 Aug. 19, 2018, 10:19 p.m. No.7320   πŸ—„οΈ.is πŸ”—kun   >>7324 >>7348 >>7367

>>7319

 

This is a neat pattern. What we're actually seeing is:

dd - d + e = 2t - 1 (for odd e)

dd - d + e = 2t (for even d)

 

When solving for t we get:

 

Odd e:

dd - d + e + 1 = 2t

(dd - d)/2 + (e + 1)/2 = t

d(d-1)/2 + (e + 1)/2 = t

 

If we treat d as the base of a triangle and add half of e (rounded up) we get the t for Nc (big N transformation for nb for a=1, b=c).

 

We also know the equation a[c + 1 - t] = n'c (shadow big N for c). This gives us:

 

c + 1 - d(d-1)/2 - (e + 1)/2

dd + e + 1 - d(d-1)/2 - (e+1)/2

dd - d(d-1)/2 + (e+1)/2

 

(For an even e it just turns into e/2)

 

What is neat here is how the formula is so similar to our a's and b's in (e, 1). And what it "means" is that if we calculate the T_d (the dth triangular number) and add half of e (rounded up) we get the t for our Nc (bigN, nb for a=1, b=c record) in (e, 1).

 

If we remove the triangle with a base of d, from the d^2 and add half of e, we get the counter part N'b (shadow big N, nb transformation for the record a=1, b=c).

 

(Trailing off a bit, but I hope this makes sense. Looking over the jargon I can't imagine anyone not familiar with this grid would be able to make sense of anything).

Hobo !!1yNgQ3NlCs ID: 6c09ed Aug. 19, 2018, 11:05 p.m. No.7321   πŸ—„οΈ.is πŸ”—kun   >>7322 >>7343 >>7367

>>7288

Hey AA. When I first saw this thing I tried to go about it like this… What is the SIMPLEST possible problem that could be solved with this thing. Not "can I break RSA" but can I do something EASY as FUCK. I really tried too see if I could find a way to use the ideas posted here to solve the easiest possible problem imaginable. 1+1=2 is too easy but what is the really simple one. What is the "Soft Lob" of Quantum Computing? If this thing is as simple as it appears then it likely could be done in one's head on a simple level.

 

Who knows, Maybe I am way off? Its the programming that keeps me from pursuing it any further. I just don't understand it and get sidetracked. Anyway, consider looking for the simplest possible solution to just make the damn thing work then expand upon that basic functionality?

 

When I travel I like to have a water bladder and a piece of black plastic. Its not pretty but it will make hot water when the sun is out. wrap the water bladder in the black plastic and put it in the sun. Easy money. Now one can easily expand up on this basic idea with pumps and hoses and circulation timers and insulated tanks and such. Hell, really smart guys have even made vacuum insulated water channels with super-flat black absorbers that will heat water to 100 degrees when the exterior temperature is below fucking zero as long as there is sun.

 

That fancy stuff is great but it is not fundamental. See if you can find the easiest problem that can be solved with this thing. Something fucking simple as hell that will be a proof of concept and then go forward from there… That may be an angle that will crack open the door.

 

I mined a little BTC in 2010. Not a lot but a bit. I have never touched my wallet but I still have it. Kinda like a bunker. I watch it daily. In the intervening decade I have learned a LOT and gone pretty far into that community in a number of ways. No one broke the blockchain by friend. Not yet. Chris is wrong on that front. No one even vaguely thinks that MAYBE it could have been broken in a different universe… Not one person thinks that. NOT ONE. This thing 'aint over yet.

 

AA, Ganon, Teach, Topol, PMA, 3Danon, Minecraft anon, Primeanon, VA, the ever-present Anonymous and more. We have all been on an adventure here to an unknown land. We have reached out into the unknown for a frontier that held promises that drew us in.

 

I was drawn in deeply by the exploration as well as the tremendous possibility for change. The world never really worked for me so change is always welcome. I hope, simply, for positive change. Others were drawn in for other reasons that belong to them alone.

 

This is the real true force that drives us all I think. Us HUMANS. We are driven by a desire to improve… We want things to be better. In the end that is what this is really all about. Making things better.

 

Well, you have all made my life far better. I am glad to have been on this journey with you all and I swear I will never forget it. Hell, I saved it all to pass it along as a great mystery to others in the future if the problem is never solved. I consider you all to be my brothers and I am proud to have walked this long and winding road into the unknown with you.

 

Remember, it is the journey not the destination. >>7288

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 Aug. 20, 2018, 7:24 a.m. No.7323   πŸ—„οΈ.is πŸ”—kun

I'm just gonna leave this here because I doubt VQC would check the EZ Bake:

 

"WhoTheFuckIs" Chris -Inner Earth -> Dewachen -> Shambhala -> Shantbhala -> Maitreya -> Maitre -> Alien HumanKind

 

This "Dante Santori" guy is interdasting.

This AHK Series is either the best biblical plot hole filler/Dungeons and Dragons premise (complete with creature book!) or there's legitimacy to it.

 

Perhaps the legitimacy is just a Sci-Fi way of explaining the machinations of the pedoraptor global sociopathic elites and their operations and machinations. A la David Icke saying they're not actual reptilians, they just lack compassion (mammalian) and use their dominant "reptilian" part of the human brain.

Blavatsky diiiiid bring it up, after all, and we all know how much of hard on for Theosophy Aleister Crowley had… though he diiiiiit have a tendency to invert and corrupt things. Maybe she was alright but her followers sucked ass, I dunno. Plus, LAM, straight up.

 

Portal to The Underworld, Portal to Another Dimension, Portal to Another Galaxy… Just because a being/entity comes through and there's a volcanic backdrop on the other side, doesn't mean they're a demon coming from Hell.

Anonymous ID: a332a2 Aug. 20, 2018, 9:45 a.m. No.7324   πŸ—„οΈ.is πŸ”—kun   >>7325

>>7320

This pattern is interesting, because the t for the nb transformation for c is located at d(d-1)/2 + (e + 1)/2, but there are values in (e, 1) that does not fit into this equation.

 

I wonder if that means this "separates" the values in (e, 1), since this does not hold true for all values in (e, 1). As in, this pattern will hold for all c, meaning a number that doesn't fit this pattern is not a c, but something else.

Anonymous ID: a332a2 Aug. 20, 2018, 11:03 a.m. No.7325   πŸ—„οΈ.is πŸ”—kun

>>7324

 

Generate from 1 to k:

2(i(i-1)/2 + (e + 1)/2) - 1

 

Each of those will be an x-value in (e, 1) which also has a corresponding record with a=1, b=i^2 + e

 

Each of these records exist in an (e, n) as a=1, b=i^2 + e. Which means they should all exist in unique n's. They will only exist in n's that have a record with a=1, meaning there are n's that is not covered by this (ones that doesn't contain a=1).

 

That should also mean it includes every a, b in column e that has a=1, b=ab, but not necessarily the n where a and b exists (I think).

 

This makes me wonder what's so interesting about the numbers that doesn't fit in these patterns in column e.

Anonymous ID: 358135 Aug. 20, 2018, 5:36 p.m. No.7331   πŸ—„οΈ.is πŸ”—kun

>>7326

it looks like the innercircle center will be at the bisector point of the 45 degree bisector of the right triangle. And thus the radius would be half the outer radius intuitively working on proving it for you as the numbers are at a 45 degree angle from x and y

Anonymous ID: 358135 Aug. 20, 2018, 5:46 p.m. No.7332   πŸ—„οΈ.is πŸ”—kun

>>7326

Ok got it - to get the center of the inner circle it takes first drawing the perpendicular to the 45 degree angle radius of outside circle. Then if you bisect the small middle triangle using the 30 and 60 degree rectangle vertexes you find the center of the inner circle.

Anonymous ID: 358135 Aug. 20, 2018, 6:31 p.m. No.7336   πŸ—„οΈ.is πŸ”—kun   >>7338

if you were looking for this innercircle the center is 29.2 by 29.2 and hypotenuse is 41.2 the radius. To prove that I would have to bisect the inner square the other direction and then drop bisectors from the 30 and 60 degree angle lines crossing that square bisector. Then to complete confirmation square the 1st quadrant of the circle to check the numbers. Like the first example.

Anonymous ID: 358135 Aug. 20, 2018, 6:42 p.m. No.7337   πŸ—„οΈ.is πŸ”—kun

so essentially you are using the sin for the y axis and cos for the x axis to move the square bisector to the origin of the inner circle and finishing the 1st quadrant square to confirm the hypotenuse of that square is the radius.

PMA !!y5/EVb5KZI ID: 284644 Aug. 20, 2018, 9:32 p.m. No.7340   πŸ—„οΈ.is πŸ”—kun

>>6972

Just posting a shortcut using the "pointer" formulas to help identify the various record sequences in any (e,n).

 

Attached pics for (1,61), (1,145), (-24,60), and (0,36) show a range of records in both positive and negative x values.

 

A new "p root" column has been added, which is calculated simply as t mod n.

 

The resulting value can be used to navigate to the first record in a sequence or as a way to identify the number if unique sequences.

 

For example, in (0,36), there are 6 sequences shown with "p root" values of 1,7,13,19,25,36.

AA !dTGY7OMD/g ID: 205cf8 Aug. 21, 2018, 2:22 a.m. No.7341   πŸ—„οΈ.is πŸ”—kun   >>7342 >>7344 >>7367

Another update on >>6774 and >>7319

 

>where the product of BigN and c are found in (e,1)

The cell where a[t]=c*BigN in (e,1) appears where x=c-d.

>and where the other value of c is in (e,1)

Another cell in (e,1) where a[t]%c==0 is where x=c+d+1. There are others, but I think it's fairly obvious why this one is probably the significant one (the x formula).

>and what information that gives us

I don't know what information that gives us yet unfortunately.

>the key that is made by column -f with the locations of c in (-f,1) and how to find x for an and a(n-1)

I haven't figured this out yet, but here are some test cases if anyone else spots anything.

 

β€”

 

7*29=203

(7,4,4) = {7:4:14:7:7:29}, f = -22, sqrt(2d) = 5

 

(7,1,95) = {7,1,18053,189,17864,18244}, f=-36100 (x=c-d)

17864/203=88 (BigN)

(7,1,109) = {7,1,23765,217,23548,23984}, f=-47524 (x=c+d+1)

23548/203=116

 

(-22,1,95) = {-22,1,17849,188,17661,18039}, e=-22 (x=c-d)

17661/203=87 (BigN-1)

(-22,1,110) = {-22,1,23969,218,23751,24189}, e=-22 (x=c+d+1)

23751/203=117

 

β€”

 

13*43=559

(30,5,6) = {30:5:23:10:13:43}, f = -17, sqrt(2d) = 6

 

a: (30,1,269) = {30,1,144199,536,143663,144737}, f=-288369 (x=c-d)

143663/559=257 (BigN)

a: (30,1,292) = {30,1,169959,582,169377,170543}, f=-339889 (x=c+d+1)

169377/559=303

 

a: (-17,1,268) = {-17,1,143639,535,143104,144176}, e=-17 (x=c-d)

143104/559=256 (BigN-1)

a: (-17,1,292) = {-17,1,170519,583,169936,171104}, e=-17 (x=c+d+1)

169936/559=304

AA !dTGY7OMD/g ID: 205cf8 Aug. 21, 2018, 2:38 a.m. No.7343   πŸ—„οΈ.is πŸ”—kun

>>7321

Nice to see you again Hobo. I don't think it's that simple. This grid is designed specifically to be used to factorize arbitrarily-large numbers. That's the only thing we know it can be used for. You can use it to find a big list of different cells. You can use it to find a bunch of variables related to any number (c) you put into it. But it's not the kind of thing you can apply a smaller problem to. It's definitely a good approach in other contexts, but not this one, unfortunately. If you made a different grid that worked in the same way (with different variables other than endxab), you could maybe create a different VQC to solve a different problem. But that wouldn't help us figure out how to use the location of a[t]%c==0 in (e,1) and (-f,1) to find the x of an and a(n-1) in (e,1) and (-f,1). I think you're right about it not being solved. I'm thinking VQC just saw a bunch of weird graphs from one of the people Topol brought over and thought it looked like an elliptic curve or something. Also, what is this?

>if the problem is never solved

The problem is already solved, in my view. Time is irrelevant. We're all here for a good reason.

AA !dTGY7OMD/g ID: 205cf8 Aug. 21, 2018, 2:51 a.m. No.7344   πŸ—„οΈ.is πŸ”—kun   >>7346 >>7350 >>7367 >>7370

>>7341

Okay, I figured out the relevance of the numbers from the other cell in (e,1) where a[t]%c==0. I wasn't sure where 116 mattered, for example, in the relevant cells for 7*29 (23548/203=116). I just realized that 88+116=204, which is c+1. It holds for the other example.

 

a: (7,1,95) = {7,1,18053,189,17864,18244}, f=-36100 (x=c-d)

17864/203=88 (BigN)

a: (7,1,109) = {7,1,23765,217,23548,23984}, f=-47524 (x=c+d+1)

23548/203=116 (c-BigN+1)

 

a: (-22,1,95) = {-22,1,17849,188,17661,18039}, e=-22 (x=c-d)

17661/203=87 (BigN-1)

a: (-22,1,110) = {-22,1,23969,218,23751,24189}, e=-22 (x=c+d+1)

23751/203=117 (c-BigN+2)

 

So in the first relevant cell in (e,1), where x=c-d, a[t]=c*BigN.

In the second relevant cell, where x=c+d+1, a[t]=c(c-BigN+1)

In the equivalent first relevant cell in (-f,1) where x=c-d, a[t]=c*(BigN-1)

In the equivalent second relevant cell in (-f,1) where x=c+d+1, a[t]=c(c-BigN+2).

 

I still don't know the relevance but it's an answer to VQC's post. So we're somehow meant to use this to find x for an and a(n-1) in (e,1) and (-f,1). For reference,

a[t] = an at (7,1,4) = {7,1,35,7,28,44}, f=-64

x=7

a[t] = a(n-1) at (-22,1,5) = {-22,1,29,8,21,39}, e=-22

x=8

Anonymous ID: a332a2 Aug. 21, 2018, 7:15 a.m. No.7346   πŸ—„οΈ.is πŸ”—kun   >>7348

>>7344

Yeah, this is a different way of looking at what I posted about regarding shadow n.

 

In your first example you have 17864/203 = 88. Here 88 is the big N of 203. Then you move t + d (or in your case x + d + 1) and you get 23548/203 = 116.

 

116 is the shadow big N of 203. And those patterns are the same as the ones I posted in >>7238 except with a perspective of x rather than t.

Anonymous ID: a332a2 Aug. 21, 2018, 7:57 a.m. No.7347   πŸ—„οΈ.is πŸ”—kun   >>7349

This is what bugs me a lot. I post tons of stuff, but no one is verifying and checking to see if it actually makes sense. It's like there is no team working here, just a bunch of people doing whatever they're doing and ignoring what everyone else is doing.

 

The patterns I've posted about the shadow n seems to be ignored and I don't get it. It seems like a pretty damn big step. There's lots of very cool patterns in it.

Anonymous ID: a332a2 Aug. 21, 2018, 8:04 a.m. No.7348   πŸ—„οΈ.is πŸ”—kun

>>7346

So the first pattern:

x = c - d

 

It's the same as

t = d(d - 1)/2 + (e + 1)/2

 

You can see it being described in >>7320

 

And x = c + d is the same as

t = d(d + 1)/2 + (e + 1)/2

 

Meaning we get big N and shadow big N by treating d as triangles. One with Triangle(d) and the other at Triangle(d + 1).

Saga ID: eca9d4 Aug. 21, 2018, 8:05 a.m. No.7349   πŸ—„οΈ.is πŸ”—kun

>>7347

Every one has his own thing, if anyone has time they will surely study your patterns. I currently am working on something of my own, and since I know a lot about it, I will choose to go down my way. It takes time to learn someone else's patterns and finds. If you have something useful, post it, if not, work on it some more. Key word here is "Improve on what you post."

 

Find, refine, post.

Anonymous ID: a332a2 Aug. 21, 2018, 8:17 a.m. No.7350   πŸ—„οΈ.is πŸ”—kun

>>7344

 

To build a bit on it.

 

The patterns I found for n and shadow n are with regards to the a[t], but if you look at them with regards of x you'll have:

 

x = d - a (x for an)

x = d + a (x for an')

 

x = b - d (x for bn)

x = b + d (x for bn')

 

Where n is normal n and n' is shadow n.

Saga ID: eca9d4 Aug. 22, 2018, 10:05 a.m. No.7358   πŸ—„οΈ.is πŸ”—kun   >>7362 >>7364

>>7334

Kinda, but I found it nevertheless, its

ab/(a+b+c)

thing is, I want to find it using only c, is it possible? sometimes two pythagorean triplets can be found using one c, for instance:

 

16 63 65

33 56 65

 

are both valid for c = 65.

Anonymous ID: 358135 Aug. 22, 2018, 3:24 p.m. No.7362   πŸ—„οΈ.is πŸ”—kun   >>7363

>>7358

No 33 56 65 is not a right triangle, not sure what you are trying to do but it might be like the rotated triangle problem I ran into.

The best way I know to solve for a non right triangle is to make a right triangle where you have overlap of angles and then calculate the difference in angles to thus calculate the non right triangle angles that will total to 180. For example the rotated angle below the hemisphere plane changed to 80 from 90 and I had to find all the angles totaling to 360 at the crossing of the right and rotated triangle to determine the 10 degree shift of the rotated triangle.

VA !!Nf9AmQNR7I ID: 13b051 Aug. 22, 2018, 5:34 p.m. No.7367   πŸ—„οΈ.is πŸ”—kun

>>7321

>>7322

Hello Hobo! Great posts! Nice to see you again :)

 

>>7341

>>7344

I'm following, AA.

 

>>7316

>>7320

Hello Isee! I've been following all your posts and trying to keep up. Thanks for all your work!

 

Here's a good read for you all!

https: //www.zerohedge.com/news/2018-08-21/internet-buzzing-after-julian-assanges-mother-implicates-seth-rich-dnc-leak

VA !!Nf9AmQNR7I ID: 13b051 Aug. 22, 2018, 6:01 p.m. No.7370   πŸ—„οΈ.is πŸ”—kun

>>7344

Hello AA. Here's a start for the list/program:

 

pick a c value from the grid. (let's start with odd x+n c values).

(prime) element shown.

(1,c) Big N.

(-f, c) Big N-1.

(e na transform).

(-f na transform).

(prime) x+n

(prime) d+n

(f-1) div 8

(f-1) mod 8

u and u+1 (to create odd x+n)

in (e,1) show (an) and (bn)

in (-f,1) show a(n-1) and b(n-1)

calc the needed movement from (e na transform) to (an) and (bn)

calc the needed movement from (-f na transform) to a(n-1) and b(n-1)

show polite numbers from (f-1) div 8 and multiply until we get a lock from rm (2d-1)

find the element in (e,1) that contains the f value which if we take SQRT(f) will give the correct x+n value.

 

There's a start. Copypasta and add more, Anons.

Anonymous ID: 358135 Aug. 22, 2018, 6:02 p.m. No.7371   πŸ—„οΈ.is πŸ”—kun

>>7369

I am looking at the Lie 8 Group and it just reminded me a lot of some of the early stuff on this board. I have no idea if it fits, just that it is a set of 246 dimensions based on a 8 dimensional spirial - rather complex

Anonymous ID: 358135 Aug. 23, 2018, 4:53 a.m. No.7377   πŸ—„οΈ.is πŸ”—kun

>>2707873 (You)

>>2707996 (You)

Geometric projection of Lie 8 group onto a higher dimensional manifold (ie hopf sphere - the Hydrogen atom electron field) will cause the scale shifts seen in the Mandelbrot set as the flat plane curls in and out on itself.

PMA !!y5/EVb5KZI ID: 52c782 Aug. 24, 2018, 5:49 p.m. No.7383   πŸ—„οΈ.is πŸ”—kun   >>7384 >>7411

>>7031

>Rather than being a pathway to a solution

Found some patterns using the odd/even triangle square formulas to create factor records between the -f and e columns that warranted a deeper look.

 

Recall that:

 

1) Odd squares are created by 1+8T(u), even squares by 4T(u) + 4T(u-1)

2) an n0 can be calculated from any estimated (x+n)(x+n), c and d values. n0 = sqrt( XPN + c ) - d;

3) the n0 can be tested for validity based on the nn + 2d(n-1) + f - 1 = XPN formula.

4) The difference between the estimated (x+n)(x+n) and the nn formula, is called the remainder 2d(n-1). Or rm2dnm1 for short.

5) A factor record exists where rm2dnm1 = 0.

 

Attached pics for c85 and c288 show 2 views of where the factor records can be found. The matrix view shows a side by side of -f and e columns and their respective calculations, while the list view is sorted by x descending. The XPN, n0, rm2dnm1 columns are explained above.

 

The new columns show the interesting patterns and relationships between -f and e.

 

"rm2dnm1 diff" is calculated as the difference between "-f rm2dnm1" and "e rm2dnm1".

 

For c85 as an example, the entry record at a matching XPN is found in the -f column at {-15:1:874:41:833:917} where x+n=42. It's related record at x-1 is {4:1:842:40:802:884}. The difference between their rm2dnm1 values is 2. i.e. the "rm2dnm1 diff" column.

 

Performing that same calculation on -f and e records x-1 apart will yield the same "rm2dnm1 diff" value all the way down to u=10 and record {4:1:222:20:202:244}.

 

The "rm2dnm1 diff" groups for c85 are 2, 4, 6, 8, 10, 12. There are several properties evident in these groups:

 

1) each group increments by 2.

2) the number of records within each group decreases.

3) the parity of the rm2dnm1 within each group is always the same.

4) a factor record will never be found when the rm2dnm1 parity is odd.

5) the parities between each group alternate. (even, odd, even, odd, etc.)

 

And finally, the "rm2dnm1 diff" value will equal 2a from the factor record x^2 + e = 2na formula. This is indicated in the "diff/2 = a[t]" column.

 

(1/2)

PMA !!y5/EVb5KZI ID: 52c782 Aug. 24, 2018, 5:49 p.m. No.7384   πŸ—„οΈ.is πŸ”—kun   >>7411

>>7383

Attached pics are a few examples to illustrate that the patterns mentioned previously hold regardless of number size.

 

For c6107, the matrix and list views show where the first group for "rm2dnm1 diff" = 2 finishes at u=763.

 

This group of records represents a jump in terms of n values from 2976 to 1450, all of which can be skipped because they will produce the same result. In this example, the last record in the group {23:1:1164349:1525:1162824:1165876} is found where x+n = u from the starting record. The range of records in this first group can be calculated depending on odd/even e and/or odd/even x+n.

 

For c9705341159, the matrix and list pics attached show the consistency in results where the prime solution record is found. Notice again the "rm2dnm1 diff" value equals 2a.

 

Being able to determine the start and end ranges for each of these "rm2dnm1 diff" groups could achieve a/4 performance if we process linearly.

 

Integrating a recursive process with appropriate jumps might be the next step to a solution.

 

(2/2)

AA !dTGY7OMD/g ID: d3023b Aug. 25, 2018, 8:22 p.m. No.7387   πŸ—„οΈ.is πŸ”—kun

Here's something to look into from the older threads. I just copied and pasted the text from Discord, if you're wondering why it's posed as a question.

 

Does anyone know why with the "there's always a cell 2n to the right" thing in each row that there seem to be arbitrary starting places? You can always start from (-1,n) and (0,n), but those aren't always the only starting places. Row 3, for example, has (-1,3) and (0,3), which do have cells 6 cells to the right, but (2,3) and (3,3) are also valid, and also have cells 2n to the right each.

AA !dTGY7OMD/g ID: d3023b Aug. 25, 2018, 9:03 p.m. No.7389   πŸ—„οΈ.is πŸ”—kun

Another thing, which Jan suggested I bring to everyone's attention:

>>761

The difference between the x value of the cells in (0,n) where c=1c^2 and where c=a^2b^2 is always divisible by 12. I verified with a couple examples myself.

AA !dTGY7OMD/g ID: f24fc0 Aug. 26, 2018, 2:40 a.m. No.7401   πŸ—„οΈ.is πŸ”—kun   >>7402 >>7404

I found something very strange with my "compare every number with every other number" program. Turns out the u of the cell in (0,n) where a=b and b=a^2b is always divisible by the b we're looking for. That is so weird.

AA !dTGY7OMD/g ID: 945834 Aug. 26, 2018, 4:28 a.m. No.7405   πŸ—„οΈ.is πŸ”—kun   >>7406

Alright, I've spent all of the last thirteen hours working on this, despite the fact that I have more non-VQC-related work to do this week than I probably had time for before I spent an entire day on this. I need a break. Seriously, please, can somebody else work on this grid pattern thread with me? Why am I the only one working on it? I'm up to around the start of RSA #7. Here's the thread >>6506

PMA !!y5/EVb5KZI ID: 284644 Aug. 26, 2018, 8:18 p.m. No.7411   πŸ—„οΈ.is πŸ”—kun   >>7415

>>7383

>>7384

Pics attached are summary views for c145, c629, c6107 and c15120, to further illustrate the groupings for rm2dnm1 between -f and e columns.

 

Each example is grouped by the "rm2dnm1 diff" value, and then filtered to only show even parities. Just looking for any obvious patterns for larger jumps.

 

Have also confirmed that if the same calculation is performed on u values above the c entry record, then the rm2dnm1 value will always equal c.

VA !!Nf9AmQNR7I ID: ff6520 Aug. 28, 2018, 7:19 p.m. No.7413   πŸ—„οΈ.is πŸ”—kun

>>7412

Nice find, Hobo!

Here's a screenshot where 0.618 is shown as part of the Phi formula. Also, VQC posted back at the beginning in response to a Golden Ratio post.

AA !dTGY7OMD/g ID: 9be96f Aug. 29, 2018, 6:53 p.m. No.7414   πŸ—„οΈ.is πŸ”—kun

I put together a program to analyse the difference between BigN and n. VQC said they were all smooth numbers ( >>7155 ), so I’ve been generating BigN-n the same way as the grid generation and taking their biggest prime factor each time. As it turns out, the largest possible biggest prime factor of the gap between BigN and n is related to i. When I set iMax to 300, the range of possible biggest prime factors for every possible BigN-n in this range is just the complete list of prime numbers from 1 to 300. When I do the same for iMax=400, it’s the list of prime numbers from 1 to 400. Interdasting.

 

I'm also curious about the distribution. The first of those numbers at the bottom is the number of BigN-ns analyzed, and the next number is the number of unique largest prime factors (which also happens to be the number of prime numbers up to i=400).

PMA !!y5/EVb5KZI ID: 284644 Aug. 29, 2018, 9:36 p.m. No.7415   πŸ—„οΈ.is πŸ”—kun

>>7411

May have stumbled across a way to find some factor records for trivial cases using only e and d (or at the very least confirmed an avenue worth exploring).

 

Pics attached are annotated matrix views showing side by side -f and e columns for c85, c145, c287, and c459. All records are included for c85 and c145, while only matching factor records are shown for the others.

 

In (e,1), we know that f = (x+n)(x+n), or sqrt(f) = x+n. And because n=1, x can always be expressed in terms of f as x = sqrt(f) - 1.

 

These small number test cases seem to indicate that there is a relationship between records where a record lower in the column can be found at d = sqrt(f).

 

For the c145 example where the starting x+n = 72, a corresponding record in (e,1) with the same x+n is found at {1:1:2592:71:2521:2665}. At this record, f = 5184 and sqrt(f) = 72.

 

Lower down in (e,1), we find the record {1:1:72:11:61:85} where d = 72, the same value as the previous record's sqrt(f). The resulting (x+n)(x+n) from this record is then used to solve the problem.

 

For c85, however, this jumping by d = sqrt(f) requires an extra iteration.

 

Starting at {-15:1:874:41:833:917}, f=1764, we calculate the lower d value as sqrt(1764) = 42 and jump down to {-15:1:42:9:33:53}, f=100. If a solution isn't found, we repeat the calculation for the next lower d as sqrt(100) = 10, and jump down to {-15:1:10:5:5:17}, f=36.

 

For c287 - similar to c145 - a solution can be found in a single calculation using d = sqrt(f) where f is the (x+n)(x+n) from the starting 1,c record.

 

And for c459, this method finds 3 out of the 4 factor records. d value jumps from 26229 -229 -> 21.

AA !dTGY7OMD/g ID: 17d6a9 Aug. 29, 2018, 11:23 p.m. No.7416   πŸ—„οΈ.is πŸ”—kun   >>7417 >>7418

The difference between BigN and the n of specific records is going to have one prime number as its largest prime factor. e.g. c=501 bign=229 n=63 gap=166 LPF=83 (gap=83*2) {17:63:22:19:3:167}

 

Multiple records in the grid have different gaps with the same highest prime factor, e.g. c=835 bign=390 n=58 gap=332 LPF=83 (8322) (51,58,12) = {51:58:28:23:5:167}

 

Since BigN and n are always the same parity, this gap will always be even.

 

Since multiple records have different gaps with the same highest prime factor, they can be grouped together, like in pic related.

The beginning record always has a=3 but b doesn’t seem predictable that I can tell.

The difference in the gap from one record to the next within a group is equal to the biggest prime factor times 2.

c=357 gap=118 BPF=59

c=595 gap=236 BPF=59 (236-118=118)

c=833 gap=354 BPF=59 (354-236=118)

c=1071 gap=472 BPF=59 (472-354=118)

c=1309 gap=590 BPF=59 (590-472=118)

 

This would only help if we could predict the BPF group any number falls within, because then we’d be able to use the way in which c scales upwards to calculate how many times we’d need to multiply the BPF by to get the correct BigN-n gap. Then we’d just take that gap away from the BigN we know how to calculate, and that would be it. Maybe there’s some way of finding that BPF.

AA !dTGY7OMD/g ID: add144 Aug. 30, 2018, 1:02 a.m. No.7419   πŸ—„οΈ.is πŸ”—kun

>>7417

This isn't true in all cases. A lot of them, but definitely not all of them (it didn't work on RSA100 for example, and I know it doesn't work on every number because records with the same gaps divisor can have different b values, obviously. Sigh)

Anonymous ID: 1e2d5d Aug. 31, 2018, 12:41 p.m. No.7425   πŸ—„οΈ.is πŸ”—kun   >>7432 >>7435

>>7424

We're turning general number sieve field theory into a straight calculation.

 

Take a[t] from (e, 1) and subtract from a in (e, 1). You'll get smooth numbers. Take d[t] from d in (f, 1) and you'll get smooth numbers. Smooth numbers are made from triangles and squares.

 

GNFS is a blind man's factorisation algorithm.

AA !dTGY7OMD/g ID: 7ddfda Aug. 31, 2018, 5:29 p.m. No.7432   πŸ—„οΈ.is πŸ”—kun

>>7425

Reading this the first time I was so confused and thought someone read "I'm up to RSA #9" as "I'm new and I'm reading the old threads" and thought they'd introduce me, but then why would they give that second clue?

 

…is this VQC giving us another crumb without his trip? They have no posting history.

AA !dTGY7OMD/g ID: ead8f7 Aug. 31, 2018, 8:37 p.m. No.7439   πŸ—„οΈ.is πŸ”—kun   >>7446

Nobody but VA seemed to give a shit on Discord or in the grid patterns thread, despite how important this seems to me, so I'll bring this here.

 

Look at this post from RSA #8 in January:

>>2858

It's from someone's private messages with VQC back when he had a Twitter account. VQC quite literally already told us what "root of d" is. It's a cell in (0,n).

"Root of d = {0, 2xd, 3xd, 2xd, d, 9xd}. All a and b of ( 0, 2xd, t) are multiple of d for all t."

Also root doesn't necessarily refer to square root. It can also refer to the root of a tree.

Anonymous ID: 64f02a Aug. 31, 2018, 8:37 p.m. No.7440   πŸ—„οΈ.is πŸ”—kun   >>7441 >>7442 >>7444 >>7445 >>7503

Hey all. I stopped keeping up with this research a while ago since I couldn't keep up with the programming aspect, but I've been browsing sporadically. So, I broke off on my own efforts autistically looking for patterns that may be helpful. So far I've identified a couple of patterns within fibonacci sequences that might be relevant. Bare with me, this is extremely autistic…

 

Im working on an infograph at the moment, but for now here's a rundown of it.

 

The first (ones) digits in the numbers of a fib sequence make up a pattern of 60 numbers which repeats ad infinitum. Within this pattern there are subpatterns: every 5th and 10th number is a 5, every 15th a 0. Every 3rd number is even. The second half (#'s 31-60) is opposite the first half, i.e. the 1st and 31st add up to 10.

 

If you break the pattern of 60 into sets of 15 more patterns emerge, 1, 7, 9, 3 and 2, 4, 8, 6. As well as a helical sort of distribution throughout the set of 60 numbers.

 

For the time being that is all that needs to be said about the first digit pattern.

 

All digit columns subsequent of the first digit pattern also have patterns which they follow. Each pattern being 5 times larger than the pattern before it. The first digit pattern is 60 numbers long, the second digit pattern is 300. Every digit in a fibonacci sequence always has a repetitious pattern. And, these patterns appear to follow rules established by the first digit pattern.

 

If you break up any of the subsequent patterns into subsets Nx numbers long where x = the length of the previous pattern, you can establish at least one that I've proven thus far rule which dictates that pattern. The second (10's) digit pattern being 300 numbers long, and the previous pattern being 60, break this pattern up into 5 columns of 60 and the rule for this pattern becomes obvious. At first glance the numbers appear highly chaotic, but surprisingly follows a very simple rule. It follows the rule of "2, 4, 8, 6" in reverse. i.e. {6,8,4,2} the 1st + 6 = the 61st, the 4th + 2 = the 64th, ad infinitum.

 

The next (100's) digit follows the same rule, but this time the sequence is shifted two spaces so that it is {4, 2, 6, 8}

 

I'm going to cut myself off for now. If you guys find this interesting and relevant let me know and I will keep going.

GAnon !jPVzzZOz2c ID: e906b3 Aug. 31, 2018, 9:04 p.m. No.7442   πŸ—„οΈ.is πŸ”—kun   >>7443

>>7440

KEEP GOING

 

This shit is crazy.

The babylonians wrote in base 60 and they also had phi, the golden ratio, the one from fibonacci, etched into stones. The fact that this is the case is actually really cool. Also you always have those people posting the cyclical digit tesla math stuff which looks like 2, 4, 8, 6 with doubling.

Anonymous ID: 64f02a Aug. 31, 2018, 10:01 p.m. No.7443   πŸ—„οΈ.is πŸ”—kun   >>7448

>>7441

>why only the first digits

Because the first digits establish a set of solid patterns and I am operating under the hypothesis that all subsequent digits amount to an abstraction of the first.

 

>>7442

Check out pics related. Like I said I am not programmer and I started running into problems with floating integers fucking up the numbers so I've been calculating everything manually. I have the first 7500 iterations of a fibonacci sequence starting with 1 calculated.

 

First pic is the infograph that Im working on. I just slapped it together so apologies if its hard to follow but this should help to visualize what I am talking about. It gets a bit difficult (as you can see with the 3rd pattern getting chopped off on the bottom cuz its huge.

 

I also made a gif to show the juxtaposition of the numbers in the first pattern. In the gif the pattern is arranged in order but it is playing through to highlight the opposite numbers and their relationship, i.e. rather than sequential order it is 1 followed by 9, 2 followed by 8, and then these numbers paired together.

 

I'll have to continue this either in a little bit or tomorrow I pushed my computer a little to hard tonight with this shit and its not playing ball right now.

GAnon !jPVzzZOz2c ID: e906b3 Sept. 1, 2018, 6:49 a.m. No.7445   πŸ—„οΈ.is πŸ”—kun   >>7447

>>7440

 

These minor patterns,

1,7,9,3 and 2,4,8,6

come from different multiplications

 

1,7,9,3, comes from multiplying repeatedly by 7

 

>>> 7*7

49

>>> 49*7

343

>>> 343*7

2401

>>> 2401*7

16807

>>> 16807*7

117649

>>> 117649*7

823543

 

If you do it by 3's then you get

3 -9 -> 27 -> 81 -> 243

3,9,7,1 which is the pattern in reverse

 

Then the 2,4,8,6 obviously comes doubling or multiplying by 2, because

22 = 4, 42 = 8, 82 = 16, 162 = 32, 32*2 = 64

2,4,8,6,2,4,…

 

Maybe we can use the fact that binary is doubling and maybe encode our number into a fibonacci number then do somethign from there.

3DAnon !!!YmM2NGExNzVlMDQ2 ID: e08c3e Sept. 1, 2018, 1:43 p.m. No.7446   πŸ—„οΈ.is πŸ”—kun

>>7439

We do give a shit, just not always have any insight to share. I've found that for e=0 you can get to all records where a and b are a multiple of n just by starting at (0, n, 1) and jumping by t = t + n. First image is using c6107's d value of 78 to get to (0, 78, 1), and then jumping.

 

Somehow this works for 2d as well, where jumping by t+78 from (0, 156, 1), the a and b values are still always divisible by 78. Next level up you could use 156 * 2 = 312 and go to (0, 312, 1+312), and a and b is always divisible by 156.

 

This diagonal movement increases c 4 times for each step, and has some really interesting patterns in the generated records, not sure what to make of them yet

Anonymous ID: 64f02a Sept. 1, 2018, 4:32 p.m. No.7447   πŸ—„οΈ.is πŸ”—kun

>>7444

>>7445

 

Actually, at the very beginning I started off looking at it in base 9 because way back when we were still trying to figure out what vqc was talking about several anons were wondering if 0 was an actual number. There were patterns but it didnt start 'making sense' until I went back to base 10.

 

>Maybe we could use the fact that binary is doubling…

That is the assumption that I am running on, that if all the rules of this pattern can be identified we would be able to use it as an abstract means of simplifying very large numbers.

 

I've ran into a problem with my theory. The 2nd pattern is 5 times the first and the third 5 times the 2nd but the 4th is not 5 times the 3rd. I made the assumption that the patterns always being 5 times larger without paying attention to the pattern of 5, 5, 0 in the first pattern. So, I have a feeling that the 4th pattern will be 10 times the 3rd. Good news is Im already at 7500, bad news is I need to get to 1500. My computer might explode.

Anonymous ID: 64f02a Sept. 2, 2018, 4:07 p.m. No.7449   πŸ—„οΈ.is πŸ”—kun   >>7450

>>7448

Interesting. Im giving it some attention right now.

>e is x, n is y

Like I said I haven't been keeping tabs here in a while which thread should I browse to get a refresher on the vqc formulas?

Anonymous ID: ebefe1 Sept. 2, 2018, 4:26 p.m. No.7450   πŸ—„οΈ.is πŸ”—kun   >>7451

>>7449

As someone here stated, it is Fermat's last theorem on steriods.

https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem

 

You understand that? create the map using a nested loop.

Read the rules!

Anonymous ID: 64f02a Sept. 2, 2018, 6:39 p.m. No.7451   πŸ—„οΈ.is πŸ”—kun   >>7452 >>7453 >>7455

>>7450

>You understand that?

insert us navy seal copypasta

 

I both do and don't understand it. I'd need to be able to play around with it to really get an appreciation for it but aside from getting numbers to print in console I have no idea how to code to get any sort of graphic output. I've been doing all my work in excel and mspaint, jej Im pretty decent with python but all documentation I've found on it is in redditese so I've never been able to quite figure out how to open a window and draw to screen.

 

Anyway, as for >>7448 it appears all odd y have a pattern x=y long and all even have a pattern x=y*2 long

Anonymous ID: 64f02a Sept. 3, 2018, 7:13 a.m. No.7454   πŸ—„οΈ.is πŸ”—kun

>>7452

It pretty much is isnt it . The odd y are growing by 2 and the even y by 4. I remember way back in like december anons were associating Lorentz transforms with vqc

Anonymous ID: ebefe1 Sept. 3, 2018, 8:43 a.m. No.7455   πŸ—„οΈ.is πŸ”—kun   >>7457

>>7451

>Anyway, as for >>7448 (You) it appears all odd y have a pattern x=y long and all even have a pattern x=y*2 long

VERY interesting, wanna be my study partner? also, can you find patterns along x? horizontally?

 

Also, for anyone doing col.1 this is a good website with a good excel file.

http://www.tsm-resources.com/alists/trip.html

Anonymous ID: ebefe1 Sept. 3, 2018, 3:41 p.m. No.7458   πŸ—„οΈ.is πŸ”—kun   >>7459

>>7457

thing is I'm pretty sure that after 499 (being d = 22 and e = 15), the data is missing. can you work with missing data? I know its hard but you can make assumptions. like first pic is e vs d to 50 incomplete, the second is how it should be.

 

vqc original code doesn't give you everything, specially numbers with high e and d that are somewhat "rough".

Anonymous ID: 64f02a Sept. 3, 2018, 3:55 p.m. No.7459   πŸ—„οΈ.is πŸ”—kun   >>7460

>>7458

Is the missing data the result of floating integer rounding? If that is the case I got around that with my fibonacci computations by taking a couple extra steps to create variables and booleans in order to add each digit in an equation individually and carry over the one if needed (like you would doing the math in your head)

 

>can you do it with missing data

Maybe. I can certainly try. Hopefully I will have time tonight to sit down and go over the equations, until I understand what I am looking at Im just playing hyper autistic connect the dots.

Anonymous ID: ebefe1 Sept. 3, 2018, 5:56 p.m. No.7460   πŸ—„οΈ.is πŸ”—kun   >>7461 >>7463

>>7459

No, I don't think it's a floating integer, the whole point of this is to use integers, we can define a number by it's d and e, d being the square root, and e being the remainder.

 

for example, take a number, 6107. It can be represented as d = 78 and e = 23.

Anonymous ID: 358135 Sept. 4, 2018, 4:23 a.m. No.7462   πŸ—„οΈ.is πŸ”—kun   >>7467

Can some please check my MATH - something is wrong.

I am digging nuclear fission verses actual documented energy in nucleus and think I found a discrepancy. The strongest and majority of force in a nucleus is the Strong Nuclear force. So trying to compare the Strong Nuclear Force Energy to the Energy of Fission.

 

Strong Nuclear Force

 

The strong force acts between quarks. Unlike all other forces (electromagnetic, weak, and gravitational), the strong force does not diminish in strength with increasing distance between pairs of quarks. After a limiting distance (about the size of a hadron) has been reached, it remains at a strength of about 10,000 newtons (N), no matter how much farther the distance between the quarks.[5]

 

Fission

 

Typical fission events release about two hundred million eV (200 MeV) of energy, the equivalent of roughly >2 trillion Kelvin, for each fission event.

 

Size of Nucleus:

The size (diameter) of the nucleus is between 1.6 fm (10βˆ’15 m) (for a proton in light hydrogen) to about 15 fm (for the heaviest atoms, such as uranium).

 

10,000/(15*10^-15)=.000000000000666 and that converts to 4.15684331565551 MeV.

 

and

 

200MeV converts to 0.0000000000320435 Newton meters.

 

There is an order difference of 4,800 times more energy in the fission reaction than the strong nuclear force.in Newton Meters and it appears 48.1 order of difference in MeV.

I must have done something wrong.

Anonymous ID: 64f02a Sept. 4, 2018, 7:38 a.m. No.7463   πŸ—„οΈ.is πŸ”—kun   >>7464

>>7460

I got two questions if you don't mind. For finding c are we limiting to the squares of integers? And for a,b is a the smallest factor or any factor smaller than b and vice verse? I can work with flash so Im writing something up.

Anonymous ID: 64f02a Sept. 4, 2018, 8:04 a.m. No.7465   πŸ—„οΈ.is πŸ”—kun

>>7464

>no fractions

Is there a reason for this or just how its being done for the moment for simplicities sake?

>floor(sqrt(c))

wouldnt d=floor(sqrt(c)) and c = (xx)-(yy) ?

Anonymous ID: 358135 Sept. 4, 2018, 4:40 p.m. No.7470   πŸ—„οΈ.is πŸ”—kun

>>7467

Well I am a newbie to your theory basically from the plots I have seen it is Lorentz transforms with a few other geographic figures in different planes. I guess I would need a solid spreadsheet to stare at for a while and see what jumped out to my geometry skills.

And that the original square to larger square minus the outer little square is basically the inner circle of the square.

Anonymous ID: e906b3 Sept. 4, 2018, 5:02 p.m. No.7471   πŸ—„οΈ.is πŸ”—kun   >>7472

12 step program

 

We admitted we were powerless over alcoholβ€”that our lives had become unmanageable.

Came to believe that a power greater than ourselves could restore us to sanity.

Made a decision to turn our will and our lives over to the care of God as we understood Him.

Made a searching and fearless moral inventory of ourselves.

Admitted to God, to ourselves, and to another human being the exact nature of our wrongs.

Were entirely ready to have God remove all these defects of character.

Humbly asked Him to remove our shortcomings.

Made a list of all persons we had harmed, and became willing to make amends to them all.

Made direct amends to such people wherever possible, except when to do so would injure them or others.

Continued to take personal inventory, and when we were wrong, promptly admitted it.

Sought through prayer and meditation to improve our conscious contact with God as we understood Him, praying only for knowledge of His will for us and the power to carry that out.

>Having had a spiritual awakening as the result of these steps, we tried to carry this message to alcoholics, and to practice these principles in all our affairs.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 Sept. 4, 2018, 5:17 p.m. No.7473   πŸ—„οΈ.is πŸ”—kun   >>7474

>>7472

I stopped going as wild as I did.

"Now do it without the drugs."

 

The bulk of all this was "sober", for me.

That's when shit got weird.

Aaaaand as usual… I'm a bit ahead of the curve and everyone kept telling me to relax and enjoy the show.

 

Buuuuuuuuuuuuut… "relax" is basically concept in my mind.

I thought I'd found it a time or two…

But I can't relax (patriots can't sleep) until this is finished.

 

Autism.

Unfinished gives me the heeby jeebies.

And I get antsy.

Finish it. Finish the thing so we can move on.

 

All Day I Dream About Serenity.

GAnon !jPVzzZOz2c ID: e906b3 Sept. 4, 2018, 7:33 p.m. No.7476   πŸ—„οΈ.is πŸ”—kun

//The VQC is a lookup table

//The question you answer is what do you do to find the lookup x for c that you gives you na or (n-1)

//On-c-e you c i-t you c[an]not un-c i[t]

 

Math in the last line:

 

on-c-e potentially c-e

c i-t

c[an] NOT (logical not)

UN-c (logical inverse of negative c)

i[t]

 

I think in the original vqc code

a = i - j

b = i + j

 

so i = (b+a)/2 = (d+n)

 

Values to maybe check

 

-c-e = -(dd+e) - e = -dd - 2e

i - t for t where i = i[t] = d+n

i[t] for various t

Inverse of c*an and -c

 

This last line wasn't an accident and there is something in it

VA !!Nf9AmQNR7I ID: 9534e2 Sept. 4, 2018, 7:34 p.m. No.7477   πŸ—„οΈ.is πŸ”—kun

>>7466

Ok, so it's c145 (I know!) but here's a working example of the jump from na transform to the correct record containing (an).

In (1,1) the na transform record has x=11

SQRT(2d) = SQRT(12 * 2) = 4 remainder 8

x=11 - 4 = 7 = correct x value at the (na) element:

{1:1:32:7:25:41}

AA !dTGY7OMD/g ID: 0cec15 Sept. 4, 2018, 10:59 p.m. No.7478   πŸ—„οΈ.is πŸ”—kun   >>7495

Hang on a minute. The x and t values in the an and a(n-1) cells are the same as the x and t values in the prime solution. VQC has said a few times that we're meant to find the x or t value of the an cell, which implies that we'd then factor an to find the solution. If we're meant to find that cell based on its x or t value, we can completely bypass it and use the x or t value to find the prime solution without needing to factor an. That renders row 1 meaningless, if x or t are the values we're looking for. So that can't be right.

AA !dTGY7OMD/g ID: 49aa11 Sept. 4, 2018, 11:41 p.m. No.7480   πŸ—„οΈ.is πŸ”—kun   >>7481 >>7484

I found an interesting property of the row 1 cells. Each one will have its own e and f values unique from the e,f pair used to find them. The difference between these e and f values is equal to either a or b in my test cases.

 

c559, 13*43, (30,5,6) = {30:5:23:10:13:43}

(30:1:6) = {30:1:75:10:65:87} f=-121, c=5655

(-17:1:6) = {-17:1:63:11:52:76} e=108, c=9607

e-|f| = 13, which is a

 

c203, 7*29, (7,4,4) = (7,4,14,7,7,29)

(-22:1:5) = {-22:1:29:8:21:39} e=35, c=819

(7:1:4) = {7:1:35:7:28:44} f=-64, c=1232

|f|-e = 29, which is b

AA !dTGY7OMD/g ID: 49aa11 Sept. 4, 2018, 11:44 p.m. No.7481   πŸ—„οΈ.is πŸ”—kun   >>7484

>>7480

Oh hey, another property of this unique pair of e and f values. You can represent a given c with its e and f values, right? Well, this unique pair of e and f values in row 1 for the an and a(n-1) cells also represent valid cs themselves. These new c values are actually divisible by a or b (depending on which the e-f/f-e gap is divisible by) and they're divisible by BigN-n. Now that is interesting.

 

c559, 13*43, (30,5,6) = {30:5:23:10:13:43}

(30:1:6) = {30:1:75:10:65:87} f=-121, c=5655

(-17:1:6) = {-17:1:63:11:52:76} e=108, c=9607

e-|f| = 13, which is a

an is divisible by e’-|f’’|

Do these e and f values produce another c?

They do: 13104, which is a(BigN-n)2*2

 

c203, 7*29, (7,4,4) = (7,4,14,7,7,29)

(-22:1:5) = {-22:1:29:8:21:39} e=35, c=819

(7:1:4) = {7:1:35:7:28:44} f=-64, c=1232

|f|-e = 29, which is b

bn is divisible by |f’|-e’’

Do these e and f values produce another c?

Yes, 2436, which is equal to b*(BigN-n)

Anonymous ID: 08d201 Sept. 5, 2018, 1:37 a.m. No.7482   πŸ—„οΈ.is πŸ”—kun   >>7483

Prostobiec's hit piece on Q.

Who in the hell are "Microchip" and "Dreamcatcher"???

 

https://twitter.com/ObiWanKEKobi/status/1037145420546428928

Part 1

 

https://twitter.com/ObiWanKEKobi/status/1037146176846548992

Part 2

AA !dTGY7OMD/g ID: 430395 Sept. 5, 2018, 2:13 a.m. No.7484   πŸ—„οΈ.is πŸ”—kun

>>7481

This thing about c being divisible by either a or b and BigN-n definitely still seems to stand. I've verified it with several other test examples.

 

>>7480

But with this thing, it seems like by pure coincidence, both of my test examples happen to be the only ones that actually make this work.

VQC !!/aJpLe9Pdk ID: 814c3d Sept. 5, 2018, 2:27 a.m. No.7485   πŸ—„οΈ.is πŸ”—kun   >>7487 >>7488 >>7489 >>7490 >>7492 >>7504 >>7506 >>7571

Any number of the form n(n-1) or multiple of it, has distinguishing properties and can be pick out of a crowd, especially and more easily if the multiple of n(n-1) is a square.

What if you multiplied each corresponding element in (-f,1) and (e,1)?

Would that series be totally predictable?

Could you quickly lookup or figure out any values that are a square multiples by two consecutive numbers like n(n-1)?

If you do this, you will quickly spot the value of t or x that is a.a.n.(n-1) at x.

This is the solution.

Multiply each entry a[t] at (e,1) by the corresponding entry at (-f,1). These pairs form a series of products.

You will quickly start to c.

Try it on some lower numbers.

Then you will be able to do it for all odd products.

Prime numbers will apear to be special.

Godspeed anons.

AA !dTGY7OMD/g ID: 430395 Sept. 5, 2018, 2:36 a.m. No.7486   πŸ—„οΈ.is πŸ”—kun   >>7491

Here's a thing based on the latest talk about the sum of two squares.

 

Since c=(d+n)(d+n)-(x+n)(x+n), the difference of two squares, you can also find another c where c'=(d+n)(d+n)+(x+n)(x+n), the sum of two squares. The difference between these c values appears to be the smaller square times two.

 

559=13x43

559=(28x28)-(15x15)

(28x28)+(15x15)=1009

1009-559=(15x15)x2

 

203=7x29

203=(18x18)-(11x11)

(18x18)+(11x11)=445

445-203=(11x11)x2

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 Sept. 5, 2018, 2:43 a.m. No.7488   πŸ—„οΈ.is πŸ”—kun

>>7485

Y'know… I finally made it through Interstella 5555.

I can't tell you how many times I attempted it, but I wasn't ready yet.

 

After Alien HumanKind…

Yeah, the story makes way more sense and I can stay awake during it.

 

Seemed a bit campy the first 1000 times I tried to watch it.

Been open in a tab for months.

Mooooooonths lol

 

https://www.youtube.com/watch?v=FGBhQbmPwH8&list=PL5wIANSfIdw_7vtJe9T5Y5WTiF2kAnZoW&index=1

AA !dTGY7OMD/g ID: 430395 Sept. 5, 2018, 2:50 a.m. No.7490   πŸ—„οΈ.is πŸ”—kun

>>7485

>Any number of the form n(n-1) or multiple of it, has distinguishing properties and can be pick out of a crowd, especially and more easily if the multiple of n(n-1) is a square.

>What if you multiplied each corresponding element in (-f,1) and (e,1)?

So if you take an and a(n-1) and multiply them together, you get a^2 and n(n-1)

c559 13*43, n=5

an=65, a(n-1)=52

65*52=3380

And apparently it will be easy to factor that number somehow.

 

>What if you multiplied each corresponding element in (-f,1) and (e,1)?

>Would that series be totally predictable?

Not that I can tell straight away but probably, if you're saying it

>Could you quickly lookup or figure out any values that are a square multiples by two consecutive numbers like n(n-1)?

So I guess this means (an)(a(n-1)) pops up somewhere, and there's an easy way to factor it. I feel like this will require a look at some of the old crumbs.

>If you do this, you will quickly spot the value of t or x that is a.a.n.(n-1) at x.

This seems like another one of those clues that we'll misinterpret several times before actually understanding it.

>you will quickly spot the value of t or x that is a.a.n.(n-1) at x.

>the value of t or x that is a.n.n.(n-1) at x

>the value of x

>at x

Huh?

AA !dTGY7OMD/g ID: 430395 Sept. 5, 2018, 3:03 a.m. No.7492   πŸ—„οΈ.is πŸ”—kun   >>7493 >>7494

>>7485

Triangle numbers are (n(n+1))/2

The triangle number before the nth triangle number is then therefore (n(n-1))/2, since n+1 becomes n and n becomes n-1.

So (an)(a(n-1)) is a^2 + 2T(n-1). If that means anything.

AA !dTGY7OMD/g ID: 430395 Sept. 5, 2018, 3:29 a.m. No.7493   πŸ—„οΈ.is πŸ”—kun

>>7492

This is a list of all (an)(a(n-1)) for the early numbers where a and b are odd and aren't 1.

18

50

54

98

108

150

162

180

242

270

294

300

338

378

450

486

500

504

578

588

648

722

Interesting to note is that some of these numbers are an odd square times 2. 18=92, 50=252, 98=492, 162=812. I'm not sure what the deal with the other numbers is.

VQC !!/aJpLe9Pdk ID: 814c3d Sept. 5, 2018, 3:45 a.m. No.7494   πŸ—„οΈ.is πŸ”—kun   >>7496 >>7498 >>7499 >>7502 >>7504 >>7511 >>7571

>>7492

If you define n-1 as k, then n(n-1) becomes k(k+1) or twice a triangular number.

So, you're looking for when a series becomes two triangles where every unit of the triangle is a square.

There is a known shortcut to do this between two limiting numbers or range.

Importantly, the search is now a calculation with a known shortcut that is at most O(log t). Think about the values in 0,1 and 0,1.

Everything that has been discussed points at this solution.

Euler? Euler? (Mispronounced)

Teach !!UgZAPoSXEk ID: 9cfa21 Sept. 5, 2018, 4:24 a.m. No.7496   πŸ—„οΈ.is πŸ”—kun   >>7500

>>7494

Thanks for these latest tips VQC.

I feel like this is so close to revealing the answer directly… I'm working on 'em now.

 

In (0,1) * (0, 1) the pattern is:

02 = 0 = 160 = 4^2 * 0^2

28 = 16 = 161 = 4^2 * 1^2

818 = 144 = 169 = 4^2 * 3^2

1832 = 576 = 1636 = 4^2 * 6^2

3250 = 1600 = 16100 = 4^2 * 10^2

5072 = 3600 = 16225 = 4^2 * 15^2

 

So each row is 4^2 * a triangle number squared (in order).

 

Going to try to extend this to other e & -f cols…

Anonymous ID: 358135 Sept. 5, 2018, 5:57 a.m. No.7497   πŸ—„οΈ.is πŸ”—kun

Have you guys considered a lot of these numbers would fit nicely into a Lie Group and would take you to the Quantum level? Manifolds flat curved and warped can then be applied, then you will discover the breaking of symmetry that Higgs did that appears to create mass. The geometry of that is important to further advancement of humanity IMO.

AA !dTGY7OMD/g ID: 077d1d Sept. 5, 2018, 12:17 p.m. No.7499   πŸ—„οΈ.is πŸ”—kun

>>7494

>Think about the values in 0,1 and 0,1.

Well, obviously you meant to mention more than just (0,1), but in the absence of that other cell, here's the clue about (0,1) from Grid Patterns:

In cell (0,1), e is zero, so all cs are perfect squares (the smaller square (x+n)(x+n) being 0). These values of c ALL appear in (0,0) but they also ALL have more than one way to arrange their factors. The factors this time produce an n value of 1. 4x4 = 16 can be arranged as 2x8 = 16, which is equal to 5x5 - 3x3. Notice that all the values of a in this cell are also each twice the value of a perfect square.

1+1 = 2

4+4 = 8

9+9 = 18

Notice that all the values of d for this cell also follow a pattern:

2x(1x2) = 4

2x(2x3) = 12

2x(3x4) = 24

 

If that other cell was meant to be (1,1), here's another potentially relevant clue from Grid Patterns:

Each value of a in cell (1,1) is also the long side of an integer right angled triangle.

Teach !!UgZAPoSXEk ID: 9cfa21 Sept. 5, 2018, 1:12 p.m. No.7500   πŸ—„οΈ.is πŸ”—kun

>>7496

One interesting thing about these a.a.n.(n-1) numbers:

  • if you add a.a.n, you'll get a perfect square

  • if you minus a.a.(n-1), you'll get a perfect square too

AA !dTGY7OMD/g ID: b621a4 Sept. 5, 2018, 2:40 p.m. No.7502   πŸ—„οΈ.is πŸ”—kun   >>7503

>>7494

>So, you're looking for when a series becomes two triangles where every unit of the triangle is a square.

I think this thing from Grid Patterns could potentially be a relevant crumb. Notice the 4, 4+4, 4+4+4 thing. I'm not sure where you'd expand this, but it's the only instance I can remember of a triangle with every unit being a square.

 

If you take a square and look for it as a b value in (0,n), the difference in n values from record to record is equal to (2x)+(4T(x)).

e.g. 9*9=81

These are the records we will get (since there can't be opposite parity between a and b as it would create an invalid n, we will only see the square of all odd numbers below 9 in a[t]):

{0:2:63:14:49:81}

{0:8:45:20:25:81}

{0:18:27:18:9:81}

{0:32:9:8:1:81}

The difference in n values can be expressed as follows:

8 - 2 = 6, 6=2+4

18 - 8 = 10, 10=2+4+4

32 - 18 = 14, 14=2+4+4+4

GAnon !jPVzzZOz2c ID: e906b3 Sept. 5, 2018, 7:49 p.m. No.7504   πŸ—„οΈ.is πŸ”—kun   >>7507 >>7511 >>7512 >>7514

>>7485

>>7494

 

>Could you quickly lookup or figure out any values that are a square multiples by two consecutive numbers like n(n-1)?

 

>Think about the values in 0,1 and 0,1.

 

Well interestingly enough in cell (0,1) the d values (also the a values in (0,0)) are these:

 

4, 12, 24, 40, 60, 84, 112, 144

 

Ignore 4 and 12 for now they're special.

 

24=2(34)

40=2(45)

60=2(56)

84=2(67)

112=2(78)

144=2(89)

 

Remember how also if you multiply any cell in column zero by a constant it is the same cell in another row

 

2 * (0,1,12,4,8,18) = (0,2,24,8,16,36)

 

So this would be a way to get any factor where the factor is divisible by 2.

 

So I thought about the values in (0,1)

 

If we think about these in 0,1 (binary) (double meaning) then we can just bit shift all these numbers over by one (or just divide by two for you newbies) and get every entry higher than 6.

 

So now we know that n(n-1)a*a MUST exist in column 0

 

This doesn't take care of the square multiple thing but I think its worth noting.

AA !dTGY7OMD/g ID: db1818 Sept. 5, 2018, 10:36 p.m. No.7507   πŸ—„οΈ.is πŸ”—kun   >>7508 >>7509 >>7510

>>7504

>So now we know that n(n-1)a*a MUST exist in column 0

>as d

Here's a test case using c559, 13*43. I would post more test cases but this is a little long. This is every cell in which d equals the (an)(a(n-1)) we're looking for. Can anyone spot a cell that we can calculate from our knowns?

 

(30,5,6) = {30:5:23:10:13:43}, f = -17, c = 559

BigN cell: (30,257,12) = {30:257:23:22:1:559}, f = -17

Cell in (e,1) where x=|f|+1: (30,1,10) = {30:1:195:18:177:215}, f = -361

c*bigN=143663, which first appears in (e,1) as a at (30,1,268) = {30,1,144199,536,143663,144737}, f=-288369

 

(an)(a(n-1)) = 3380

The cell in (0,n) where d=(an)(a(n-1)):

(0,2852721,1690) = {0:2852721:3380:3378:2:5712200}, f=-6761, c=11424400

(0,1424672,1689) = {0:1424672:3380:3376:4:2856100}, f=-6761, c=11424400

(0,710649,1687) = {0:710649:3380:3372:8:1428050}, f=-6761, c=11424400

(0,567845,1686) = {0:567845:3380:3370:10:1142440}, f=-6761, c=11424400

(0,282240,1681) = {0:282240:3380:3360:20:571220}, f=-6761, c=11424400

(0,216333,1678) = {0:216333:3380:3354:26:439400}, f=-6761, c=11424400

(0,139445,1671) = {0:139445:3380:3340:40:285610}, f=-6761, c=11424400

(0,110889,1666) = {0:110889:3380:3330:50:228488}, f=-6761, c=11424400

(0,106496,1665) = {0:106496:3380:3328:52:219700}, f=-6761, c=11424400

(0,53792,1641) = {0:53792:3380:3280:100:114244}, f=-6761, c=11424400

(0,51597,1639) = {0:51597:3380:3276:104:109850}, f=-6761, c=11424400

(0,40625,1626) = {0:40625:3380:3250:130:87880}, f=-6761, c=11424400

(0,25281,1591) = {0:25281:3380:3180:200:57122}, f=-6761, c=11424400

(0,18720,1561) = {0:18720:3380:3120:260:43940}, f=-6761, c=11424400

(0,13689,1522) = {0:13689:3380:3042:338:33800}, f=-6761, c=11424400

(0,7865,1431) = {0:7865:3380:2860:520:21970}, f=-6761, c=11424400

(0,5733,1366) = {0:5733:3380:2730:650:17576}, f=-6761, c=11424400

(0,5408,1353) = {0:5408:3380:2704:676:16900}, f=-6761, c=11424400

(0,1664,1041) = {0:1664:3380:2080:1300:8788}, f=-6761, c=11424400

(0,1521,1015) = {0:1521:3380:2028:1352:8450}, f=-6761, c=11424400

(0,845,846) = {0:845:3380:1690:1690:6760}, f=-6761, c=11424400

(0,117,391) = {0:117:3380:780:2600:4394}, f=-6761, c=11424400

AA !dTGY7OMD/g ID: 1780ba Sept. 5, 2018, 11:34 p.m. No.7509   πŸ—„οΈ.is πŸ”—kun

>>7507

From (0,18720,1561) to (0,117,391), the difference between every value and any other value of that same variable is divisible by a.

 

This is the series of numbers the gap between the n values divided by a is equal to (from highest n to lowest n):

387, 448, 164, 25, 288, 11, 52, 56

Here's the same thing for t:

3, 7, 5, 1, 24, 2, 13, 35

Same thing for x:

6, 14, 10, 2, 48, 4, 26, 70

Same thing for a (but (next-current)/a since a gets larger as n decreases):

6, 14, 10, 2, 48, 4, 26, 70

Same thing for b:

780, 910, 338, 52, 624, 26, 130, 182

AA !dTGY7OMD/g ID: 1780ba Sept. 6, 2018, midnight No.7510   πŸ—„οΈ.is πŸ”—kun   >>7511

>>7507

We can find this d if we can find any c value. In (0,n), it'll be 11424400, and since e is zero, this is the minimum c value for our d. The maximum will be 11431160, since this is (3381*3381)-1 and after this point d=3381. So if we can find any c value within these two c values, we'll have a d value that is equal to the (an)(a(n-1)) we want.

Anonymous ID: 4e3773 Sept. 6, 2018, 8:51 a.m. No.7512   πŸ—„οΈ.is πŸ”—kun   >>7513 >>7514

>>7504

So I was looking into patterns in (0, 1) and found that our n(n-1)aa will exist in (0, 1) along a multiplier. I was looking in to (0, n) where a = n(n-1)aa and noticed the pattern. It appears that our n(n-1)aa will always exist in column 0 at the row of triangle of n, that is (0, n(n-1)/2).

 

This also means that in (0, 1) there is an a[t] = (n(n-1) * n(n-1) * a * a)/2

Anonymous ID: 4e3773 Sept. 6, 2018, 8:52 a.m. No.7513   πŸ—„οΈ.is πŸ”—kun

>>7512

To clarify:

> I was looking in to (0, n) where a = n(n-1)aa

 

I was just looking for patterns in (0, i) (for some i 0) where a[t] = n(n-1)aa.

 

We're running out of single-letter variables.

Anonymous ID: 4e3773 Sept. 6, 2018, 9:02 a.m. No.7514   πŸ—„οΈ.is πŸ”—kun   >>7515

>>7504

>>7512

 

Since we can multiply and a[t] in (0, 1) with any number and it will exist in some row, we can multiply our a[t] = (n(n-1) * n(n-1) * a * a)/2 with 2 and it will occur in (0, 2) as a[t] = 2 * (n(n-1) * n(n-1) * a * a) / 2.

 

In (0, 3) it is a[t] = 3 * n(n-1) * n(n-1) * a * a / 2

In (0, 4) a[t] = 4 * n(n-1) * n(n-1) * a * a / 2 etc.

 

And again in (0, 5), (0, 6) etc.. It should exist in every single row in column 0.

 

Wow, that is kind of mindblowing. I think I'm starting to see why VQC thinks column 0 is so special.

Anonymous ID: 4e3773 Sept. 6, 2018, 9:12 a.m. No.7516   πŸ—„οΈ.is πŸ”—kun   >>7517

>>7515

 

And the t in (0, 1) for a[t] = n(n - 1)^2 * a^2 / 2 is equal to: n(n-1) * a / 2. Meaning the t in (0, 1) is the triangle of n multiplied by the factor of a. So we "scale" the triangle of n by a?

Saga ID: 24884d Sept. 6, 2018, 5:20 p.m. No.7519   πŸ—„οΈ.is πŸ”—kun

Column0 : https://pastebin.com/kyeughcR

 

Row 1 : https://pastebin.com/VeYNdTTV

 

Old research material, hope it helps. It's good to review.

Also sorry for using t = 2x, "or in my case, n", that was before I knew it's correct substitution.

AA !dTGY7OMD/g ID: 73e121 Sept. 6, 2018, 11:32 p.m. No.7520   πŸ—„οΈ.is πŸ”—kun   >>7531

>>7495

What do you mean elaborate? I don't see how anything I said was hard to understand. If we find x or t from the na cell, we've found the x or t from the prime solution cell. That means, if we're meant to find x or t, the na cell is irrelevant. Which means there's something weird going on.

 

>>7503

Considering it didn't come up in my post, I wouldn't have a clue.

 

>>7511

Thanks anon. Are you one of the namefags posting without your name?

AA !dTGY7OMD/g ID: 92669d Sept. 7, 2018, 10:14 p.m. No.7524   πŸ—„οΈ.is πŸ”—kun   >>7525 >>7526 >>7527

I've been looking into this crumb a bit:

"The values of a in the first cell where e=1 contains ALL factors for odd numbers that are the sum of two squares"

I haven't found anything groundbreaking (that I'm aware of) but I did find some interesting patterns. I'll dump numbers in this post and then explain some things in some replies. The numbers on the left are the sequence of a values in (1,1). The numbers on the right are the factors of these numbers.

 

1

5

13

25 5

41

61

85 5 17

113

145 5 29

181

221 13 17

265 5 53

313

365 5 73

421

481 13 37

545 5 109

613

685 5 137

761

841 29

925 5 25 37 185

1013

1105 5 13 17 65 85 221

1201

1301

1405 5 281

1513 17 89

1625 5 13 25 65 125 325

1741

1861

1985 5 397

2113

2245 5 449

2381

2521

2665 5 13 41 65 205 533

2813 29 97

2965 5 593

3121

3281 17 193

3445 5 13 53 65 265 689

3613

3785 5 757

3961 17 233

4141 41 101

4325 5 25 173 865

Archive Anon Anon !dTGY7OMD/g ID: 92669d Sept. 7, 2018, 10:20 p.m. No.7525   πŸ—„οΈ.is πŸ”—kun   >>7527

>>7524

The first most obvious thing is that when a number is the sum of two squares and it's odd, it will only be divisible by other numbers that are also the sum of two squares and odd. So if you have a c value and it's in an e column where e is a square (because this means it's the sum of two squares, based on another crumb), its factors will be in columns where e is a square too. I'll post more in a while.

Archive Anon Anon !dTGY7OMD/g ID: 92669d Sept. 7, 2018, 11:26 p.m. No.7526   πŸ—„οΈ.is πŸ”—kun   >>7527

>>7524

The second useful thing to mention, that I probably should have mentioned first, is that every value of a in (1,1) is the sum of two squares and is odd. VQC's crumb said that these values "contain ALL factors for odd numbers that are the sum of two squares", but they are also themselves odd numbers that are the sum of two squares.

 

Third thing: while these a values are each odd sums of two squares, sum odd sums of two squares are missing. Here's the first ten odd sums of squares: 1, 5, 13, 17, 25, 29, 37, 41, 45, 53. Here are the first ten a values in (1,1): 1, 5, 13, 25, 41, 61, 85, 113, 145, 181. So, as you can see, while a in (1,1) are all odd square sums, some of the odd square sums are missing, such as 17, 29, 37, 45 and 53.

 

Fourth thing: the sequence of a values starts at 1 and increases by 4(x-1) where x is the place in the sequence. In other words, 1, +4=5, +8=13, +12=25, +16=41, etc. This follows the pattern of sums of squares that are odd in that they are all equal to a multiple of 4 plus 1.

 

Fifth thing: these odd numbers that are the sum of two squares appear as factors of others in the series in a pattern. Where a number appears as a factor, it will appear as a factor again the number of numbers away that is equal to itself. If that was worded confusingly, take this example: where 5 appears as a factor, it will appear again 5 numbers in the sequence away, then 5 away again, and so on. It will also always appear once between these times. This seems to hold for all of them. For 5, it appears two away, then three away (this second one being 5 away from the origin), and this pattern repeats. When 13 appears as a factor, it appears 8 away, and then appears again 5 away, and so on. Where 17 appears as a factor, it appears 4 away, and then appears again 13 away, and so on. For 25, it's 18 and 7. For 29, it's 12 and 17. This seems to hold for all numbers that are odd and the sum of two squares.

Archive Anon Anon !dTGY7OMD/g ID: 92669d Sept. 7, 2018, 11:41 p.m. No.7527   πŸ—„οΈ.is πŸ”—kun

>>7524

>>7525

>>7526

In summary, the crumb "the values of a in the first cell where e=1 contains ALL factors for odd numbers that are the sum of two squares" is true because all of these a values are themselves odd numbers that are the sum of two squares, and because every number that is odd and the sum of two squares can only be divided by other numbers that are odd and the sum of two squares. I'm not sure if every c value in the columns where e is a square are the sum of two squares, but every c value that is the sum of two squares does appear in these columns. If you have a c value that is the sum of two squares, it will appear as an a value in (1,1) and its factors will appear in (1,1).

Archive Anon Anon !dTGY7OMD/g ID: c72d24 Sept. 8, 2018, 1:45 a.m. No.7528   πŸ—„οΈ.is πŸ”—kun   >>7529

I've been thinking about this

>>7466

>what do you do to find the lookup x for c that gives you na or (n-1)

>find the lookup x

>as opposed to the actual x

I'm thinking maybe the x we're looking for isn't actually the correct x. Maybe it's a value that we can substitute for x to find na or (n-1) as one of the other values. This lookup x would probably need to be in our list of knowns (or stuff we can find that we haven't paid attention to). More evidence for it not actually being the correct x is the thing I've been saying about the cell in (e,1) where a[t]=na being irrelevant if you find the correct x for it since it's the same x as the solution record. If we're looking for a different x value that gives us na in a different way (probably based on (an)(a(n-1))), we would still be using (e,1) and (f,1), and we would also still be looking for an x value related in some way to na, so it would still fit all the criteria.

 

So if we had a different, incorrect x value, and we applied it to our knowns, how would that produce na, (n-1) or (an)(a(n-1))? Well, if we are keeping our knowns the same, that would mean we'd find it as a, b or n. This doesn't seem to really make any sense, though. If it turned up as a or b, c would need to be divisible by it, which, in the case of semiprimes, isn't possible. If it turned up as n, a and b would need to change, which, again, in the case of semiprimes, isn't possible. So this means we're probably applying it to numbers that aren't directly related to our given c.

 

I think maybe we're looking for the x value from a cell where a=aa and b=n(n-1), or something similar to that. Maybe, since n can be either greater than or smaller than a, and since b always has to be greater than a, whether the n we're looking for is greater or smaller than the a we're looking for determines whether we're looking for na or (n-1), since VQC mentioned both.

Archive Anon Anon !dTGY7OMD/g ID: c72d24 Sept. 8, 2018, 2:03 a.m. No.7529   πŸ—„οΈ.is πŸ”—kun   >>7530

>>7528

If this applies to semiprimes, a will be odd. That means aa will be odd. In order to be a valid difference-of-two-squares cell, a and b have to have the same parity. That means n(n-1) would have to be odd. But since it's a number multiplied by itself minus 1, whether n is odd or even, the other one will have the opposite parity. When a number is multiplied by another number with the opposite parity, you get an even number. That means, in this case, a would be odd and b would be even. That means it doesn't exist in the grid. I still think we're looking for an incorrect x value that somehow gets applied to a known (or multiple) and produces either (an)(a(n-1)), na or (n-1) somehow. But I guess it's not going to be in the cell I was thinking of.

Anonymous ID: a332a2 Sept. 8, 2018, 7:11 a.m. No.7530   πŸ—„οΈ.is πŸ”—kun

>>7529

Well if you remember a bit back, I ranted a lot about the shadow n. Every record has a "main" record and the shadow record. The shadow record is a valid record, but with values shifted by 2d, this is also true for the x. In the shadow record, x is negative and is equal to the "original" x - 2d.

 

Not sure if this is what we're looking for, but if we find the "shadow" x, then we could use that to find the "correct" x.

Anonymous ID: a332a2 Sept. 8, 2018, 8:48 a.m. No.7532   πŸ—„οΈ.is πŸ”—kun

Not sure I'm heading down the correct path (or if I know where I am heading), but I'm trying to think about the aan(n-1) and what it means.

 

We know n(n-1) is two triangles and a singular triangle is defined as n(n-1)/2. This is also the sum from 1 to n.

 

So two triangles is the same as:

2*(1 + 2 + .. + n)

or

2 + 4 + 6 + .. + 2n

 

Essentially the sum of even numbers from 2 to 2n.

 

Let's ignore that for now and pretend our number is really aan(n-1)/2, that is, we're only dealing with a single triangle.

 

Then what we're really doing is:

 

aa(1 + 2 + 3 + .. + n)

 

Or

 

aa + 2aa + 3aa + … + naa)

 

Which is summing the square up to n times. For our case, since we're dealing with two triangles, we instead have:

 

2(aa + 2aa + 3aa + 4aa + … + naa)

 

or

 

2aa + 4aa + 6aa + … + 2naa

 

Not sure where I'm heading with it, though. I was thinking that maybe we could coherence information from this.

AAA !dTGY7OMD/g ID: 91c416 Sept. 8, 2018, 5:14 p.m. No.7536   πŸ—„οΈ.is πŸ”—kun   >>7547 >>7548 >>7567

When VQC took us through nn+2d(n-1)+f-1 and made a coloured diagram, he said "the colours in the MAP will be the KEY to GUIDE you through the process of how the algorithm works." As I was going through RSA #10 for the Grid Patterns thread the other day I noticed this post and that nobody had said anything about it. Well, consider this. This first picture is the one he posted at the time. It has several specific colours. Picture 2 (confusingly named "colours 1.png") is another picture he posted a couple months later in RSA #12. It has very similar colours. The colours are most likely linked, I think.

Anonymous ID: 358135 Sept. 9, 2018, 2:58 a.m. No.7540   πŸ—„οΈ.is πŸ”—kun   >>7541

at about the 18 min mark in that video the is fundamentally showing dimensional shifts and how they shadow in 2D - personally i never knew that was the exponent value until now.

Anonymous ID: 358135 Sept. 9, 2018, 3:09 a.m. No.7541   πŸ—„οΈ.is πŸ”—kun

>>7540

At about 19:30 he is explaining that the circle of i (imaginary plane) is the circle equivalent to the Y axis. That i is an axis that is Y equivalent shadowed onto 2 D.

Anonymous ID: a332a2 Sept. 9, 2018, 5:23 a.m. No.7545   πŸ—„οΈ.is πŸ”—kun   >>7546 >>7550

>>7535

Note though, that bn and b(n-1) occur in different indexes of t (a[t] = b(n - 1) in f, a[t + 1] = bn in e). So they won't appear when we do the naive multiplication.

 

That might be the "key", but I haven't seen any obvious patterns there yet.

AA !dTGY7OMD/g ID: 31e34d Sept. 9, 2018, 2:20 p.m. No.7550   πŸ—„οΈ.is πŸ”—kun

>>7548

That means a lot VA. I would have been putting this much work into it whether or not other people were too but I'm glad I'm not just talking to myself here.

 

>>7547

He said the thing about colours being part of the key before he posted anything about colours, so I only just noticed it a couple days ago. I don't think anyone's looked at this yet. So it might mean we use the -f column to find 2d(n-1) and use d^2 in (0,n), or something like that.

 

>>7545

We don't seem to have even found all that much about (an)(a(n-1)) yet to begin with, so I'm just thinking over ideas.

GAnon !jPVzzZOz2c ID: e906b3 Sept. 11, 2018, 9:51 a.m. No.7559   πŸ—„οΈ.is πŸ”—kun   >>7560 >>7561

>>7547

What happened last night? I posted like 4 posts with regards to the tetrahedral numbers and their locations in the grid. Also the numbers of the form t(t+h)/2 locations in the grid. I guess it was sort of a mass shitpost. Did BO delete it?

 

Also this post right here that I'm quoting has a different trip but I remember posting that yesterday. Is this a timeline shift or something? Am I the only one??

 

Also AA made a bunch of posts that got delete too. WTF

AA !dTGY7OMD/g ID: 5e8992 Sept. 11, 2018, 2:09 p.m. No.7561   πŸ—„οΈ.is πŸ”—kun   >>7563

>>7559

>>7560

Yeah, as you noticed, you posted those in the Grid Patterns thread. You do know that I'm the BO, too, right? That's how I got my name. It would be a bit weird if I deleted a bunch of my own posts. You can see every mod action here. I haven't done anything for two weeks. https://8ch.net/log.php?page=1&board=vqc

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 08d201 Sept. 11, 2018, 3:34 p.m. No.7562   πŸ—„οΈ.is πŸ”—kun

i'm still curious as to what it means to look at this in binary…

https://oeis.org/search?q=A181891

as if there's a simpler way of representing the "binary" so that it's not 8brazillion units long… or like… there's a simple, significant pattern in the records that "once you see it, you can't unsee it", sorta thing.

 

Doing my "intuitive/vaguely remembering" schtick… this came to mind:

 

"Palindromic Prime" (I saw the word Palindrome, and I wondered if any was is be prime.

https://en.wikipedia.org/wiki/Palindromic_prime

 

https://oeis.org/A117697

What caught my eye….

The Palprimes are palindromic in binary, also, not that they logically wouldn't be, but I don't speak that language enough to know how it "should" work in the first place.

 

11, 101, 111, 10001, 11111, 1001001, 1101011, 1111111, 100000001, 100111001, 110111011, 10010101001, 10110101101, 11000100011, 11001010011, 11011111011, 11100100111, 11101010111, 1001100011001, 1001111111001, 1010001000101

AA !dTGY7OMD/g ID: 527f48 Sept. 11, 2018, 6:37 p.m. No.7564   πŸ—„οΈ.is πŸ”—kun

>>7563

I originally made the board to archive everything when we were still on /cbts/, so VA started calling me "Archive Anon". As Q would say, "the more you know".

GAnon !jPVzzZOz2c ID: e906b3 Sept. 11, 2018, 6:46 p.m. No.7565   πŸ—„οΈ.is πŸ”—kun   >>7566 >>7573

>>7563

Confirmed.

 

>>7552

>>7553

>>7554

 

So I took these patterns and I wrote some code to figure out where these were. Here is a summary of my findings.

 

For any odd number s, the series is defined by this function:

f(i) = i*(i+s) / 2

The first, for s=1, is the triangular numbers: 1, 3, 6, 10, 15, 21

The second for s=2, is 2, 5, 9, 14, 20, 27

Here is a table of the values horizontally.

 

What I found is that If your shift is s, an odd number, then

f(1) = 1*(1+s)/2

This value is found in the A[t] value where X[t] = 1 at (e,n)=(1+s,1).

 

If you increase to f(2) the new (e,n) would be (1+2s,1) with x=2.

 

Generally for any shift s:

for any x:

let f(x) = x*(x+s)/2

The f(x) is equal to the a value at the entry (1+x*s, 1, x)

 

Moreover, if you look at this sequence of records, the d values are equal to the sequence with the shifts + 2.

Example, if I'm looking at A[t] values equal to x(x+3)/2, then the D[t] values for those same records are gonna be equal to x(x+5)/2.

 

Code output related

GAnon !Nx57Pyux3E ID: e906b3 Sept. 11, 2018, 7:27 p.m. No.7566   πŸ—„οΈ.is πŸ”—kun   >>7568 >>7573

>>7565

So if we want to find our c value as an A[t] value in row n=1, then we could find it in 3 places [that I know of for now], maybe 6 if we can use negative s too complex for this point in time.

 

We would need c = i*(i+s)/2 for some i and s

2c = ii + is

 

This is a little weird but let's do it with c=145.

 

2c = 290 = 2529

10 * 29 = i*(i+s) if i = 10 and s = 19

 

Then this would put it at (e,n,x) = (1+si, 1, i) = (1+1910, 1, 10) = (191,1,10)

[191, 1, 155, 10, 145, 167]

 

Or we could go to this:

 

2c = 290 = 2529

 

=558 = i(i+s) if i = 5 and s=53

 

(e,n,x) = (1+5*53, 1, 5) = (1+265, 1, 5)

[266, 1, 150, 5, 145, 157]

 

Or:

 

2c = 290 = 2529

 

= 2145 = i(i+s) if i=2 and s = 143

 

(e,n,x) = (1 + 2*143, 1, 2) = (287, 1, 2)

[287, 1, 147, 2, 145, 151]

 

Could we triangulate with these or something??

GAnon !Nx57Pyux3E ID: e906b3 Sept. 11, 2018, 7:49 p.m. No.7568   πŸ—„οΈ.is πŸ”—kun

>>7566

But of course the only one we could access at first would be the third one (e,n,x) = (2c-3, 1, 2), so maybe it would be worthwhile to check out the negative value.

 

2c = i*(i+s)

i=c

i+s = 2 =c+s=2 => s=2-c (negative value)

 

Then of course this would appear at (e,n,x) = (1+s*i, 1, i)

= (1 + (2-c)c, 1, c) = (1 + 2c - cc, 1, c)

 

for c=145 it would be:

 

[-20734, 1, 290, 145, 145, 437]

AA !dTGY7OMD/g ID: d7d16e Sept. 11, 2018, 8:21 p.m. No.7570   πŸ—„οΈ.is πŸ”—kun

>>7567

Of course it isn't to scale. It's just a mock up VQC put together to show the relationship. It's not meant to be to scale, so you can't base equations on the proportions of the sections in the picture. Also the e portion is broken up and spread around. All of the green parts make up e. e isn't always 1. That wouldn't make any sense. Have you read all the old threads yet?

PMA !!y5/EVb5KZI ID: 02cf6e Sept. 11, 2018, 9:33 p.m. No.7571   πŸ—„οΈ.is πŸ”—kun   >>7572 >>7576

>>7485

>>7494

Thanks for these latest hints!

 

Pics attached for c145, c287, c288, and c6107 are an attempt to further explore the relationship between the product of (-f,1).a and (e,1).a and the corresponding factor record's aan(n-1) value.

 

Relevant columns are as follows:

 

  • "a diff": difference between (e,1).a and (-f,1).a.

  • "d diff": difference between (e,1).d and (-f,1).d.

  • "aa product": (e,1).a * (-f,1).a

  • "div 2": "aa product"/2. (abbreviated as aa/2 in this post)

  • "ti": largest triangle base from aa/2 as calculated from the inverse triangle formula.

  • "ti rm": remainder from aa/2 less the largest triangle.

  • "square multiples": possible square factors of aa/2.

 

The aa/2 value is being analyzed because aan(n-1) = aa * 2T(n-1) for each factor record. Therefore we can isolate aa and T(n-1).

 

Each possible "square multiple" has it's own triangular breakdown. See the c6107 example for 4426800 that shows possible T formulas plus remainders for the 1, 2, 4, 5, 10, and 20 squares.

 

Might be going too deep here, but the objective is to see if there is a way to calculate the "aa product" for any factor record by understanding it's components.

 

Other observations on this data:

 

1) The "a diff" column matches the factor record's a value.

2) The "d diff" column can be used to navigate from the starting x value to the solution x:

 

If "d diff" is even, then starting x - "d diff" == solution x.

If "d diff" is odd, then starting x - "d diff" - 1 == solution x.

PMA !!y5/EVb5KZI ID: 02cf6e Sept. 11, 2018, 9:55 p.m. No.7572   πŸ—„οΈ.is πŸ”—kun

>>7571

There are a few limited cases where a factor record can be found directly from the triangle breakdown of the aa product.

 

Pics attached are for c119, c189, c275, c377, c495, and c629.

 

These numbers represent a small portion of integer sequence A181890 (oeis.org/A181890). (thanks SAGA for pointing this out)

 

In these cases, n-1 from the entry record is a perfect square, and the a value in the factor record is sqrt(n-1). Notice the matching ti and "div 2" columns.

GAnon !Nx57Pyux3E ID: e906b3 Sept. 12, 2018, 9:59 a.m. No.7573   πŸ—„οΈ.is πŸ”—kun   >>7574

>>7565

>>7566

 

So all of these equations are of the form f(x) = x(x+2h+1)/2 for any h.

Basically if x is even, then x+2h+1 is odd, but we can still divide it by two because of the x. If x is odd, then 2+2h+1 will be even, so that is divisible by two, so either way the product is divisible by two.

 

For the tetrahedral numbers, 1, 4, 10, 20, …, they have the formula:

f(x) = x(x+1)(x+2) / 6

 

This is similar to above, but instead of dividing by 2 we divide by 6.

Therefore we need one of x, x+1, and x+2 to be divisible by 2 and another (or the same) to be divisible by 3.

 

So basically if we want to do the same analysis for tetrahedral-type numbers, then the equation must be of the form:

 

x(x+6g+1)(x+6h+2)/6 for any g,h.

 

My lunch break is almost over so I'll try and find these in the grid after work.

GAnon !Nx57Pyux3E ID: e906b3 Sept. 12, 2018, 3:22 p.m. No.7574   πŸ—„οΈ.is πŸ”—kun

>>7573

Nevermind. The 4,10,20 is NOT the tetrahedral numbers in (4,1), because the next number is 34 not 35. This kind of throws a whole wrench in my idea so I'm going to go back to looking at the other sequences

VA !!Nf9AmQNR7I ID: c0ec86 Sept. 12, 2018, 9:05 p.m. No.7575   πŸ—„οΈ.is πŸ”—kun

Hello Anons!

Just found something very interesting.

In (-f,1) and (e,1) columns the difference between a values increases by 1,3,5,7, etc moving up and down from the element(s) containing the (-f na transform) and (e na transform). Here's some data from c145.

 

(-f a values) (e a values) (Delta)

20 25 5

38 41 3

60 61 1 (na transform record)

86 85 1

116 113 3

150 145 5

188 181 7

PMA !!y5/EVb5KZI ID: 02cf6e Sept. 12, 2018, 9:30 p.m. No.7576   πŸ—„οΈ.is πŸ”—kun   >>7584

>>7571

Have found a way to calculate a valid aa factor in (0,n) starting from the (-f,1) and (e,1) aa product.

 

Pics attached for c361, c961, c21025, and c144 show the previously explained aan(n-1) view, as well as the triangle breakdowns for each possible square multiple.

 

Recall that the aa product can be represented as aa * 2T(n-1) and potentially has multiple valid aa and T formula solutions.

 

For c361, as an example, the "aa product" is 26082 with possible square factors 1^2, 3^2, and 9^2. In the case of the 9^2 square multiple, the triangle formula can be represeted in terms of our starting x value of 18.

 

Triangle portion = T(x-1) + (x/2 - 1) = T(17) + 8 = 161

 

And solving for a possible aa factor:

 

26082 = aa * 2 * (T(17) + 8)

aa = 26082 / 322 = 81

a = 9

 

This similarly works for the other odd examples c961 and c21025, and finds that largest possible aa factor.

 

In the case of even c144, the methodolgy is the same, with a slight adjustment to the triangle formula:

 

Triangle portion = T(x-1) + (x/2 - 2)

 

So given the aa product from (-f,1) and (e,1), we are now able to determine the largest possible aa factor by calculating the triangle and it's remainder from x.

Anonymous ID: 358135 Sept. 14, 2018, 5:14 p.m. No.7578   πŸ—„οΈ.is πŸ”—kun

>>7577

possibly two close circle rotations around x and Y and all the others are diagonal at a lower scale - maybe. The first circle rotation is about 8 degrees and then is spins into 3d and warps into ellipses is what I see.

Anonymous ID: 358135 Sept. 14, 2018, 5:17 p.m. No.7579   πŸ—„οΈ.is πŸ”—kun

I know this is off topic as it is about my discovery that even physics say 99 percent of mass is not matter but binding energy. But I need a better math fag than I am. I basically want to figure out the energy per meter of a proton. Here is my feeble attempt:

 

How big is the proton?

Because protons are not fundamental particles, they possess a physical size, though not a definite one; the root mean square charge radius of a proton is about 0.84–0.87 fm or0.84Γ—10βˆ’15 to0.87Γ—10βˆ’15 m.

 

Mass of Proton

Mass

1.672621898X. 10βˆ’27 kg[1]

938.2720813(58) MeV/c2[2]

 

the speed of light =

299 792 458 m / s

 

E=mc^2

EProton= 1. 672621898 X10^-27 Kg* (299 792 458 m / s)^2

= 1. 672621898e-27 Kg* 8.987551787e+16

=1.50327759289610516349668072e-10 Kgm/s

 

How big is the proton?

Because protons are not fundamental particles, they possess a physical size, though not a definite one; the root mean square charge radius of a proton is about 0.84–0.87 fm or0.84Γ—10βˆ’15 to0.87Γ—10βˆ’15 m.

Thus the Energy per meter area is:

The energy of Proton / the area of the proton

1.50327759289610516349668072e-10 Kgm/s / 0.85e-15m=

= 176856.18739954178394078596705882 Kg/s

 

Option 2:

Ratio of area of proton to c^2=

0.85e-15 m / 8.987551787e+16 m/s=

=9.4575254768431856157937707803052e-33 seconds

This is the area of energy the proton contains.

 

If you then divide the EProton by the ratio of area of proton to c^2=

15895041431120167484788.273385791 kgm

VA !!Nf9AmQNR7I ID: 23f1bb Sept. 14, 2018, 9:50 p.m. No.7583   πŸ—„οΈ.is πŸ”—kun

Hello Anons! Working on diagrams, found a better offset layout which clearly shows how the u and u+1 dimensions form 8 equal triangles surrounding one unit. Maybe it's old news, but I had fun making it, lol ;)

for c6107, T(41) * 8 + 1 = 861 * 8 + 1 = 6889

PMA !!y5/EVb5KZI ID: 9a0997 Sept. 14, 2018, 10:24 p.m. No.7584   πŸ—„οΈ.is πŸ”—kun   >>7604

>>7576

Have taken this a bit further, and found a way to calculate all triangle portions of the aan(n-1) formula for the initial factor record for any odd n-1 in (0,n).

 

Pics attached for c21025 (145^2) and c104329 (323^2) show the aan(n-1) derivations for all factor records, and then all square multiples in terms of triangles for the first factor record.

 

The revised formulas are:

 

aan(n-1) = aa * 2T(n-1) = aa * 2(T(u) + remainder)

 

where u = m * x - 1

and remainder = m * x / 2 - m^2

m is just a multiplier, but it turns out to work inversely to each "a" square multiple.

 

For c21025, as an example, there are 12 possible values that represent the aa square multiple of aan(n-1). 1,2,3,4,6,8,9,12,18,24,36,72.

 

Using the above formulas, and using m=4 as an example, we can find an "a" factor as:

 

53742528 = aa * (T(4144 - 1) + (4144/2 - 4^2)) = aa * (T(575) + 272)

aa = 53742528 / 165872

a = 18

 

And for each possible value of a, a * m will have the same value of 72, hence the inverse relationship between a and m.

 

Therefore, to find all valid square multiples, we can simply start at the lowest m = 1, find the largest "a" value, and then iterate halfway upwards.

 

This isn't a solution to our general factoring problem, as it doesn't yet appear to provide a way to jump to any other factor record for our starting c.

 

It does, however, show that there is a way to calculate both the triangle base and remainders to enable extracting factors from larger products.

Anonymous ID: a332a2 Sept. 14, 2018, 11:50 p.m. No.7585   πŸ—„οΈ.is πŸ”—kun

So I've been looking a bit into our grid, thinking more about what we actually are looking at and I was reading up on arithmetic progression, geometric progression and series in general.

 

Our grid is just a bunch of series with different offsets and triangles and squares, right.

 

So I started to play a bit with sequences and did some analysis using that way of thinking about our (e, n) and I found what I believe is the general equation for generating the nth record in (e, n).

 

It's remarkably simple, in fact so simple I'm surprised we haven't noticed it before. First I want to clarify on our n's. We have two types of n, one that occurs as a in (e, 1) and one that occurs as a factor of a in (e, 1). The first one I refer to as primary n, or root n, while the second one is a sub-primary n, or factor n. The first one will always have a=1 in (e, n), thus following the rules of the grid (a[t] = an in (e, 1), where a[t] = 1n for some t). The second one will not, it will have a "start" record with a=k (for some k, factor of a[t] = kn).

 

We know the equations for (e, 1). They are either the sum of half of e + 1 if e is odd added with either a product of triangles or squares.

 

odd: a[t] = (e/2 + 1) + 4t(t-1)/2

even: a[t] = e/2 + 2tt

or, if we want to limit our selves to purely triangles:

a[t] = e/2 + 2(t(t-1)/2 + t(t+1)/2)

 

The pattern for primary n's, or root n's is:

a = 1

b = 1 + 2(n + x)t + 4n(t(t-1)/2)

d = x + 1

 

while for sub-primary n's it is:

a = k

b = k + 2(n + x)t + 4n(t(t-1)/2)

d = k + x

 

In the first pattern, the x refers to the x in (e, 1) where it first occurs (which is also the same the cell in (e, n) where a = 1).

 

For the second pattern, the k is an offset, some factor where a[t] = kn in (e, 1). Here x is also the same as for when the n first occurs as a factor in (e, 1).

 

Now this also makes me think about VQC's hints. We know using -f and e, we're supposed to find an offset (in the second pattern you can think of k as an offset) and we know by multiplying -f and e we will end up with aan(n-1). That means, if we multiply the product of -f and e with 2, we will have one part of our equation for calculating records in (e, aa).

 

That is, k + 2(n + x)n + 2aan(n-1). The latter part will be known (although only in the sense that it exists as a product of a[t] in e and -f, but not which product).

Anonymous ID: a332a2 Sept. 15, 2018, 11:12 a.m. No.7586   πŸ—„οΈ.is πŸ”—kun   >>7587

I've been playing a bit with the idea from VQC about multiplying -f and e, but I trailed of a bit and found something funny.

 

If you imagine a number line with d's as poles (GA talked about this before) then our c exists between dd and (d + 1)(d + 1). It will exist at dd + e and at (d+1)(d+1) - (2d + 1 - e).

 

That is the variable we call f, I decided to take a look at c from "below" our d, that is d-1. I have called it g, simply because I needed a name. In g we will have n+1 instead of n-1. This means if we were to multiply e and g, there is a result that is equal to aan(n+1) instead of aan(n-1).

 

It's just a different point of view, but I decided to just play around with g and f out curiosity and I decided to ignore e and just play with g and f. The idea was that we know that in -f we have a[t] = a(n-1) and at g we will have a[t] = a(n+1). Multiplying these two rows will give us aa(n-1)(n+1). But I trailed off again and decided to play with addition. This means at some t we will add a(n-1) + a(n+1). This should result in 2an. However, I did a mistake in my function and instead of adding a[t] from -f with a[t] from g, I ended up skewing it and added a[t] from f with a[t + 1] from g. The result surprised me, as it generated the list that exists in (4e, 2) instead. The funny thing is that in (4e, 2) we have (at least) two sequences. One sequence will contain c and the other is the result of this addition.

 

I'm sure it can be explained easily using algebra, I haven't looked into it yet, but I was just surprised and found it very interesting. Not sure if this is a blind alley or not, but I figured I should share.

Anonymous ID: a332a2 Sept. 15, 2018, 11:46 a.m. No.7587   πŸ—„οΈ.is πŸ”—kun

>>7586

Meh, I think it's a blind alley. I looked into the algebra and it boils down to the difference of g and f. The series created from it won't always match 4e either, although the pattern inside of it should still be valid.

 

As in we have two sequences, one with a[t]=c and the other where an should appear at some t.

VQC !!/aJpLe9Pdk ID: 9f8800 Sept. 15, 2018, 1:36 p.m. No.7588   πŸ—„οΈ.is πŸ”—kun   >>7589 >>7590

How do you stop releases being scrubbed from the internet?

Can you write it to the blockchain?

Can you write it to many?

Would they take out all crypto to scrub?

Journey to the Centre of the Earth

VA !!Nf9AmQNR7I ID: d4b724 Sept. 15, 2018, 2:20 p.m. No.7589   πŸ—„οΈ.is πŸ”—kun

>>7588

Senpai! Nice to see you. Can you give a hint on how to correctly use the f-1 div 8 mods in combination with the f-1 div 8 multiples? I'm puzzling over the diagrams now.

Anonymous ID: a332a2 Sept. 15, 2018, 9:30 p.m. No.7590   πŸ—„οΈ.is πŸ”—kun

>>7588

> How do you stop releases being scrubbed from the internet?

Only way to do that while minimizing the risk of scrubbing would be to expose it to as many people as possible in one big go, or to the right people who could then help spreading it. If enough people are spreading it faster than they can scrub it, it's out there.

 

> Can you write it to the blockchain?

Yup, might take multiple transactions depending on what you're releasing. Releasing the whole thing vs just enough equations to solve the problem.

 

> Can you write it to many?

Yes, should be able to write it to any blockchain that supports extra data. But you would still be limited to the amount of data pr. transaction you could store.

 

> Would they take out all crypto to scrub?

Most definitely. At least try.

PMA !!y5/EVb5KZI ID: 9a0997 Sept. 16, 2018, 7:56 p.m. No.7604   πŸ—„οΈ.is πŸ”—kun   >>7632

>>7584

Turns out there is a way to calculate these triangle bases in terms of n-1.

 

Continuing with the c21025 example, attached tree shows a path to each of the triangle bases in aa2T(u) starting from u = n-1 = 10367.

 

The u value for each subsequent node is calculated either as (previous u - 1)/2 for blue nodes or (previous u - 2)/3 for orange nodes.

 

The denominator for each formula when multipled up the tree will equal the corresponding "a" square multiple.

 

For example, from (863-1)/2 = 431, the denominators are 2 x 3 x 2 x 2 = 24.

 

It is also possible to calculate this u value of 431 directly from the starting n-1 as (10367 - 23)/24 = 431. Other examples are indicated by curved lines.

 

The formula used in this example to calculate the triangle base u value from n-1 is:

 

((n-1) - (m-1)) / m

 

where m is any factor of aa.

 

A more generalized formula, that requires further testing, is:

 

((n-1) - (n-1 % m)) / m

PMA !!y5/EVb5KZI ID: 52c782 Sept. 18, 2018, 3:17 p.m. No.7632   πŸ—„οΈ.is πŸ”—kun   >>7634

>>7604

>A more generalized formula, that requires further testing:

 

Attached pic is a test of the u mod formula to verify the triangle base calculations from valid known factors.

 

Starting from the third factor record for c1020100, (0,9216,481) = {0:9216:1010:960:50:20402}, each known factor represents a square multiple "a" value in aan(n-1).

 

The formula for each cell is:

 

( u - mod( u, p ) ) / p

 

where p is a valid factor of aa shown in the header row, and u is the triangle base starting position in the "u" column.

 

Green cells indicate an exact match to a valid u value.

Yellow cells indicate a mismatch by 1.

 

For the case of yellow cells, the correct formula would be: ( u - mod( u, p ) ) / p - 1, though not yet sure when that -1 rule applies.

GAnon !Nx57Pyux3E ID: e906b3 Sept. 20, 2018, 7:18 p.m. No.7642   πŸ—„οΈ.is πŸ”—kun

>>7636

 

>Think col 0

>Think col 1

>Think -1

 

Column 0 and column 1 are specified here.

Think -1. He doesn't say column -1 so I'd say they mean row -1, which as all the negative squares in it. And if you notice from all these n=-1 entries there are little projections of them that map going from (-d^2, -1) to (-d^2 + 2, 0), (-d^2 + 4, 2), …. (-d^2 + 2i, i)

 

So maybe we could start from these -1 entries with the right d (or d and d+1 because our f record) and check something like in between these entries in the column. Obviously since one of d or d+1 will be odd and the other will be even, one will get cast into the col 0 and one will go into col -1. (I thought it was gonna be 1 so it would make more sense with this crumb but whatever) maybe this could be a path into using the center columns.

GAnon !Nx57Pyux3E ID: e906b3 Sept. 20, 2018, 9:41 p.m. No.7643   πŸ—„οΈ.is πŸ”—kun   >>7658

So if you draw a square and a smaller square inside of it, you can tilt the inner square a bit and then it makes four identical right triangles around the edge. One length of the triangle is the base of the big square, (d+n), then from the corner it goes into the middle corner of the tilted one like pic related. These are all the squares for products where the factors differ by 2. Then there is 4 and 10. Now I don't know if this is completely related, but it is a completely unique shape to every product. Also it looks like they start off moving then sort of set into place but approach a limit. Idk. You could also see that this is just two rectangles also and you could pull this out and fractal it up

Anonymous ID: c01508 Sept. 21, 2018, 7:47 p.m. No.7645   πŸ—„οΈ.is πŸ”—kun   >>7658

I have 2 good leads that are coming to me right now.

 

  1. I was thinking about the asymmetry thing.

And 0.

And if 0 is a number.

And symmetry.

And vqc's euler? euler?.

And e.

And how the background temperature of the universe is e.

Instead of absolute 0.

Weird.

Cosmic Microwave Background they call it.

Or maybe they call it something else now.

Which is directly connected to free energy and white noise and the casimir effect and all that fun stuff.

A force that can be tapped into for seemingly infinite energy.

It's the noise that's left over when everything else is quiet.

It's the ghost in the machine.

The noise in the lines when there shouldn't be anything there.

We're getting rid if it like it's something bad, when we could be tapping into it.

A background frequency that's always there.

And where is it coming from?

Seemingly everywhere.

Propogating in all directions.

Or maybe you could call it directionless?

You can also think of it like a temperature.

So what is that temperature?

It happens to be unfucking believably close to the value of e.

The background temp of this universe is e. Instead of 0.

 

So my thinking is this - What if 0 isn't the center? What if the center of all numbers is e? How sexy does that feel with all this talk of asymmetry? Because to me it feels perfect.

 

  1. Number 2 isn't as cool, but I'm wondering what patterns would be noticed in base e. Some anons were talking about changing base to notice patterns (9 and 2 were suggested). I think base e would be perfect if there is a change from 10.

 

And

  1. Number 3 because I just noticed it. And this is gonna really blow your minds because I'm about to casually state one of the most profound things in mathematics, physics, spirituality, religion, philosophy and life in general:

 

e is God.

 

I'm gonna let you think about that for a second.

It probably won't connect until I give you the next piece:

 

phi is Satan.

 

phi is the law of the real. phi is the lord of this physical realm. WHO DO THEY WORSHIP? Why the SPIRALS? Covered in GOLD? What's it MEAN? (kek)

 

To put it in different terms:

 

e is the SPIRIT and phi is the ANIMAL

 

Also - here is a .pdf (yeah I know people don't usually like pdfs) that you guys probably understand more than me.

>Relationship between irrational constants Phi and e - viXra

 

Also Also -

>What is the connection between euler's number, phi (golden ratio) and pi (3.14…)???

<https://www.researchgate.net/post/What_is_the_connection_between_eulers_number_phi_golden_ratio_and_pi_314

>I've been struggling with this question for a while and I'm wondering if you may help me with any insight/references? My intuition tells me that it is growth that links these irrational numbers…growth from negative to positive matter. The link between these numbers may possibly lead to an explanation of the origin of matter, or 'substratum'. Any help….?

>An elegant solution to this question is:

phi = eipi/5 + e-ipi/5

 

I give to you the key.

I know of no other nerds that are more perfect for this task.

 

Godspeed.

Anonymous ID: c01508 Sept. 21, 2018, 7:53 p.m. No.7646   πŸ—„οΈ.is πŸ”—kun

Sorry for the link, the < should have broken it I had thought.

I may never get credit for this, but a rare pepe would be appropriate. Nothing good on this machine.

Anonymous ID: 64f02a Sept. 22, 2018, 7:25 a.m. No.7658   πŸ—„οΈ.is πŸ”—kun   >>7659 >>7660 >>7661

>>7643

This is interesting, what would be the dimensions of 1st pic related?

 

>>7645

Also interesting. Im the anon that was sperging about fibonacci sequences earlier in the thread. I managed to get to the next set of patterns and I noticed something. First off the 1st pattern is 60 numbers long, the 2nd is 300, the 3rd is 1500, I had assumed the 4th would be 7500 but that was incorrect, it is 15,000 numbers long. Im working on getting to the next pattern but it seems that it too is a 10 fold increase, as ive gotten to the 75,000th iteration of the fibonacci sequence and it didnt repeat as predicted at 75,000.

 

Anyway. in an attempt to make sense of such large groups of numbers I tried organizing them all into columns of as long as the 2nd prior pattern. (so with the 1500 pattern I arranged it into columns of 60 and with the 15,000 pattern I arranged it into columns of 300 numbers. I then assigned each number a color. 2nd pic related is the 15,000 number long pattern and 3rd pic related is the 1,500 number long pattern. First off, wew that is quite the image that it creates. Second off, look at how similar they are. the 1,500 number long pattern looks like a much lower resolution version of the same image.

 

For a refresher on where I'm getting these patterns from, the first (60) pattern is the first digit spot in a fibonacci sequence, after 60 numbers the singles digit of said fibonacci sequence will repeat forever. The 300 number long pattern is the 10's digits, 1500 the hundreds and the 15,000 pattern is the thousands.

 

So far my theory that they will always repeat holds true.

Anonymous ID: 64f02a Sept. 22, 2018, 7:41 a.m. No.7659   πŸ—„οΈ.is πŸ”—kun

>>7658

the first picture of the pattern is kinda hard to see the way it maximizes in browser, heres it zoomed out a bit and broken into two images (couldnt get the whole image into two screencaps its missing the center portion of it) for phone lurkers.

GAnon !Nx57Pyux3E ID: e906b3 Sept. 22, 2018, 6:40 p.m. No.7661   πŸ—„οΈ.is πŸ”—kun   >>7678

>>7660

>>7658

 

So the dimensions of this boils down to a quadratic equation. If you have the inner square as x+n and the outer as d+n, then d+n would be the hypotenuse. You would also see that the little extra length (on the far right and far left) would be the same distance, lets call it w for no specific reason.

 

This function screenshot returns the length and angle

Anonymous ID: a332a2 Sept. 24, 2018, 9:22 a.m. No.7662   πŸ—„οΈ.is πŸ”—kun

I'm not sure what I'm actually doing. I've been looking into the VQC thinking "lookup" table and I was playing around with x=d in column 0, row 1 for even d's. I noticed a pattern for BigN - n, but I don't know if it's anything usable. It doesn't solve anything, just a pattern.

 

Anyway, if you take an (e, n) and create the cell in (0, 1, x=d). The a will relate to the bigN - n, but not in any obvious way. What I've looked at is creating the smooth number BigN - n, then subtracting it from (0, 1, x=d).a, which gives another number. If you do this for a record, it will generate a new sequence of numbers that correspond to another column where a=(0, 1, x=d).a - (BigN - n).

 

Example:

a=7, b=37, c = 259, d = 16

 

BigN = 114,

n = 6

114 - 6 = 108

 

Column 0, row 1 x=d: {0:1:144:16:128:162}

128-108 = 20.

 

The next record in the sequence of (3, 6) is a=37, b=91, c=3367, d=58. This has:

BigN = 1626

n = 6

1626 - 6 = 1620

 

{0:1:1740:58:1682:1800}

1682 - 1620 = 62.

 

These two numbers will exist in (15, 6): {15:6:35:15:20:62}.

 

This appears to hold true for all records.

 

At first glance it could look like it is (e + 2n, n), but it isn't true for all the records. If you look at the other sequence in (3, 6) you will have one record that has a=19, b=61. So it could also look like 19 + 1, 61 + 1, but again, that doesn't work for all the records. Sometimes it is an increase of something else.

Anonymous ID: a332a2 Sept. 24, 2018, 9:29 a.m. No.7663   πŸ—„οΈ.is πŸ”—kun   >>7664

Another interesting thing regarding n's and shadow n's and smooth numbers.

 

Create both Big N's for a record (Normal and shadow). Subtract the smooth number from these two. It will give you a new set of numbers that will exist in (e, k) (for some k) where x + n = d.

 

Example:

 

a=7, b=37, c=259, d=16

BigN = 114, shadow BigN = 146

Smooth number (114 - 6) = 108.

 

114 - 108 = 6, 146 - 108 = 38

{3:7:15:9:6:38}

 

7+9 = 16

 

Another example:

a=61, b=131, n=7, d=89, e=70

 

BigN = 3907, shadow BigN=4085

Smooth number (3907-7) = 3900

 

3907 - 3900 = 7

4085 - 3900 = 185

{70:61:35:28:7:185}

 

61 + 28 = 89.

 

Meaning there is a record in column e, where d = x+n that is related to our own record. In this record the d is equal to x+n.

Anonymous ID: a332a2 Sept. 24, 2018, 9:35 a.m. No.7664   πŸ—„οΈ.is πŸ”—kun   >>7665

>>7663

> Meaning there is a record in column e, where d = x+n that is related to our own record. In this record the d is equal to x+n.

 

Just to clarify a bit

 

Here a is the n for the record a=61, b=131 while 185 is the shadow record. That means a=n, b=shadow n exists in (e, a).

These two records are the "normal" record and the shadow record:

{70:7:89:28:61:131}

{70:185:-89:-150:61:131}

 

This is the record that is created when you remove the smooth number from BigN and shadow BigN, it is also the record where a=n and b=shadow n.

{70:61:35:28:7:185}

Anonymous ID: a332a2 Sept. 24, 2018, 10:42 a.m. No.7666   πŸ—„οΈ.is πŸ”—kun   >>7670

In some cases the a's in column e will match the parity of x in column -f. If you highlight, or "mark" these columns in -f you will see that they are spaced out by 2n - 1 (The difference between square n^2 - (n-1)^2 for some n). For smaller values, if you highlight the center between these marked cells (since 2n-1 is odd there will be a center piece) it tends to contain our factors. Might just be a side-effect of using smaller numbers, though.

Anonymous ID: a332a2 Sept. 25, 2018, 8:27 a.m. No.7670   πŸ—„οΈ.is πŸ”—kun   >>7671 >>7673

>>7666

The product of the two n's (normal n and shadow n) for our cell (a, b) is equal to (x + n)^2 + e.

 

Given the cell in column e for (a, b) for some a and b. This cell will have n, but there will also exist a shadow n' for it. The product of n and n' is equal to the (a, b)s cells (x + n)^2 + e.

 

The cell in column e for (n, n') will have x + n = d from the cell of (a, b).

 

This gives us: n * n' = (x + n)^2 + e

Anonymous ID: a332a2 Sept. 25, 2018, 9:22 a.m. No.7673   πŸ—„οΈ.is πŸ”—kun

>>7670

To build a bit on this.

 

We have the cell: {3:6:58:21:37:91}

This has another cell with negative d (shadow cell?): {3:122:-58:-95:37:91}

 

The records for where a=6, b=122 is:

{3:37:27:21:6:122}

{3:91:-27:-33:6:122}

 

In these cells, x + n = 58 (d). The product of these records are 732 (6 * 122) which is (x + n)^2 + e (6 + 21)^2 + 3. Meanwhile those records will have (x + n) as d. (6 + 21 = 27).

 

This means we can find the triangular base for the shadow cells, but we'll only know e and x+n.

Anonymous ID: a332a2 Sept. 25, 2018, 11:26 a.m. No.7674   πŸ—„οΈ.is πŸ”—kun   >>7675

I'm not entirely sure what to call this, but I noticed something when I was looking at 4c. In my head I've been thinking of it as a kind of record transformation. Take a record and multiply it by 4 (you can think of it as by dividing the 4 into 2 by 2, multiplying a by 2 and b by 2.).

 

The reason I'm calling it a transformation is because we skew or transform the patterns in (e, 1). Take a column e and row 1. Multiply it by 4. In (4e, 4) you will find the same values, but this time the patterns are skewed, as in, the sequences / chains within this record no longer matches the ones we have in (e, 1).

 

Example, look at (3, 1). Multiply it by 4 and we get (12, 4).

 

{3:1:3:1:2:6} {12:4:4:2:2:14}

{3:1:9:3:6:14} {12:4:12:6:6:26}

{3:1:19:5:14:26} {12:4:24:10:14:42}

{3:1:33:7:26:42} {12:4:40:14:26:62}

{3:1:51:9:42:62} {12:4:60:18:42:86}

{3:1:73:11:62:86} {12:4:84:22:62:114}

{3:1:99:13:86:114} {12:4:112:26:86:146}

{3:1:129:15:114:146} {12:4:144:30:114:182}

{3:1:163:17:146:182} {12:4:180:34:146:222}

{3:1:201:19:182:222} {12:4:220:38:182:266}

{3:1:243:21:222:266} {12:4:264:42:222:314}

 

Here we can see the sequences have changed. In (3, 1, 1) we have a=2, b=6 followed by (3, 1, 2) where a=6, b=14. This forms a single chain / sequence within (3, 1). In (12, 4) however, we can see that we now have transformed (3, 1) into two sequences. If we were to multiply (12, 4) again by 4 the pattern would repeat (transforming yet again).

 

If we were to multiply (e, 1) by n^2 (For example n=6 where a=7, b=37 exists) we would end up with a record that connects / pairs all the an, bn transformations in (e, 6).

 

{3:6:16:9:7:37} =an = 42, bn = 222 => {108:36:96:54:42:222}

{3:6:58:21:37:91} =an = 222, bn = 546 => {108:36:348:126:222:546}

{3:6:124:33:91:169} an = 546, bn = 1014 ={108:36:744:198:546:1014}

 

I haven't looked enough into this yet, there's so many damn patterns to try and understand, but I suspect that multiplying any (e, 1) with a square will result in (e * k^2, k^2) records to "skew" the original record, generating multiple sequences.

Anonymous ID: a332a2 Sept. 25, 2018, 11:28 a.m. No.7675   πŸ—„οΈ.is πŸ”—kun

>>7674

Forgot to mention, 108 is also the smooth number belonging to record a=7, b=37, c=259. We have n = 6 and BigN = 114 with 114 - 6 = 108. I believe I've seen patterns before, but forgot to write up about the smooth numbers within a column running in a pattern similar to the other patterns we're seeing.

 

Also, note that there are two sequences in (3, 6) and both of them are "paired" in (108, 36). Example a=1, b=19, n=6. This gives an = 6, bn = 114, which is at (108, 36, 2).

Anonymous ID: d4c3ed Sept. 28, 2018, 1:31 p.m. No.7684   πŸ—„οΈ.is πŸ”—kun   >>7686

I've been looking a bit more into movements. One thing I noticed was cells at (e + nn, n). In those records the d-values will be equal to the a-values in (e, n) and the x-values will be equal to (x + n) while the a's (and thus b's) will be equal to (d + n) values.

 

Since we don't know n, this isn't usable, but I've been thinking more about the fractal nature VQC has been talking about. I'm wondering if we should be trying to draw some figures that attempt to match up with what our fractal actually looks likes.

 

Having said that, anyone actually looked into the mandlebrot set? I was reading up on it (Only quickly between breaks), but the damn thing is screaming VQC. f_c(z) = z^2 + c. That's pretty much what we're doing, except instead of dealing with complex numbers or float numbers we're working with integers.

 

I've been thinking that we will either extend the grid (to support float / complex numbers) or there is a deeper connection (Is our fractal of mandlebrot nature?).

Anonymous ID: d4c3ed Sept. 28, 2018, 1:37 p.m. No.7685   πŸ—„οΈ.is πŸ”—kun

So we have (e * nn, n) and (e + nn, n), both of which extend or transform the records in some way.

 

(e * nn, n) will extend the sequence (Essentially increase the number of sequences in our row) while (e + nn, n) will transform / modify our row by exchanging values of d, x and a with other values. I'm going to try and think more about what this actually means and try and formulate a better description of what actually is happening. I'm struggling a bit with visualizing what I'm doing, so I would like to try and step back a bit to get a better understanding of why (e + nn, n) and (e * nn, n) works the way they do.

 

As for (e * nn, n) we will see repetitions of (1, 1) in every (nn, n) as 1 * nn = nn, meanwhile this explains the repetitions of (0, 1) in (0, nn).

Anonymous ID: d4c3ed Sept. 28, 2018, 1:41 p.m. No.7686   πŸ—„οΈ.is πŸ”—kun   >>7687

>>7684

Regarding the fractal / mandelbrot idea I'm wondering if that's actually our fractal. As in, (0, 1) or (1, 1) is the "base" mandelbrot set, as in the grand picture and why VQC has been telling us to use bigger numbers. Because by doing so we'll start seeing the repeated numbers.

 

I'm just spitballing, so I'm probably way off, but I can't help but shake the fact that our equation for c (d^2 + e) is exactly the same as the one for mandelbrot (z^2 + c).

Anonymous ID: d4c3ed Sept. 28, 2018, 1:53 p.m. No.7687   πŸ—„οΈ.is πŸ”—kun

>>7686

In the mandelbrot set you have a term called "escaping" in which a point that is "escaping" means that the point is not within the boundaries of the fractal (If I grasped it correctly). I wonder if prime numbers are the ones that "escape".

Anonymous ID: 64f02a Sept. 28, 2018, 8:02 p.m. No.7688   πŸ—„οΈ.is πŸ”—kun   >>7689

Fibonacci autist here.. This might be really fucking important. I need to test something. I need a couple of large sub-prime numbers that I do not know the factors to. Fucking pronto.