AA ID: 72180a RSA #14 Sept. 28, 2018, 8:25 p.m. No.7690   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.

Anonymous ID: 72180a Sept. 28, 2018, 8:32 p.m. No.7691   πŸ—„οΈ.is πŸ”—kun

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

AA ID: 72180a Sept. 28, 2018, 8:44 p.m. No.7692   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

RSA #13 β€”β€” https://archive.is/Fblcs

AA !dTGY7OMD/g ID: 72180a Sept. 28, 2018, 8:52 p.m. No.7693   πŸ—„οΈ.is πŸ”—kun

The fact that we're almost a year into this and we still don't have an easily accessible pastebin for making new threads with is a bit aggravating. https://pastebin.com/VkaTJcX1

ID: b129f9 Sept. 28, 2018, 10:17 p.m. No.7695   πŸ—„οΈ.is πŸ”—kun   >>7696

I apologize for necesitating an emergency bake. I was busy.

 

In a clearer format (the one I keep) (also, fixed some typos):

post 1, glossary - https://pastebin.com/gWV7vySY

post 2, rules - https://pastebin.com/cE195ptk

post 3, code - https://pastebin.com/heweAtWs

Anonymous ID: 83b459 Sept. 29, 2018, 12:22 p.m. No.7697   πŸ—„οΈ.is πŸ”—kun   >>7699

The Dimension equals 8?

Q=D5=Minecraft Avalanche

VQC=8D=Mathcraft Avalanche, starting with a single particle of water.

 

Difference of 3 Dimensions.

 

E8 H2O

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Sept. 29, 2018, 12:31 p.m. No.7698   πŸ—„οΈ.is πŸ”—kun   >>7699

Aaaaand now to officially BLESS THIS BREAD!

8=D… wheeeeeere to go with this…

8==D….

8===D…

 

https://en.wikipedia.org/wiki/Eight-dimensional_space

 

https://en.wikipedia.org/wiki/Euclidean_distance

 

https://en.wikipedia.org/wiki/Coxeter–Dynkin_diagram

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Sept. 29, 2018, 12:37 p.m. No.7699   πŸ—„οΈ.is πŸ”—kun   >>7700

>>7698

>>7697

Golly… Pic 1 looks a might familiar…

 

Moving on…

https://en.wikipedia.org/wiki/Negative-dimensional_space

 

https://en.wikipedia.org/wiki/Zero-dimensional_space

 

https://en.wikipedia.org/wiki/Polish_space

 

https://www.thefreedictionary.com/Hypertriangle

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Sept. 29, 2018, 12:52 p.m. No.7701   πŸ—„οΈ.is πŸ”—kun

>>7700

Also duh, I'm jumping ahead.

We're on 14.

 

Which woulda been perfect for 7700, cuz 7+7=14, and then 7x7=49 and 7^7=823543

 

Anyway.

14th prime is 43

 

Forty-three is the 14th smallest prime number. The previous is forty-one, with which it comprises a twin prime, and the next is forty-seven. 43 is the smallest prime that is not a Chen prime. It is also the third Wagstaff prime.[1]

 

43 is the fourth term of Sylvester's sequence, one more than the product of the previous terms (2 Γ— 3 Γ— 7).[2]

 

43 is a centered heptagonal number.[3]

 

Let a0 = a1 = 1, and thenceforth an =

1

/

n βˆ’ 1

(a02 + a12 + … + an βˆ’ 12). This sequence continues 1, 1, 2, 3, 5, 10, 28, 154… (sequence A003504 in the OEIS). a43 is the first term of this sequence that is not an integer.

 

43 is a Heegner number.[4]

 

43 is the largest prime which dives the order of the Janko group J4.

 

43 is a repdigit in base 6 (111).

 

43 is the largest natural number that is not a (original) McNugget number.[5]

 

43 is the smallest prime number expressible as the sum of 2, 3, 4, or 5 different primes:

 

43 = 41 + 2

43 = 11 + 13 + 19

43 = 2 + 11 + 13 + 17

43 = 3 + 5 + 7 + 11 + 17.

When taking the first six terms of the Taylor series for computing e, one obtains

 

βˆ‘

i

=

0

5

1

i

!

=

163

60

=

2

+

43

60

,

 

\sum _{{i=0}}^{{5}}{\frac {1}{i!}}={\frac {163}{60}}=2+{\frac {43}{60}},\

which is also five minus the fifth harmonic number.

 

Every solvable configuration of the Fifteen puzzle can be solved in no more than 43 multi-tile moves (i.e. when moving two or three tiles at once is counted as one move).[6]

 

McNugget Number

GAnon !Nx57Pyux3E ID: b542af Sept. 29, 2018, 10:48 p.m. No.7702   πŸ—„οΈ.is πŸ”—kun   >>7703

So if you're in the discord you may have heard about this but I made a little app for you guys to use to navigate the map. I think it can be really useful for those of us who don't like to calculate or make an excel grid every time. It is fairly easy to use as well. It is all in one python file I'm not sure if you need to run it from a terminal but I would do that if I were you. Before you run the file make sure your console window is big enough I couldn't figure out how resize it through python.

 

Essentially the top left is the map and you can highlight cells based on whether they have the same d x or a value as your starting cell. You can navigate through this map however you want. If you highlight a cell and it has records, they will show up on the right window with other ones in the same cell. You can navigate up and down these records. If you press 'i' while highlighting one, it will get sent down to the bottom detail row. It will stay there until you set a new starting record or delete it. If you press 'h' while highlighting a record that will become the new starting one. I'd recommend typing !help to learn how it works and just play around with it

 

To create records type 'abd 5 29 12' or the similar with enx or end or others. Screencaps are here

 

code:

https://pastebin.co m/745cFABb

GAnon !Nx57Pyux3E ID: b542af Sept. 29, 2018, 10:54 p.m. No.7703   πŸ—„οΈ.is πŸ”—kun

>>7702

>7702

Also theres definitely room for optimization. I tried to generate the records in a dictionary and then load them so I didn't have to iterate through x values for every pixel in the map but I couldn't get it for some reason and I wasn't concerned about it really so I just generate every pixel. Someone could easily fix that. I plan on making improvements and adding more features I just think this is at a pretty solid state so I want to send it out. Also don't fuck with pressing j or k on a record in the n=0 row. It will crash the thing because I index with x and x is all the same with that so it never finds the next x and I dont feel like adding a whole nother indexing set for n=0 so just understand the patterns there.

Anonymous ID: b1d934 Sept. 30, 2018, 6:47 a.m. No.7704   πŸ—„οΈ.is πŸ”—kun   >>7705 >>8091

I've been spending some more time looking into the smooth numbers. I've noticed that given (e, n) the smooth numbers for those records (bigN - n) exists in (e - 2*n, 1). If e is odd, then the smooth numbers will exist as d's. If e is even then smooth number will appear as a. If you jump to (e + 2n, n) the smooth values of those records will match the a's or d's in (e, 1) at t for bigN. Essentially, given a record a, b we get bigN. If e is odd then the d at a[t] = aBigN (e, 1) will be a "smooth" number for the jump (or a + 1, b + 1) in (e + 2n, n).

 

The smooth numbers of (0, 1) exist in (-2, 1) at t = 1, 4, 9, 16…

The smooth numbers of (1, 1) exist in (-1, 1) at t = 1, 4, 9, 16…

 

It also appears that the t's for the smooth numbers in (e, 1) exist in (e - 1, 2). Although that is just a preliminary result. I haven't had much time to look into that result.

Anonymous ID: b1d934 Sept. 30, 2018, 6:53 a.m. No.7705   πŸ—„οΈ.is πŸ”—kun   >>7706 >>8091

>>7704

The patterns aren't always [1, 4, 9, ..] but offsets.

 

Take (7, 1). The first 10 smooth numbers are:

[2, 5, 10, 17, 26, 37, 50, 65, 82]

 

Here the relationship with squares is 2 + 2, 5 + 4, 10+6, 17 + 8.

 

(8, 1) has first 10 smooth numbers: [4, 8, 14, 22, 32, 44, 58, 74, 92] which has the pattern 4 + 0, 8 + 1, 14 + 2, 22 + 3, 32 + 4 …

Anonymous ID: b1d934 Sept. 30, 2018, 7 a.m. No.7706   πŸ—„οΈ.is πŸ”—kun   >>7707 >>8091

>>7705

Take cell (3, 6). Moving to the left will put us in (3 + 2*6, 6) =(15, 6).

 

The first 10 rows in (3, 6) is:

 

and their smooth numbers are:

{3:6:4:3:1:19}

{3:6:16:9:7:37}

{3:6:34:15:19:61}

{3:6:58:21:37:91}

{3:6:88:27:61:127}

{3:6:124:33:91:169}

{3:6:166:39:127:217}

{3:6:214:45:169:271}

{3:6:268:51:217:331}

{3:6:328:57:271:397}

[0, 108, 540, 1620, 3780, 7560, 13608, 22680, 35640]

 

While the first 10 rows in (15, 6) are:

{15:6:17:9:8:38}

{15:6:35:15:20:62}

{15:6:59:21:38:92}

{15:6:89:27:62:128}

{15:6:125:33:92:170}

{15:6:167:39:128:218}

{15:6:215:45:170:272}

{15:6:269:51:218:332}

{15:6:329:57:272:398}

{15:6:395:63:332:470}

 

and their smooth numbers are:

[129, 579, 1683, 3873, 7689, 13779, 22899, 35913, 53793]

 

Note that (15, 6) has "shifted" one record down, meaning the first cell in (3, 6) (a=1, b=19) is gone.

 

Each of the smooth numbers in (15, 6) are the d's at (3, 1) where a[t] = BigN.

 

The big n for a=7, b=37 is 114 and the d at that location = 129 and exists at t = 8.

 

This means that the t's in (3, 1) where d = smooth number for (15, 6) are: [8, 17, 29, 44, 62, 83, 107, 134, 164]

Anonymous ID: b1d934 Sept. 30, 2018, 7:04 a.m. No.7707   πŸ—„οΈ.is πŸ”—kun   >>8091

>>7706

Minor typo, pasted the records along with the smooth numbers, but you'll see the difference.

 

The smooth numbers for (3, 6) (0, 108, …) exists in (3 - 2*6, 1) =(-9, 1) at t's: [2, 8, 17, 29, 44, 62, 83, 107, 134]

 

Which we can see occur at the same location in (3, 1) as the smooth numbers do for (15, 6).

PMA !!y5/EVb5KZI ID: 4fbd64 Oct. 2, 2018, 8:37 p.m. No.7708   πŸ—„οΈ.is πŸ”—kun

Continuing with the aan(n-1) work from the previous bread >>7584, have found an alternate way to calculate the triangle base and remainders in some cases for more than the first factor record.

 

Drawing from these VQC hints, though not sure if related and/or relevant.

 

>>7636

>Think -1.

 

>>7639

>The key is how many squares make up the remainder.

 

All aan(n-1) calculations for (-f,1) and (e,1) will have at least 1 square multiple of aa for the first factor record, and 2 or more for subsequent factor records. Idea being that if we can understand how those triangles and remainders are calculated, we would be able to jump to another record.

 

Attached pics for c255025 (101^2 * 25^2) show the first 3 factor records where a[t] is 1, 5, and 25.

 

The "u calc" column is calculated relative to (n-1) as follows:

 

if a a[t]: (n-1) - (((a - a[t]) * (n-1))/a) - 1

if a < a[t]: (n-1) + (((a[t] - a) * (n-1))/a) + 1

if a = a[t]: always (n-1)

 

where "a" is a valid square multiple in aan(n-1), and a[t] is the "a" value from the factor record.

 

For triangle bases u < (n-1), the a a[t] formula seems to be relatively accurate as reflected in the "u diff" column for various test cases in both (0,n) and (1,n).

 

When u (n-1), however, something else starts to creep into the calculation as can been seen in the a[t]=25 example. Larger a[t] values will have a larger discrepancy, but not yet sure how these are calculated. An "(x)" in the "u calc" column indicates an incorrect value.

 

The triangle remainders where a[t] is 1 and 5 can also be calculated in terms of (a-a[t]) as shown in the "rm/(a-a[t])" and "sqrt" columns. Again showing the inverse relationship between the "aa" square multiple and the triangle remainder formula. More discrepancies also appear in calculating these remainders as a[t] increases and for u (n-1) as can be seen in the a[t]=25 example.

AA !dTGY7OMD/g ID: e77daa Oct. 2, 2018, 9:34 p.m. No.7709   πŸ—„οΈ.is πŸ”—kun   >>7710 >>7711 >>7712 >>7721

This has completely destroyed my life. It didn't hit me how terrible it was until today, but fitting in enough time to study everything I've studied about it over the past 3 months or so and doing Grid Patterns completely by myself has led to my friends thinking I'm a useless, selfish cunt, my sleep schedule being ruined to the point of it probably affecting me long-term, my university grades tanking, my diet becoming very unhealthy, and in general my mood being very negative. I'll try to stop this from turning into a wall of text and just say that I'll still be around to do mod duties, but unless something big happens very soon, I'm possibly going to have to take an extended break, for my own good.

Anonymous ID: b1d934 Oct. 3, 2018, 7:26 a.m. No.7710   πŸ—„οΈ.is πŸ”—kun

>>7709

Hey man, that's understandable. You need to take care of yourself. You've done a massive job and I'm very impressed (and a bit guilty for not participating as much in the pattern thread). I hope you'll straighten things out.

Anonymous ID: d02d29 Oct. 3, 2018, 7:57 a.m. No.7711   πŸ—„οΈ.is πŸ”—kun

>>7709

Ive found that in my studies I always wind up in that same position after about 3 months or so. Also, think of it like a video game. Ever play a game and you just cant get past one part? But then if you put the game down and come back after a couple of days you beat that level on your first attempt? Take a break. Sometimes we need to get out and see some trees in order to get the inspiration we need to make any progress.

 

It's easy to want to be 'the one' to figure it all out but we are racing against them, not each other. When I figure it out while you are taking a break I'll include you in a footnote somewhere towards the back of my dissertation :^)

VA !!Nf9AmQNR7I ID: 94c848 Oct. 3, 2018, 9:47 a.m. No.7712   πŸ—„οΈ.is πŸ”—kun   >>7721

>>7709

Hello AA, I can relate. I was spending so much time on this that my family life and health was suffering. I'm still thinking and working every day, but I've had to dial back my involvement a bit and make sure the other parts of my life are getting the focus and attention they need. Also, I had to ask for help to quit drinking which was fueling my late night obsessive working till early hours and wrecking my sanity and health. I have 37 days sober today.

 

I think I can speak for everyone here when I say Thank You for all your work over the last few months. Take the break you need to put your life back in balance, my friend. Love ya!

ID: b129f9 Oct. 3, 2018, 1:40 p.m. No.7713   πŸ—„οΈ.is πŸ”—kun

I've taken a few steps on formulating Shor in terms of and linking it to the language of the VQC and I've found that the possible values in the most important step of Shor (period finding) are actually related to N-n.

 

The possible period values found in calculating Shor's algorithm with random m values from 1 to c-1 all originate from the prime factorization of N-n.

 

An example, c=253, (the number we are looking at here is the second number in each bracket, which is the period value arrived at when the first number in the bracket is used as m in Shor).

 

https://pastebin.com/beY9hPx0

 

N-n = 110 = 2 * 5 * 11 which means we expect to find that a combination of these 3 factors results in every possible period value, and that is what is found in the data. 55, 22, 10, 5, and 110 are period values in the data and they are all products of a combination of 2, 5, and 11.

 

You find the same thing for other examples (it takes a lot of computing to calculate these examples)

 

6107, N-n = 2940: (in the attachment, I had to put it into a pdf because 8chan doesn't allow text files)

 

2940 = 2^2 x 3 x 5 x 7^2, so we expect every period value (second number in the brackets) to have a combination of 2, 3, 5 and 7 as a factor (note, it still counts if it's missing one of these factors from the prime factorization because it can be written with the factor it's missing to the power of 0)

 

145, N-n = 56: https://ghostbin.com/paste/2cg75

123, N-n = 40: https://ghostbin.com/paste/fqu8u

287, N-n = https://ghostbin.com/paste/7te9u

 

So the first noted connection between Shor and the vqc happens to be N-n and smooth numbers.

ID: b129f9 Oct. 3, 2018, 9:39 p.m. No.7714   πŸ—„οΈ.is πŸ”—kun   >>7715

A concept you all might find interesting, from the General Number Field Sieve (it is pretty simple if you know what a smooth number is.)

 

Let's say we're going to look at 3-smooth numbers, numbers which at most have 3 as a prime factor. We can represent all of the 3-smooth numbers as evaluations of:

 

2^x * 3^y = 3-smooth number

 

where the exponents determine which number it is. For example, 12 = 2^2 * 3^1

 

We can represent this in a different notation called exponent vectors (the premise of the notation is simple, don't let that scare you).

Since we know we're looking at 3-smooth numbers, we only need to know the values of the exponents to represent a given 3-smooth number.

2^2 * 3^1 becomes (2, 1). (I haven't confirmed a connection here, but does this look familiar?)

 

So, our 3-smooth numbers become sets of exponent values. E.g, (3,1) = 24. We can also define multiplication of these "exponent vectors." Since a pair of values now represents a number, we can simplify multiplying these pairs by adding their exponent values instead of calculating the actual multiplication.

 

(4,2) = 144

(4,1) = 48

 

(4, 2) times (4, 1) = (4+4, 1+2) = (8, 3) = 6912 = 14448 (This also defines squaring a 3-smooth number, as in (2,1) times (2,1) = (4,2) = 1212.

 

Incidentally (or maybe not), any exponent vector that only contains even exponent values happens to be a perfect square. As in, (4,2) = 12*12, (4,4) = 36^2, (6,2) = 24^2, and so on. (This type of notation is used in GNFS to find one of the nontrivial squares (i^2 or j^2) which, interestingly enough happen to be equal when you are operating in GF(c) where c = ab, a field modulo c. When you are operating in GF(c) (you may have seen Galois fields in my ECC explanations), i^2 = j^2 (mod c) and i^2 - j^2 = 0 (mod c) (because they are equivalent to c, when it's modulo c then it's 0 since c%c=0).

 

So there you have it. Smooth numbers in Shor, smooth numbers in GNFS in a relatively simple introduction, and N-n as a smooth number.

ID: b129f9 Oct. 3, 2018, 9:53 p.m. No.7715   πŸ—„οΈ.is πŸ”—kun

>>7714

Any k-smooth number can be represented this way, this paper introduces the concept with 19-smooth numbers.

 

An example of this type of notation with a number we're familiar with:

 

N-n for 145 = 56 = 2^3 * 7^1

 

So if we define that as a 7-smooth exponent vector with 2 inputs it'd be (3, 1). It's also not the difference of two squares so it isn't found in the grid, at least in positive e and positive n.

 

As an example, all of the period values for 145 can be represented as one of these exponent vectors.

2 = (1, 0)

4 = (2, 0)

14 = (1, 1)

28 = (2, 1)

 

The same is true for all N-n from what I have seen so far. All of the period values can be represented in the exponent notation for the prime factorization of N-n, as further proof that N-n's smoothness determines all possible period values evaluated in Shor's algorithm for c.

ID: b129f9 Oct. 4, 2018, 9:07 p.m. No.7718   πŸ—„οΈ.is πŸ”—kun

>>7717

Follow up paper, I like this guy:

 

His paper includes finding a variable that is equivalent to d or d+1. Then calculating square's difference from c, which is either e or f.

 

Someone should tell him we simplified his work.

ID: b129f9 Oct. 4, 2018, 9:13 p.m. No.7719   πŸ—„οΈ.is πŸ”—kun

for future reference on our variable names (we are working on formulating these papers in the language of the VQC)

 

d = floored sqrt(c) = sqrt(c)//1 = sqrt(c) with no decimals

 

g = sqrt(c)

 

So we'll call the actual square root of c with the decimals g

ID: b129f9 Oct. 4, 2018, 9:29 p.m. No.7720   πŸ—„οΈ.is πŸ”—kun

When we formulated this first paper (A-Quick-Way-to-Factor-Large-Semiprimes.pdf) in the terms of the VQC, we found that the VQC language simplifies the approach greatly and requires less steps.

 

In this first screenshot I have here, Kurwzeg's "Nf(N)" (a polynomial equation with c as the input multiplied by c again, in number theory c is N and pq is ab) is equal to 2*i or 2(d+n) or a+b.

 

In essence this paper is saying Kurzweg has developed a method of factorization that is equivalent to searching for the value of a+b or 2(d+n). We find that we can greatly simplify his work using our equations.

 

The equivalent equations in the terms of the VQC:

 

c = ab

c = i^2 - j^2

 

His Nf(N) is equivalent to 2*i, so, if we iterate for 2i in the same manner he does ("using MAPLE math program," screenshot two), we can derive the factors by

 

i = 2i / 2

j = sqrt(i^2 - c)

 

a = i - j

b = i + j

Anonymous ID: 7e317b Oct. 4, 2018, 11:24 p.m. No.7721   πŸ—„οΈ.is πŸ”—kun

>>7712

>>7709

 

Its amazing that it has come this far. Been following this for a while with a friend, but I really gotta say that this whole thing here is problematic and everyone here should take a break honestly. It says a lot that Chris would let it go this far. Simple as that.

 

That said, this is the most fascinating experiment in human nature I have ever seen in my life. Someone could get a PHD from this place without ever even doing a math problem to be honest. Simply analyzing the people and their interactions and such would be enough. I encourage you to all just maybe come here once a week or so AT MOST. Also, I encourage Chris to move things along. Things have gone too far imo.

ID: b129f9 Oct. 4, 2018, 11:38 p.m. No.7722   πŸ—„οΈ.is πŸ”—kun   >>7723

Some formulae for deriving Kurzweg's variable k (from solution values j and i so far), in (A-New-Method-for-Factoring-Large-Semiprimes.pdf)

 

if (c%6) == 1 {

k = ( ( sqrt( (2j)^2 + 4c ) - (c+1) ) / 36) * -1

}

 

if (c%6) == 5 {

k = ( ( sqrt( (2i)^2 - 4c ) - (c-1) ) / 36 ) * -1

}

 

in the program "calc," they'd be written as (in the same order as above for the cases of which to use):

 

define kF(c, j) = (-1(sqrt((2j)^2 + 4*c) - (c+1)))/36

define kG(c, i) = -1((sqrt((2i)^2 - 4*c) - (c-1))/36)

 

examples of k values (examples c's from the paper):

kF(34417, 72) = 945

 

kF(21428053, 594) = ~594964.444

kF(21428053, 594)*36 = 21418720

 

kG(106577, 369) = ~2950.889

kG(106577, 369) * 36 = 106232

 

kG(732010841, 29229) = 20333020

 

In his paper,

G = (2i)^2 = 2(d+n)^2

F = (2j)^2 = 2(x+n)^2

ID: b129f9 Oct. 4, 2018, 11:47 p.m. No.7723   πŸ—„οΈ.is πŸ”—kun

>>7722

So, in the second paper, Kurzweg's second factorization method went from being a search for 2i to being a more informed search for 2(d+n)^2 or 2(x+n)^2 based on the remainder of c%6.

 

Interestingly, (c-1)/6 or (c+1)/6 is always in the ring of Z, which is number theory speak for it's always a whole integer, and which one it is is based on whether c%61 or c%65, respectively.

ID: b129f9 Oct. 5, 2018, 12:37 a.m. No.7724   πŸ—„οΈ.is πŸ”—kun

Also, an idea on what the point of the remainder tree was:

We were told to repeatedly divide each term by 2 until it was no longer a multiple of two. What does that make? A list of odd numbers, which also isolates the primes.

 

So , hypothesis:

with smooth numbers you are looking at the primes that make up a number by multiplication and with the tree you are looking at the primes in the remainders.

 

The tree wasn't the solution, it was a hint.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Oct. 5, 2018, 2:32 p.m. No.7725   πŸ—„οΈ.is πŸ”—kun

Sooooo….

 

Considering what's happening with Linux recently…

Since 2015…

 

It'd be a hell of a plot twist if Chris was non-other than Linus himself…

 

And this is his planned "Fuck you, I'll just make another one! I've done it before!" to the SJWs trying to ruin Linux…

 

And this is how the New One gets Open Sourced.

 

That'd be fuckin' stellar.

GAnon !Nx57Pyux3E ID: b542af Oct. 7, 2018, 10 a.m. No.7728   πŸ—„οΈ.is πŸ”—kun

Hey I just made a walkthrough of some of the grid patterns in the grids. Pic related. I hope this works as an ad for my program too. I also hope we can navigate through the grid on with these patterns.

ID: b129f9 Oct. 7, 2018, 10:44 p.m. No.7729   πŸ—„οΈ.is πŸ”—kun   >>7730 >>7731 >>7732

Kurzweg's 3 papers all have confusingly similar names, so I'm going to give them distinct names:

 

(A-Quick-Way-to-Factor-Large-Semiprimes.pdf) = the 2i paper, the 2(d+n) paper (because that's the value he's searching for, in his terms, Nf(N))

(A-New-Method-for-Factoring-Large-Semiprimes.pdf) = the 2i^2 and 2j^2 paper, because this paper contains a description of an iterative search for the values of 2i^2 and 2j^2 in our terms

(A-New-Approach-to-Factoring-Semiprimes.pdf) = the d, e and f paper, because the variables d, e and f all show up in this paper (I'll show how in a moment).

 

Next I'm going to outline his paper (the d e and f paper) in terms of our equations. I'm going to use greek letters for his variables that don't exactly map to ours because we've almost taken up the entire alphabet, and he introduces a lot of different variables.

 

A NEW APPROACH TO FACTORING SEMI-PRIMES:

 

step 1:

g = sqrt(c)

g is a variable we've defined before, this one includes the decimals

 

step 2:

Ξ± = the closest whole integer to g (d or (d+1))

 

step 3 - realize that there are two unknown values which add and subtract from Ξ± to make a and b:

 

a = Ξ± - Ξ²

(where Ξ² = x or x+1)

 

b = Ξ± + Ξ³ (this isn't a y (why), this is a Ξ³ (gamma))

(where Ξ³ = (x+2n) or (x+2n-1))

 

step 4 - which become, if we rearrange:

Ξ± * (Ξ³ - Ξ²) - Ξ²*Ξ³ = c - Ξ±^2

 

In our terms,

Ξ²*Ξ³ = x * (x+2n)

or

Ξ²*Ξ³ = (x+1) * (x+2n-1)

 

Here's where it gets fascinating:

c - Ξ±^2 = e or f

 

step 5, solve for (Ξ²*Ξ³) and (Ξ³ - Ξ²):

 

Ξ²*Ξ³ = Ξ΄ + Ξ΅ (where Ξ΄ and Ξ΅ are an unknown integers)

Ξ³ - Ξ² = (c - Ξ±^2 + Ξ΄ + Ξ΅)/Ξ±

Ξ΄ < Ξ±

 

step 6, eliminate Ξ³ from the equations and set Ξ΄=c - Ξ±^2 (I'm not sure of the exact path of this mathematical jump so I'll figure it out later if it is important):

Ξ²^2 + Ξ²*(Ξ΅ / Ξ±) - (Ξ΄ + Ξ΅) = 0

 

step 7, solve for Ξ²:

Ξ² = -(Ξ΅ / (2Ξ±)) + sqrt( (Ξ΅ / (2Ξ±))^2 + (Ξ΄ + Ξ΅) )

 

step 8, since we know Ξ΅ / (2*Ξ±) must be an integer (because it's defining Ξ² and a = Ξ± - Ξ²), we can get to the value that we need to search for to factorize:

let Ξ΅ / (2*Ξ±) = ΞΆ

 

Ξ² = -ΞΆ + sqrt( ΞΆ^2 + 2Ξ±ΞΆ + Ξ΄)

 

step 9, rearrange for factors:

 

a = (Ξ± + ΞΆ) - sqrt( ΞΆ^2 + 2Ξ±ΞΆ + Ξ΄ ) = (Ξ± + ΞΆ) - Ξ·

b = (Ξ± + ΞΆ) + sqrt( ΞΆ^2 + 2Ξ±ΞΆ + Ξ΄ ) = (Ξ± + ΞΆ) + Ξ·

letting Ξ· = sqrt( ΞΆ^2 + 2Ξ±ΞΆ + Ξ΄ )

 

step 10:

Ξ± and Ξ΄ are known from c ( Ξ΄ = c - Ξ±^2 and Ξ± is the closest integer to g (d or d+1)), so now it just becomes finding a value of ΞΆ that makes Ξ· evaluate to a whole integer.

 

next, Kurzweg solves for ΞΆ and defines the search as between 0 and 1 (decimals), but I don't think that is the right direction to go with this.

ID: 3cab20 Oct. 8, 2018, 6:22 p.m. No.7731   πŸ—„οΈ.is πŸ”—kun

>>7729

We don't have to calculate g to determine whether g is closer to d or (d+1), we just have to see whether e or f is smaller. If f < e then Ξ± = (d+1), if f e then Ξ± = d

 

Kurwzeg has been treading on our territory since 2013 (most recent paper on factorization from what I've seen so far was in 2015).

ID: 3cab20 Oct. 9, 2018, 12:01 a.m. No.7737   πŸ—„οΈ.is πŸ”—kun

Ulrich H Kurzweg factorization papers list (to eliminate confusion) (they are all describing different algorithms and approaches):

 

November 1, 2012 - (Factorizing-Semiprimes-with-the-Integer-Spiral.pdf)

 

August 2013 - the search for 2(x+n)^2 or 2(d+n)^2 paper (A-New-Method-for-Factoring-Large-Semiprimes.pdf)

 

Christmas (December?) 2013 - the search for 2(d+n) [N(f(N)] paper (A-Quick-Way-to-Factor-Large-Semiprimes.pdf)

 

January 2015 - d, e and f paper, (A-New-Approach-to-Factoring-Semiprimes.pdf) (hereby renamed the zeta search paper since it is a search for the value of a variable I've renamed to zeta ΞΆ for our purposes)

 

August 2015 - Factorizing with Q primes (Factoring-Large-Semiprimes.pdf)

 

April 7, 2016 - (UNIVERSAL-CURVE.pdf)

 

Who knows where he's gotten by now.

ID: b129f9 Oct. 11, 2018, 1:03 a.m. No.7741   πŸ—„οΈ.is πŸ”—kun   >>7742

Kurzweg contacted.

 

Location of forum and of us was not revealed, only mathematical concepts shared in order to prove we are serious and are making enough progress (even to the point of improving on current research).

 

Pursuit of more of his research and his expected interest in our research was made clear.

GAnon !Nx57Pyux3E ID: b542af Oct. 11, 2018, 8:46 p.m. No.7743   πŸ—„οΈ.is πŸ”—kun   >>7744 >>7745 >>7747

>>7742

Made a new python script to look at records. It has way more features.

 

Now you can:

  • view d, x, a, c, (d+n), (x+n), f, t, grids. (f grids look cool they're like a spiral)

  • track and highlight records

  • mark any cell on the grid

  • view all e,n,d,x,a,b,c,f,x+n,d+n in the list view on the right

 

I highly recommend it. If it doesn't run on your computer or something I could help you out with the settings or something.

 

https://pastebi n.com/aZGz2RHF

GAnon !Nx57Pyux3E ID: b542af Oct. 11, 2018, 8:50 p.m. No.7744   πŸ—„οΈ.is πŸ”—kun   >>7745

>>7743

start it in a terminal and make sure the window is big enough or resize the settings or change the terminal font size. If you want to learn type !help basic

Then from there there will be a tutorial type path to go !help commands or !help map

Really cool tool

VQC !!cJ17mYdx.I ID: 0e79d3 Oct. 12, 2018, 3:30 p.m. No.7746   πŸ—„οΈ.is πŸ”—kun   >>7790 >>7795 >>7803 >>8015 >>8276

It is a lookup via the x values in row 1.

It is about finding how to make the lookup with column -f and e.

Something about the product of squares and triangles. I(t) stands out. Once you c it, you cannot un-c (i)t.

In The End, it is a lookup.

No more secrets.

Sneakers.

GAnon !Nx57Pyux3E ID: b542af Oct. 12, 2018, 4:56 p.m. No.7749   πŸ—„οΈ.is πŸ”—kun   >>7801

>>7747

 

x does x values, shift +x shows -x values. I noticed they were always the same. I thought about consolidating them to one button but I figured people should know.

 

Here is the record

 

9, 17, 6, 5, 1, 45

 

If we go along the n=1 row for -x of our original record, we get this pattern on the bottom row.

 

c = 65, 84, 105, 128, 153, 180, 209, 240, 273, 308, 345, 384

change in c = 19, 21, 23, 25, 27, …+2, +2, +2,

at e = 9, the same e as the original record, the value is 153.

 

153 - 128 = 25

180 - 153 = 27

so I guess the derivative at that point is 26.

 

Otherwise e increments by 2, n doest move, d increments by 1 and so does a and b.

t, f, and x+n seem to be constant (d+n) increments by 1.

 

 

 

Also people there are still some bugs I'll fix them.

Anonymous ID: 930093 Oct. 13, 2018, 1:39 a.m. No.7751   πŸ—„οΈ.is πŸ”—kun   >>7792

Just a quick look:

 

c287

X = 15, N = 128

(X+N) = 143, U = 2556, (X+N)^2 = 20449

(X-N) = 113, U = 56, (X-N)^2 = 12769

f = 2, e = 31

 

(31, 1, 7) -{31:1:113:13:100:128}

(31, 1, 8) -{31:1:143:15:128:160}

 

(-2, 1, 8) -{-2:1:111:14:97:127}

(-2, 1, 9) -{-2:1:143:16:127:161}

 

|

 

(31, 1, 0) -{31:1:15:-1:16:16}

(31, 1, 1) -{31:1:17:1:16:20}

 

(-2, 1, 1) -{-2:1:-1:0:1:-1}

(-2, 1, 2) -{-2:1:3:2:1:7}

 

287 = 7 * 41, x = 9, n = 8

(x+n) = 17, u = 8, (x-n)^2 = 289

(x-n) = 1, u = 1, (x-n)^2 = 1

 

The patterns say, look, I, the solution am here! But can we bring it out?

GAnon !Nx57Pyux3E ID: b542af Oct. 13, 2018, 9:53 a.m. No.7789   πŸ—„οΈ.is πŸ”—kun

>>7752

>>7753

>>7754

>>7755

>>7756

>>7757

>>7758

>>7759

>>7760

>>7761

>>7762

>>7763

>>7764

>>7765

>>7766

>>7767

>>7768

>>7769

>>7770

>>7771

>>7772

>>7773

>>7774

>>7775

>>7776

>>7777

>>7778

>>7779

>>7780

>>7781

>>7782

>>7783

>>7784

>>7785

>>7786

>>7787

>>7788

 

Made a big string of posts on my grid thread. I think it's interesting that the get 7777 was the introduction of the f grid. Not planned.

I hope you guys enjoy and learn!!

AA !dTGY7OMD/g ID: 8b4129 Oct. 13, 2018, 11:45 p.m. No.7794   πŸ—„οΈ.is πŸ”—kun   >>7801

>>7793

This is explainable with algebra. If d=t, depending on the parity of e, 2t will be within 1 of 2d. The square root of a number is very close to the square root of itself plus or minus 1.

AA !dTGY7OMD/g ID: 8b4129 Oct. 13, 2018, 11:57 p.m. No.7797   πŸ—„οΈ.is πŸ”—kun   >>7798 >>7801

Since I couldn't find any triangle visualization tools written in Java, I've been putting one together myself. It isn't finished. If anyone did already write one I must have missed it. Nobody will probably use this so it doesn't matter. Here it is, anyway. https://pastebin.com/3nDs75dy It relies on this GridCell object https://pastebin.com/HEC87GAQ

AA !dTGY7OMD/g ID: 8b4129 Oct. 14, 2018, 5:25 a.m. No.7799   πŸ—„οΈ.is πŸ”—kun   >>7801

>>7798

With colours now (that was fucking horrendous to get my head around): https://pastebin.com/4CE3tirj

I just wasted another 8 hours straight doing something that probably isn't going to solve the thing. I haven't even eaten dinner and it's already past midnight. I have a problem, and I think the only way I'm ever not going to keep doing this is either if I quit altogether or if we finally solve it. I don't know which of those is going to come sooner than the other.

PMA !!y5/EVb5KZI ID: 4fbd64 Oct. 14, 2018, 12:13 p.m. No.7802   πŸ—„οΈ.is πŸ”—kun

Was reviewing (0,n) in terms of squares and triangles.

 

Pics attached for c144 are the (0,n) entry record, it's na record in (0,1), and an annotated view of (0,1) in terms of the a and d formulas.

PMA !!y5/EVb5KZI ID: 4fbd64 Oct. 14, 2018, 12:34 p.m. No.7803   πŸ—„οΈ.is πŸ”—kun   >>7804

>>7746

Expanded the search into -x for the aan(n-1) hint.

 

c145 and c287 examples attached show how the aan(n-1) calculation for each factor record in positive x equals the product of (-f,1) a[t] and (e,1) a[t].

 

In negative x, however, these values differ as shown in the highlighted rows.

 

We could force these values to be equal by adjusting the x offset between the -f and e columns by 1. For example in c 145, aligning the x=-7 record with x=-8 instead of x=-6 would result in an aa product of 500 and match to the factor record.

 

This would require a rethinking of how -x records align between the -f and e columns.

 

On the other hand, this difference in aa product vs aan(n-1) from the factor record is correct and offers additional information to be analyzed.

PMA !!y5/EVb5KZI ID: 4fbd64 Oct. 14, 2018, 1:01 p.m. No.7804   πŸ—„οΈ.is πŸ”—kun

>>7803

Condensed output attached for c145, c287, and c6107 showing just the relevant factor and -x factor records.

 

Highlighted "d diff" columns represent difference between d values in -f and e columns for the solution factor records.

 

Add both highlighted values together and you will get 2x.

 

Which is the same "d diff" as the starting c record and it's -x offset.

 

For c145, where starting x=11.

 

c records "d diff" = 0 + 22 = 22

prime records "d diff" = 4 + 18 = 22

 

Similar for other examples, and makes the case that the aa product and aan(n-1) differences in -x are correct.

AA !dTGY7OMD/g ID: 8b4129 Oct. 14, 2018, 5:34 p.m. No.7812   πŸ—„οΈ.is πŸ”—kun   >>7873

I've been avoiding talking about this because I know I'm not going to get an answer that actually explains jack shit but what was the point of the rm2dnm1 thing from RSA #10? n0 is somewhere between f/8 and 1. We want some number between them that is greater than n and less than x+n. That divided f value is a staircase number. You put it into the rm2dnm1 function and it gives you a number that isn't 0 because it isn't going to equal the correct u value. We never seemed to do anything with that value aside from checking whether it was zero. What was the point of that entire thing? If n0 is never going to equal n, we're never actually going to use that function for the sake of checking that a number gives a remainder of 0 and is thus u. So we do want the remainder, and we do want a specific value from the staircase number, but what are we even using them for?

AA !dTGY7OMD/g ID: 8b4129 Oct. 14, 2018, 6:25 p.m. No.7813   πŸ—„οΈ.is πŸ”—kun   >>7814 >>7815 >>7817

Now you put a and b in through terminal and you can iterate through all cells with the same x+n. For even n odd x, it takes one away from n (to make it an odd square in the center) and adds the difference to f.

Triangles - https://pastebin.com/Kjm6jhZE

GridCell - https://pastebin.com/KRg2QTH9

AA !dTGY7OMD/g ID: 8b4129 Oct. 14, 2018, 6:41 p.m. No.7815   πŸ—„οΈ.is πŸ”—kun   >>7817

>>7813

>>7814

Max x+n is 649, by the way, because I needed to make it all actually fit on the screen so the minimum button size is one pixel. I don't know if that's big enough for what we're doing but VQC has never elaborated about what "big enough" is. It takes a very long time to load this since it has to make a maximum of 421201 squares and figure out whether they're fs or not (which was pretty complex, and works in a while loop iterating f by 8 each time). It's not very useful for iterating a and b at that scale (a-b=(x+n)*2 so to get the next (x+n) you just add or subtract one to a and b). In fact I haven't successfully waited through one with an x+n of 649. It might take a few hours. I don't know. x+n of 135 took a little over 10 seconds.

VA !!Nf9AmQNR7I ID: 9f86d9 Oct. 14, 2018, 7:29 p.m. No.7816   πŸ—„οΈ.is πŸ”—kun

>The best and most pragmatic advice I can offer is to use known numbers at scale as well as what you are doing.

>Reconcile an earlier post with a HUGE drop in BTC.

>Everything is connected.

 

And theeen you see that we're looking for the element in (e,1) where x + (e a value) - (-f a value) = original d value.

for c145 here's the calc:

original d value = 12

(e na transform) x value = 11

at x = 7, we have the following calc: x= 7 + (25) - (20) = 12 = original d value.

which gives us d = x + a, 12 = 7 + 5

 

So my idea is to use the original d value as the target that we're trying to make x+(e a)-(-f a) add up to. Have we already done this lads? Just thinking the clue through again and working it through.

AA !dTGY7OMD/g ID: 8b4129 Oct. 14, 2018, 8:14 p.m. No.7817   πŸ—„οΈ.is πŸ”—kun   >>7818

>>7815

>>7813

Here's a version that displays the (d+n)(d+n) square beside it in the fashion of VQC's diagrams. It doesn't work with very big numbers, or numbers that are quite far apart. That's because of how much larger (d+n)(d+n) usually is than (x+n)(x+n) (pic two is a good example of why it might not be super useful for bigger numbers (note that this is before I scaled them the same)). I might turn it into a bitmap generator at some point (I would be super grateful if someone else did but I know nobody will). So while this working with big numbers would obviously be ideal, it seems like the only way we'd get it working is with bitmaps. But I'm not sure.

 

(x+n)(x+n) and (d+n)(d+n) visualization in Java - https://pastebin.com/TV4Ud7j7

AA !dTGY7OMD/g ID: 8b4129 Oct. 14, 2018, 9:03 p.m. No.7818   πŸ—„οΈ.is πŸ”—kun   >>7819 >>7820

>>7817

Now you can view the na transform and bign cells (if their x+ns are odd), and go back to the original. https://pastebin.com/rBEFuU05

Would definitely not recommend the bign button for large numbers (because of what I explained in that quoted post about (d+n)(d+n)) unless you remove the (d+n)(d+n) functionality, but even then it depends on the (x+n)(x+n) size.

AA !dTGY7OMD/g ID: 8b4129 Oct. 14, 2018, 9:45 p.m. No.7827   πŸ—„οΈ.is πŸ”—kun   >>7828

>>7825

>>7824

>>7826

"when something happens that justifies the amount of time I've put into this" != proof. If I didn't think this was real, why would I have spent all that time? That would be fucking moronic. What I mean is finding the solution. That would justify me destroying a friendship and completely neglecting university because I was spending too much time doing Grid Patterns.

{file.ext}BrotherofEbot**┬┬ Β―\_(ツ) ID: 83b459 Oct. 14, 2018, 9:50 p.m. No.7828   πŸ—„οΈ.is πŸ”—kun   >>7829

>>7827

I remember friends… be glad you only have VQC to sink into.

You want the solution… but you haven't earned it and we're not at a place where we need to have it finished yet.

 

What are you going to school for?

Something that will be meaningless when this is successful?

Even if you're going for something that will still exist… I'm an artist.

You have way more of an advantage by doing this, hopefully.

{GR8TST.GE}GrampaEbot***(γƒŽΰ² η›Šΰ² )γƒŽε½‘β”»β”β”» ID: 83b459 Oct. 14, 2018, 9:55 p.m. No.7830   πŸ—„οΈ.is πŸ”—kun

I'm only as good as what I have to work with.

Want me to send you money for a whore?

I'd… I might throw some cash at that, but I need proof.

No Proof = Refund.

Anonymous ID: 43930d Oct. 14, 2018, 10:55 p.m. No.7831   πŸ—„οΈ.is πŸ”—kun   >>7833 >>7841

c6107

{23:2976:78:77:1:6107} (23, 2976, 39)

{23:1:18641629:-6107:18647736:18635524} (23, 1, -3053)

{23:1:18653843:6107:18647736:18659952} (23, 1, 3054)

 

{-134:1:18647657:-6108:18653765:18641551} (-134, 1, -3053)

{-134:1:18659873:6108:18653765:18665983} (-134, 1, 3055)

 

Ξ± = 18641607

Ξ² = 9322268

Ξ³ = 18647657

Ξ΄ = 18647712

 

Ξ³ - Ξ± = 6050

Ξ΄ - Ξ³ = 55

Ξ³ - Ξ± - c = -57

 

Ξ± - (-134, 1, -3053)[b] = 56

Ξ± - (-134, 1, -3053)[a] = -12158

(Ξ± - (-134, 1, -3053)[a])/2 = -6079

Ξ± - (23, 1, -3053)[b] = 6083

Ξ± - (23, 1, -3053)[a] = -6129

Ξ± - (23, 1, -3053)[b] - Ξ± - (23, 1, -3053)[a] = -46

 

6107[x] = 47

Anonymous ID: 43930d Oct. 14, 2018, 10:58 p.m. No.7832   πŸ—„οΈ.is πŸ”—kun   >>7841

c34117

{261:16875:184:183:1:34117} (261, 16875, 92)

{261:1:581950858:-34117:581984975:581916743} (261, 1, -17058)

{261:1:582019092:34117:581984975:582053211} (261, 1, 17059)

 

{-108:1:581984790:-34118:582018908:581950674} (-108, 1, -17058)

{-108:1:582053026:34118:582018908:582087146} (-108, 1, 17060)

 

Ξ± = 581950598

Ξ² = 290983839

Ξ³ = 581984790

Ξ΄ = 581984713

 

Ξ³ - Ξ± = 34192

Ξ΄ - Ξ³ = -77

Ξ³ - Ξ± - c = 75

 

Ξ± - (-108, 1, -17058)[b] = -76

Ξ± - (-108, 1, -17058)[a] = -68310

(Ξ± - (-108, 1, -17058)[a])/2 = -34155

Ξ± - (261, 1, -17058)[b] = 33855

Ξ± - (261, 1, -17058)[a] = -34377

Ξ± - (261, 1, -17058)[b] - Ξ± - (261, 1, -17058)[a] = -522

 

34117[x] = 75

 

This hasn't been ironed out, as one can see.

Anonymous ID: dc77e5 Oct. 14, 2018, 11:13 p.m. No.7834   πŸ—„οΈ.is πŸ”—kun   >>7835 >>7838

I give you guys presents but I'm not sure anyone opens them.

The only one who takes me seriously is VQC.

I guess it's because I connect dots in a non-linear fashion.

And because I have the barest understanding of the math you guys do.

You guys are amazing.

Seriously. You guys are smart as hell.

But it takes all kinds to complete the puzzle.

I need you.

And you need me.

My "crumbs" will stand the test of time.

Of that I am certain.

My mathematical intuition has never failed me.

I have the box, but not the key.

I cannot open them by myself.

So here are more presents for those anons who wish to open them.

May you find some inspiration!

 

#1.

>https:// medium.com/@thoughttheory/lets-see-how-deep-the-rabbit-hole-goes-9af5f9896941

=

>http:// archive.is/64L2h

#2.

>https:// www.academia.edu/7489568/The_pythagorean_relationship_between_Pi_Phi_and_e

AA !dTGY7OMD/g ID: 8b4129 Oct. 14, 2018, 11:58 p.m. No.7838   πŸ—„οΈ.is πŸ”—kun

>>7834

>I give you guys presents but I'm not sure anyone opens them.

>I have the barest understanding of the math you guys do.

I don't mean to be patronizing, but, I mean, duhhhh. I don't know who you are because your IP hash only has three posts associated with it, but chances are if you're complaining about nobody reading your posts because they aren't the same kind of math that the rest of us are doing, it's because you haven't done the following:

 

(a) read all of the threads so you understand the grid (it's not difficult, it's just time consuming - I was going to work on a simple guide in June (before that too) but nobody ever wanted to help me so it never happened). VQC explained it pretty well at the beginning of the first thread. We're finding a and b, the factors of c, c being any arbitrary number of your choice. c has a bunch of variables associated with it. You can grid these sets of variables with e as the x axis and n as the y axis, and they create infinite sets. Relationships between cells of variables (different a and b pairs) can be used to factor any arbitrarily large number. Understanding the multitude of patterns will lead us to the solution. Go read the threads. You don't turn up to a calculus class with a chemistry textbook and pass the class, man.

 

(b) figure out just how related your math is to the grid - if VQC is taking you seriously, maybe it's important stuff, but if you don't even understand the grid, how do you actually know it's related? Intuition is a strong thing, but you have to back your shit up with facts, anon. If we're all talking about (x+n)(x+n) and you're trying to relate the golden ratio to the Mandelbrot set, we're in completely different worlds. Apply it to the concepts we're all studying or we're not going to have a use for it (at least with our current objective).

 

(c) now that you'll know how the math we're doing works, you'll have the terminology and the context to apply your math to it in such a way that you can explain where you're coming from. We're working explicitly on what VQC is trying to explain to us. If you post something that seems like it doesn't have anything to do with that (like weird graphs with lots of geometric shapes in seemingly arbitrary places like that one anon), obviously you're not going to get very much attention. If it is related, explain in terms that we understand (i.e. with our math). Let's say you found out that f divided by d is equal to a (that's definitely not true but it's an example). Let's say you found it because you're super into geometry and you were messing around with the grid and found this weirdly shaped line. If you didn't understand the grid (which you said you don't), you'd be telling us "hey guys, there's this weird shape here and you really need to know about it" then all you're doing is showing us a weird line. What are we meant to do with that? We've seen lots of weird lines and none of them got us anywhere. If you understood the grid, you would be able to say "hey guys, f/d=a, isn't that neat". Then we'd actually know what you're talking about.

 

tl;dr: read the threads

Anonymous ID: 3cab20 Oct. 15, 2018, 6:15 p.m. No.7845   πŸ—„οΈ.is πŸ”—kun

I've got a new idea and it's simpler than what I was working on earlier. It's putting the remainder tree and the grid together and it also unifies the "the solution is recursive" and "the grid does all the work for you" hints.

Anonymous ID: 3cab20 Oct. 15, 2018, 6:28 p.m. No.7847   πŸ—„οΈ.is πŸ”—kun

you start at the odd column (it's either f or e), where t=-(d+N-1) then you take sqrt(2d) and it equals c, then you take sqrt d and it gives you another value and then you set that value as x and it will send you down the column, and then you take sqrt(d) again and use that as x, then keep doing that, sometimes it'll have you switch columns

Anonymous ID: 3cab20 Oct. 15, 2018, 8:37 p.m. No.7850   πŸ—„οΈ.is πŸ”—kun   >>7851

It can't factorize large semiprimes yet. I think it's missing something.

 

Interestingly it instantly terminates when you enter a prime, ie it takes one step to terminate on known rsa number factors. So if I can find the missing piece of this it could be a prime test too.

AA !dTGY7OMD/g ID: 8b4129 Oct. 15, 2018, 9:07 p.m. No.7851   πŸ—„οΈ.is πŸ”—kun   >>7852

>>7850

Without seeing your code, it doesn't take into account

>(e,1)

>(0,n)

>triangles

>diagonals

etc so if you generate triangle diagrams and output the respective (e,1) cells, just as an example, during your process, it might display some patterns.

AA !dTGY7OMD/g ID: 8b4129 Oct. 15, 2018, 10:05 p.m. No.7853   πŸ—„οΈ.is πŸ”—kun

>>7852

Well for a start, if you're starting in whichever column is odd, chances are it's different for odd e and even e, so you might just have to add one or divide by two somewhere. Just in general, you aren't utilizing every concept. While (0,n) and (e,1) might not be used in one particular solution, VQC said the three solutions he's aware of involve either (0,n) or (e,1) or diagonals. So as much as you could potentially find another solution, I think it's far more likely that you've found something that you're meant to use in conjunction with the other concepts. So maybe if you run your algorithm and every time you calculate a new cell you also find that cell's (e,1), for example, it'll have some important value in it, or its triangle configuration will have something in common with the solution cells, or something. My point is to start analyzing this at a deeper level. One other idea is that you could work backwards. If you're taking a cell and setting the square root of its d to x, you could take xx and xx+1 as potential d values in the cell that leads to the solution cell.

ID: 3cab20 Oct. 15, 2018, 11:21 p.m. No.7854   πŸ—„οΈ.is πŸ”—kun

Here's the code.

 

https://pastebin.com/sEse8cxm

try exploring setting x = f or f-1 (depending on whether the column is even and odd), and try exploring setting x to root of f and root of e. Also, we should figure out what a diagonal is.

 

Also, I have no idea how to mix 0,n into this.

ID: b129f9 Oct. 16, 2018, 12:35 a.m. No.7855   πŸ—„οΈ.is πŸ”—kun

There was a "ghost in the machine," so to speak, as to the reason why some of the records are invalid in the screenshots, but the code has been corrected to reflect the original idea of the movement from (-(odd column), 1, -(d+N)) to (col, 1, -(d+n). Also, it wasn't terminating in one step, I just forgot to print out the movement, but Chris has already said an instant prime test will be included with the solution.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Oct. 16, 2018, 12:38 a.m. No.7856   πŸ—„οΈ.is πŸ”—kun

>>7800

Grazie grazie.

I told Olga Vishnevsky that I'd promote her whenever I could.

 

I was almost tempted to hit up my The Professor to give him a update.

Buuuuut he'll find out eventually.

 

After all, someone needs to teach the Maths + Arts class we were talking about before this all started.

GAnon !Nx57Pyux3E ID: b542af Oct. 16, 2018, 4:16 p.m. No.7857   πŸ—„οΈ.is πŸ”—kun

Slightly off topic, but I think that he chose these variables for a reason. I think that the N in supposed

to represent the natural numbers. I say this because for e=0, you can multiply any record by a constant and it will represent another record.

 

On a side note, is this the offset we're looking for?

 

Notice how for the records e = 85, it is the origin of the parabola, where in the f values, the origin of the parabola is in e=84.

They're offset by one

ID: b129f9 Oct. 17, 2018, 1:21 a.m. No.7858   πŸ—„οΈ.is πŸ”—kun

VQCGUI revision 3, includes source

https://anonfile.com/37C6mej1b4/VQCGUI-0.0.3_7z

 

try navigating from a[t] = N record to a[t] = na record, you will see amazing patterns in the values on the right.

 

sqrt(2d) has an intricate relationship with the x values.

 

(yes I know there's a typo, too late to recompile).

ID: b129f9 Oct. 17, 2018, 9:10 a.m. No.7860   πŸ—„οΈ.is πŸ”—kun

I will come back to clarify the other by the end of the week but briefly. Using the values of RSA 100 just as a large number example is a suggestion. In column e, the values at (e,1) for each element includes a value for a[t] which is an. The value bn is at a[t+n]. At (-f,1) the value a(n-1) is at the same value x in that cell as "an" for (e,1), the value b(n-1) at cell (-f,1) is one element less than bn at (e,1). This difference is key. It is not the only key, as you are seeing. I have found three, not including yours. Those three are in row 1, column zero and the side by side diagonal cells from the origin. There may be infinite keys.

 

Two sets of equations that run together and then merge, lock and key.

ID: b129f9 Oct. 17, 2018, 11:55 a.m. No.7864   πŸ—„οΈ.is πŸ”—kun

//using the diagonal to find pairs that are one row apart

 

c1541

e=20, f = 59

sqrt(e) = 4, sqrt(f) = 7

sqrt(f) - sqrt(e) = 3

 

2d = e + f - 1

 

{20:732:39:38:1:1541} i[t] = 771; j[t] = 770; sqrt(d[t]) = 6; sqrt(2d[t]) = 8; f[t] = 59

{-59:731:40:39:1:1541} i[t] = 771; j[t] = 770; sqrt(d[t] = 6; sqrt(2d[t]) = 6; f[t] = 140

 

140 - 59 = 81

difference between f[t] in (e,N, t) and (e, N-1, t) = 81

 

{20:6:39:16:23:67} = 1541; i[t] = 45; j[t] = 22; sqrt(d[t]) = 6; sqrt(2d[t]) = 8; f[t] = 59;

{-59:5:40:17:23:67} = 1541; i[t] = 45; j[t] = 22; sqrt(d[t]) = 6; sqrt(2d[t]) = 8; f[t] = 140;

 

140 - 59 = 81

sqrt(140) - sqrt(59) = 4

difference between f[t] in (e,n, t) and (e, n-1, t) = 81

 

{20:1:770:38:732:810} i[t] = 771; j[t] = 39; sqrt(d[t]) = 12; sqrt(2d[t]) = 17; f[t] = 1521; c[t] = (d+N)^2 - d^2

{-59:1:770:39:731:811} i[t] = 771; j[t] = 40; sqrt(d[t]) = 27; sqrt(2d[t]) = 39; f[t] = 1600; c[t] = (d+N-1)^2 - d^2

 

1600 - 1521 = 79

sqrt(1600) - sqrt(1521) = 1

 

{20:1:154:16:138:172} f[t] = 289; c[t] = (d+n)^2 - d^2

{-59:1:132:17:115:151} f[t] = 324; c[t] = (d+n-1)^2 - d^2

 

324 - 289 = 35

sqrt(324) - sqrt(289) = 1

ID: b129f9 Oct. 17, 2018, 2:12 p.m. No.7865   πŸ—„οΈ.is πŸ”—kun

I'm making the most progress working backwards, seeing that it's not about going from c to n, it's about going from d and e to n and N.

 

"Two sets of equations running together."

ID: b129f9 Oct. 17, 2018, 2:35 p.m. No.7866   πŸ—„οΈ.is πŸ”—kun

since f=(x+1)^2 at (e,1), f increases in squares by t of the opposite parity of e. if e is even, f increases in odd squares, and if e is odd, f increases in even squares.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Oct. 19, 2018, 2:21 a.m. No.7869   πŸ—„οΈ.is πŸ”—kun   >>7870

From the Discord:

 

how much effort do you really need to put into a tree?

they kinda… just… do on their own… y'know?

you can go all bonzai if you like, or train 'em to make bridges…

but a tree's gonna do it's do whether you do your do or not

it's like a…. fractal… binary… self-replicating algorithm.

 

so where do you plant the seed?

let's say you have an effect or decision in this

just… y'know… find yourself in such a position

you have the seed, and the ground… plus the environment

but let's say this is a tree that can grow anywhere and will adapt to every situation

where do you want it?

Mmmmmmmminecraft.

 

Basically, you could put a palm tree on the top of mount everest.

er… no

it would take the form of whatever it takes to survive right there

you're not like… GOD all of a sudden

you get that seed into the ground, the Tree will do its do… however it HAS to do and ONLY that way way in such an EXTREEM environment

the "tree" at the bottom of the marianas trench has gotta look nutz

ooooor the one on mars. or jupiter. or the sun. or a comet.

Just a matter of getting The Seed there.

 

Ham Fisted Terraforming

so whaaaaat is the Seed "plugged into"?

a server? an id? a portfolio of accounts? a control panel of some sort?

math is sexy, but in this analogy… what is the vqc and what is the encryption? what's the seed, ground, and… is the VQC the tree? The Ground?

Is The Grid the ground that we put The Seed into to see The Tree produced by the seed?

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Oct. 19, 2018, 3:01 a.m. No.7870   πŸ—„οΈ.is πŸ”—kun

>>7869

Or is it like this?:

The Grid is the The Environment.

The Ground/Fertile Soil is the VQC.

The Seed is the Encrypted… Thingy.

Put The Seed into Ground to see The Tree.

The End is a Forest of Trees.

 

Maybe the "Leaves" are the "Remainders" or something.

VA !!Nf9AmQNR7I ID: bd5068 Oct. 20, 2018, 12:31 p.m. No.7876   πŸ—„οΈ.is πŸ”—kun   >>7877 >>7879

>>7873

AA, this is the best I summary I have from my work on the (f-1)/8 method:

 

How to find the two adjacent Triangle numbers: Take the SQRT of (f-1)/8. This will be the larger of the two Triangle numbers we need, let's call it T2. The smaller staircase number, T1, is simply the T1- 1. Here's an example. For c6107, (f-1)/8 = 16. SQRT 16 = 4. So our first staircase number T2 is 4, and T1 is 3. T2 + T1 = 10 + 6 = 16.

 

My question is what to do with the remainders. We have remainder from (f-1) mod 8, and sometimes we also have an additional remainder from SQRT ((f-1)/8). So we end up with our two staircase numbers, and at least one remainder to deal with, sometimes two remainders.

 

The idea is that those two triangle numbers multiply, and that multiples of them will equal our correct (x+n) square. The cool thing about this is that as you work outward from the center of the x+n square, the pattern grows in a very orderly fashion.

 

I could use some help trying to figure out how to handle the two sets of remainders, but VQC said that we had come VERY close to completing the pattern, which would then lead to the grid shortcut.

VA !!Nf9AmQNR7I ID: bd5068 Oct. 20, 2018, 12:40 p.m. No.7877   πŸ—„οΈ.is πŸ”—kun

>>7876

>The smaller staircase number, T1, is simply the T1- 1

Typo here lads. Should read:

The smaller staircase number, T1, is simply T2- 1. (the next smallest triangle number)

VA !!Nf9AmQNR7I ID: bd5068 Oct. 20, 2018, 12:48 p.m. No.7878   πŸ—„οΈ.is πŸ”—kun   >>7879

>>7873

Here's one almost complete, but still trying to figure out the edges like VQC asked. Not sure how the remainders play into completing the edges. There must be a way that fills the x+n square perfectly, once we know how to correctly use or dismiss the remainders. Oops, I see a symbol in there, wonder where all those ancient cultures got that good luck symbol from?

AA !dTGY7OMD/g ID: 8b4129 Oct. 20, 2018, 4:19 p.m. No.7879   πŸ—„οΈ.is πŸ”—kun   >>7880 >>7881

>>7876

>>7878

What does that picture with the mini yellow and green triangles represent? Is it c? All I've ever seen going through past threads is the picture itself and not an explanation, so I haven't been taking note of it since I have no idea what it is. It doesn't look like (x+n)(x+n) because of the whitespace. But if it's c, I haven't ever seen anyone break it down into smaller parts that aren't dd+e.

VA !!Nf9AmQNR7I ID: 6fe22c Oct. 20, 2018, 9:54 p.m. No.7880   πŸ—„οΈ.is πŸ”—kun   >>7882

>>7879

Hey AA! The diagram is the area of (x+n)^2. The two triangle numbers are represented by yellow (T2) and green (T1). The blue units are the remainders from (f-1) mod 8.

 

I think the key is adding the two bases of the staircase triangle numbers together, so in this case it's 4+3=7. For c6107, the correct u is 42. So in this example, 7 * 6 = 42. So the correct u value is a multiple of 7. In this case, u and u1 are 42 and 41, which add to 83, which is the correct x+n value. The correct area of this x+n square is 83^2 = 6889

 

I think the rm(2d-1) comes into play here by being able to give us a lock on the correct multiple of u that gives us a perfect square, 6889.

 

Here's the breakdown from VQC:

>>6185

>The solution to this problem introduces a new form of algebra where two concurrent forms of equations run side by side and then merge.

>Take some time to think about what that means.

>Have you seen this approach before?

>What could it look like?

>Many new solutions often seem obvious in hindsight.

>In fact many new designs seem to simplify in many varied approaches to design.

>Does it seem obvious in hindsight that in order to solve a multivariate equation, that something new but similar to what we have always done, would be the solution? Just taken in a new direction? Expanded thinking.

>The two sets of equations take the problem and simplify it. Together they handle the "lock and key" nature of the problem/solution, particularly when c is divisible by 1, c, and two other prime numbers, such as in RSA.

>The two forms of equations that merge together handle staircase numbers where the base of one staircase, is one unit longer than the other.

>The analogy to Fermat's Last Theorem. Two objects that seem to be completely unrelated were proved to be the same object.

>Two seemingly different 'fields' will be used side by side and merged to create an elegant solution. Again, this can only be done in reverse, using the assumption that P=NP. I only saw at the end of over seven years work. Anons have got much further in six months than I would have. I would have walked away in frustration back in the day but you anons here have been amazing.

>Take some time to think what this will look like and how the diagrams (especially animated) might show this.

>I think things will start moving quickly.

>This will be new mathematics. It will make more sense than how this problem has been approached up until now.

VA !!Nf9AmQNR7I ID: 279dad Oct. 20, 2018, 11:32 p.m. No.7881   πŸ—„οΈ.is πŸ”—kun   >>7882 >>7888 >>7922

>>7879

Here's the steps for a test program:

start with c

get d, e, and f

calc (f-1)/8

calc SQRT((f-1)/8) = base T2

define base T1 as T2-1 = T1

T1 + T2 = potential u value

iterate potential u value looking for lock using rm(2d-1) formula, basically looking for a perfect square.

hopefully find a lock for the correct (x+n)^2 value.

 

Basically the correct u value should be a multiple of the combined bases of T2 and T1. Remainders may need to be accounted for as well, not sure yet.

GAnon !Nx57Pyux3E ID: 6feebe Oct. 21, 2018, 3:07 p.m. No.7886   πŸ—„οΈ.is πŸ”—kun   >>8080

>>7885

Heres all the info I could figure out for the A grid. It looks like at most points there are at least like 5 dimensions to work into. Horizontal movement, diagonal, parabolic (x2), parabolic again if at the origin, then increasing t values for a given cell. I want to do these for every grid. The e+2a, n+1 movement is common though so I wanted to do the A grid first.

ID: b129f9 Oct. 21, 2018, 4:43 p.m. No.7891   πŸ—„οΈ.is πŸ”—kun

Also, 2 values of c at (e,1) (a different idea)

 

2 values of c in both columns in -x and +x

 

145:

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

{1:1:10368:-145:10513:10225} (1, 1, -72)

{1:1:10658:145:10513:10805} (1, 1, 73)

 

{-24:1:10500:-146:10646:10356} (-24, 1, -72)

{-24:1:10792:146:10646:10940} (-24, 1, 74)

 

34117:

{261:1:581950858:-34117:581984975:581916743} (261, 1, -17058)

{261:1:582019092:34117:581984975:582053211} (261, 1, 17059)

 

{-108:1:581984790:-34118:582018908:581950674} (-108, 1, -17058)

{-108:1:582053026:34118:582018908:582087146} (-108, 1, 17060)

 

rsa100c:

{61218444075812733697456051513875809617598014768503:761302513961266680267809189066318714859034057480651323757098846024549192056136964455981270339102876:39020571855401265512289573339484371018905006900194:39020571855401265512289573339484371018905006900193:1:1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139} (61218444075812733697456051513875809617598014768503, 761302513961266680267809189066318714859034057480651323757098846024549192056136964455981270339102876, 19510285927700632756144786669742185509452503450097)

{61218444075812733697456051513875809617598014768503:1:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306382919000431748126203278565987540269446653259579647945903160561369722164795555272230598393346522221773:-1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306384441605459670659563814184365672906876371327694609326591818469864302287758814225128252393697214227912:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306381396395403825592842742947609407632016935191464686565214502652875142041832296319332944392995830215636} (61218444075812733697456051513875809617598014768503, 1, -761302513961266680267809189066318714859034057480690344328954247290061481629476448827000175346003069)

{61218444075812733697456051513875809617598014768503:1:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306385964210487593192924349802743805544306089395809570707280476378358882410722073178025906394047906234051:1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306384441605459670659563814184365672906876371327694609326591818469864302287758814225128252393697214227912:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306387486815515515726284885421121938181735807463924532087969134286853462533685332130923560394398598240192} (61218444075812733697456051513875809617598014768503, 1, 761302513961266680267809189066318714859034057480690344328954247290061481629476448827000175346003070)

 

{-16822699634989797327123095165092932420211999031886:1:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306384441605459670659563814184365672906876371327694609287571246614463036775469240885643881374792207327717:-1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006140:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306385964210487593192924349802743805544306089395809570668259904522957616898432499838541535375142899333857:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306382919000431748126203278565987540269446653259579647906882588705968456652505981932746227374441515321579} (-16822699634989797327123095165092932420211999031886, 1, -761302513961266680267809189066318714859034057480690344328954247290061481629476448827000175346003069)

{-16822699634989797327123095165092932420211999031886:1:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306387486815515515726284885421121938181735807463924532048948562431452197021395758791439189375493591339997:1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006140:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306385964210487593192924349802743805544306089395809570668259904522957616898432499838541535375142899333857:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306389009420543438259645421039500070819165525532039493429637220339946777144359017744336843375844283346139} (-16822699634989797327123095165092932420211999031886, 1, 761302513961266680267809189066318714859034057480690344328954247290061481629476448827000175346003071)

Anonymous ID: 47ad18 Oct. 22, 2018, 12:53 a.m. No.7901   πŸ—„οΈ.is πŸ”—kun   >>7906 >>8508

a prime number appears once (2^(1-1)) in a column.

 

a product of two primes appears twice (2^(2-1)) in a column.

 

a product of three primes appears four (2^(3-1)) times in a column.

 

a product of four primes appears eight (2^(4-1)) times in a column.

 

a product of five primes appears sixteen (2^(5-1)) times in a column.

Anonymous ID: 398ad7 Oct. 22, 2018, 1:43 a.m. No.7903   πŸ—„οΈ.is πŸ”—kun   >>7904 >>7905

Introducing the grid and the diagonals then moving to a series of odd x+n hints was not pointless. It's crucial the final step of the journey is solved organically and not given away.

 

The x+n hints are one solution and will lead to the grid solution.

 

It will all make sense in hindsight.qGncp

Anonymous ID: abe23c Oct. 22, 2018, 9:25 a.m. No.7907   πŸ—„οΈ.is πŸ”—kun   >>7909 >>7910

>>7906

I've been suspecting for a while that VQC has actually been dropping hints for all three keys, mixed in with each other. I believe the smooth numbers is part of one key, the x+n triangle is part of another and that some of his other hints belong to the third part.

Anonymous ID: abe23c Oct. 22, 2018, 9:33 a.m. No.7908   πŸ—„οΈ.is πŸ”—kun

I've been busy with life for a while, but I did have some realizations the past few weeks.

 

I can't believe I missed the sqrt(2d), it's obvious now, but it's how you find the triangular base of d. Take a triangle n(n-1)/2 = k. To find the triangular base you do 2k = n(n-1), then the square root will be sqrt(2k) = sqrt(n(n-1)). Floor this number and you will have (n-1).

Anonymous ID: 473c10 Oct. 22, 2018, 8:43 p.m. No.7915   πŸ—„οΈ.is πŸ”—kun   >>7916 >>7921 >>7922 >>7928 >>7948

c = 6107

d = 78

e = 23

f = 134

 

(d+n)^2 {

d^2 = 6084 //first square in structure

 

f is divided into two pieces, 67+67

e is divided into two even pieces with one left over, 11+11+1^2

d^2 + e + f = 6141 //next square in structure

 

(n^2 - 1) + last piece of e = n^2 //next square in structure

 

d^2 + e + f + (n^2 - 1) + 2d(n-1) = (d+n)^2 //the large square

}

 

(x+n)^2 {

f is divided into two even pieces with two pieces, one and one left over

 

(n^2 - 1) + one of the leftover pieces of f = n^2 //first square in structure

(n^2 - 1) + f + 2d(n-1) = (x+n)^2 //the smaller square

}

GAnon !Nx57Pyux3E ID: 6feebe Oct. 23, 2018, 3:52 p.m. No.7917   πŸ—„οΈ.is πŸ”—kun   >>7918

>>7906

>>>7906

>This is interesting too. Look at xx + e = 2na

>If you set a = 1, then xx + e = 2c because n = c

 

>>7913

 

Sometimes this is the best way to view it. It's in a grid for a reason right? I threw out my old pc and lost that pic and I missed it sorely

VA !!Nf9AmQNR7I ID: 9f86d9 Oct. 23, 2018, 8:45 p.m. No.7922   πŸ—„οΈ.is πŸ”—kun   >>7923

>>7915

Nice work on this diagram, Anon!

 

Alright, I have an idea lads.

In the linked c6107 diagram, (n-1) = 35

Let's derive the staircase numbers, as I explained here: >>7881

 

(f-1)/8 = 16, T2 = 4 , T1= 3, T1+T2 = 7

The 2d(n-1) rectangles are 2730 + 2730, or 2 * 78 * 35

35 = (n-1)

T1 + T2 * iterate = correct (n-1)

7 * 5 = 35

So (f-1)/8 gives us the staircase triangle number bases to add together, and then multiply out looking for a lock. It will scale properly upward, even with huge c values.

Obviously we need to test some other examples, but deriving the T2 + T1 staircase numbers from f , and then using it to iterate/solve for (n-1) works in this c6107 example.

ID: b129f9 Oct. 24, 2018, 7:23 p.m. No.7932   πŸ—„οΈ.is πŸ”—kun   >>7933

Was watching some videos and reading some literature on the Riemann Zeta function today. It's really fascinating stuff. There's a lot of calculus and number theory behind it, but basically if the Riemann Zeta hypothesis is true, then you can calculate the exact number with no error of how many primes there are beneath a given integer.

 

You can already do this with the more accurate version of the prime number theorem and a formula that gives you the amount of error (calculated with the zeros of the Riemann Zeta function), but nobody is sure it's correct because it depends on the Riemann Zeta hypothesis.

Anonymous ID: e2aefa Oct. 25, 2018, 1:19 a.m. No.7939   πŸ—„οΈ.is πŸ”—kun

I was really shitfaced last night and I could have sworn there is a pattern of n being a factor of the a values in the cell at e,1 and that n-1 had a pattern of being a factor of the a values in the cell at -f,1

ID: b129f9 Oct. 25, 2018, 6:31 a.m. No.7943   πŸ—„οΈ.is πŸ”—kun

>>7942

If you look at the patterns of n in a[t] you find pairs of elements of the n we're looking for multiplied by different a's and b's for a different, smaller c, that are separated by n for that c's elements

ID: b129f9 Oct. 25, 2018, 8:27 a.m. No.7946   πŸ—„οΈ.is πŸ”—kun   >>7947

These particular patterns may be coincedence but this is the approach I'd like to be investigated:

 

c21428053:

N=10709398

n=38

 

So we want to find a way to find multiples of 38 in e,1 (work backwards).

In e,1 if we look for n and its multiples, we do find

 

{412:1:318:14:304:334} a = n*8 t=8

{412:1:470:22:448:494} a = n*13 t=13

8*13 = 104

104's n value = 5 and these elements are 5 elements apart.

 

So we find our n times a different set of a and b for a different c with elements that are that c's n apart at e,1.

Anonymous ID: abe23c Oct. 25, 2018, 1:10 p.m. No.7948   πŸ—„οΈ.is πŸ”—kun   >>7949

So it's been a while since I posted much, I've been busy looking over patterns in the grid, but with >>7915 something hit me. What if we've been doing this wrong. We're trying to triangulate (x+n)^2, but we only have f and 2d remainders right? What if we could increase it? What if we could add more information to our data?

 

I've attached an image that shows my idea. If we connect (d + n)^2 with (x + n)^2 at nn, we have a new square that contains more information that we have. The equation for this square is:

 

(d + n + x)^2 = nn + dd + 2d(n - 1) + 2d(n - 1) + f + 2 + f - 2 + 2dx

 

This can be simplified to:

 

nn + dd + 2d(2n + x - 2) + 2f + 1

 

In this "upper" square the dd will cover the entire outside of the triangle, meaning the triangle base will be smaller than the u of dd. Essentially, we have a new square and an upper limit of the triangle base of (x+n).

 

Another thing is that our dd square will now occupy one part of the new square, while xx will occupy the opposite side.

 

I've also been a bit huffing about the fractal VQC has been talking about, but now I think I'm starting to see it. If we combine the squares as in the attached image we can now see what could potentially be a fractal. I'm not entirely sure how one would map it / graph it, but it's a start.

 

I do feel like something clicked, but I'll be honest and say that we've all had that feeling before.

 

A legend for the image:

 

In the upper left corner you see the dd square with the f/2d extension on the side (n=6) along with the nn square shared between (d+n)^2 and (x+n)^2.

In the lower right corner you see xx and the extended f/2x around it.

 

As for what we're trying to do, which is find (x+n)^2, this will exist inside of this (d+n+x)^2 with nn in the center, think of it like our original (x+n)^2 = nn + 2d(n-1) + f - 2 equation, except this is inside of dd + 2d(n-1) + 2dx + f + 2.

AA !dTGY7OMD/g ID: 8b4129 Oct. 26, 2018, 12:50 a.m. No.7951   πŸ—„οΈ.is πŸ”—kun

Getting back into it a bit. I found a pattern in regards to f/8 being less than or greater than the x+n triangle base u. It seems that there are specific j values which always seem to create an f/8 value which is greater than u. In this image, the numbers are the j values which, together with every i within a range of i values, create a cell with an f/8 greater than its u. The i constraints are the blue text beside each block.

 

>>7950

HOW DO I TURN CAPS LOCK OFF?

Another joke answer would have been to say UTF-8. I don't know if you're VQC or someone else but I don't see how this question is meant to help other than in hindsight.

Anonymous ID: 2344ac Oct. 26, 2018, 3:56 a.m. No.7957   πŸ—„οΈ.is πŸ”—kun   >>7961 >>8015 >>8027

Some random food for thought:

 

c=ab

2 inputs

infinite solutions

infinite wrong solutions

1 right solution - two primes

encode solution not as a number

but as a pattern, algorithm, approach towards the number

ID: b129f9 Oct. 26, 2018, 10:48 p.m. No.7970   πŸ—„οΈ.is πŸ”—kun

What do we get when we sort numbers by their distance from a square instead of as a line?

 

2, 5, 10, 17, 26, 37, 50, 65, 82, 101

3, 6, 11, 18, 27, 38, 51, 66, 83, 102

4, 7, 12, 19, 28, 39, 52, 67, 84, 103

5, 8, 13, 20, 29, 40, 53, 68, 85, 104

6, 9, 14, 21, 30, 41, 54, 69, 86, 105

7, 10, 15, 22, 31, 42, 55, 70, 87, 106

8, 11, 16, 23, 32, 43, 56, 71, 88, 107

9, 12, 17, 24, 33, 44, 57, 72, 89, 108

ID: b129f9 Oct. 26, 2018, 11:59 p.m. No.7972   πŸ—„οΈ.is πŸ”—kun

[iterating d for e=31], relevant c=287

32

35

40

47

56

67

80

95

 

na = 56

 

so e,1 in our grid analyzing every value of d in d^2 + e for us. understanding why the grid is configured the way it is is a good exercise.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Oct. 27, 2018, 3:49 p.m. No.7979   πŸ—„οΈ.is πŸ”—kun   >>7989

>>7977

145=64+81

145=8^2 + 9^2

So… Polite Squares? Or would there need to be a third?

 

7,8->113

8,9->145

 

7,8,9->194

8,9,10->245

 

7,8,9,10->294

8,9,10,11->366

 

Then what?

AA !dTGY7OMD/g ID: 8b4129 Oct. 27, 2018, 7:04 p.m. No.7986   πŸ—„οΈ.is πŸ”—kun   >>7987 >>7995 >>8011 >>8062 >>8076

VQC, please read this post and respond to it

As far as I can tell. I've finished getting all of the patterns and concepts together in the Grid Patterns thread >>6506 here. I have a couple things I wanted to ask about.

 

For whatever range of reasons, none of the other anons have been helping, and none of them have been proofreading or checking that I'm correct and that I haven't missed anything. That means it's probably likely that I missed something or that maybe something is incorrect. I know you of all people will know if anything is right or if anything important has been missed. There's a lot of information in that thread, but would there be any chance you might have the time to go through and check everything?

 

Secondly, in your explanations of (x+n)(x+n)=nn+2d(n-1)+f-1, you introduced us to n0 and the guess triangle base made of f/8 or f/(8*something). As far as I could tell going through the threads, you never actually seemed to do anything with either of these values. If n0=n and guess base=u, you can construct the (x+n)(x+n) square, but you even said yourself that we know n0 won't equal n and we know the guess base won't equal u. After explaining that, you stopped talking about n0 altogether, and only mentioned the guess base a bit cryptically. So what was the point of those variables?

 

Thirdly, with the grid patterns thread done, do you have any newer advice as to what we should be looking into? Is there anything specific (as in looking at this concept and that concept at the same time) that we should be doing that we can now that everything's in one place?

ID: b129f9 Oct. 27, 2018, 11 p.m. No.7987   πŸ—„οΈ.is πŸ”—kun

>>7986

I'm thinking fractals. That has to be what we are missing here. It's a way you can turn that massive search space into nothingβ€”because the same patterns that apply to the massive apply to the small, in a fractal.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Oct. 28, 2018, 8:51 a.m. No.7992   πŸ—„οΈ.is πŸ”—kun   >>7993

Riemann Zeta + Mandelbrot

http://www.dhushara.com/DarkHeart/RH2/RH.htm

 

Riemann Zeta + Non-Mandelbrot Set Fractals

ttp://primepatterns.tumblr.com/tagged/Riemann-zeta-function

AA !dTGY7OMD/g ID: 8b4129 Oct. 28, 2018, 5:40 p.m. No.8002   πŸ—„οΈ.is πŸ”—kun   >>8003 >>8006

>>8001

I'll "calm my feckin' tits" when getting a giant two-month-long set of work done doesn't get a response of "you have more work to do". I'll calm down when I'm not the only one who does that work (I'm not the only one who is meant to work on it by the way, don't know if anyone realizes that). I'll calm down when suggesting that we work on something together for once rather than hiding off in our own little research corners completely ignoring everyone else doesn't get a response (this is a direct quote) of "who cares" on Discord. This is ridiculous. Give me one good reason not to ditch this board and all of you lazy pricks and figure this out by myself, Topol. I might as well considering how little interaction there is between any of us that isn't just mindless chatting on Discord. I'm over being told to calm down when every time I do anything useful here I either get completely ignored or I get this bullshit.

ID: b129f9 Oct. 28, 2018, 7:47 p.m. No.8005   πŸ—„οΈ.is πŸ”—kun   >>8007

>>8004

You're going about this as if you've earned the solution, as if you deserve more. You're too obsessed with the endpoint to the point of putting your friends down just because they don't work on what you want them to.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Oct. 28, 2018, 7:49 p.m. No.8006   πŸ—„οΈ.is πŸ”—kun   >>8008

>>8002

You're just being unnecessarily aggressive.

Also, I not Jan and the PMA/VA-duo working on this more often than not.

Sure… Minecraft sorta kinda fucked off for whatever reason…

But to claim that you're the only one doing anything is asinine.

 

You're lashing out. Calm yer tits.

You haven't done this on your own and left to your own devices you wouldn't do it, anyway.

 

I'm not saying you're not "reasonably frustrated with this whole affair".

But you're not some sort of "uniquely frustrated" where you should be throwing these fits and snapping at mufuggaz.

AA !dTGY7OMD/g ID: 8b4129 Oct. 28, 2018, 7:59 p.m. No.8007   πŸ—„οΈ.is πŸ”—kun

>>8005

>you're going about this as if I've earned the solution

How am I doing that? Topol told me to calm down. I told him why I'm not calm. Everything I'm saying is backed up with facts. I'm not just having a big emotional rant. These are genuine issues (to varying degrees).

>point 2: nobody helped me with grid patterns

Now if everyone explicitly told me "we don't think this is useful, maybe you should do it yourself", maybe I wouldn't mind so much. You actually said multiple times that you would help me with it (so did a couple of the others) and yet you didn't do anything. So this isn't me putting people down because they don't do what I want them to. This is people saying they will do something and never doing it.

>point 3: nobody is working together

I'm not the only one who has complained about this. Several other anons have. Isee was one. I think he was complaining about it on Discord. I suggested we work together on something for once and the only response I got was "who cares". That's a thing that happened. Do you think that's a useful outlook for everyone to have in regards to working together?

 

Now how am I acting entitled? Are any of the things I said objectively incorrect?

AA !dTGY7OMD/g ID: 8b4129 Oct. 28, 2018, 8:03 p.m. No.8008   πŸ—„οΈ.is πŸ”—kun   >>8009

>>8006

When did I say I was the only one doing anything? I said I was the only one working on one thing in particular. And then when I said I was done, someone told me to do more work (that they could do themselves, might I add). Also, I can admit I did snap that one time (that was generally a shitty period) but the only snapping I've done today is calling people lazy pricks. Nothing else I'm saying isn't based purely around facts. They might be angry facts, but they're facts.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Oct. 28, 2018, 8:08 p.m. No.8009   πŸ—„οΈ.is πŸ”—kun   >>8010

>>8008

Sweet, you've acknowledged you snapped at mufuggaz.

Took Chris 7 years to do this…

And yeah, when you finish one thing and there's more that needs to be done… what kind of response do you expect?

"Great job on that one thing! Your job here is done. Please join us in the clubhouse to rest on your laurels!"

 

Like I said, your not unreasonably frustrated.

But you're not uniquely frustrated and we don't deserve to be the butt of your anger and frankly, neither does Chris.

 

I'm not saying don't be human.

But if you want a word battle, let's duke it out in the EZ Bake.

AA !dTGY7OMD/g ID: 8b4129 Oct. 28, 2018, 8:13 p.m. No.8010   πŸ—„οΈ.is πŸ”—kun   >>8011

>>8009

>And yeah, when you finish one thing and there's more that needs to be done… what kind of response do you expect?

When it's something VQC told all of us to do, and it therefore isn't just my job, I don't expect a response of "there's more work to do, go do more work".

>But you're not uniquely frustrated and we don't deserve to be the butt of your anger

All I'm doing is pointing out things that I think are problems. If there's a bit of anger in there, it isn't for the sake of making people upset.

>and frankly, neither does Chris.

Would you like to point out where I got angry at Chris? Because as far as I'm aware I didn't do that.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Oct. 28, 2018, 8:16 p.m. No.8011   πŸ—„οΈ.is πŸ”—kun   >>8012

>>8010

There you go saying you're the only one doing something again…

And you've seen my try to herd the cats into playing with the same ball of yarn.

You're dealing with maths-people.

It's complex.

 

We get it. You're upset. Quit lashing at us.

 

Yeah: >>7986 <<< there you are being pissy and making entitled commands.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Oct. 28, 2018, 8:38 p.m. No.8013   πŸ—„οΈ.is πŸ”—kun   >>8015

>>8012

Jesus, you sound like you're new here!

I've made bomb ass points before that he didn't acknowledge at the time.

IF he mentions something I've worked on outside of saying he likes my style… it's when it's coming into play.

 

For starters… let's look at another pattern.

Have you NOT noticed "someone" ip hopping and making posts?

Sometimes they respond to someone, and sometimes they seem to be in their own little world…

Haaaaave you not figured out what's going on with that, yet?

You are the only one here who can see the ip hashes, after all.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Oct. 28, 2018, 8:44 p.m. No.8014   πŸ—„οΈ.is πŸ”—kun

>>8012

Basically, you'll get a lot farther with this if you treat VQC like Q.

 

I've posted things that Q didn't acknowledge, only to give the nod to someone else a hundred breads later.

 

But can you make demands of Q? No.

Should you throw a tantrum and threaten to nuke the place if Senpai doesn't give you the attention you feel you deserve? No.

Pic related.

 

You've blown off some steam.

Do you feel better?

Can we all go back to playing nice?

AA !dTGY7OMD/g ID: 8b4129 Oct. 28, 2018, 8:44 p.m. No.8015   πŸ—„οΈ.is πŸ”—kun   >>8016

>>8013

No offense to whoever posted these if they're new and they're just firing off whatever ideas come to them, but is there anything we can actually use any of these posts for?

>>7950

>>7957

>>7988

>>7990

I don't think VQC posted any of those. They're the kinds of musings we all had at the beginning, which makes me think it's a newfag with a dynamic IP or a VPN. When VQC has crumbs, he posts more specific stuff like >>7746 this. Whatever the case, I really don't see how I was acting pissy towards VQC.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Oct. 28, 2018, 8:47 p.m. No.8016   πŸ—„οΈ.is πŸ”—kun   >>8017

>>8015

Aaaaah you little faggot.

Why'd you leave out the one with the clue in the e-mail?

 

And let's Occam's Razor this…

It would take an incredibly versed maths-troll to pull some convincing shit out of their ass that we'd "fall for".

 

Look at the pattern.

Here's another example: Look at how ebot does their schtick.

<;3=

ID: b129f9 Oct. 28, 2018, 9:01 p.m. No.8019   πŸ—„οΈ.is πŸ”—kun   >>8020

Okay, but if the goal is to rethink our approach to mathematics as a way of rewriting it, then why would any musing be unwelcome? None of them are things that have been completely explored

AA !dTGY7OMD/g ID: 8b4129 Oct. 28, 2018, 9:11 p.m. No.8020   πŸ—„οΈ.is πŸ”—kun

>>8019

Maybe read my post again because I never said they were unwelcome. I said I don't think VQC posted them because they're musings and VQC posts more specific hints about variables.

PMA !!y5/EVb5KZI ID: 4fbd64 Oct. 28, 2018, 9:24 p.m. No.8021   πŸ—„οΈ.is πŸ”—kun

>>7988

>Re: 38! + 1

38! + 1 = 523022617466601111760007224100074291200000001

 

Is a semi-prime with factors 14029308060317546154181 Γ— 37280713718589679646221

 

Interesting write-up and sample code at https://codegolf.stackexchange.com/questions/8629/fastest-semiprime-factorization

PMA !!y5/EVb5KZI ID: 4fbd64 Oct. 28, 2018, 9:29 p.m. No.8022   πŸ—„οΈ.is πŸ”—kun

>>7988

>c = 15105

>123 = floor_sqrt(-14883 + 244i)

 

Relevant records:

 

entry c: (221,7431,61) = {221:7431:122:121:1:15105} = 15105; f=24;

na: (221,1,61) = {221:1:7552:121:7431:7675} = 57032925; f=14884

 

Rewriting the floor_sqrt in terms of our variables:

 

123 = floor_sqrt( -(f-1) + 2di )

 

where f=14884, and 2d = 244

 

shows a connection to the d value in the (-f,n) space at:

 

(-24,7430,62) = {-24:7430:123:122:1:15105} = 15105

Anonymous ID: 2344ac Oct. 29, 2018, 10:34 a.m. No.8023   πŸ—„οΈ.is πŸ”—kun   >>8027

Imagine playing chess.

Every piece has a specific pattern/permutations in which it can move/exist.

Solution to the game is check mate - a specific pattern amongst some pieces.

Superimpose piece patterns to get all possible patterns for all pieces.

On every move discard those patterns that dont lead to check mate.

 

Play the game.

At the begining there is a starting pattern amongst all pieces.

And a superposition of all possible patterns.

As the game progresses and moves are made, solution space shrinks.

Everytime some cell gets occupied by some piece, solution space shrinks.

On every move evaluate if this specific pattern (can) leads towards check mate.

 

Does it make sense?

Can we apply it to our problem?

Pieces are equations,

Check mate is c=ab,

Patterns are grid patterns?

Row 0 is a superposition of all possible patterns at every step?

Everytime a cell gets occupied/calculated….???

 

Just some thoughts, maybe it helps someone.

Will probably work on the analogy further as I can see some resemblance.

Anonymous ID: 2344ac Oct. 29, 2018, 12:05 p.m. No.8024   πŸ—„οΈ.is πŸ”—kun   >>8026 >>8027

Without any information or sense of what the best move is, of what the best pattern to follow is an algorithm would still reach the solution-check mate, but by randomness.

 

So, how do we evaluate which pattern in any moment is the best, optimal?

ID: b129f9 Oct. 29, 2018, 12:48 p.m. No.8027   πŸ—„οΈ.is πŸ”—kun   >>8030

>>8024

>>8023

>>7957

It is the unity and enumeration of all grid patterns.

 

>Enumerate the rules.

>Win.

 

>The solution to this problem introduces a new form of algebra where two concurrent forms of equations run side by side and then merge.

>The two sets of equations take the problem and simplify it. Together they handle the "lock and key" nature of the problem/solution, particularly when c is divisible by 1, c, and two other prime numbers, such as in RSA.

>The two forms of equations that merge together handle staircase numbers where the base of one staircase, is one unit longer than the other.

 

Two threads in a program, one starts with c=ab, the other starts with c=1c. The equations run independently and encompass all grid patterns, then eventually merge.

 

c=ab {

d = sqrt(c)

e = c - d^2

f = 2d+1-e

n = (a+b)/2-d

x = d-a

(d+n)^2 - (x+n)^2 = c

(d+n) + (x+n) = b

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

 

a[e, 1, t] = na

b[e, 1, t] = na+2(x+1)

a[-f, 1, t] = (n-1)a

b[-f, 1, t] = (n-1)a+2(x+2)

}

 

c=1c {

d = sqrt(c)

e = c - d^2

f = 2d+1-e

N = (c+1)/2-d

X = d-1

(d+N)^2 - (X+N)^2 = c

(d+N) + (X+N) = c

(d+N) - (X+N) = 1

 

a[e, 1, T] = N

b[e, 1, T] = N+2d

a[-f, 1, T+1] = N-1

b[-f, 1, T+1] = N-1+2(d+1)

}

 

The equations keep enumerating patterns until they merge.

ID: b129f9 Oct. 29, 2018, 1:22 p.m. No.8030   πŸ—„οΈ.is πŸ”—kun   >>8031 >>8033

>>8027

cont'd

 

c = ab {

[1] the entry (e, 1, t+pna) where p is any multiplier of na will contain a multiple of na in a[t]

in other words, na will be a factor of a[t] in that element

[2] the entry (-f, 1, t+1+p(n-1)a) where p is any multiplier of (n-1)a will contain a multiple of (n-1)a in a[t]

}

 

c = 1c {

[1] the entry (e, 1, T+pN) where p is any multiplier of N will contain a multiple of N in a[t]

in other words, N will be a factor of a[t] in that element

[2] the entry (-f, 1, T+1+p(N-1)) where p is any multiplier of N-1 will contain a multiple of N-1 in a[t]

[3] a[-f, 1, T+1+1(N-1)] = N*c

a[-f, 1, T+1+1(N-1)] / a[-f, 1, T+1] = c

}

 

They will merge when enough patterns are enumerated. Continue enumerating these patterns -reach solution.

 

A question to answer (to define the equivalent pattern for the ab thread). In (-f, 1, t+1) where a[t] = (n-1)a, what defines the multiple of (n-1)a that appears at (-f, 1, t+1+(n-1)a)?

Anonymous ID: 47d6d1 Oct. 31, 2018, 6:10 p.m. No.8037   πŸ—„οΈ.is πŸ”—kun

>>8036

From a lurker, brainlet anon. Thank you for attempting to ennunmerate all of the hints in an organized manner. Hell, you’ve kept the board followable.

The chans are a breeding ground for disorganized but intuitive thought processes. You have been fighting against its natural state. Take care of yourself. That goes for all of you.

Anonymous ID: 266807 Nov. 1, 2018, 12:23 a.m. No.8040   πŸ—„οΈ.is πŸ”—kun   >>8042 >>8058 >>8063 >>8079 >>8512

A staircase number is an (odd) number composed of one number added to the number less than it.

 

To find the staircase numbers for an odd value, we add one to the value then divide it by two. Then the number is equal to the result added to the result minus one.

Calculating this for the number we started with, would make two familiar values. Adding two side by side (t aligned) x values in e,1 and -f,1 would also make a staircase number.

 

When na is odd, what do you do to find an identifier for the element where the x is equal to one of the staircase numbers for na? And if na is even, for n-1?

ID: b129f9 Nov. 1, 2018, 2:02 p.m. No.8043   πŸ—„οΈ.is πŸ”—kun   >>8044

Equation threads cont'd:

 

c = ab {

[3] a[-f, 1, (t+1)+(n-1)a] = (2d+1)[t+1] * (n-1)a

 

a[-f, 1, (t+1)+(n-1)a] / a[-f, 1, t+1] = (2d+1)[t+1]

}

 

c = 1c {

//since c is equal to 2d+1 of the na transform record at T+1 in cell -f,1, c appears as our multiple of (N-1) when we go (N-1) elements down

[3] a[-f, 1, T+1+1(N-1)] = N*c

 

a[-f, 1, T+1+1(N-1)] / a[-f, 1, T+1] = c

}

 

//a factor of 2d+1 of t in -f,1 is added when we go na cells down

ID: b129f9 Nov. 1, 2018, 2:06 p.m. No.8044   πŸ—„οΈ.is πŸ”—kun

>>8043

Clarification:

(notation is being pushed to its limits here), pattern [3] applies to going (n-1)a and N-1 elements down from the -f,1 transform record where a[t] = (n-1)a or N-1

 

t of this record is written t+1 since it is one element below the na transform record in e,1, which we derive from x or X using the t from x formula

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Nov. 1, 2018, 3:12 p.m. No.8047   πŸ—„οΈ.is πŸ”—kun   >>8051 >>8056 >>8061

>>8045

>T-24hr.

~10pm UTC…

>Blackout in the Red Room

10 days of Darkness?

Red Room… Twin Peaks?

RR…

>Highlander 2

I thought there could only be 1?

 

Hmmmm… also, considering what's going on with Assange and Ecuador…

I wonder what hands are being played right now…

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Nov. 1, 2018, 3:28 p.m. No.8054   πŸ—„οΈ.is πŸ”—kun

>>8052

Well, if chillin' ain't required:

 

Highlander 2 is subtitled:

The Quickening.

 

If Podesta is scheduled to be picked up 11.3…

 

Might be a good time to start the 10 days of Darkness.

 

Get all the bad actors while they're acting to do something against the elections.

 

And don't forget, 2 rallies, each day, Nov 3rd and 4th.

 

That's lots of insulated, secure airtime on AF1, especially on the 3rd from D.C. to Montana to Florida.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Nov. 1, 2018, 3:30 p.m. No.8055   πŸ—„οΈ.is πŸ”—kun

>>8053

"The Other"… or maybe M… that you?

 

I remember chatting in the open waters about pokin' VQC to bring him back to wrap this up while ago.

 

Timing… timing…

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Nov. 1, 2018, 3:57 p.m. No.8056   πŸ—„οΈ.is πŸ”—kun   >>8057

>>8047

>T-24hr.

>Blackout in the Red Room.

>Highlander 2.

 

Another Red Room…

Here's the JFK Edition:

https://www.jfklibrary.org/asset-viewer/archives/JFKWHP/1960/Month%2006/Day%2016/JFKWHP-1960-06-16-A?image_identifier=JFKWHP-KN-C16087

 

https://en.wikipedia.org/wiki/Red_Room_(White_House)#Kennedy_refurbishment

 

Curious that the only image of the Kennedy Edition on Wikipedia is of the Funeral Reception…

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Nov. 1, 2018, 4:05 p.m. No.8057   πŸ—„οΈ.is πŸ”—kun

>>8056

Granted… dimension bending…

Also a Villa of Mysteries reference?

 

"What really made Vesuvius explode? Did something go wrong in the Villa of Mysteries?"

 

https://www.archaeology.org/issues/124-1403/features/1813-pompeii-saving-the-villa-of-the-mysteries

AA !dTGY7OMD/g ID: 8b4129 Nov. 1, 2018, 5:44 p.m. No.8062   πŸ—„οΈ.is πŸ”—kun

Before I take a break properly I suppose I should apologize for getting angry and calling you all lazy pricks. That's quite obviously not true. I stand by most of the rest of what I said, but it's inconsequential at this point, so I promise I won't mention it again. All I'll say is that >>7999 was pretty unnecessary.

 

>>8048

Would you mind responding to this (or telling me you aren't going to respond)? >>7986 The red text was just to get your attention since you miss questions semi-often (whether intentional or not).

AA !dTGY7OMD/g ID: 8b4129 Nov. 1, 2018, 5:56 p.m. No.8063   πŸ—„οΈ.is πŸ”—kun

One last thing: based on >>8040 this, when you take the cells in (e,1) and (-f,1) where the two x values add together to make na, b-a in the (-f,1) cell is na+3 and b-a in the (e,1) cell is na+1. This is only tested on odd na so far.

 

Here are some test cases:

 

Find the element/cell where the x is equal to one of the staircase numbers for na/(n-1)a

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

(-17:1:6) = {-17:1:63:11:52:76} e=108

na=65 (so where x=32 and 33)

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

(-17,1,17) = (-17:1:569:33:536:604} e=1120

b-a=604-536=68, b-a=593-527=66, na=65

 

(13:7:7) = {13:7:26:13:13:53} f=-40

(-40:6:8) = {-40:6:27:14:13:53} e=13

na=91 (so 45+46)

(13,1,23) = {13:1:1064:45:1019:1111} f=-2116

(-40,1,24) = {-40:1:1084:46:1038:1132} e=2127

b-a=1132-1038=94, b-a=1111-1019=92, na=91

 

(13:13:7) = {13:13:20:13:7:59} f=-28

(-28:12:8) = {-28:12:21:14:7:59} e=13

na=91

(13,1,23) = {13:1:1064:45:1019:1111} f=-2116

(-28,1,24) = {-28:1:1090:46:1044:1138} f=2151

94, 92, 91

 

(69:5:10) = {69:5:62:19:43:91} f=-56

(-56:4:11) = {-56:4:63:20:43:91} e=69

na=215 (107, 108)

(69,1,54) = {69:1:5866:107:5759:5975} f=-11664

(-56,1,55) = {-56:1:5912:108:5804:6022} f=11767

218, 216, 215

 

It might be useful to look into d[t]-d and a[t]-a in these cells, as well as comparing them to the cells where cBigN and cshadowN show up in (e,1) and (-f,1), and where x=f-1.

AA !dTGY7OMD/g ID: 8b4129 Nov. 1, 2018, 8:05 p.m. No.8070   πŸ—„οΈ.is πŸ”—kun   >>8071 >>8074

Man, I've been angry for a while, huh? I've just been reading through older things. I'm really sorry everyone. I've had a shitty year and I didn't realize I was taking it out on you guys. This is the last I'll post about this whole thing, I swear. Don't mind me. Back to math.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Nov. 1, 2018, 8:53 p.m. No.8072   πŸ—„οΈ.is πŸ”—kun

Did you mean "Blackout 42?"

Led me to some topologies…

 

https://en.wikipedia.org/wiki/Bus_network

https://en.wikipedia.org/wiki/Star_network

https://en.wikipedia.org/wiki/Tree_network

https://en.wikipedia.org/wiki/Hypertree_network

 

Teehee.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Nov. 1, 2018, 10:52 p.m. No.8073   πŸ—„οΈ.is πŸ”—kun

From an anon in the open waters:

 

"check this out anons..might be relevant?

 

At the start of the summer term at Meadowbank School for Girls, a prestigious prep school in England, there is no reason for Miss Bulstrode, the popular but aging headmistress, to believe that the challenges facing her will be more than the occasional irate or inebriated parent. She scarcely listens when Mrs Upjohn, a parent, recognizes someone that she sees from her wartime days in the intelligence service. But there is a killer at the school who does not wait long to strike."

VQC !!cJ17mYdx.I ID: a26807 Nov. 1, 2018, 11:48 p.m. No.8076   πŸ—„οΈ.is πŸ”—kun   >>8079

>>7986

The n0 base is a way to calculate through iterations the correct value of n or n-1. The iterations required are no bigger in complexity than Big Oh for the root of c.

Well done with the patterns.

It will happen when you come back to it.

From nowhere, in a moment when you decide not to give up.

Never give up.

I expect there is a very good reason why you were chosen to bring this to the world. That is why you are doing the pattern heavy lifting.

The patterns will give you a lookup.

When you see it, you won't unsee it.

I believe in you 100%

VA !!Nf9AmQNR7I ID: 0a3773 Nov. 2, 2018, 6:20 a.m. No.8079   πŸ—„οΈ.is πŸ”—kun

>>8076

Hello VQC! Nice to see you.

The n0 base and triangle staircase numbers are derived from f, so they scale with every c value, correct?

 

On another thought: For the grid shortcut, does our program/algorithm need to be able to generate the elements surrounding the na transform in e,1 and -f,1?? The na transform lands us right in between (an) and (bn) in e,1 and a(n-1) and b(n-1) in -f,1. The movement from the na transform to the correct (an) and (bn) t or x values is very similar to the formula you described in your IP hopping post >>8040. The movement to (an) and (bn) is definitely two adjacent staircase numbers, and for a(n-1) and b(n-1) it happens as well. For c145 those numbers are 2 and 3 in e,1.

>To find the staircase numbers for an odd value, we add one to the value then divide it by two. Then the number is equal to the result added to the result minus one. Calculating this for the number we started with, would make two familiar values. Adding two side by side (t aligned) x values in e,1 and -f,1 would also make a staircase number.

>When na is odd, what do you do to find an identifier for the element where the x is equal to one of the staircase numbers for na? And if na is even, for n-1?

GAnon !Nx57Pyux3E ID: 566843 Nov. 2, 2018, 2:31 p.m. No.8080   πŸ—„οΈ.is πŸ”—kun

>>8077

 

Hey VQC. Huge fan

Should I keep doing the grid stuff I'm doing?

 

>>8039

>>7886

 

I can look further into the factoring method if we should be doing that. I think there is a lot to be learned from these grids and if its as versatile as you say then it could be useful. On the otherhand the first discovery might be so mindblowing that anything else we do to utilize the grid would be trivial. I don't want to be wasting time.

Anonymous ID: 2344ac Nov. 4, 2018, 7:08 a.m. No.8083   πŸ—„οΈ.is πŸ”—kun   >>8084 >>8089

So… what are your thoughts on how The Grid relates to RSA?

 

:

RSA revolves around Eulers PHI function, phi(N), pic related

phi has all primes on its diagonal and a fractal pattern inside a triangle

Our c is somewhere in the fractal, related to a,b who are predictably on the diagonal

Every point in a fractal is self-similiar

We are trying to find a pattern that ties our c to all its self-similiar points in the fractal

a and b are hidden in the pattern

 

Something about triangles, squares and fractals?

The Grid reverses the operation

It starts with some input 'c' and applies a square pattern on phi untill perfect squares(primes) are found???

Diagonals of prime squares?

The Grid is built for a specific version of RSA, specific constants

 

It might be easier to concentrate all our efforts on one specific path if we discuss how we SEE the problem.

Anonymous ID: abe23c Nov. 5, 2018, 5:37 a.m. No.8089   πŸ—„οΈ.is πŸ”—kun

I was playing around with the Euler Totient. Based on the triangle from >>8083 I wanted to see how it looked like as a square. Not sure if this is the fractal we're looking for, but never the less it looks quite pleasing. You can see how there are squares and triangles pretty much everywhere inside of it. I really don't think this is where we are moving with regards to our fractal, but it was fun reading about the function and playing a bit with it.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Nov. 5, 2018, 10:03 p.m. No.8094   πŸ—„οΈ.is πŸ”—kun

>>8093

Hello coinkydinkle first time poster anon!

Welcome to our corner of the circle.

Please, check out the other topics in the catalog and remember…

 

Math is only EVERYTHING; so don't fuck up.

ID: b129f9 Nov. 5, 2018, 11:23 p.m. No.8099   πŸ—„οΈ.is πŸ”—kun

fractals:

 

Mandelbrot: recurse over f(z) = z^2 + c over and over for random complex points and color them based on whether they stay bounded or tend to infinity

 

Remainder tree: recursively calculate c = d^2 + e

Anonymous ID: abe23c Nov. 6, 2018, 8:35 a.m. No.8103   πŸ—„οΈ.is πŸ”—kun   >>8104 >>8105

I feel like an idiot again, it seems like every time I "get" part of what we do I end up feeling like an idiot.

 

Anyways, the smooth numbers are (a - 1)*(b - 1)/2. That is, BigN - n = (a - 1)(b - 1)/2.

Anonymous ID: e12188 Nov. 7, 2018, 7:45 p.m. No.8110   πŸ—„οΈ.is πŸ”—kun   >>8135

 

n values for the first 28 triangle numbers (from ColumnKeys function)

 

T(1) = n/a

T(2) = [1]

T(3) = n/a

T(4) = n/a

T(5) = [1, 5]

T(6) = [1, 7]

T(7) = [3]

T(8) = [4]

T(9) = [1, 3, 17]

T(10) = [1, 21]

T(11) = n/a

T(12) = n/a

T(13) = [1, 37]

T(14) = [1, 3, 9, 43]

T(15) = [1, 3, 7, 21]

T(16) = [8, 24]

T(17) = [1, 15, 65]

T(18) = [1, 17, 73]

T(19) = n/a

T(20) = n/a

T(21) = [1, 5, 25, 101]

T(22) = [2, 112]

T(23) = [10, 54]

T(24) = [3, 11, 59]

T(25) = [1, 17, 145]

T(26) = [2, 6, 42, 158]

T(27) = n/a

T(28) = n/a

Anonymous ID: e12188 Nov. 7, 2018, 7:53 p.m. No.8111   πŸ—„οΈ.is πŸ”—kun   >>8206 >>8207

>>7007

It returns pairs of n and n-1.

For example, if we input -24, 1 it will return

 

[[4, 5], [60, 61]]

 

since -f and e combined make the columns unique to c. Can this function be reconstructed from a description of how it behaves?

Anonymous ID: e12188 Nov. 7, 2018, 8:02 p.m. No.8112   πŸ—„οΈ.is πŸ”—kun   >>8113

This is a diagonal, since it describes pairs of n that are one (row) apart. How do column -f and e when used together at once materialize the solution, since the diagonal and the patterns they make when used in tandem are unique to c?

 

It will all make sense in hindsight.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Nov. 7, 2018, 8:34 p.m. No.8113   πŸ—„οΈ.is πŸ”—kun

>>8112

I do my darnedest to remind the Nerds to post here, I swear.

 

Side note, if there is "space between squares", is there space between trangles?

What about space between Trangles AND Squars?

Anonymous ID: 2bd1d5 Nov. 7, 2018, 9:36 p.m. No.8114   πŸ—„οΈ.is πŸ”—kun   >>8115

https:// curvaturasvariantes.com/2018/06/12/galois-theory-hodge-conjecture-and-riemann-hypothesis-visual-geometric-investigations/

https:// curvaturasvariantes.com/2018/05/25/about-prime-numbers-symmetries/

<3

Anonymous ID: 0c3c6f Nov. 9, 2018, 8:36 p.m. No.8119   πŸ—„οΈ.is πŸ”—kun   >>8122 >>8123 >>8125

1) Draw a circle.

2) Pick a c value which is the sum of squares (c values where e=1,4,9,16,25..)

3) Set the radius of the circle equal to the square root of c (this calculation includes decimals).

4) Every integer point the circle passes through will be a valid sum of squares that makes c. (e.g the circle for sqrt(145) passes through (12,1) and 145 = 12^2 + 1^2)

 

A c example equal to the sum of squares was chosen for a reason.

Anonymous ID: 0c3c6f Nov. 9, 2018, 8:42 p.m. No.8120   πŸ—„οΈ.is πŸ”—kun   >>8122 >>8125

This is because the formula for a circle reduces to the sum of squares when the circle's centerpoint exists at the origin (a and b in the formula = 0). The cell where e and n = 0 also functions as an origin.

GAnon !Nx57Pyux3E ID: 566843 Nov. 10, 2018, 8:37 a.m. No.8125   πŸ—„οΈ.is πŸ”—kun   >>8127

>>8119

>>8120

>>8121

 

https://www.m ath.hmc.edu/funfacts/ffiles/20008.5.shtml

 

>Which whole numbers are expressible as sums of two (integer) squares?

 

>Here's a theorem that completely answers the question, due to Fermat:

>A number N is expressible as a sum of 2 squares if and only if in the prime factorization of N, every prime of the form (4k+3) occurs an even number of times!

 

>Examples: 245 = 577. The only prime of the form 4k+3 is 7, and it appears twice. So it should be possible to write 245 as a sum of 2 squares (in fact, try the squares of 14 and 7). But because 7 appears only once in 42=237, it is impossible to write 42 as the sum of two squares.

 

>A corollary of this fact is that every prime of the form (4k+1) can be written as the sum of two squares.

 

 

So since 145 is a sum of 2 squares so 145 = 4k+1 for k=36 = 1212 + 11

 

I think I learned in number theory way back when a way to get these squares for a number. Not sure though.

 

We know for a fact that it is coprime, so there are two factors. If one is equal to 4k+3, then the other must also be equal to 4k+3 for different k's because it can be written as a sum of two squares. If one factor is equal to 4k+1, then the other must also, because 0 is for this sake an even number.

 

So if a number is a sum of two squares, it is either

 

c = (4k+1)(4j+1) or (4k+3)(4j+3) for some j and some k

 

Lets look at the formula (d+n) = (b+a)/2 and (x+n) = (b-a)/2,

 

Suppose a=4k+1 b=4j+1,

d+n = (4j+1 + (4k+1))/2 = (4j + 4k + 2)/2 = 2(j+k) + 1

x+n = (4j+1 - (4k+1))/2 = (4j+1 - 4k - 1)/2 = 2(j-k)

 

for c=145 = 4(36) + 1

a = 5 = 4(1) + 1 ==k=1

b = 29 = 4(7) + 1 ==j=7

 

(d+n) = 2(j+k)+1 = 2(7+1)+1 =17

(x+n) = 2(j-k) = 2(7-1) = 12

 

Now suppose a=4k+3 and b=4j+3

 

d+n = (4j+3 + (4k+3))/2 = (4(j+k) + 6) / 2 = 2(j+k) + 3

x+n = (4j+3 - (4k+3))/2 = (4(j-k))/2 = 2(j-k)

 

For c = 253 = 4(63) + 1

a = 11 = 4(2) + 3 ==k = 2

b = 23 = 4(5) + 3 ==j = 5

 

(d+n) = 2(j+k) + 3 = 2(2+5) + 3 = 17

(x+n) = 2(j-k) = 2(5-2) = 6

 

This is super cool because every number that is the sum of two squares has (x+n) = 2(j-k).

 

Its almost as if it maps all the numbers from 2k+1 to 4k+1 in such a way.

Like for the 145 example,

a = 4(1) + 1

b = 4(7) + 1

 

Call the 1 and 7 the core components of the number. You can add and subtract these, and when you do you generate an odd and an even number

For these situations d+n is the odd component. I'd imagine when c=4k+3 that is the reverse but same type of stuff.

 

(d+n) = 2(7+1) + 1 = 17

(x+n) = 2(7-1) = 12

 

The same type of thing holds for c=253, but the leaf (odd number) is 3 instead of 1

a = 4(2) + 3

b = 4(5) + 3

 

(d+n) = 2(5+2) + 3 = 17

(x+n) = 2(5-2) = 6

 

Getting anywhere??

GAnon !Nx57Pyux3E ID: 566843 Nov. 11, 2018, 10:01 a.m. No.8127   πŸ—„οΈ.is πŸ”—kun   >>8128

>>8125

 

If we look in here

 

Pic 1 can show you a few things. The core element of b is always the sum of the core for d+n and x+n. Also the core for a is always the difference of d+n and x+n.

 

Whats interesting is that for these 4 numbers (a,b,d+n,x+n), the pattern always holds no matter what base you're in. Also another thing is that the leaf numbers of d+n and x+n always add up to the leaf number of b.

GAnon !Nx57Pyux3E ID: 566843 Nov. 11, 2018, 12:03 p.m. No.8129   πŸ—„οΈ.is πŸ”—kun

>>8128

 

If a number c is a sum of two squares, then it is of the form 4k+1. Here, the 4 is the multiplier, the k would be the core and the 1 would be the leaf. I'm proposing that the multiplier can be any power of 2 and the k can be any number and then the leaf would then have to be a number less than the multiplier.

 

If a square c is of the form 4k+1, then according to the link I posted it's prime factorization must have primes of the form 4k+3 occurring an even number of times [here, 4 would be multiplier, k core and 3 leaf]. Since we know c is coprime (2 prime factors) and that they are odd, the only two types of numbers they can be are 4k+1 and 4k+3. Since the factors of the type 4k+3 must occur an even number of times and there are only 2 factors, the even numbers are either 2 or 0, so either both factors are 4k+3 for different k's or they are both 4k+1 for different k's. Keep in mind, this is only for c = 4k+1.

 

Knowing this, I algebraically looked into the d+n and x+n values in this format through the formulas:

d+n = (b+a)/2

x+n = (b-a)/2

 

I found that if

 

a = (4)(k) + L

b = (4)(j) + L

 

For L=1 or 3 and since b>a, j>k

 

Then you can find the d+n and x+n values like this

 

d+n = (2)(j+k) + L

x+n = 2(j-k)

 

So instead of doing all the calculations you could look at it like this:

>Divide the multiplier by two

>Take sum and difference of core elements

>Decrement L by one for the difference

 

I'm thinking we might be able to take this formula and do other things with it, but I am still looking. For instance what if we increase all the powers? Then does the function map to a different calculation? If you increase the multiplier to greater than the initial number, then the entire number would just be a leaf because the core would be 0. Obviously every single number can be written as another number.

 

15 = 16(0) + 15 = 8(1) + 7 = 4(3) + 3 = 2(7) + 1 = 1*(15) + 0

 

Look here, the leaf goes 15->7->3->1->0

and the core goes 0->1->3->7->15

 

They reverse but the power decreases.

 

I think we should look into this

GAnon !Nx57Pyux3E ID: 566843 Nov. 11, 2018, 12:49 p.m. No.8130   πŸ—„οΈ.is πŸ”—kun

Another thing we can use this type of stuff for is Legendre Symbols.

 

The legendre symbol (I'll denote it as (a//p)) is:

0 if a is divisible by p

1 if a is a quadratic residue mod p

-1 if a is not a quadratic residue mod p

 

If a IS a quadratic residue mod p, then there exists a number M where MM - a = kp for some k.

Or the square of M is a units more than a multiple of p.

We could make a a negative number, which would switch the equation to MM + a = kp.

Then its a square plus a number equals a product of 2 numbers (one of which is necessarily prime), which is our entire problem.

 

If you look further into these numbers, you'll see things where this type of math is used.

 

If you let a=-1, then the legendre symbol (a//p) can be calculated directly through this calculation:

(a//p) = 1 if p = 4k+1

(a//p) = 3 if p = 4k+3

 

If you let a=-2, then

 

(a//p) = 1 if p = 8k + 1 or 8k + 3

(a//p) = -1 if p = 8k + 5 or 8k + 7

 

If you let a=-3, then

 

(a//p) = 1 if p = 12k + 1 or 12k + 7

(a//p) = -1 if p = 12k + 5 or 12k + 11 (3 or 9 dont seem to show up)

 

a = -4

 

(a//p) = 1 if p = 8k + 1 or 8k + 5

(a//p) = -1 if p = 8k + 3 or 8k + 7

 

a = -5

 

(a//p) = 1 if p = 20k + {1, 3, 5, 7, 9}

(a//p) = -1 if p = 20k + {11, 13, 17, 19} (15 doesnt seem to show up)

 

a = -6

 

(a//p) = 1 if p = 24k + {1, 5, 7, 11}

(a//p) = -1 if p = 24k + {13, 17, 19, 23}

 

Basically it looks like you can check divisibility without actually calculating anything except the mod of 4 * the number. The -4 case is unique though I'll look more into that. There may be numbers missing from these but generally speaking the ideas is right. I'll write a script later to more easily calculate this stuff and verify it.

PMA !!y5/EVb5KZI ID: e42aae Nov. 11, 2018, 1:14 p.m. No.8131   πŸ—„οΈ.is πŸ”—kun   >>8133

>>7990

>c=i^2+ij-ji-j^2

 

Thanks for this perspective.

 

Using ij = (d+n)(x+n) as the mid-point between the large and small squares, went looking for a way to either directly calculate a factor or reduce the iterative search space for finding the solution small square.

 

Pics attached explore the ij = (d+n)(x+n) "middle" rectangle for various test cases, and includes a square/triangle breakdown for c6107 as an example of how factors could be extracted from this middle area.

 

The analysis is based on the formulas (DPN - ij) and (ij - XPN), where DPN = (d+n)^2 and XPN = (x+n)^2.

 

Reason for this approach is that the a[t] values from the na transform (-f,1) and (e,1) starting records are either very close to or an exact match with the (ij - XPN) or (DPN - ij) values. (see c145 as an example).

 

The "gcd" and "sqrt" columns provide an alterative way to find the solution "a" values and are calculated as:

 

gcd = gcd(c,(ij - XPN))

sqrt = sqrt((DPN - ij) - (ij - XPN))

 

Both formulas return the solution "a" value, except for c363 where the gcd result is a^2.

 

Additional testing has found that the following gcd formulas also work:

 

a = gcd(c,(ij - XPN))

a = gcd(c,(DPN - ij))

 

b = gcd(c,(ij + XPN))

b = gcd(c,(DPN + ij))

 

Furthermore, the sqrt formula (DPN - ij) - (ij - XPN) can be simplified to aa=(d-x)^2. Indicating again the need to find the correct "x" value.

 

One way to take advantage of these formulas would be to somehow calculate or iterate to the (ij - XPN) value. The last 2 columns using the min_a variable were an attempt to isolate and understand these gaps (2d+1 keeps appearing).

 

Alternatively, this analysis may lead to an improved iterative search estimated x+n starting position based on the na (e,1) a[t] value.

 

For c6107, that starting position would be x+n = sqrt(2976) = 54, and reduces the iterative search space by half (see pic attached).

PMA !!y5/EVb5KZI ID: e42aae Nov. 11, 2018, 4:05 p.m. No.8135   πŸ—„οΈ.is πŸ”—kun   >>8163

>>8110

Pic attached is for triangle numbers between 500 and 530 and includes variables previously hinted for the ColumnKeys function.

 

The n/a columns represent c mod 4 = 2 numbers that don't exist in the grid. These require c/2 to find valid entries.

 

n values of 22 and 23 have been highlighted to show perhaps similar behaviors that needs to be explored further.

 

Interesting that in some cases the sqrt(2d), sqrt(f), or sqrt(e) values come very close to a correct n.

VA !!Nf9AmQNR7I ID: 296e69 Nov. 11, 2018, 6:34 p.m. No.8136   πŸ—„οΈ.is πŸ”—kun

Hello Lads! I'm reviewing crumbs looking for traction. We haven't figured out yet how 2d and f limit the possible (n-1) values. Based on VQC's instructions in the attached pic, we split off one (n-1) value in the x+n square formula. The formula looks like this after being modified:

 

nn+(2d-1)(n-1)+f+(n-1)-1

 

I was just thinking about a new way to examine the mods.

Let’s take familiar c values, and break them down in two ways:

 

Method 1: Using the (f-1)div 8 method, showing all mods as well.

Then let’s have columns with 2d-1 and f for that c.

 

Method 2:

Then let’s have columns breaking down each part of nn+(2d-1)+f-1 into div 8 and mod 8.

 

Goal is to find the connection between 2d, f, and the n or (n-1) values that give a correct lock for the x+n square. We need to see how EVERY piece contributes to completing the x+n square.

To make it work, I think we need medium/large c values, like in the millions range.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Nov. 12, 2018, 2:54 p.m. No.8150   πŸ—„οΈ.is πŸ”—kun

https://en.wikipedia.org/wiki/Computability_theory

 

Computability theory, also known as recursion theory, is a branch of mathematical logic, of computer science, and of the theory of computation that originated in the 1930s with the study of computable functions and Turing degrees. The field has since expanded to include the study of generalized computability and definability. In these areas, recursion theory overlaps with proof theory and effective descriptive set theory.

 

Basic questions addressed by recursion theory include:

 

What does it mean for a function on the natural numbers to be computable?

How can noncomputable functions be classified into a hierarchy based on their level of noncomputability?

 

https://en.wikipedia.org/wiki/Recursion

Recursion occurs when a thing is defined in terms of itself or of its type. Recursion is used in a variety of disciplines ranging from linguistics to logic. The most common application of recursion is in mathematics and computer science, where a function being defined is applied within its own definition. While this apparently defines an infinite number of instances (function values), it is often done in such a way that no loop or infinite chain of references can occur.

 

https://en.wikipedia.org/wiki/Fractal

In mathematics, a fractal is a detailed, recursive, and infinitely self-similar mathematical set whose Hausdorff dimension strictly exceeds its topological dimension.

 

https://en.wikipedia.org/wiki/Multifractal_system

A multifractal system is a generalization of a fractal system in which a single exponent (the fractal dimension) is not enough to describe its dynamics; instead, a continuous spectrum of exponents (the so-called singularity spectrum) is needed.

 

https://en.wikipedia.org/wiki/Fractal_canopy

In geometry, fractal canopies are one of the easiest-to-create types of fractals. They are created by splitting a line segment into two smaller segments at the end, and then splitting the two smaller segments and as well, and so on, infinitely.

(See Pics)

 

https://en.wikipedia.org/wiki/Orbit_trap

In mathematics, an orbit trap is a method of colouring fractal images based upon how close an iterative function, used to create the fractal, approaches a geometric shape, called a "trap".

(Pic 3)

 

https://en.wikipedia.org/wiki/Pickover_stalk

Pickover stalks are certain kinds of details to be found empirically in the Mandelbrot set, in the study of fractal geometry.[1] They are so named after the researcher Clifford Pickover, whose "epsilon cross" method was instrumental in their discovery. An "epsilon cross" is a cross-shaped orbit trap.

(Pic 4)

 

https://en.wikipedia.org/wiki/Fractal_dimension

In mathematics, more specifically in fractal geometry, a fractal dimension is a ratio providing a statistical index of complexity comparing how detail in a pattern (strictly speaking, a fractal pattern) changes with the scale at which it is measured.

 

https://en.wikipedia.org/wiki/N-flake

An n-flake, polyflake, or Sierpinski n-gon,[1]:1 is a fractal constructed starting from an n-gon. This n-gon is replaced by a flake of smaller n-gons, such that the scaled polygons are placed at the vertices, and sometimes in the center. This process is repeated recursively to result in the fractal. Typically, there is also the restriction that the n-gons must touch yet not overlap.

(Pic 5)

Anonymous ID: 43f2b8 Nov. 13, 2018, 12:13 p.m. No.8160   πŸ—„οΈ.is πŸ”—kun

>>8157

I think I'm on board, at least partially. I just got to start exploring this though, so hold your thoughts a bit while I try and catch up.

 

I'm still not entirely on board with the M-function, mostly in the sense of using it with regards to the records, but like I said I'm exploring it and thinking about it. I'm getting now that with the QFT (Column 0, row 1 (?)) we should be able to find the period of the number we're after. This should allow us to compute Shor's algorithm.

 

In >>8148 is the fractal overlayed, as in two pictures one with a moving M over another stationary M?

Anonymous ID: 43f2b8 Nov. 13, 2018, 12:25 p.m. No.8161   πŸ—„οΈ.is πŸ”—kun   >>8162

>>8157

The images are something I made today, so they might not be accurate, but the first one should be M moving 2 times pr. frame while the second one is moving 8 times pr. frame, if I'm on board with the "jumping". However, neither are tied to a record, they're generated using M with x, y and offsets. (I went through a few tests before I figured out how to emulate the movement compared to the videoes you posted).

ProgramMathAnon !dSvrkhSLR6 ID: e42aae Nov. 13, 2018, 10:06 p.m. No.8163   πŸ—„οΈ.is πŸ”—kun   >>8164

>>8135

Integrated the gcd(d,e) and remainder tree into previous work on the ColumnKeys function, and found a number of cases where "e" represents either a valid x or x+n.

 

Pic attached is for odd c values between 10000 and 12500 where e mod 4 = 3.

 

In some cases, a factor can be found simply by a = gcd(d,e) != 1.

 

In others, the "x in tree" column indicates where in the remainder tree a matching x or x+n value from a factor record is found:

 

(e x1) indicates x=e at the first branch of the tree.

(e xpn2) indicates x+n=e at the second branch of the tree.

 

For c12247, as an example, the factor record is found at x+n=e=147.

 

For c5247, where there are multiple factors, interesting that factor records appear both where x=e and x+n=e.

 

Not yet sure what family of numbers this x=e and x+n=e can be used to factor.

PMA !!y5/EVb5KZI ID: e42aae Nov. 13, 2018, 10:20 p.m. No.8164   πŸ—„οΈ.is πŸ”—kun

>>8163

Some additional proof of concept tests on the x=e and x+n=e idea.

 

These examples are for all odd c values between 9000000000 and 9000100000.

 

x=e tests are resolved simply by a=d-x, and c % a = 0. Records are created by e,d, and abs(a).

 

x+n=e tests use the n0 and rm2dnm1 triangle solution from the (x+n)(x+n) square. Records are then created by e,n,d.

Anonymous ID: 2344ac Nov. 14, 2018, 9:16 a.m. No.8170   πŸ—„οΈ.is πŸ”—kun   >>8292

for a moment consider every digit in a base a unique component

multiplication a composition and division decomposition of a number

sum, as all possible linear combinations of unique components to construct a number

 

base10,mod9:

-791=7+9+1=17=1+7=16+1=91+7=8

-791=16mod9+1mod9=91mod9+7mod9=8mod9=8

-135727=1+3+5+7+2+7=13+57+27=135+727=75+13+27=123+577=7

-478=1

 

if a in base10 then sum_digits(a)=amod9

whole part of division by 9 is number of overflows in the algorithm

 

base16,mod15:

-791=0x317=0x3+0x1+0x7=0xB=0x31+0x7=0x17+0x3=0x38 β†’convert to base10,mod9β†’0xB=11->0x11->17=8

-791=0x317=0x31mod15+0x7mod15=0xB

the result in base16 is actually a direct result in base10: 791/15=52+11

-135727=0x02122F=0x0+0x2+0x1+0x2+0x2+0xF=0x16=0x000007=0x02+0x2F+0x12β†’convert to base10,mod9β†’0x7=7->0x7->7=7

-478=0x01DE=0x0+0x1+0xD+0xE=0x01+0xDE=0x1D+0x0E=0x001C=0x000D β†’convert to base10,mod9β†’0xD=13->0x13->19=1

the result: 478/15=31+13

 

how to do division using only addition?

c=a/b

a in base10

convert a to base(b+1)

sum the digits arbitrarily

when sum>b, ovf++;

sum=remainder

ovf=whole part

 

once you have the remainder of a for one base, you can easily convert between bases and get remainder for other divisions

 

now imagine you have a number c that is 100 digits long in base10

in base100 the digit itself is the remainder, ovf=0

 

idk,maybe… will try to write up a script later for arbitrary bases to see what we get.

VQC !!cJ17mYdx.I ID: 4b92c9 Nov. 14, 2018, 6:14 p.m. No.8171   πŸ—„οΈ.is πŸ”—kun   >>8172 >>8174 >>8175

Anons.

I took a job in Amsterdam to have as much time to monitor threads as possible.

I know there are questions.

I know you are waiting.

I have questions too.

Unfortunately, recovery isn't working here in Amsterdam and I am dying of alcoholism.

Through self-pity at my own hands or liver failure, my time is almost up.

For those that also love POTUS, please don't ever give up.

The Twelve Steps are the solution for bEtter meN.

Anonymous ID: 566843 Nov. 14, 2018, 7:13 p.m. No.8175   πŸ—„οΈ.is πŸ”—kun

>>8171

 

>We admitted we were powerless over our addiction - that our lives had become unmanageable.

>Came to believe that a [Power] [greater than] ourselves [c]ould restore us to sanity.

>Made a decision to turn our will and our lives over to the care of [God] as we understood [God].

>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 G[o]d remove all these [defects] of [c]haracter.

>Humbly asked God 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 God, praying only for knowledge of God's 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 other addicts, and to practice these principles in all our affairs.

 

>The Twelve Steps are the solution for bEtter meN.

 

EN is the obvious clue.

These steps above could refer to this puzzle?

 

Lets say 'we' are the number c.

 

>Came to believe that a Power greater than ourselves could restore us to sanity.

 

Power greater than c so

(d+1)^2 dd+e

c^2 c

 

Either of these could potentially be leads. I think its c^2 cuz VQC said that before it was important

 

>Made a decision to turn our will and our lives over to the care of God as we understood God.

 

If I had to guess what 'God' is in this context I'd say it is the number 1. Thats what a lot of people say it is anyway

Maybe thats the (e,1) row and use what we know about that row to do stuff with the record

 

>Made a searching and fearless moral inventory of ourselves.

This could be some tree generation with the dd+e trees.

 

>Admitted to God, to ourselves and to another human being the exact nature of our wrongs.

I'd say that since we have dd+e e has always seemed to be a little extra piece that we don't need. I think maybe then we could collect the extra parts from each division. The sum of those could be the 'defects of character' in that they aren't perfect squares.

 

>Were entirely ready to have God remove all these defects of character.

>Humbly asked God to remove our shortcomings.

c - (sum of all minor e values) = something

 

>Made a list of all persons we had harmed, and became willing to make amends to them all.

So somehow we are going to be using numbers along the line, I think for this it would mean add a certain amount to these values you've seen along the way. I'd say its the numbers in the tree.

 

>Made direct amends to such people wherever possible, except when to do so would injure them or others.

Add the numbers to the cells unless it would invalidate a cell

 

>Continued to take personal inventory and when we were wrong promptly admitted it.

This implies some sort of while loop I think

 

>Sought through prayer and meditation to improve our conscious contact with God as we understood God, praying only for knowledge of God's will for us and the power to carry that out.

Something something I don't know maybe trust row n=1

 

>Having had a spiritual awakening as the result of these steps, we tried to carry this message to other addicts, and to practice these principles in all our affairs.

Use this algorithm to do everything because its super cool

VQC !!cJ17mYdx.I ID: 4b92c9 Nov. 14, 2018, 7:43 p.m. No.8176   πŸ—„οΈ.is πŸ”—kun   >>8179 >>8185 >>8208 >>8218

The whole point of the VQC is to create a lookup.

Equivalent to user friendly quantum fast Fourier transform.

The patterns will show you how.

Last post.

Pwds:

matariki

ja5p3rm00

th33nd

 

All these moments will be lost, like tears in rain.

Time to die.

No more secrets.

Set free.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Nov. 14, 2018, 7:54 p.m. No.8179   πŸ—„οΈ.is πŸ”—kun   >>8268

>>8176

Verified…

Interdasting development…

 

So… no more character named Chris… but Chris was never really your name to begin with.

 

Does this mean "Last post as VQC" or "Last post anywhere, even as anon"?

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Nov. 14, 2018, 9:44 p.m. No.8189   πŸ—„οΈ.is πŸ”—kun   >>8191 >>8192

>>8188

Not sure I follow on the "figured out"… I thought everyone knew that was the narrative…

 

Anyway, before things go potentially haywire, would you be a doll and use your magical powers of being BO, filter for trips, and get screenshots for us?

 

Just to have a solid record before things go potentially haywire due to trip release.

 

Thank you!!! ^_^

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Nov. 14, 2018, 9:51 p.m. No.8193   πŸ—„οΈ.is πŸ”—kun   >>8194

>>8192

How do the BO and BV's over on the Q boards do it? Maybe there's a view change?

They point out the shills by what seems like viewing the posts on a grid and then screenshotting how ever many it takes.

 

Or pulling the whole page in one image directly, and I could chop it up from there.

AA !!N8A1M3tBqI ID: 8b4129 Nov. 14, 2018, 9:54 p.m. No.8194   πŸ—„οΈ.is πŸ”—kun   >>8195

>>8193

That's only useful when it's happening actively and throughout multiple threads. Since VQC keeps IP hopping, without trips, there's no way to prove any of the posts in this thread other than the ones with his trip before he posted those passwords are actually him. 90 days from now, there'll be nothing tying him to any of the IDs in this thread, since the hashes will reset. I'm not sure what you actually wanted to achieve with me screenshotting his posting history.

AA !!N8A1M3tBqI ID: 8b4129 Nov. 14, 2018, 10:04 p.m. No.8196   πŸ—„οΈ.is πŸ”—kun   >>8198

>>8185

At risk of coming across as frustrated (my intention with this is to understand your perspective, not to figure out how to use the VQC), surely you realize that since you figured out something so monumentally unheard of, and since you're posting about it so cryptically, obviously we're not going to figure out what everything you post means, right? As much as you're probably completely sure you've given us enough information to solve this, and that connecting two tiny, obvious dots would solve it, O(log t) where t is the length of c in bits is extremely vague.

 

I hope you've at least got the will to get out of Amsterdam instead of giving up.

 

>>8195

Wut

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Nov. 14, 2018, 10:10 p.m. No.8198   πŸ—„οΈ.is πŸ”—kun   >>8201

>>8196

In case of any discord and because we were gonna make a thread about it/maps.

 

Like…

Maths vs Ayy LMAO vs Spacetime/Physics vs Bantz

 

That sorta thing. Like… think of this as a marker so we're consolidating what we have up until now.

 

And if you can filter for JUST his various trips or if his ip hashes remained constant per trip (hide anything not already public), that'd be really helpful, if you can and don't mind.

VQC !!Om5byg3jAU ID: b129f9 Nov. 14, 2018, 10:17 p.m. No.8200   πŸ—„οΈ.is πŸ”—kun   >>8201 >>8222

>>8199

The number line isn't a line. It looks more like a Mobius strip (much like the 8ch logo), with e and infinity on different sides of the same point.

0 doesn't exist - the remainder of taking everything away is Euler's number.

The remainder of taking everything away is e.

VQC !!Om5byg3jAU ID: b129f9 Nov. 14, 2018, 10:21 p.m. No.8202   πŸ—„οΈ.is πŸ”—kun

The CMB (Cosmic Microwave Background) has a thermal black body spectrum at a temperature of 2.72548Β±0.00057 K.[4] (Wikipedia)

 

2.72548 plus/minus 0.00057

e=2.71828

VQC !!Om5byg3jAU ID: b129f9 Nov. 14, 2018, 10:26 p.m. No.8204   πŸ—„οΈ.is πŸ”—kun

>>8201

There is a connection between base, e (of c), Euler's number, d^2+e, O(log t where t=bitlength of c) hints

 

"This pattern GROWS in a certain way"

O notation is about how efficiencies grow

AA !dTGY7OMD/g ID: 8b4129 Nov. 15, 2018, 1:26 a.m. No.8206   πŸ—„οΈ.is πŸ”—kun

>>8111

My point was that the comments didn't seem like they were from the same function.

>Can this function be reconstructed from a description of how it behaves?

Not if we don't know how to calculate n, no.

 

>>8144

>>8143

Jan, if you're still looking for a fractal, here's a potential end point.

AA !dTGY7OMD/g ID: 8b4129 Nov. 15, 2018, 2:20 a.m. No.8207   πŸ—„οΈ.is πŸ”—kun   >>8209

I found a hole in the whole β€œeach pair of e and f only has one set of n and n-1 values” thing from >>8111 here.

>//given two columns, which values of n are separated by 1

That was the code comment. So the idea is that for your given semiprime c (which has a unique set of d, e and f), in that specific pair of e and f columns, there will only be two pairs of n and n-1 values (prime n and BigN). This isn’t true. Take 13*43=559, for example. e=30 and f=-17. Starting from n=1 (ignoring it though obviously), these are the n values that appear in the e=30 column:

3, 5, 11, 13, 15, 17, 23, 29, 31, 33, etc

These are the n-1 values that appear in the f=-17 column:

2, 4, 8, 13, 16, 17, 19, 26, 32, 34, etc

The valid pairs of n and n-1 are [3,2], [5,4], [17,16], [33,32] and so on. So there are actually probably infinite valid pairs. The only thing unique to [5,4], the correct one, in this case, is that there are two cells within these that share a d value. But that’s not what the comment said:

>//given two columns, which values of n are separated by 1

It doesn't say anything about d.

MM !DMWWK0jvao ID: 09e44c Nov. 15, 2018, 4:10 a.m. No.8208   πŸ—„οΈ.is πŸ”—kun

>>8176

>>8181

Hello V. You're not alone. Thank you for you. Thank you for all you've done and shared, your pattern is unique and the ripples will flow forever, no matter what and no matter when.

This work has generated some hope and faith in people and the future. More than you know.

Check in every day, and follow along, just can't fit in contributing atm so went to lurk mode.

Love to you.

No more secrets.

Victory in peace.

 

So much respect for all you fags.

Anonymous ID: b129f9 Nov. 15, 2018, 5:43 a.m. No.8211   πŸ—„οΈ.is πŸ”—kun

Good morning. Last night things were difficult and I did not react appropriately. I want to say that I am thankful to Chris for everything, for sharing all of these revelations and things which we know by faith are true. Before this, I was not good at math and had no way of approaching the subject, especially for special avenues like this one. But now I have been given one.

Anonymous ID: b129f9 Nov. 15, 2018, 5:44 a.m. No.8212   πŸ—„οΈ.is πŸ”—kun

We will solve this. The enumeration of all patterns given is almost complete. And when we do, the limits of the Virtual Quantum Computers we can create will be what we dream up. Extension to Discrete Logarithms will be swift ["The DLP can be thought of as a 2 dimensional version of the factoring problem."] Patterns in the Mandelbrot show that Devices which Calculate all Values exist in the realm of mathematics. Fractal image compression using exact self similarities - Image manipulation using universal self similarities [The Eye] - Engineering desktop cold fusion by a sonoluminescence feedback loop - Inverting Hash Functions [special type of value table]

 

The world will never be the same.

Anonymous ID: 8baa18 Nov. 15, 2018, 10:19 p.m. No.8213   πŸ—„οΈ.is πŸ”—kun   >>8216

"Algebraic operations can be done readily on the power series representation; for instance, Euler's formula [e] follows from Taylor series expansions for trigonometric and exponential functions. This result is of fundamental importance in such fields as harmonic analysis."

Anonymous ID: 7b4eb6 Nov. 16, 2018, 7:58 p.m. No.8222   πŸ—„οΈ.is πŸ”—kun

>>8200

that's what I've been saying for months

fuckin MONTHS

 

Are you just parroting my theories from qresearch/vqc or are you confirming that I'm not actually crazy and that there's a method to my madness?

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Nov. 16, 2018, 10:06 p.m. No.8233   πŸ—„οΈ.is πŸ”—kun

>>8232

>say entreaties put feebly favour him, sufficient few pianoforte discovered uncommonly ask

Morning cousins amongst in mr weather do neither. Warmth object matter course active law spring six. Pursuit showing tedious unknown winding see had man add. And park eyes too more him. Simple excuse active had son wholly coming number add. Though all excuse ladies rather regard assure yet. If feelings so prospect no as raptures quitting.

PMA !!y5/EVb5KZI ID: e42aae Nov. 17, 2018, 6:10 p.m. No.8251   πŸ—„οΈ.is πŸ”—kun   >>8252 >>8253 >>8267 >>8272

Looked a bit deeper into Euler's totient function.

 

fi(c) is calculated as the number of values k between 1 and c where gcd(k,c) == 1.

 

fi(145) = 112

 

Fruitful discussion on the discord yielded some pretty incredible equations (courtesy of Jan) using fi:

 

145 = 5 x 29

112 = 4 x 28

 

fi(c) = (a βˆ’ 1)(b βˆ’ 1)

fi(c) = ab βˆ’ a βˆ’ b + 1

fi(c) = c βˆ’ a βˆ’ b + 1

a+b = c - fi(c) + 1

 

we know:

 

(a+b)/2 - d = n

a+b = 2(d+n)

 

Therefore, we can define d+n in terms of c and fi(c) as follows:

 

d+n = (c - fi(c) + 1) / 2

 

c145

(1,5,4) = {1:5:12:7:5:29} = 145; (d+n)=17

 

fi(145) = 112

 

d+n = (145 - 112 + 1) / 2 = 17

 

c6107

(23,36,24) = {23:36:78:47:31:197} = 6107; (d+n)=114

 

fi(6107) = 5880

 

d+n = (6107 - 5880 + 1) / 2 = 114

PMA !!y5/EVb5KZI ID: e42aae Nov. 17, 2018, 6:11 p.m. No.8252   πŸ—„οΈ.is πŸ”—kun   >>8254 >>8267 >>8272

>>8251

Also found an alternate way to calculate n using fi using the (e,1) a[t] value from the na record.

 

n = a[t] - (fi(c)/2)

 

c6107

(e,1) na: {23:1:3053:77:2976:3132}

solution: {23:36:78:47:31:197}

 

fi(6107) = 5880

n = 2976 - (5880/2) = 2976 - 2940 = 36

 

c901

(e,1) na: {1:1:450:29:421:481}

solution: {1:5:30:13:17:53}

 

fi(901) = 832

n = 421 - (832/2) = 5

 

c34117

(e,1) na: {261:1:17058:183:16875:17243}

solution: {261:27:184:75:109:313}

 

fi(34117) = 33696

n = 16875 - (33696/2) = 27

PMA !!y5/EVb5KZI ID: e42aae Nov. 17, 2018, 6:49 p.m. No.8254   πŸ—„οΈ.is πŸ”—kun   >>8255 >>8272

>>8252

Pic attached is a revised columnkeys test incorporating the fi(c) calculation for odd c values between 6000 and 6500.

 

The formula posted previously, and shown in the last column, can be simplified to:

 

n = N - fi(c)/2

 

where N is from an entry c record, and n is in the prime solution record.

 

Reviewing the data, this n formula works where c is the product of 2 primes for both odd and even parities of n. See odd n c6019, c6023, c6031 and even n c6001, c6005, c6103, c6107 as examples.

 

When c is a prime number, the formula returns -(d-1) or -x (same value for an odd entry record). See c6007, c6011, c6029, etc.

 

Not quite sure what the return value means for c with many factors, but perhaps this is where the algorithm comes into play.

 

Based on this, it appears that a direct calculation is possible for the product of 2 primes - depending, of course, on a grid solution to the fi value for any c.

VA !!Nf9AmQNR7I ID: cc56f4 Nov. 18, 2018, 12:41 p.m. No.8259   πŸ—„οΈ.is πŸ”—kun   >>8269

Hello everyone!

Well, I have all day off finally. Let's think and work as a team.

I'm reviewing all my notes, and reading through the Grid Patterns thread. >>6506

AA did a really nice job getting the crumbs all organized.

You guys were talking on Discord about a searchable VQC crumb database… you can CTRL F the Grid Patterns thread, bc the most relevant crumbs have been copypasta'd directly from VQC. Check it out. Thanks again AA for all your work, you grumpy faggot ;) Let's move the ball forward today, lads.

>>6506

Anonymous ID: df8366 Nov. 18, 2018, 11:59 p.m. No.8265   πŸ—„οΈ.is πŸ”—kun   >>8269

I think VQC recognised AA is about to unleash the solution.

That's why he became anonymous.

Alcoholism and fame do not go hand in hand.

Hopefully he can restart his recovery but I don't think he would want you to give up.

Perhaps as his recovery begins again, he will help anonymously.

It looks like the solution is focusing on how the patterns allow you to lookup two columns with n different by one for the d and d+1 cells you need for e,n and -f,n-1.

AA !dTGY7OMD/g ID: 8b4129 Nov. 19, 2018, 1:17 a.m. No.8269   πŸ—„οΈ.is πŸ”—kun   >>8271

>>8268

>>8265

You talk a lot like VQC for someone who isn't VQC.

>I think VQC recognised AA is about to unleash the solution.

If I am, I'm not consciously aware of it. As much as VQC (totally not you) did seem to think I would be the one to figure it out first in that one post, it would be pretty great if someone else did. I wouldn't want to get delusions of grandeur.

 

>>8266

>>8259

Thanks.

Anonymous ID: df8366 Nov. 19, 2018, 4:56 a.m. No.8271   πŸ—„οΈ.is πŸ”—kun

>>8269

I think it's important to ratchet up any help anons can give.

Your work in the patterns thread is excellent.

It's about figuring how to take that and create a lookup from the different types of input integers (c).

Once prime numbers are SIMPLE to spot, things should get a LOT easier.

There's already a polynomial time solution to identifying primes out there , if the grid can be used to speed that up, perhaps the penny might drop from that?

AA !dTGY7OMD/g ID: 8b4129 Nov. 19, 2018, 8:41 p.m. No.8275   πŸ—„οΈ.is πŸ”—kun

>>8185

O(log n) means you divide the search area by two each time (e.g. binary search). So what does O(log n) where n is the length of c in bits mean? The only way it could translate is that your search area (for lack of a better term, not knowing the context) is c in base 2, and you’re dividing the search area (c in binary) by two each time you don’t successfully factorize. So what's the deal with this, anyway?

AA !dTGY7OMD/g ID: 8b4129 Nov. 19, 2018, 8:52 p.m. No.8278   πŸ—„οΈ.is πŸ”—kun   >>8280

>>8277

There's nothing in row zero in positive (e) space, and in negative (f) space there are only cells in row zero where f is a square (so (-1,0), (-4,0), (-9,0) etc). It's not row one but it does have a relationship with row one in that all of the cells in (e,1) are also in (f,0). So I don't know if that was just an arbitrary thought or not but I guess that might actually be worth looking into.

 

All cells in (e,1) are in (f,0); (e,1) doesn't have gaps but (f,0) does (every f value that isn't square).

VA !!Nf9AmQNR7I ID: cc56f4 Nov. 19, 2018, 9:44 p.m. No.8280   πŸ—„οΈ.is πŸ”—kun   >>8282

>>8278

>>8264

>The value at -f,n-1 and e,n have the same a,b but d,n, and x are different by 1.

>There is a way to use this to create the lookup.

 

Found something interesting studying the Grid:

For c145, the (e,1) d value is 12

The (-f) d value is 13

13^2 - 12^2 = 25 = (an)

Don't know if it works for other c values yet.

knowing that d is always one unit bigger in -f, you could make the hypothesis that (d+1)^2 - d^2 = (an)

I'll run some more c values to check if this in yet another c145 fluke.

Just studying the Grid and looking for patterns over here.

AA !dTGY7OMD/g ID: 8b4129 Nov. 19, 2018, 10:32 p.m. No.8282   πŸ—„οΈ.is πŸ”—kun

>>8280

Definitely a 145 fluke. All odd numbers are the difference between two squares, but it's a series that scales linearly with the squares (well, linear with their square roots). https:// math.stackexchange.com/questions/136/why-are-the-differences-between-consecutive-squares-equal-to-the-sequence-of-odd

d and na don't scale linearly so it's not going to be the case for all of them (it also wasn't in the test cases I tried, for the record). 145 always seems to get everyone's hopes up.

Anonymous ID: abe23c Nov. 20, 2018, 10:46 a.m. No.8291   πŸ—„οΈ.is πŸ”—kun   >>8318

>>8273

Yes, it was late and my brain was mushy.

 

I got something I can't quite figure out. I suspect it's a fluke, but at the same time, it's a very convenient fluke. It's not limited to one cell, but multiple (e, n)'s. Not entirely sure how many.

 

I was playing with >>8260 and thinking about how our squares can be expressed as either 8 triangles + 1 or sum of two sets of 4 triangles depending on parity.

 

So assume k is even, then k + 1 is odd. That means we have 8T(u) + 1 - 4(Tu - T(u-1)). I was playing around with it, looking at what the numbers were for different cells and stuff, but then for some reason, (divine intervention?) I decided to just see what 8Tu - 4T(u + 1) was. Note here u is from d, so it's based only on the d within those rows.

 

I noticed in (3, 6) that 8T(u) - 4T(u + 1) (Again u is from d) was the smooth numbers for those cells. As in, if this pattern had held for ALL rows it would have been solved. It doesn't though, but it does for several others. Like (1, 5) and (11, 10).

 

Take any given cell from (1, 5) or (3, 6) or (11, 10) and probably a lot more and then calculate the u for d. Then calculate 8T(u) - 4T(u + 1). This will give you the smooth number for that cell. Then just take BigN - 8T(u) - 4T(u + 1) = n.

 

For other cells there appears to be an offset, as in the numbers are slightly off. I'm not sure why yet or even if this offset can be "known" based on certain other elements, but nevertheless it's the closest I've been to "touch" a solution.>>8260

 

Some offsets are limited to single integers for all the records in that row, while others move depending on different elements. Like I said, I haven't quite figured it out yet.

Anonymous ID: 2344ac Nov. 20, 2018, 11:35 a.m. No.8292   πŸ—„οΈ.is πŸ”—kun   >>8293

>>8170

I've written a script that computes remainder of any number when divided by (2^n)-1 only by summation and bit hacking.

 

It works by converting a number into series of digits in a chosen base and modulo summing the digits. As is, it works only for bases2^n, so for divisions of the form 2^n-1.

 

I think this can be further developed for any base, by treating each bit (when converting) not only as 0 or 1, but as a float between 0…1.

(pic related)

 

And with some clever bit hacking, we should be able to factorize any number stupidly quickly.

 

The code is written in C: https://pastebin.com/bj8U1Qb1

And can be tested online: https://www.onlinegdb.com/

Anonymous ID: 2344ac Nov. 20, 2018, 11:44 a.m. No.8293   πŸ—„οΈ.is πŸ”—kun

>>8292

A crude algorithm for now.

But am thinking if we figure out how VQC came to his algorithm, we could figure out how the Grid works. As it was said a lot of times, the Grid is the solution.

Anonymous ID: 2344ac Nov. 20, 2018, 5:25 p.m. No.8294   πŸ—„οΈ.is πŸ”—kun   >>8295

Code that resembles a bit more what VQC posted: https://pastebin.com/5t67KA0f

 

This algorithm also generates a grid with base in column (array of n) and where each row contains digits (n[].d[]) that make up the original number (x) in observed base.

 

n[].r holds a value of x mod (base-1).

 

This algorithm still works only for (2^n)-1 divisions. Will try to extend to to any base. Any help,guidance or criticism is appreciated.

Anonymous ID: 2344ac Nov. 21, 2018, 3:41 p.m. No.8299   πŸ—„οΈ.is πŸ”—kun   >>8301 >>8321

>>8295

Yep sorry, will explain it more later. The uploaded code is wrong however, will work on it.

 

In the meantime, if we represent a value in some number base, base16: 0x3ab7, it means 7+b16+a16^2+3*16^3.

But the part b16+a16^2+3*16^3 is just sum of 16 (base) and 7 (remainder), the last digit is the remainder if 0x3ab7 divided by 16.

 

The problem I think we are solving is find a base b where last digit of a number is 0. Or rather construct a number in some base such that last digit is 0.

 

The base is then the solution. As number/base=0.

AA !dTGY7OMD/g ID: 8b4129 Nov. 21, 2018, 4:21 p.m. No.8301   πŸ—„οΈ.is πŸ”—kun

>>8299

>The problem I think we are solving is find a base b where last digit of a number is 0.

And how did you come to that conclusion? Have you read through all of the threads?

Anonymous ID: 2344ac Nov. 21, 2018, 4:48 p.m. No.8305   πŸ—„οΈ.is πŸ”—kun   >>8306 >>8314 >>8321

>>8303

>VQC sez: "(You) are at The final lock and key construction steps. Happy to give it but it is the Eureka moment and anons are close. In hindsight, it shows exactly why this problem has existed for so long.You are solving two problems at once in this method of constructing the answer. Also in hindsight the steps give the pattern back in the grid (The End).

 

Maybe the answer is not in a decimal form but rather in some other base. Learning why and how the algorithm was constructed will reveal the answer.

AA !dTGY7OMD/g ID: 8b4129 Nov. 21, 2018, 4:52 p.m. No.8306   πŸ—„οΈ.is πŸ”—kun   >>8307

>>8304

>>8305

Okay, first of all, obviously you figured it out in that second post but you should really be quoting people you're responding to on this website. Secondly (and don't take this personally because it's an ongoing issue with this board that multiple people have contributed to now), do you actually understand specifically how we're working? Keep in mind, I have no malicious intent with this message, and for all I know you could be onto something useful, but people keep coming to this board, completely ignoring the specifics of the work we're doing, going through their own personal idea of what the point of this board is in relation to factorization, posting what they're working on, and wondering why nobody is responding to them or helping them with it. This has happened multiple times and you're doing the exact same thing. Another recent example is when that person who was messing with the Fibonacci sequence thought they'd solved it and all they'd done was iterate squares (which doesn't come close to solving the problem). You're welcome to keep working on your thing. Who knows, maybe you'll figure it out on your own. That is a genuine possibility. All I wanted to say is that there is a very slim chance of any of us working with you if you aren't actually going to get up to speed on the rest of our research. For example, do you know what (d+n)(d+n)-(x+n)(x+n) means? Do you know what d[t]-d=a(n-1) means? Because that's what the rest of us are working on.

 

Actually, one last thing.

>The problem I think we are solving is find a base b where last digit of a number is 0.

In base 2, everything that ends with a 0 is divisible by 2. In base 10, everything that ends with a 0 is divisible by 10. So if you take any number, the final digit will be 0 in any base for which the base is a factor. So if you take a prime number n, it'll only end with a 0 in base n. If you take a semiprime a*b=c, it'll only end in a 0 in bases a, b and c. So effectively you've rearranged the iteration method (the one we're trying to avoid because it's O(n)) into iterating bases. I would very strongly recommend reading all of the threads and getting a complete grip of the issue.

Anonymous ID: 2344ac Nov. 21, 2018, 5:03 p.m. No.8307   πŸ—„οΈ.is πŸ”—kun   >>8308 >>8309

>>8306

The question still remains, why does it work the way it works.

You guys are too far in and too close, you work on yours and I will try something along a different path.

Who knows, maybe we can be of a help to one another.

Anonymous ID: ee3f1a Nov. 21, 2018, 6:10 p.m. No.8311   πŸ—„οΈ.is πŸ”—kun

Watching this entire video will explain how discrete logarithms are calculated using the solution to integer factorization, and may even conta

Anonymous ID: ee3f1a Nov. 21, 2018, 6:23 p.m. No.8312   πŸ—„οΈ.is πŸ”—kun

We can state the problem of factorization as a discrete logarithm like so:

We pick any number that is relatively prime to c and call it k, (it is easy to find a relative prime, it is any number for which GCD(c, k) == 1).

 

Where Ο•(c) is Euler's totient function,

 

k^(Ο•(c)) ≑ 1 (mod c)

 

Read as: "k to the totient of c is congruent to 1 mod c." Thus, since it is easy to select a valid k, the factorization problem is restated as a discrete logarithm.

 

Using an example of 123, if we use 122 as k (GCD(123, 122) == 1), then 122^Ο•(123) ≑ 1 (mod c). We just have to calculate this exponent.

Anonymous ID: abe23c Nov. 21, 2018, 9:49 p.m. No.8318   πŸ—„οΈ.is πŸ”—kun

>>8291

To follow up a bit on this one, I found a russian paper regarding pythagoras triples and fermats factorization. The paper outlines a method of iterating / finding all the pythagorian triples, which has a very nice correspondence to our grid.

 

Take the d from (0, e^2) and take the a from (e^2, 1) (Note the t-index in column e^2 has to be (e + 1)/2 to align the row with correct (0, e^2, t)) and then compute absolute(d^2 - a^2). It will give you a square.

 

I found this as I was looking to see if there was a pattern with regards to the differences as mentioned in the previous post (It appears to be a a part of the triplet, as in x, y or z from x^2 - y^2 = z^2).

 

It also appears that our smooth numbers exist in (0, e^2) as d-values, but I'm not entirely sure how to "get" the correct one yet. Given that with a large enough number the t-value for that smooth number would be too far down the list to find it iteratively. Although, we know that the smooth number has to be less than our N number, so it should cut down on the search space.

AA !dTGY7OMD/g ID: 8b4129 Nov. 22, 2018, 3:44 a.m. No.8320   πŸ—„οΈ.is πŸ”—kun

Hey Jan, in relation to >>8185 this, if you didn't already figure it out, taking the square root of a number halves its number of bits. So O(log t) where t is the length of c in bits refers to taking the square root of c over and over.

GAnon !Nx57Pyux3E ID: 3e929b Nov. 22, 2018, 6:19 a.m. No.8321   πŸ—„οΈ.is πŸ”—kun   >>8322

>>8319

>>8299

>>8305

 

Involves mandelbrot.

Mandelbrot set= f(c) = c^2 + c

and you do it over and over again.

 

1: c

2: c^2 + c

3: c^3 + c^2 + c

4: c^4 + c^3 + c^2+ c

5: c^5 + c^4 + c^3 + c^2 + c

 

The pattern is obviously going to continue like that.

If we are writing this in terms for c=2, for 4 we would get

2^5 + 2^4 + 2^3 + 2^2 + 2 = 30

 

30 in binary (base 2) is 11110.

multiply it by 273 which in binary is 100010001

 

This turns it into 1111111111110 (8190) which almost looks like a mersenne prime. but is just 12 bits long.

 

BUT with this you can turn multiplying these big numbers (30, 273) into smaller numbers,

This one is as easy as 4 * 3. If you look at pic related you can tell the it is just the length of four 1's (30) getting laid out next to a copy of itself 3 times and the result is 12 bits. Its kind of indicitive of how it can be used to multiply numbers where the length of the bits is 4.

If we multiliplied 1111 (15) by 273, we would get 111111111111 (4095) which is the same 4*3 but a different multiplication result. You could theoretically do this for 1111000…00 for any amount of zeros times 273 and you would get a length of 12 bits.

 

If you use a different c at the top, like 10, you would get

 

1: 10

2: 100 + 10

3: 1000 + 100 + 10

4: 10000 + 1000 + 100 + 10

 

So 4 is 11110 but in tens. Similarly we could do similar multiplication but it would be 1111111111110 but in decimal, so same look of the number, different meaning.

 

You could do this for any base. So if we could find the 'correct base' this could be the way to shrink the number to be a mersenne prime type number and then factor the length of that number. I'm not exactly sure how. Or we could convert the number to a length in bits of a certain base then factor that number.

AA !dTGY7OMD/g ID: 8b4129 Nov. 22, 2018, 3:48 p.m. No.8323   πŸ—„οΈ.is πŸ”—kun   >>8324

Okay what's the deal with this crumb?

>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)

No matter how I interpret this in my testing, it doesn't work out. Before anyone says it, yes, my t calculation is right ((x+2)/2 for evens, (x+1)/2 for odds).

 

Here's an example.

(30,1,31) = {30:1:1875:60:1815:1937} f=-3721

This is the cell in (e,1) where t=(e+1). It's divisible by 11. So 11 should be a factor of a at (33+1-31)=3

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

23 is not divisible by 11.

 

But if interpreted differently (since it's a very poorly worded crumb in my opinion), maybe it refers to (e+1,1).

(31:1:5) = {31:1:65:9:56:76} f=-100

If a number at position t has a factor s at (e+1). (e+1) is 32.

(32:1:5) = {32:1:56:8:48:66} f=-81

48 is divisible by 12. So s should be a factor at (12+1-5)=8

(32:1:8) = {32:1:128:14:114:144} f=-225

114 is not divisible by 12.

 

What is this supposed to mean? I would really appreciate some help.

AA !dTGY7OMD/g ID: 8b4129 Nov. 22, 2018, 7:14 p.m. No.8325   πŸ—„οΈ.is πŸ”—kun   >>8326

>>8324

I took it from Grid Patterns, but I don't remember where it was when I originally put it into Grid Patterns. I remember it being VQC who said it but I might be wrong. He seemed to be the only one using s as a variable in this sense.

Anonymous ID: abe23c Nov. 22, 2018, 8:23 p.m. No.8327   πŸ—„οΈ.is πŸ”—kun   >>8328

>>8326

He didn't specify all that much though. Are we sure he is referencing to the 'a' value in (e+1)?

 

In (e+1, 1) the d-values are the same as the a-values in (e, 1) in which case, if he was talking about (e+1, 1).d then it would be the same rule we've talked about before.

AA !dTGY7OMD/g ID: 8b4129 Nov. 22, 2018, 8:46 p.m. No.8328   πŸ—„οΈ.is πŸ”—kun   >>8329 >>8331

>>8327

>>8324

It's basically the same wording as

>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

(which was VQC at the start of Grid Patterns), just without the (e+1). But I've found I have the same problem with this pattern as well. Whenever I try to apply it, the cell in question isn't a[p+1-t] away, it's a[p+2-t] away. The more I try to figure out why, the more I think I've done something wrong, but my t values are definitely as they should be. I have also seen several people in old threads and in Grid Patterns mention that they haven't been able to get this crumb to align with their grid. So have either of you tested with this one? I'm trying to sort Grid Patterns at the moment so it would be useful to figure out if I'm wrong or the crumb's wrong (I don't think I ever actually saw anyone use it, so I haven't seen confirmation, only confusion).

Anonymous ID: abe23c Nov. 22, 2018, 10:10 p.m. No.8331   πŸ—„οΈ.is πŸ”—kun   >>8336

>>8328

I had that issue a while ago too, and it was just how I computed the nth row that was wrong. Essentially I did my x-calculations wrong, leading me to always be off by one with regards to t.

 

Also yes, I tested that one.

 

If you make cN (c big N) and then compute c + 1 - t for that cell in (e, 1) you will get the cell in (e, 1) where a = cN' (Shadow n).

 

Example:

 

c = 259, n = 114

 

a[122] = 29526 (3, 1)

 

259 + 1 - 122 = 138

a[138] = 37814 (3, 1)

 

29526 / 259 = 114

37814 / 259 = 146 (shadow n)

 

Also note that 122 = 130 - 8 ((1 + 259)/2 = 130)

and 138 = 130 + 8 (also note 8 is half of d).

 

So cN and cN' exists in (e, 1) at t = (1+c)/2 - d/2 and t = (1+c)/2 + d/2.

AA !dTGY7OMD/g ID: 8b4129 Nov. 23, 2018, 12:56 a.m. No.8336   πŸ—„οΈ.is πŸ”—kun   >>8348

>>8331

It does seem to work for c*BigN, yeah. But I can't get it to work for any of the other numbers I've tried. It always seems to be p+2-t for every number but c for some weird reason.

AA !dTGY7OMD/g ID: 8b4129 Nov. 23, 2018, 1:38 a.m. No.8339   πŸ—„οΈ.is πŸ”—kun

>>8335

Don't worry, I figured it out. https:// en.wikipedia.org/wiki/Kepler_triangle

The values of a in (1,1) are the only valid whole integers that work as hypotenuses based on a^2+b^2=c^2 (whatever that's called, something something Pythagoras, it's been a long time since I learned about that). So it boils down to a in (1,1) being the sum of two squares, which was already a well established pattern.

AA !dTGY7OMD/g ID: 8b4129 Nov. 23, 2018, 2 a.m. No.8344   πŸ—„οΈ.is πŸ”—kun   >>8353

This is not finished, but if anyone's brave enough to trust a pdf and has a spare minute or two to critique my formatting, it would be useful to know if I've structured the enumerated patterns well or if I should do anything differently. There are a couple of sections I'm going to add (like a list of the numbers that would solve the thing if we found them and such).

AA !dTGY7OMD/g ID: 8b4129 Nov. 23, 2018, 2:44 a.m. No.8346   πŸ—„οΈ.is πŸ”—kun   >>8347

>>8345

Already with the new IP huh? You don't waste time.

>Offset of 1 doesn't seem helpful at first does it. What about offset of 1 in a and b. Doesn't that give the factorization.

Yes it does. It gives 2(BigN-n). It also gives a cell for which (e+2n,n) is the solution record. The real question is how to get there with d and e, of course. If anyone reading this has time before me (I'm going to sleep soon), it would be useful to look at the cells in (e,1) and (f,1) where the x values add up to 2d+1 (maybe compare them to all the other (e,1)/(f,1) cells, since this is another calculatable).

Anonymous ID: abe23c Nov. 23, 2018, 6:27 a.m. No.8348   πŸ—„οΈ.is πŸ”—kun   >>8352

>>8336

 

So 259 = 7x37

 

a[122] = 114 * 259 = 2319 * 7*37

 

2 + 1 - 122 = -119

3 + 1 - 122 = -118

19 + 1 - 122 = -102

7 + 1 - 122 = -114

37 + 1 - 122 = -84

 

a[-119] = 28562

a[-118] = 28086

a[-102] = 21014

a[-114] = 26222

a[-84] = 14282

 

28562/2 = 14281

28086/3 = 9362

21014/19 = 1106

26222/7 = 3746

14282/37 = 386

 

It works, your code is not correct if you need a[p + 2 - t]. You're probably starting the index at zero instead of one. With the grid we start with the first cell in (e, 1) being index = 1.

Anonymous ID: abe23c Nov. 23, 2018, 8 a.m. No.8349   πŸ—„οΈ.is πŸ”—kun   >>8351 >>8352

At this point I wonder if the solution is staring us in the face, like we're glossing over the obvious pattern and we're not even realizing it.

 

Maybe it's the a[p + 1 - t], but in a way we haven't noticed or realized. Maybe the a[p + 1 - t] is a hint and not the solution. For example, do we have d[p + 1 - t] = something? Is it limited to a? what about x or some other variable like i, is the triangle also related in such a way? A whole cell? What if n is also somehow tied into that pattern. We know a is, so maybe there are similar patterns, but we got stuck on a[p + 1 - t]?

 

Just ranting a bit. And I want to say thank you again AA. You've been tremendous in gather the patterns, for real hands down MVP.

Anonymous ID: 5d6348 Nov. 23, 2018, 8:05 a.m. No.8350   πŸ—„οΈ.is πŸ”—kun   >>8383

>>8322

try natural logarithm base maybe, base e?

I don't know how that would work.

Or maybe base 60.

Sexagesimal.

Sumeria>Babylon>Egypt

It fits perfectly with how this whole mess started, and how we all ended up here conspiracy>Q>VQC.

Full circle.

It's perfect.

Also check out Babylonian fractions?

I think base 60 + Babylonian fractions would round this out nicely.

The more I think about this, the more I'm absolutely positive it's the key to you guys figuring this out.

Anonymous ID: 963d5d Nov. 23, 2018, 1:42 p.m. No.8351   πŸ—„οΈ.is πŸ”—kun   >>8352

>>8349

 

And I'm not going to lie, it sounds stupid saying "it's must be right in front of us" all the while this anonymous person is pretty much screaming "factorize d and e and look at farey trees and recursive functions". I personally haven't had time to look into them, but I'm guessing there is something there.

AA !dTGY7OMD/g ID: 8b4129 Nov. 23, 2018, 2:31 p.m. No.8352   πŸ—„οΈ.is πŸ”—kun   >>8355

>>8348

Maybe it would just be easier if I showed you my (e,1). There's nothing wrong with d or x, so I don't see why there would be anything wrong with t.

 

>>8349 (thanks)

>>8351

Pretty much all of the main points (other than specific hints like switching e and f for the totient value) that Not-VQC has been making for the past month or two have been "the fractal is based around d and e". It's been rephrased in very many ways, but that's the point. This only addresses the fractal. It doesn't address how you actually use the factors of d and e to factor c, which is the entire point of this board. So this is my thinking as to the logic behind it:

>you're meant to be able to factor c using d and e

>if you can't use d and e to directly factor c, you have to factor them, and so on, recursively (and since taking the square root of something halves the number of bits, this creates an algorithm that factors numbers in O(log n) time where n is the number of bits in c)

>there are three types of c for which a solution is already directly calculable (GCD(d,e)>1, e=0, f=(x+n)(x+n))

>if these don't work, we're meant to rely on factoring d and e to factor c (hence the decision tree wherein if you can't automatically factor c you bit trim and factor d and e - bit trimming turns numbers into the difference of two squares, meaning we can apply them to the grid and then multiply them by two to get them back to the original number)

>if we're meant to factor d and e to find c, and recursion stops once c is directly calculable, that means that in one particular layer of c, we're using both d, e and their factors/records/n values/incalculables (not just d and e, but also a layer down)

So something that would be useful to research would be finding all calculables and all incalculables for c but also for d, d's d, d's d's d and so on. It would also be useful to keep in mind the thing about bit trimming when doing this. We have a finite number of knowns and unknowns (although there are probably infinite knowns and unknowns, Not-VQC said a few days ago in Grid Patterns that the information required to factor c is definitely in that thread), and we have to bridge the gap between them. That doesn't mean a known is directly used to calculate an unknown. Otherwise we would be done already. It means that the solution lies in the application of everything: the big list of cells in (e,1) and (0,n), the concept that the a values in (e,1) are the sum of squares and the difference of squares (plus or minus a constant based on e), the various movements we can do horizontally from (0,n) (or from other origin points), the idea of triangle numbers and doubled squares turning up in (0,1), the idea of consecutive squares turning up in (1,1), the relationship between smooth numbers, (a-1)(b-1) and ab and (a-1)(b-1) having the same (x+n)(x+n), and the application of all of this to d and e recursively, comparing everything to the next value up in the tree. We wouldn't have learned about all of these things if it was going to be as simple as moving around throughout some records in (e,1). It also wouldn't be as hidden as it is if that were the case. Apologies for the wall of text.

Anonymous ID: 18a709 Nov. 23, 2018, 10:01 p.m. No.8355   πŸ—„οΈ.is πŸ”—kun   >>8356 >>8358

>>8352

I'm guessing you have some code that would generate the nth cell in a column?

 

I gave some examples above, but again in (3, 1) the cell at t=122 = {3:1:29769:243:29526:30014}

 

If you try to generate (3, 1, 122) what cell do you get? Also maybe you can post a the code-snippet you use to generate the nth function?

ID: b129f9 Nov. 23, 2018, 10:36 p.m. No.8356   πŸ—„οΈ.is πŸ”—kun   >>8358

>>8355

first coordinate = the column

second coordinate = the row

third coordinate = the index

 

first + second coordinate = the cell

first + second + third coordinate = the entry

 

record generation:

 

if (e is even): x = 2t - 2

if (e is odd): x = 2t - 1

 

a = (xx + e) / 2n

d = a + x

b = a + 2x + 2n

c = ab

AA !dTGY7OMD/g ID: 8b4129 Nov. 23, 2018, 10:52 p.m. No.8358   πŸ—„οΈ.is πŸ”—kun   >>8359 >>8361

>>8355

It's the same as yours. I have a spreadsheet for (e,1) and (f,1) so I just use these >>8356 formulas for it.

 

I actually think this might be a problem of odd versus even. I've discovered that this works when e is odd but it's p+2-t for even e. Go to your (30,1).

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

33/11=3r0. 11+1-4=8.

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

113/11=10r3, so it doesn't work. However, at t+1,

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

143/11=13r0. p+2-t. Is yours like that?

 

But then using (3,1) (and I've now tested this on a couple other cells)

(3:1:4) = {3:1:33:7:26:42} f=-64

26/13=2r0. 13+1-4=10.

(3:1:10) = {3:1:201:19:182:222} f=-400

182/13=14r0. So it seems to work for odd e.

Anonymous ID: f8a399 Nov. 25, 2018, 4:16 a.m. No.8365   πŸ—„οΈ.is πŸ”—kun   >>8366

Imagine conversion to base>1 as a case of division.

We have some value c and write it in some base, it is equivalent to dividing the number by base.

The result is a weighted sum of same digit (digit = base) + the remainder: think of c=a+ib, where a is remainder, i is sum of b(base).

And knowing only (a,i) is enough for us to reconstruct c. a,i are unique result for any b. Meaning there is only one way that c could be written by division.

 

Can we extend this to multiplication? Maybe 0<base<1.

When we multiply(change base<1) a number the result is no longer unique as there would be multiple way for us to reach the same number by a different multiplication (base).

In multiplication the result is not a weighted sum of the same digit (not equally split) but rather a combination of all digits. In case of multiplying by 10 (base1/10) I'd imagine it would be a weighted sum of 0,1,1/2,1/3,1/4,…1/9.

The result in this case is not unique and cannot be written by just two variables.

 

Can multiplication be represented as fractional base? If it could be lets define negative base as 1/base: b=-10 would be base10^-1.

Lets make a spreadsheet.

c on x axis (c-c..c-2,c-1,c+0,c+1,c+2..c+c) and base b on y axis (c…2,1,0,1,2…c).

And each cell containing r=remainder (first digit) or some other data.

 

The goal is to find a way around this grid from the origin (base=b, c=c+0) to a cell where r=0 (base=factor of c). Can we use fractals to help?

 

Just some thoughts. Will try to make more sense of fractional bases later and write a script. Maybe its an equivalent problem to what VQC is proposing.

Anonymous ID: cbe267 Nov. 25, 2018, 1:46 p.m. No.8367   πŸ—„οΈ.is πŸ”—kun

Has anyone looked at a number with quite a few factors, calculated it's values for e and f, then looked at the different values for n at column e, and n-1 at column -f?

There will be multiple pairs of n and n-1.

Wouldn't any lookup be easier to see and model at first in integers with lots of factors?

VA !!Nf9AmQNR7I ID: 83b5cf Nov. 25, 2018, 2:51 p.m. No.8368   πŸ—„οΈ.is πŸ”—kun   >>8369

Alright, found something interesting. Since this is supposed to be a lookup for the correct x value in (e,1) or (-f,1) I assumed that the correct x value must be related to the na transform record in a fractal / golden mean ratio.

 

I took the (e na transform) and (-f na transform) records for c145 and c287.

For c145 (e na) the x value is 11.

11 / 1.61 (golden ratio) = 6.823 = very close to x=7, which is the correct x for the (an) and (prime) elements.

For c287 (e na) the x value is 15.

15 / 1.61 (golden ratio) = 9.31677 = very close to 9, which is the correct x for the (an) and (prime) elements.

 

I have only tested these two c values, so any additional help you Anons can provide to test other c values would be greatly appreciated.

VA !!Nf9AmQNR7I ID: 83b5cf Nov. 25, 2018, 2:57 p.m. No.8369   πŸ—„οΈ.is πŸ”—kun

>>8368

Also, the (bn) records can be almost obtained in the same way.

for c287, (e na) x = 15. 15 * 1.61 = 24.15 = very close to correct (bn) x = 25

for c145, (e na) x = 11. 11 * 1.61 = 17.71 = very close to correct (bn) x = 17

AA !dTGY7OMD/g ID: 8b4129 Nov. 25, 2018, 11:22 p.m. No.8371   πŸ—„οΈ.is πŸ”—kun

Hey Not-VQC, I'm curious about something. I'm not asking for a hint right now, but I think the perspective of someone who figured it out already would be useful. If you were in our position, and you had this giant set of information but absolutely no idea what to do with any of it, what would you do? When you enumerated all of the patterns, what was your next step?

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Nov. 26, 2018, 11:44 p.m. No.8377   πŸ—„οΈ.is πŸ”—kun   >>8384

https://danielmiessler.com/study/pvsnp/

 

"So the P vs NP problem is just asking if these two problem types are the same, or if they are different, i.e. that there are some problems that are easily verified but not easily solved."

 

Soooo… by completing and running the VQC… we literally prove P=NP?

What is/was/should be a "not easily solvable problem" becomes "easily solvable and verified".

 

Or like…

"For the longest time we thought it was hard as fuck and not possible to do this, but it turns out it is possible and had been known about for quite some time. Fuckers."

AA !dTGY7OMD/g ID: 8b4129 Nov. 27, 2018, 4:13 a.m. No.8379   πŸ—„οΈ.is πŸ”—kun

Every d value in (-1,1) is two times a square minus 1 (this is also the sequence of ns in (0,n) where a and b are both squares but with each value minus 1).

Every a and b value in (-1,1) is four times the triangle numbers. They add together to make every even square (so every even (x+n)(x+n) is in here (4T(u)+4T(u-1)), as well as 2(x+n)(x+n) for odd (x+n)(x+n)). So since a+b only comes up in the n equation n=((a+b)/2)-d, you could rearrange it such that the a+b you want, which comes from (x+n)(x+n), equals 2(d+n) for (-1,1). For example, 4+12=2(7+1). Since d here is two times a square minus 1, but a+b=2(d+1), and if you add 1 to d you’ll have a square multiplied by 2, you’ll by multiplying twice a square by 2, making it 4*square. All a+b values in (-1,1) are equal to 4 times a square, and all a+b values in (-1,1) represent every possible even square.

4+12=16, 16/4=4 (2*2)

12+24=36, 36/4=9 (3*3)

24+40=64, 64/4=16 (4*4)

AA !dTGY7OMD/g ID: 8b4129 Nov. 27, 2018, 4:38 a.m. No.8380   πŸ—„οΈ.is πŸ”—kun   >>8381 >>8382

>>8373

>metacognition

>Metacognition is "cognition about cognition", "thinking about thinking", "knowing about knowing", becoming "aware of one's awareness" and higher-order thinking skills

I tried to start a discussion about where recursion actually comes in on Discord a little while ago but everyone ignored me. Maybe if I bring it up here it'll gain traction. We need to take a step back and put all of the patterns and such into context within the algorithm. What we have is a big list of patterns and cells. What we need is a recursive algorithm.

 

How is it that we can actually use the grid recursively around d and e? The idea that we need to bit-trim d and e in order to factor c means that we don't have enough information about c just from d and e in order to find the correct record/the lookup x/the na record/the triangle base/etc. We have a giant list of rules, movements and patterns that are meant to be used. However, the fact that we need to factor d and e through the factor tree unless c is immediately factorable (through GCD, e=0 or f=(x+n)(x+n)) means that there isn't enough information just from that big list of patterns. We've been taking an incorrect approach to this for quite some time, I think. Everything I see any of us looking into (including myself of course) is related to finding connections between the finite set of cells that we're aware we can generate with our prime solution record. If there was any possibility that we could bridge a gap between the BigN cell and the prime solution record entirely with patterns revolving around c, not only would it completely bypass the bit-trim factor tree, but it wouldn't be recursive at all. It would be O(1). Where c is immediately factorized is the recursion termination point. We factor d and e recursively (that's what VQC said a bunch of times).

 

So what is the missing link between a factorization of d and e and a factorization of c? Looking into anything else isn't going to solve it; it's just going to expand our list of patterns. This idea of metacognition is actually extremely useful right now (thanks Chris).

Anonymous ID: 125cd1 Nov. 27, 2018, 8:26 a.m. No.8381   πŸ—„οΈ.is πŸ”—kun   >>8386 >>8388

>>8380

Just to nitpick it would be O(sqrt) as we do compute the sqrt of c to get d. That is the upper limit in the case you're describing.

 

I see the bit trimming as a means to simplify. When we trim bits we simply remove 2 as a factor, leaving us with factors beyond 2. I'm not sure why, but the number 2 appears to be "special" in the grid. Several times I've seen patterns where I expect the result to contain 2x2x…, but instead contain 2x… (One less factor of 2 than I expected).

 

I've said it before, but I suspect that VQC has been giving us hints related to all the keys he knows. The recursive one I suspect is one method. The triangle one is another and I suspect the (n-1)naa is a third.

 

How factorizing d and e solves it, I'm not sure. We know that if gcd(d, e) != 1, then we have the d = kq (for some k and q) and e = ko (for some k, o) which gives us: kqkq + ko =k(qq + o). I wonder if, when solving d and e we will end up with something similar that we can then use to "figure it out". It's the figure it out part that's hard and recusing over d and e doesn't give me any good ideas about how to solve it.

 

As for the subject, metacognition, I'm wonder what do we have? We have a huge grid, that exists as a lookup and we know that column 0 (row 1?) is supposed to operate as the QFT (Quantum Fourier Transformation), but just writing this doesn't make me understand it.

 

The grid shows us how numbers can be grouped together as the difference of two squares, and how this two squares are related to the integer and the integers factors.

 

We have patterns inside the grid for knowing how a cell is related to other cells in different columns. We can move in many ways, by shrinking or increasing the largest square we remove from the number. By transforming the number, but so far we haven't actually gotten a solution.

Anonymous ID: 125cd1 Nov. 27, 2018, 8:52 a.m. No.8382   πŸ—„οΈ.is πŸ”—kun

>>8380

I'm ranting a bit I know, but I'm trying to think loud.

 

If we assume the recursive method is the solution, then I'm wondering more about the patterns we spend time on. Are they then irrelevant? Or are they irrelevant for the problem at hand, but useful once you understand the grid? As in, will the grid be used further to solve other problems?

 

Based on the description of the recursive method, we only need to care about d and e. We generate a tree where we branch out from d and e, but for each node we only store d and e (and c?). However, we know that doing so doesn't solve for all d and e. As in, gcd(d, e) doesn't solve much, we still have branches that aren't solved, and I think I've seen "trees" where none of the nodes are solved using just that method. Which means, we're missing something.

 

Assume we're building a tree, where each node contains three values, d, e and c. c is the input number of the node, d and e are the d and e-values of the c-value. This tree is what the recursive algorithm gives us, but we aren't done yet.

 

What then would be the next logical step? How would you connect this tree to the grid? We have an initial c, we have a tree. We can move recursively down the tree and then what?

 

To digress a bit, I remember in one of the first threads, VQC talked about multiplying our c number with other numbers. As a means to create a column that is predictable and that we "control". From this column we could coax out the factors of c, did we ever to much with this? Was it just something to get us interested? Knowing now what we know, is that a possibility?

GAnon !Nx57Pyux3E ID: 566843 Nov. 27, 2018, 9:42 a.m. No.8383   πŸ—„οΈ.is πŸ”—kun

>>8350

 

If we look at any number base, the 1's place is 1, the next place is b, the next is bb, then bb*b etc. Obviously, you can't really have non-integers as bases and e=2.718 is a decimal.

In a way, though, the fibonacci numbers are like base e. because F(n) * phi ~= F(n+1), which is similar to how, in base 10 for example, if you have 7 * 10 = 70.

 

Sexagesimal again obvious pointer to fibonacci numbers. (great site btw)

 

https://www.mathsisfun.com/numbers/fibonacci-sequence.html

 

If you look at the last digit of fibonacci numbers it repeats every 60. The tens repeats every 300, etc.

 

>Full circle

Implication we need to use a circle.

>It's perfect

Perfect circle

>I think base 60 + Babylonian fractions would round this out nicely

>round this out

Either another reference to the roundness of a circle or has to do with rounding the numbers.

>Absolutely positive its the key

Absolute value of something

GAnon !Nx57Pyux3E ID: 566843 Nov. 27, 2018, 9:59 a.m. No.8385   πŸ—„οΈ.is πŸ”—kun

I gotta go back to work, but I thikn we could do something with this notation

 

If we have an number like:

 

'100001000' in fibonacci notation,

we could rewrite it as

'100000110'

or

'011000110'

'002100110'

'000320110'

'000053110'

etc.

 

Like any number can be turned into the two next to it.

I think that if we can make this a constant digit then it must be divisible by that digit at the very least.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Nov. 27, 2018, 11:19 a.m. No.8386   πŸ—„οΈ.is πŸ”—kun

>>8378

in bold: 1+2+3+4+5+6+7… etc.

 

>>8381

I've tried explaining the Limits I see, c/2 and √c, with my Masonic Square'n'Compass but it only gets remembered every now and then.

 

>>8384

Factoring a Google length RSA Semi-Prime into its Co-Primes isn't "NP-hard"? Well that's gay.

AA !LF1mmWigHQ ID: 8b4129 Nov. 27, 2018, 3:29 p.m. No.8388   πŸ—„οΈ.is πŸ”—kun   >>8389

>>8381

>it would be O(sqrt)

O(sqrt) is O(log n) where n is the length of c in bits, which is what VQC has always said. Taking the square root of something halves its number of bits.

 

>I see the bit trimming as a means to simplify.

>I suspect that VQC has been giving us hints related to all the keys he knows. The recursive one I suspect is one method. The triangle one is another and I suspect the (n-1)naa is a third.

If the recursive solution was one of several methods, there would be one O(log n) method and several O(1) methods. The only way anything VQC has said makes any sense is if every method relies on factoring d and e. That means we have to know the factors of d and e to solve. That means there isn't enough information purely from c to solve.

Anonymous ID: abe23c Nov. 27, 2018, 8:47 p.m. No.8389   πŸ—„οΈ.is πŸ”—kun   >>8390

>>8388

No it wouldn't be O(1). So far, everything involved requires us to calculate d which would require O(sqrt) (or O(log n) as you put it).

 

If we need to go recursive and we need to factor d and e, does that mean the other approaches VQC has talked about in the past (triangle, -f and e … ) are just approaches that won't work (or will, but with a too high complexity)? Or do you think it all ties in together?

AA !LF1mmWigHQ ID: 8b4129 Nov. 27, 2018, 9:34 p.m. No.8390   πŸ—„οΈ.is πŸ”—kun

>>8389

If we only calculate d and e for one c, it isn't O(log n). It's O(log n) if we calculate d and e continually, because in the worst case we would halve the number of bits log n times (but we wouldn't always have to do that). If we were only meant to calculate d and e for one c, it would be O(sqrt), which is different, and definitely not what VQC said.

>If we need to go recursive and we need to factor d and e, does that mean the other approaches VQC has talked about in the past (triangle, -f and e … ) are just approaches that won't work (or will, but with a too high complexity)? Or do you think it all ties in together?

My theory is that we use those approaches in conjunction with factored d and e. We wouldn't have learned those things for no reason. My theory is that we use a fully factored d and e in order to apply those approaches that we currently can't use. So the cells and numbers for 5*29=145 could be used to factor other numbers for which d=145, maybe.

ID: b129f9 Nov. 27, 2018, 10:10 p.m. No.8391   πŸ—„οΈ.is πŸ”—kun

Good insights, AA. Since factorizing d and e enable the factorization of c, then that also means that the patterns in the factorization of d apply to every number inbetween d^2 and (d+1)^2.

Anonymous ID: 609e7d Nov. 28, 2018, 12:24 p.m. No.8397   πŸ—„οΈ.is πŸ”—kun   >>8398 >>8399

>>8396

AA and Jan, since we're not that many people left still discussing things I'm asking you two a bit more directly. Have we identified any of these two equations VQC has talked about? I know we have a lot of equations and he has talked in the past about two branches of math meeting, solving two equations in tandem(?)/parallel only to have them meet and combine, which will become a new thing.

 

My initial though is that he might be referring to the multiplication of -f and e. Since we know they will not be of the same parity, we then also know that they will have a different set of equations for the a-values. If f is even then a's are 2 squares - f/2, and a-values in e will be 4 triangles + (e + 1)/2. If f is odd then a's are 4 triangles - (f - 1)/2 and e will be 2 squares + e/2. These are two "different" equations, meeting (in the exclusive middle?).

 

But how would those spiral?

 

Also to note, I'm asking because I'm not sure, but I personally doubt that we're talking about the equations for 'a' alone here.

Anonymous ID: 609e7d Nov. 28, 2018, 12:37 p.m. No.8398   πŸ—„οΈ.is πŸ”—kun

>>8397

Based on previous posts and work we've done, I would assume that one of the equations is nn + 2d(n - 1) + f - 1. Or PMA's suggestion of nn + (2d - 1)(n - 1) + n - 1 + f - 1.

 

Assuming that is true, and we're going to combine the triangular solution with the recursive function, then we will end up with a tree and parts of that equation. We'll know 2d, f and -2. Somehow, this is going to be connected to the tree and the grid.

Anonymous ID: 566843 Nov. 28, 2018, 8:08 p.m. No.8401   πŸ—„οΈ.is πŸ”—kun   >>8402 >>8404 >>8405

Heres some examples for shrinking the fibonacci notation down. First is for full notations.

Next is if you keep it as 1's and 0's. I think that is the better route to take but idk why I say that. It seems that the most sparse it will be is always 101010101 etc and the most dense is 11111111. Basically the rule of this when you collapse it is that it is 1s broken up by 0s and there are never two zeroes touching, because if there were then you could populate them with the next higher digit. If you have a streak of 1's (aka a sum of consecutive fibonacci numbers) then it is as follows

 

>https://math.stackexchange.com/questions/833231/the-sum-of-n-consecutive-fibonacci-numbers

Sum of (F(1) .. F(k)) = F(k+2) - 1

 

So the sum from j to k would be:

[F(k+2) - 1] - [F(j+2) - 1] = F(k+2) - F(j+2)

 

I don't know where you'd fit in the negatives. Maybe make a second number. Basically I'm thinking we collapse it until we can't anymore then trim out the streaks by removing the negative element, then keep collapsing until its just a string of 1's then do something with the negative

GAnon !Nx57Pyux3E ID: 566843 Nov. 28, 2018, 8:09 p.m. No.8402   πŸ—„οΈ.is πŸ”—kun   >>8404

>>8401

This could also be like 12 steps

>take deep inventory of self

>take out the bad (negatives)

>do something with it at the end (give to friend or something)

>maybe thats the original number

>idk im sleepy

GAnon !Nx57Pyux3E ID: 566843 Nov. 29, 2018, 9:54 a.m. No.8405   πŸ—„οΈ.is πŸ”—kun

>Admitting powerlessness over the addiction and our lives had become unmanageable

Powerless AKA not a difference of squares. AKA 1 dimensional.

If our lives were manageable then we'd have two factors for it

 

>Believing that a higher power (in whatever form) can help

I think this higher power is phi and in specific this big spiral.

 

>Deciding to turn control over to the higher power

I think this would be to represent ones self as a sum of fibonacci numbers (none consecutive)

For example 3*5 = 15 would be 13 + 2 = 1000100 = (13)(0)(0)(0)(2)(0)(0)

 

>Taking a personal inventory

[Interesting how 145 = 144 + 1 and 144 and 1 are both fibonacci numbers]

I think this is to look at your own bits within the number.

(this is just my idea for now of what this entails)

1000100

0110011

0101111

This streak of four 1's is 1+1+2+3 = 7

The 1 on the left is 8

7+8 = 15

>>8401

Sum of consecutive fibs = F(1) + … + F(k) = F(k+2) - 1

15 = 0101111 = 0100000 + 0001111

and 0001111 = 0100000 - 1

15 = 0100000 + 0100000 - 1

15 = 0111000 - 1

15 = 0110110 - 1

15 = 0110110 - 1 = 0110000 + 0000110 - 0000001

0000110 = 0010000 - 0001000

0110000 = 10000000 - 1000000

so

15 = 10000000 + 0010000 - [ 0001000 + 1000000 - 0000001 ]

15 = 10010000 - 1001001

15 = 10001100 - 0110111

15 = 10001011 - 0110111

etc..

 

>Admitting to the higher power, oneself, and another person the wrongs done

Something something get the negative amount and give it to the fibonacci

15 = 1000100 = 1111111 - 0111011 Maybe this value is the negative.

 

>Being ready to have the higher power correct any shortcomings in one’s character

Correction of shorcomings would be the result of this calculation whatever that may be

 

>Asking the higher power to remove those shortcomings

Subtract that amount from you

Anonymous ID: 708ece Nov. 29, 2018, 4:49 p.m. No.8406   πŸ—„οΈ.is πŸ”—kun

Have we considered using "Imaginary Phi"?

https:// math.stackexchange.com/questions/1851698/imaginary-golden-ratio

 

>Some joke about radicalization and flying complex planes into buildings

Anonymous ID: cbe267 Nov. 30, 2018, 5:53 a.m. No.8408   πŸ—„οΈ.is πŸ”—kun

Is it possible that if we multiply c by a certain number, say q, and get the remainder e, given that we can calculate -f, we can force the resulting -f and e columns to create our lookup?

Thinking outside the box, what if actually multiplying c by certain value or values, forces the result to be where we want, to make the lookup easier.

We could theorectically do it in two steps, getting information from the first product, qc, and introduce a second factor, v, again forcing the result to give us a deterministic result for e and -f. This product, vqc, could then be used to "triangulate" a lookup somehow?

The chosen numbers for v and q may depend on the type of c we are using but I would speculate they would be from a limited set…

This approach would still be O(log l) or less, where log l is the length of c in bits because the largest operation remains finding a square root and remainder. It doesn't matter if we need to do it more than once, the overall complexity in Big Oh remains the same.

VA !!Nf9AmQNR7I ID: 01587d Dec. 1, 2018, 10:36 p.m. No.8420   πŸ—„οΈ.is πŸ”—kun   >>8421 >>8425

For c6107, interesting that our u1 and u2 values are 41 and 42, since our f derived polite numbers = 3+4=7

The total x+n dimensions are 83 * 83 = 6889.

4 rectangles of 41 * 42 +1 = 6889

It’s the (n-1) pattern playing out at 2 levels simultaneously.

Maybe a fluke, just brainstorming over here.

c6107 f= 134

f div 8 = 16

Sqrt(16) = 4

Sqrt(16)- 1 = 3

4+3=7. These are our two polite triangle number bases added together.

7 * unknown + 1 = correct (x+n)^2

For this example it turns out to be 984.

VA !!Nf9AmQNR7I ID: 01587d Dec. 1, 2018, 10:40 p.m. No.8421   πŸ—„οΈ.is πŸ”—kun

>>8420

This should be blazing fast at computer calc speed! since the two polite triangle bases come from f, it should scale upwards with increasing c values.

All we have to do is check each iteration to see if it's a perfect square.

This is the two equations merging together or running alongside each other. (i think)

VA !!Nf9AmQNR7I ID: 174e7d Dec. 2, 2018, 10:29 a.m. No.8423   πŸ—„οΈ.is πŸ”—kun   >>8424

>>8422

Farey Tree Anon, I studied that Plimpton 322 article. Hilarious that all this probably started as a project cost analysis for some engineer or business manager ~6000 years ago. Like "Shit guys we gotta figure out the dimensions of this ramp some we know how much limestone to order from the quarry for this 1000 cubit length ramp. Go figure this shit out, we're on the King's payroll anyways, and we don't want too many slaves dying hauling extra rock from the quarry if we're gonna make our deadlines. Tell that fucking nerdy kid who sits around thinking all day to figure it out. He's a fucking weirdo, oh well, he's great at drawing triangles all day."

Thus was born the First Autist, lol.

Anonymous ID: 5d6348 Dec. 2, 2018, 11:49 a.m. No.8424   πŸ—„οΈ.is πŸ”—kun   >>8425

>>8423

Lol. It is funny isn't it, how everything seems to connect. Maybe history can provide the perspective needed to finish this thing.

I found this short read to be extremely good.

https:// www.google.com/url?sa=t&source=web&rct=j&url=http://vixra.org/pdf/1407.0062v1.pdf&ved=2ahUKEwjkpbXY84HfAhWMwMQHHVioDawQFjAQegQIBRAB&usg=AOvVaw1mZHbw_xNX0CbQ4bx-tmQA

 

Why Base 60 by JF Sweeney if you want to google the link yourself

VA !!Nf9AmQNR7I ID: 174e7d Dec. 2, 2018, 11:01 p.m. No.8426   πŸ—„οΈ.is πŸ”—kun   >>8429

>>8425

This was why VQC told us we would be estimating n0 when we used the (f-1) div 8 method. We’re looking for a way to use f to construct (x+n)^2

Correction: this is one of the ways to do it. There are multiple correct methods.

I’m just working on the one that I understand the mostοΏΌ.

Can we try it out on the RSA100 Numbers we have all the solutions for?

Essentially I'm postulating that (x+n)^2 is a multiple of f or one of its roots, so a big difference from just (x+n).

The factor derived from f * iteration forms / fills the area of the square.

Odd (x+n) formula is f * unknown + 1 = (x+n)^2

Even (x+n) formula is f * unknown = (x+n)^2 (possible)

Anonymous ID: 366f62 Dec. 5, 2018, 10:08 a.m. No.8429   πŸ—„οΈ.is πŸ”—kun   >>8430

>>8426

Since the two polite triangle bases come from f, it should scale upwards with increasing c values.

All we have to do is check each iteration to see if it's a perfect square.

This is the two equations merging together or running alongside each other. (i think)

Equation 1 is: SQRT(c + (x+n)^2) - d = n

Equation 2 is: f factor * iteration + 1 = a perfect square? If not, don't run it through equation 1.

Anonymous ID: a4498f Dec. 6, 2018, 11:54 p.m. No.8431   πŸ—„οΈ.is πŸ”—kun

Enumerating e in a given row n is a way of enumerating every possible n value in an intelligent way.

 

If we swap n and x coordinate wise, we can enumerate x values the same way we did n values. The horizontal gap in every other cell highlights the fact that only every other x value is defined in a cell. (Hence, cells have either odd or even x values).

 

What type of movement would this be?

AA !LF1mmWigHQ ID: 8b4129 Dec. 7, 2018, 2:12 a.m. No.8432   πŸ—„οΈ.is πŸ”—kun   >>8433

I wrote a program (in Java) that creates the factor tree and generates every relevant cell and value that we've studied throughout the tree for each c/d/e/d's d/d's e/etc. You need the GridCell class I wrote for it to work (also below). I didn't bit-trim throughout the tree. I just divided each number by two until it was the difference of two squares (so it has evens that are divisible by 4 and odds). It also isn't done with BigInteger. I'll have to rewrite both programs with BigInteger at some point because it isn't completely functional without BigInteger (for example, generating the cells in (e,1) and (f,1) where cBigN and c(BigN-1) creates integer overflows). Either way, here it is. My thought is that it'll be quite useful to look for patterns that occur between all relevant information for c and all relevant information for d and e.

 

everything.java - https://pastebin.com/mbH8eZ7s

GridCell.java - https://pastebin.com/Cn36RwPA

Anonymous ID: a3777e Dec. 7, 2018, 1:27 p.m. No.8433   πŸ—„οΈ.is πŸ”—kun   >>8435 >>8438 >>8454 >>8476

>>8432

Got a present for you AA anon.

Where a value, c first appear in row 1 at e for the value a[1], t=1?That is a column e.

Where does it next appear in row 1 at a different e for the value a[t], t=2?

And so on back towards the origin.

Every column that c appears in has what in common?

How many columns do you need to create a lookup?

What specific information does that set of columns build up?

We know a and b are values of n in EVERY one of these columns because c is in the first cell of each column for a value of a[t].

Once you 'c' it, you cannot unsee it.

Godspeed anon.

Anonymous ID: a3777e Dec. 8, 2018, 12:49 a.m. No.8436   πŸ—„οΈ.is πŸ”—kun

>>8435

This was never about me.

It's about You.

It's about us.

The larger a number c, the more columns it is in, the more information about c is in the number c, the easier the lookup.

The single most important takeaway is that you now have a method that gives you more information about c the larger it is, not less.

Important.

Anonymous ID: a3777e Dec. 8, 2018, 12:51 a.m. No.8437   πŸ—„οΈ.is πŸ”—kun

One of the most important prayers that exist is when two or more gather in His name, that they spare a moment for all still suffering anon both inside and outside the rooms regardless of time sober.

AA !LF1mmWigHQ ID: 8b4129 Dec. 8, 2018, 6:03 a.m. No.8438   πŸ—„οΈ.is πŸ”—kun   >>8441

>>8433

>Got a present for you AA anon.

>Where a value, c first appear in row 1 at e for the value a[1], t=1?That is a column e.

If c is semiprime and the difference of two squares, it'll be odd. Any odd number c appears as a[1] at (2c,1), since for even e in (e,1) a[t]=e/2+2(t-1)(t-1), and a[1] is e/2+2(1-1)(1-1)=e/2. So for example c145 appears as e[1] at (290:1:1) = {290:1:145:0:145:147} f=-1. It'll also appear as a[1] at (2c-1,1), since for odd e in (e,1) a[t]=(e-1)/2+(t-1)(t-1)+tt, and a[1] is (e-1)/2+(1-1)(1-1)+(1*1)=(e-1)/2+1. (289:1:1) = {289:1:146:1:145:149} f=-4

 

To simplify: since all a values in (e,1) are twice a square or the sum of consecutive squares (based around t) plus either e/2 or (e-1)/2, but we're choosing t, we can just take 2(t-1)(t-1) and (t-1)(t-1)+tt away from c to find e/2 or (e-1)/2 and therefore calculate the columns in which c will appear as an a value in (e,1).

 

>Where does it next appear in row 1 at a different e for the value a[t], t=2?

2(2-1)(2-1)=2, and 145-2=143, and then 1432=286, so it'll appear in (286:1:2) = {286:1:147:2:145:151} f=-9. (2-1)(2-1)+(22)=5, 145-5=140, and (140*2)+1=281, so it'll appear in (281:1:2) = {281:1:148:3:145:153} f=-16

 

>And so on back towards the origin.

Where c145 appears as a[t] in (e,1) with t increasing from 1 upwards, to begin with for even e is 290, 286, 274, 254, 226, 190, etc. -4, -12, -20, -28 etc. The gap between cells increases by 8 each time for even e. For odd e, 289, 281, 265, 241, 209, 169, etc. -8, -16, -24, -32, etc. The gap between cells also increases by 8 each time, but the first started at 4 and this one starts at 8.

 

>Every column that c appears in has what in common?

They're all in (e,1) so they follow the (e,1) patterns of every factor of this particular number having a valid n in this column. So in every column that follows the above pattern, there'll be a valid cell at the a and b we're trying to find. There are probably other things but I'm assuming that's your point.

 

>How many columns do you need to create a lookup?

That depends on what other information is required for the lookup, and we don't know that.

Anonymous ID: 2bc45b Dec. 8, 2018, 10:22 p.m. No.8440   πŸ—„οΈ.is πŸ”—kun

(-238, 1, 12) = {-238:1:145:22:123:169}(-286, 1, 13) = {-286:1:169:24:145:195}(574, 1, 2) = {574:1:291:2:289:295}(-1798, 1, 31) = {-1798:1:961:60:901:1023}(-5742, 1, 68) = {-5742:1:6241:134:6107:6377}

AA !LF1mmWigHQ ID: 8b4129 Dec. 9, 2018, 2:55 a.m. No.8442   πŸ—„οΈ.is πŸ”—kun

I've rewritten my program that outputs everything from Grid Patterns (unless I missed something) down the recursive d/e factor tree for BigInteger. Pic related is a small section of example output. You need the following two java files:

https://pastebin.com/mvrvxVr4

https://pastebin.com/NjGYV8ip

VA !!Nf9AmQNR7I ID: bf321c Dec. 9, 2018, 4:11 p.m. No.8443   πŸ—„οΈ.is πŸ”—kun

Hello Anons! I noticed that the Polite Triangle Bases for c6107 = 7 and (n-1) = 35 = 7 * 5.

Is this just a fluke?

For c145, PTB = 1+0=1 and (n-1) = 4 = 1 * 4

"A few more iterations and you have the Grid related short cut!!!"

Let's run a few other c values a bit larger and see if the PTB's are factors of (n-1)

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Dec. 10, 2018, 5:23 p.m. No.8448   πŸ—„οΈ.is πŸ”—kun

https://ventrellathing.wordpress.com/2018/08/16/very-large-numbers-are-not-numbers-infinity-does-not-exist/

 

Inside you find:

 

http://www.divisorplot.com

Anonymous ID: abe23c Dec. 13, 2018, 1:46 a.m. No.8450   πŸ—„οΈ.is πŸ”—kun   >>8451

>>8449

 

Almost, you're missing half of them. So we can reason a simple method of finding the e's without the need for brute forcing by just using algebra.

 

We have two equations, one for even e and one for odd e.

 

even e:

2(t - 1)(t - 1) + e/2 = a

odd e:

2t(t-1) + (e+1)/2 = a

 

What we're talking about now is solving for a, which will be our c.

 

2(t - 1)(t - 1) + e/2 = a

4(t - 1)(t - 1) + e = 2a

e = 2a - 4(t - 1)(t - 1)

 

2t(t-1) + (e + 1)/2 = a

4t(t-1) + e + 1 = a

e = 2a - 4t(t-1) - 1

 

Example:

 

We want to find which e's a[1] = 145. There's two of them.

 

290 = 2145 - 40*0

289 = 2145 - 41*0 - 1

 

Since a[1] is a bit bland, let's find the e where a[3105] = 145

 

2145 - 43104*3104 = -38538974

2145 - 43105*3104 - 1 = -38551391

 

And the records in the cells:

{-38538974:1:6353:6208:145:12563}

{-38551391:1:6354:6209:145:12565}

 

You can also use this to find columns where c exists in a specific n (for some e).

Say we want to find the columns where there is an a[t] = 145 in (e, 5).

 

25145 - 4t(t-1) - 1

25145 - 4(t-1)(t-1)

 

This takes advantage of the fact that if 5*145 exists in (e, 1) then 145 exists as an a in (e, 5) (At the same time 5 exists in (e, 145)).

Anonymous ID: abe23c Dec. 14, 2018, 5:41 a.m. No.8453   πŸ—„οΈ.is πŸ”—kun

>>8451

Yeah. It's kind of hard to verify that it generates the accurate records. I'm not 100% sure since the numbers are so big, but it appears to work for smaller numbers, so I think it's fine. I included the first records for RS100: https://pastebin.com/nart63j2

 

One thing I noticed while checking with regards to smaller numbers (so it could be a fluke) is that all the e's where c exists in row 1 are also numbers that are part of phi(2c) with a few exceptions (ie a few numbers that divide c). I was looking into it to see if that meant we could take advantage of this to factorize, but that is still no dice given the amount of values to iterate over for a large number.

Anonymous ID: abe23c Dec. 14, 2018, 7:37 a.m. No.8454   πŸ—„οΈ.is πŸ”—kun

I was thinking about >>8433 and why we're looking at c at a[1], a[2].. and I started to think about these hints from a while ago:

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

> When is the second time it appears?

 

I decided to see how many E's squared numbers appear in. The number of e's per square correlate with the Beatty sequence (A001951). That is, 11 appears in column 1, 22 appears in 2 columns, 33 appears in 4 columns… The number of e's that the first 72 squares have match the numerical sequence of Beatty sequence for floor(nsqrt(2)).

 

I did a quick check against the first 5000 squares and the first 5000 values of floor(n*sqrt(2)) and it appears to check out.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Dec. 14, 2018, 11:08 a.m. No.8457   πŸ—„οΈ.is πŸ”—kun   >>8458

>>8456

>72 + 73=145

>27+28+29+30+31=145

>10+11+12+13+14+15+16+17+18+19=145

 

77+78=155

29+30+31+32+33=155

11+12+13+14+15+16+17+18+19+20=155

 

(jump 5) 82+83=165

(jump 2… yup) 31+32+33+34+35=165

12+13+14+15+16+17+18+19+20+21=165

 

—–

Let's start with 145 as our base

 

You build from the bottom, so you start with your lowest value set.

GCD or HCF is the middle number in the middle value set.

Top value set is your sum/2, and then you make 'em whole numbers by adding remaining .5 to one of 'em.

 

145 -155 -> 165 (avg + 5, sum +10)

29 -31 -> 33 (HCF +2)

10 -11 -> 12 (base +1, 10 numbers {thus sum increases by 10 each time})

 

So your sum that you're working with will shift by n, where n=number of values being summed (in your lower value set)

 

I don't know if this holds, but the middle value set has half the amount of numbers as the lower set, so n/2=n'

 

Which makes me wonder if that only works because 2 is the biggest chunk you can take out at once…

 

Like, if you had n=9 if you'd then n/3=n' and have 3 numbers at the top, but then that would only work if the number isn't prime or comprise of co-primes.

Does it have to be an odd sum? I'll play with it.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Dec. 14, 2018, 11:23 a.m. No.8458   πŸ—„οΈ.is πŸ”—kun   >>8459

>>8457

With 9 numbers instead of 10

10+11+12+13+14+15+16+17+18=126

add 2 to the previous highest value of the mid-set and work backwards

31+2=33: 33+32+31+30=126

63+63=126

42+42+42=126… actually… it does this for every divisor:

  • The factors of 126 are 1,2,3,6,7,9,14,18,21,42,63,126;

 

But if you were presented with the just the 9 numbers, it looks like you'd take the highest value from the low-set and subtract 1, double the result, and subtract another 1 from that, so:

18-1=17

17*2=34

34-1=33, and work down from there.

 

Though, I don't know if that holds in the higher ranges.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Dec. 14, 2018, 12:25 p.m. No.8459   πŸ—„οΈ.is πŸ”—kun   >>8460

>>8458

10+11+12+13+14+15+16+17=108

35+36+37=108 (108/3=36)

54 + 54=108 (108/2=54)

 

Hmmmmmmrmrmrmrmrmrmrmmmmmmm….

On 3 step polite 126:

41+42+43=126

Midstep is lowest non-2 factor when even?

I need more coffee.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 83b459 Dec. 14, 2018, 4:28 p.m. No.8460   πŸ—„οΈ.is πŸ”—kun

>>8459

20+21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37=513/18 (28.5 midpoint)

83+84+85+86+87+88=513/6 (Β±.5 midpoint)

170+171+172=513/3 (Β±1 midpoint)

256+257=513/2 (Β±.5 midpoint)

 

How to scale a polite number?

pick however many steps you want… like… 6? and then you can produce the politeness

Anonymous ID: b129f9 Dec. 14, 2018, 10:42 p.m. No.8463   πŸ—„οΈ.is πŸ”—kun

This is what occurrences of b[t] = c look like.

 

c145{289:1:144:-1:145:145} (289, 1, 0) c = 21025 f = 0{281:1:142:1:141:145} (281, 1, 1) c = 20445 f = 4{265:1:140:3:137:145} (265, 1, 2) c = 19865 f = 16{241:1:138:5:133:145} (241, 1, 3) c = 19285 f = 36{209:1:136:7:129:145} (209, 1, 4) c = 18705 f = 64{169:1:134:9:125:145} (169, 1, 5) c = 18125 f = 100{121:1:132:11:121:145} (121, 1, 6) c = 17545 f = 144{65:1:130:13:117:145} (65, 1, 7) c = 16965 f = 196{1:1:128:15:113:145} (1, 1, 8) c = 16385 f = 256{-71:1:126:17:109:145} (-71, 1, 9) c = 15805 f = 324{-151:1:124:19:105:145} (-151, 1, 10) c = 15225 f = 400{-239:1:122:21:101:145} (-239, 1, 11) c = 14645 f = 484

Anonymous ID: b129f9 Dec. 15, 2018, 11:45 p.m. No.8472   πŸ—„οΈ.is πŸ”—kun

>>8471

a[1] = c

a[2] = c

a[3] = c

…

 

The "n is the same thing as a" remark starts to make sense when you realize that c at a[t] functions as an an value and a OR b can be n, since t+a and t+b both have the same factor

Anonymous ID: fd1528 Dec. 18, 2018, 2:26 p.m. No.8476   πŸ—„οΈ.is πŸ”—kun   >>8478

>>8433

>Every column that c appears in has what in common?

The first valid n (above 1) in said columns is always prime when c is coprime

c3637 (2, 3, 5 and 7 filtered)(7258,19)(7078,3539)(7018,11)(6598,3299)(5158,2579)(4358,2179)(4138,2069)(3674,11)(2918,1459)(1798,29)(874,19)(218,109)c6107 (n = 2, 3 and 5 filtered)(12214,31)(12178,6089)(11638,11)(11314,5657)(9298,4649)(8614,59)(7858,3929)(5158,2579)(4114,11)(2998,1499)...

 

When c has three factors, the first valid n in the e=c*2 column is always one of the factors.

c135336263417 (392956576089)(270672526834,1)(270672526834,3929)c447791149 (1356576089)(895582298,1)(895582298,13)c6380123599 (56930833637)(12760247198,1)(12760247198,569)c240520532961377359853213671 (1233232127918610991654346153)(481041065922754719706427342,1)...gap...(481041065922754719706427342,123323)

 

Not a direct calculation but haven't found any values where the above isn't true yet.

AA !LF1mmWigHQ ID: b00cdb Dec. 19, 2018, 10:11 p.m. No.8477   πŸ—„οΈ.is πŸ”—kun

Is there anything we can actually do with the gaps aside from them being a stopper for any (e+2n,n) movements? "Not VQC" said "lookup in gaps" but if there aren't any cells we can generate what are we even meant to look at when it comes to the gaps?

Anonymous ID: b129f9 Dec. 19, 2018, 11:36 p.m. No.8478   πŸ—„οΈ.is πŸ”—kun

>>8476

Let's say we have a way to find the first valid n in a column efficiently.

 

Assuming your assertion is true for all semiprime c, we can make c the product of three primes where the a value that we want is the smallest valid n value by picking the closest prime to d (perhaps by the nextProbablePrime() function in Java), since d a in a semiprime. Thus, calculating the first valid n value would be equivalent to factoring.

Anonymous ID: b129f9 Dec. 19, 2018, 11:38 p.m. No.8479   πŸ—„οΈ.is πŸ”—kun

Making this prime that is known to be bigger than a as the third factor of c, qc, the a value we want would be the smallest factor of c making the lookup reveal itself more easily.

AA !LF1mmWigHQ ID: b00cdb Dec. 21, 2018, 4:02 p.m. No.8480   πŸ—„οΈ.is πŸ”—kun   >>8482

I've updated my program that shows you everything down the factor tree. Instead of just creating a giant text dump, you can tell it what you want to see specifically (so you could just analyze the an cells for example, rather than having to look for them in a giant 6-page wall of text with every other pattern). It'll also work with bigger numbers now (probably not RSA-sized, but if the highest prime factor is maybe 7 digits long it should only take 80 seconds at most - this is because it looks for all ab pairs down the tree). Since we're on 8ch, just so nobody gets paranoid, I'll just put it here using pastebin, but I can put a jar somewhere if anyone wants.

 

everything.java (contains main) β€” https://pastebin.com/XfG1SwYN

BigCell.java (also required) β€” https://pastebin.com/309hK39c

AA !LF1mmWigHQ ID: b00cdb Dec. 22, 2018, 5:24 a.m. No.8481   πŸ—„οΈ.is πŸ”—kun

The x values in the (e,1) and (f,1) cells where a[t]=BigN and BigN-1 add together to make 2d-1 (so if you took the next cell up (t+1) in one of those you’d get 2d+1 too since x iterates by 2).

AA !LF1mmWigHQ ID: ec6c10 Dec. 22, 2018, 7:12 p.m. No.8482   πŸ—„οΈ.is πŸ”—kun   >>8483

I found a link between the 2(x+n)2(x+n) square and the cell in (1,1) where d is equal to the n value from (0,n)'s a=aa b=bb cell. The polite triangle bases from 2(x+n)2(x+n) add together to give the x value in the (1,1) cell. I think you can link it back to the original (x+n)(x+n) with algebra. Once I figure that out I'll post more. This "analyze everything at once" program has already proven to be very useful.

 

By the way, if anyone's planning on using my program >>8480 here, you'll want to change the subtraction on lines 553, 554, 581 and 582 to an addition (whoops).

AA !LF1mmWigHQ ID: ec6c10 Dec. 22, 2018, 7:30 p.m. No.8483   πŸ—„οΈ.is πŸ”—kun

>>8482

The x value (which is the 2(x+n)2(x+n) bases added together) is actualy 2(x+n)-1 from the original (e,n) cell. So the x value in the cell in (1,1) where d is the n value from (0,n)'s a=aa b=bb cell is actually also 2(x+n)-1. That's the algebra thing I was talking about.

AA !LF1mmWigHQ ID: ec6c10 Dec. 22, 2018, 10:01 p.m. No.8484   πŸ—„οΈ.is πŸ”—kun

For even e, the t value in the cell in (e,1) where a[t]=BigN is equal to half the t value of the Root of D cell. For odd e, it's 2t+1.

Anonymous ID: 02496e Dec. 26, 2018, 4:32 a.m. No.8488   πŸ—„οΈ.is πŸ”—kun   >>8490 >>8492

I noticed this a few days ago, but given the holidays I haven't spent time on it.

 

Generate the e's for a number, example 259. The odd e's for that 259 appear as a[t] in (e, 1) are:

[517, 509, 493, 469, 437, 397, 349, 293, 229, 157, 77]

 

I decided to play with them for no specific reason, but I mapped the numbers through a getU function and I got the following:

[258, 254, 246, 234, 218, 198, 174, 146, 114, 78, 38]

 

What stands out is 146 and 114 (and 78). The Big N for 259 is 114 and the Big Shadow N is 146. I didn't expect this, but I checked with a few other numbers and it seems to hold for those as well. Another thing, 78 is also a value in (3, 259) but belongs to a different chain. Anyone looked into this before? Anyone knows why this seems to occur?

AA !LF1mmWigHQ ID: 2c2fb2 Dec. 27, 2018, 12:51 a.m. No.8489   πŸ—„οΈ.is πŸ”—kun   >>8490 >>8491

The x value of the cell in (1,1) where d is the n value from (0,n)'s a=aa b=bb cell is equal to b-a-1 (the b and a from the solution cell, not this cell).

 

Example:

a=3, b=177, n from (0,n) a=aa b=bb = 15138

(1,1) d=aabbn = (1,1,87) = {1:1:15138:173:14965:15313}, f=-30276, c=229159045, u=87, i=15139, j=174

x from this cell = 173

173=b-a-1=177-3-1

This seems to apply to all cells.

Anonymous ID: abe23c Dec. 27, 2018, 10:14 a.m. No.8490   πŸ—„οΈ.is πŸ”—kun

>>8489

Just in case this wasn't known, I believe that n (0, n where a=aa, b=bb) is equal to 2(x + n)^2.

 

As for >>8488 I still haven't quite figured this out. It's almost like it is an index for n-values for a given number.

Anonymous ID: abe23c Dec. 27, 2018, 11:10 a.m. No.8492   πŸ—„οΈ.is πŸ”—kun   >>8493

>>8488

I'm still not entirely sure what I'm looking at, I need to think some more about it, but I noticed something.

 

Big N and shadow Big N seems to always appear as a u for the value of e that c exists in, but the value they exist in also seems to act like a map.

 

Take 259 for example, it has Big N = 114 and shadow N = 146.

 

The getU simply calculates the u-value of the numbers, so by doing it in reverse we can do sqrt(8(114115/2) + 1) = 229 and sqrt(8(146*147/2) + 1) = 293. The records for these two are:

 

{4:100:15:14:1:229}

{4:130:17:16:1:293}

 

The two records for 259 are:

{3:114:16:15:1:259}

{3:146:-16:-17:1:259}

 

So we can see a method of moving between these records. 100 + 14 (n + x for 229) = 114 and 130+16 = 146 (n + x for 293). This also means we could move across e's and n's on a diagonal (But this does not conserve the c-value).

 

For example {3:114:16:15:1:259}, we can now move diagonally by doing 114 - 1415/2 = 9 and 3 + 1415/2 = 108. That means at e = 108, n = 9 exists. It should be noted that those numbers got my heart to jump. 108 is the smooth number for 259 (114 - 6) and 9 is the x-value of a=7, b=37, but a quick reality check against other numbers slowed it down. This also works moving "up" with n and "down" on the e.

 

114 + 15 = 129 (e=2)

129 + 14 = 145 (e=1)

145 + 13 = 162 (e=0)

Anonymous ID: abe23c Dec. 27, 2018, 2:13 p.m. No.8493   πŸ—„οΈ.is πŸ”—kun

>>8492

Fyi I wouldn't think of this as what VQC has talked about when he says "diagonally". I'm not sure if it's a fluke, or exists because of how numbers are connected or what, but it allows for navigation / moving around cells.

 

For example:

 

You can start at (5, 1, 1) and end up at (0, 8, 1) moving through these records:

{5:1:4:1:3:7}

{4:2:4:2:2:10}

{3:2:2:1:1:7}

{2:3:3:2:1:11}

{1:5:4:3:1:17}

{0:8:5:4:1:25}

 

I don't know if this movement makes any sense. I might be going down a new rabbit hole that isn't going to span out.

AA !LF1mmWigHQ ID: 2c2fb2 Dec. 27, 2018, 3:41 p.m. No.8494   πŸ—„οΈ.is πŸ”—kun   >>8499

>>8491

There are a lot of things you can do with the (1,1) cell. VQC said it's the most important cell in the grid at one point. Specific cells in (0,1) and (1,1) with a[t] added to e/2 or (e-1)/2 (depending on parity) give you an and bn, for example.

VA !!Nf9AmQNR7I ID: 4877aa Dec. 29, 2018, 8:40 p.m. No.8496   πŸ—„οΈ.is πŸ”—kun

Hello Anons!

Anyone got ideas they’re working on?

I’m still working on how square numbers and triangle numbers can be broken down into the polite triangle bases and then reassembled to solve the (x+n)^2 area. We had both successes and failures on the test AA helped me out with.

Thanks btw AA! οΏΌ

I’m also compiling and running java files now from my Mac terminal, and doing an online java class.

I can edit the files in txt format and then recompile too. It feels fun to learn new skills!οΏΌ

Thanks @everyone for all your work and dedication to this project.

I’m stoked to be here.

AA !LF1mmWigHQ ID: 2c2fb2 Dec. 30, 2018, 5:17 a.m. No.8497   πŸ—„οΈ.is πŸ”—kun   >>8509

The t value of the (1,1) d=aabbn cell for a=1 b=c is equal to the d value of the (f,1) cell where (e,1) and (f,1)’s polite x values add together to give 2d+1 (I’m not sure if this works for all parities in the same way but it worked for multiple cases).

e.g. c=1427

The cells in (e,1) and (f,1) where x+x=2d+1=75

(-17,1,19) = {-17:1:713:37:676:752}, f=1408, c=508352, u=19, i=714, j=38

(58,1,20) = {58:1:789:38:751:829}, f=-1521, c=622579, u=19, i=790, j=39

[1, 1427]

(1,1) d=aabbn = (1,1,713) = {1:1:1016738:1425:1015313:1018165}, f=-2033476, c=1033756160645, u=713, i=1016739, j=1426

d=713, t=713

Anonymous ID: 1ea46d Dec. 30, 2018, 10:55 a.m. No.8498   πŸ—„οΈ.is πŸ”—kun   >>8501 >>8502 >>8564

Here is a tip, if i(t) has not been done before.

Take products, c, that have odd x+n. And even n.

Take one from x+n.

8 triangles.

Create a table.

What does f and 2(n-1) contribute?

Because the tips of all 8 triangles are from (n-1)(n-1) - 1.

Find the pattern from large to small.

There are h families!

Avoid the 13th step of AA.

Anonymous ID: abe23c Dec. 30, 2018, 11:02 a.m. No.8500   πŸ—„οΈ.is πŸ”—kun   >>8508

>>8499

Lovely to see another anonymous poster (Been a while since we've seen you)!

 

I've been playing with t a[t] = c patterns and I made a few images. Each show the progression of the pattern as more factors are added.

 

Prime numbers don't have much in the sense of interesting patterns. As you increase the number of factors the interleaving patterns increase.

 

Note, white is c, black is 1 and the other colors are either prime factors or sub-compositional numbers.

AA !LF1mmWigHQ ID: 2c2fb2 Dec. 30, 2018, 1:42 p.m. No.8501   πŸ—„οΈ.is πŸ”—kun   >>8503

>>8499

>>8498

You said factoring d and e allows for the factoring of c. I've put together a program that outputs every known and unknown (unless I missed something) down the factor tree. I've been looking for something in a given d and e's unknowns that somehow shows up in c's unknowns but there are so many possible combinations of numbers. Could you maybe give a hint as to where we should be looking, or even just what concepts we should focus on in this context?

VA !!Nf9AmQNR7I ID: df3148 Dec. 30, 2018, 7:47 p.m. No.8502   πŸ—„οΈ.is πŸ”—kun

>>8498

Hello Senpai!

>Take products, c, that have odd x+n. And even n.

>Take one from x+n.

>8 triangles.

>Create a table.

 

I've included an updated sheet soon including f and 2(n-1) and how they contribute to the 8 triangles. PMA and I spent a lot of time working to understand this! This is the lock and key, isn't it? I'm excited to work on this again.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 81cdd5 Dec. 31, 2018, 12:11 p.m. No.8507   πŸ—„οΈ.is πŸ”—kun

>>8506

you already had me at 47, but at least I finally had my question answered as to "what kinda trangle we lookin' fer whenever we're trangulatin'?" ^_^

Anonymous ID: abe23c Dec. 31, 2018, 4:30 p.m. No.8508   πŸ—„οΈ.is πŸ”—kun

>>8500

I'm guessing this isn't much noteworthy stuff then? Either way, to give some background. Each row of pixels represent column e_t where e is a column where c exists as a[t]. The first row is where a[1] = c, second row is a[2] = c etc.

 

The color grid represents gcd-value between a[t] and c. So in the first (very tiny) image we have 5 (but the same pattern holds for all primes) while the second is two primes. Black is gcd(a, c) = 1, white is gcd(a, c) = c. The rest are randomly assigned based on gcd(a, c) != 1 / gcd(a, c) != c.

 

Given 2 primes you have 3 values, a, b and ab (Note this relates to >>7901). For three prime factors you have a, b, c, ab, ac, bc and abc.

 

The length of the row (width) is equal to c, so each image represents a grid of c * c. The problem, as I see it with regards to this structure is aligning them.

 

You can also see that it is in reality just 8 triangles, each equal and it generates a square pattern where there are multiple squares depending on the number of factors. Each type of square represents a factor.

 

What is very neat though is the beauty of the patterns it creates. It's stupidly slow, even when multi threaded and partially generated since it involves so many different values. It could probably be more optimized and my machine isn't the fastest.

AA !LF1mmWigHQ ID: 2c2fb2 Dec. 31, 2018, 10:53 p.m. No.8515   πŸ—„οΈ.is πŸ”—kun

>>8514

You might be onto something. I've checked with most c values where d=23 and it appears that for most of them the t value for one of the unknown (1,1) cells appears as a d value in either (e,1) or (f,1). I think a few of them didn't but I forgot which, and I also didn't check what happened when you added the x values together. I've look into it properly and report back later.

AA !LF1mmWigHQ ID: 2c2fb2 Jan. 1, 2019, 1:28 a.m. No.8516   πŸ—„οΈ.is πŸ”—kun   >>8517

Turns out the t value of the (1,1) cell where d is the n value from (0,n)'s a=aa b=bb cell was actually j or (x+n) all along. I don't know if anyone figured that out already. (1,1,j)'s d value is the n value from (0,n)'s a=aa b=bb cell. It also appears that j turns up as a d value in (e,1) or (f,1) sometimes. I'm just looking into it some more.

AA !LF1mmWigHQ ID: 2c2fb2 Jan. 1, 2019, 3:33 a.m. No.8518   πŸ—„οΈ.is πŸ”—kun

>>8514

I looked at all of the odd c values with a d of 23 (529 to 575). I took the t value of the cell in (1,1) where d is equal to the n in (0,n)’s a=aa b=bb cell, and tried to find it as a d value in either (e,1) or (f,1).

 

Every time you have a=1 b=c, t from (1,1) shows up as d in (e,1) and (f,1).

Example:

c575 = (46,265,12) = {46:265:23:22:1:575}, the relevant (1,1) cell is (1,1,287), and 287 appears as d in (46,1,12) and (-1,1,12). This works for all a=1 b=c.

 

For solution cells, it gets a bit weird. t for at least one of the relevant (1,1) cells (so solution j or solution (x+n)) appears as a d value in (e,1) for most of them. I didn't notice anything in particular that could be used to calculate this cell, but it's a possibility.

 

Some of them (c535, c551, c575, c567) have it turn up in (f,1).

 

Occasionally (c539, c549), the t value of the cell in (0,1) where a is equal to the n value from (0,n)’s a=aa b=bb cell (which is actually the t value we were already looking for plus 1) appears as a d value in (e-1,1).

 

Some of them (c543, c545, c565, c573) just don’t turn up. Who knows why.

 

So while j definitely turns up as a d value in (e,1) quite often, it doesn't seem to follow any stable patterns, considering it just kinda doesn't happen sometimes.

VA !!Nf9AmQNR7I ID: 341dcd Jan. 6, 2019, 5:39 p.m. No.8520   πŸ—„οΈ.is πŸ”—kun

>>8499

Hmmm. (1,1) is one of the most important cells. So I started there looking for patterns. Interesting find for a *b values moving diagonally in the Grid from (1,1). Found the following combos:

(4,4) a * an

(9,9) BigN * c

(25,25) an * c

(49,49) a * c

Anonymous ID: 1c551a Jan. 11, 2019, 12:19 p.m. No.8524   πŸ—„οΈ.is πŸ”—kun   >>8525

So, I've been working on this for a while trying to attack it from different angles. I've identified a couple of rules governing the factoring of numbers, Im not sure if this is new stuff or not… For every number (except for the square of every second even number) has a perfect square number larger than it that when subtracted from that larger perfect square number results in a smaller perfect square. The root of that larger perfect square is always the average of the sum of the factors of the starting number. The root of the smaller perfect square is always the distance between the average of the factors and the factors themselves.

 

For instance:

7 *11 = 77

7+11=18

18/2=9

9*9=81

81-77= 4

9-2=7

9+2=11

 

This works with all odd numbers and every other even number. And it also always seems to work with any number that is the product of two primes.

 

Unfortunately, its just as hard to predict the average of the two factors as it is to predict the factors themselves. But, I noticed something else:

 

For almost every number there is only 1 larger number that is 1) a perfect square and 2) the difference between the two is also a perfect square. So, I've been thinking that since every subsequent square number is the root of the next lower square * 2 +1… I think it should be possible to plot an intercept course.

 

Sort of like the old 'police pursuit' physics problem but with the diffence being, the two cars travel a constant speed, and the distance "between stops" is what changes. Going off of the example of 77 above, car 1 starts at 77, car 2 starts at 0. If they travel the same speed and only stop at square numbers then the next time they stop at the same time is when car 1 is at 81 and car 2 is at 4.

 

 

TLDR; is there a formula that could be used to plot an intercept course where speed is constant, the 2 objects in motion make stops where the distance between stops increases by a known variable and the point of interception is when they both stop at the same time?

AA ID: 2c2fb2 Jan. 17, 2019, 10:57 p.m. No.8528   πŸ—„οΈ.is πŸ”—kun   >>8529

>>8526

It was destined to stagnate a bit at least. There are still daily conversations on Discord. The board isn't dead but obviously when all we have to go on is metaphors and vagueness it gets a bit difficult.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 81cdd5 Jan. 22, 2019, 10:07 p.m. No.8533   πŸ—„οΈ.is πŸ”—kun

>Be God

>Have a child in mind.

>Only one man at a specific time can produce the one specific sperm that will make that child a thing.

>Put/encrypt that specific child into a woman

 

sperm-small prime

unfertilized egg-large prime

fertilized egg-semiprime

Gestation-same process for every RSA situation.

Birth-The child/packet/message/file/whatever

 

Cutting the umbilical cord - Going from .999… to 1 (e?)

First Breaths - Priceless.

VA !!Nf9AmQNR7I ID: 291f8b Jan. 27, 2019, 10:18 p.m. No.8538   πŸ—„οΈ.is πŸ”—kun

Damn my intuition bells are going off right now Lads! What I’ve found in using my method is that when you begin building up the factors of the (x+n)^2 -1 area, you first arrive at a perfect square that is a potential match for Potential x+n or P(x+n). D, e, and f all determine what the first perfect square will be, and then

 

The big idea is finding the first perfect square built using the root of f.

And then using multiples of it.

F limits the possible correct values.

It’s a 2 step process. All larger x+n values can be built from the first perfect square.

 

Once you find the first perfect square, the other possible values are just multiples.

So PTB for x+n=6889 = 2(sqrt ((f-1)/8))-1 = 7

First match is 7 * 5 +1 = 36

Sqrt(36) = 6 = potential x+n

 

Here's the perfect square creation I'm talking about. This is building the x+n square using 4 rectangles (or 8 triangles) +1

Step 1: find the first perfect square.

PTB = 7

7 * 5 +1 = 36

potential triangle base value(u) is 6

potential triangle side value (u-1) is (6-1)=5

 

Step 2: correct x+n is a multiple of the correct value above ^^^

Formula is: u * (u-1) * 4 +1 = potential x+n

  1. 6 * 5 *4 +1 = 121, sqrt(121)=11

  2. 12 * 11 * 4 +1 = 529, sqrt(529)=23

  3. 18 * 17 * 4 +1 = 1225, sqrt(1225)=35

  4. 24 * 23 * 4 +1=2209, sqrt(2209)=47

  5. 30 * 29 * 4 +1= 3481, sqrt(3481)=59

  6. 36 * 35 * 4 +1= 5041, sqrt(5041)=71

  7. 42 * 41 * 4 +1= 6889, sqrt(6889)=83 = correct x+n value

 

Step 3: use the quadratic to check/verify each iteration above ^^^ resulting in an integer answer (no remainder)

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

 

I'll run it on some more c values, I know this is c6107.

Just trying to find an idea that makes sense first, and then test it out.

 

Pretty cool to do this from memory with a calculator and pad. I love this problem. I think about it in the morning when I wake up, and it pops into my mind during the work day. I haven't posted much lately bc I had nothing new, but maybe this idea is worth exploring.

GAnon !Nx57Pyux3E ID: 566843 Jan. 29, 2019, 5:57 p.m. No.8546   πŸ—„οΈ.is πŸ”—kun

>>8526

I still lurk. I got sidetracked during the holidays. Still do work on this though. Noticed something recently. thanks to

>>8535

I was looking at (1,1).

 

In (1,1) the values for x and a are tied to a pythagorean triple,

(x, a-1, a) so that aa = xx + (a-1)(a-1). Then I couldn't really follow what d was getting to, but I think this might be a way to tie together unrelated squares, because xx, aa, (a-1)(a-1), d*d, are all related to eachother in some way in these equations. Granted there isn't much but forgive me.

 

Another thing I saw is that if you look at the records in row 1, all of their 'ones' place values repeat over time, no matter what base you are using. For instance, if we have a base equal to 45, then all of the values in column A % 45 will repeat over time. For instance, if we use base 60, then for every even e, the a's repeat every 30, then if we use an odd e, it repeats every 15. Hovever, if our base is odd, it will always repeat at that interval. Also within the repeating the same pattern occurs backwards so theres half as many options.

 

For base divisible by 4:

Even e:

A[t] % base = A[t+base/4] % base

Odd e:

A[t] % base = A[t+base/2] % base

 

Base divisible by 2:

A[t] % base = A[t+base/2] % base

 

Other base:

A[t] % base = A[t+base] % base

 

We could then iterate through the first t records in (e,1) because those would have the correct a%t values,

then maybe we could use modular arithmetic to solve it from there. Maybe mix in the -f entry to get two equations. Maybe deliberately use a base that is divisible by 4 so that d will be even and -f will be odd, and you'll have two different bases to use in the modular equation.

 

 

Lets use base 60 for c=145

We know that A[t] = na for some value in (1,1), and that the values A[t]%60 for this are pic related:

(1, 5, 13, 25, 41, 1, 25, 53, 25, 1, 41, 25,13,5,1,1,5,13,…)

We calculated all of these and there are a bunch of repeats so we filter them out

now our options are na == 1, 5, 13, 25, 41, 53 (mod 60)

 

Second pic related is -24.

If we look at (-24, 1) = (-f, 1), then there is an A[t] = n(a-1).

Now all the options are :

(6, 20, 38, 00, 26, 56, 30, 8, 50, 36, 26, 20, 18, 20, 26, 36, 50, 08, …

which can be reduced to n(a-1) == 6, 8, 18, 20, 26, 36, 38, 50, 56

 

We can check every permutation which would be 6*9 attempts = 54 iterations, and each time you'd do this.

Lets say I'm testing n(a-1) 18, and na 41 (mod 60)

 

na - n(a-1) == 41 - 18 (mod 60)

na - na + n == 23 (mod 60)

n == 23 (mod 60)

 

So I guess it would always turn out to be just getting the difference of all these numbers, which isn't too intense of a calculation. Of course once we get n (mod 60) it's not like we have n or anything. We might be able to narrow it down further with other equations and known variables but idk. Just thought this was neat and wasn't sure if anyone else had posted it.

AA ID: 27241f Feb. 3, 2019, 1:02 a.m. No.8557   πŸ—„οΈ.is πŸ”—kun   >>8558 >>8561

>>8556

I was being sarcastic. You could apply the colour thing to any set of numbers so it doesn't really matter. It looks like this is for 5x31. It doesn't really add any new information aside from highlighting (-1,n). We were already told to do that, but when I looked into (-1,n) I didn't find any unique patterns that changed anything, and it had the same patterns as the other negative square columns ((-4,n), (-9,n), etc). We were already told to do that colour diagram thing too, early last year. He said that the colours were significant. The only other time he used these same colours was in his triangle diagrams, but I don't remember being able to find any useful connections personally (plus nobody found anything originally (GA made some example diagrams but I don't remember which thread this was)).

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 81cdd5 Feb. 3, 2019, 1:20 a.m. No.8558   πŸ—„οΈ.is πŸ”—kun

>>8557

So you can explain 3 black segments, one gray segment, 2yellow highlights, the green segments, the purple segments, the 2 same-purple highlights separating green segments, and the 2 blue highlights?

VA !!Nf9AmQNR7I ID: 01587d Feb. 3, 2019, 8:44 p.m. No.8559   πŸ—„οΈ.is πŸ”—kun   >>8561

2d(n-1)+f-1= 2xn+ x^2

F and d limit possible n and (n-1) values.

Let’s work on this!

Is it a graph line where only certain integers work?

More about eliminating values than solving?

The algorithm is ancient.

According to VQC

How to combine multiple variable equations to run side by side.

a new form of algebraic solution

Underlying fractal patterns

Triangles and squares

And how they combine

Limiting possibilities

Factors contained within f.

(D+n)^2 - (x+n)^2=c

Simple and clean.

So much complexity and simplicity combined.

Mods and patterns

(Of mods)

Factoring each square

Multiple possibilities for perfect squares in (x+n)^2

Geometric patterns in the Grid

Geometric patterns in the 8Tu+1

Leading to a Grid shortcut based on understanding of the geometric patterns.

How do we factor the two squares (d+n)^2 and (x+n)^2 ?

Using only c d e f ?

I'm reworking equations over here.

I have too much time invested here to give up.

And I'll keep going till we solve this shit.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: 81cdd5 Feb. 4, 2019, 6:47 p.m. No.8560   πŸ—„οΈ.is πŸ”—kun

I bet these folks would SUPER appreciate it/be terrified if someone were to open that wallet up and give everyone their $190 million back…

 

https://www.marketwatch.com/story/crypto-exchange-customers-cant-access-190-million-after-ceo-dies-with-sole-password-2019-02-04

 

Crypto exchange customers can’t access $190 million after CEO dies with sole password

Anonymous ID: 566843 Feb. 5, 2019, 7:30 p.m. No.8561   πŸ—„οΈ.is πŸ”—kun   >>8564

>>8557

That was a tkinter app I made that would do it for any records. Never found much from it. Then again I never really used it because I saw the flaws and I made a better app. I think we could use the coloring to highlight these cells but then view them from a different perspective. I'm stil working on a program that will let us view the cells from any angle. AKA originally the axes are (e,n) but I'm trying to make something that will enable use to view from any axes and highlight any cells.

 

>>8559

>using only c d e f

 

I think that this could be the key. He probably chose those letters for a certain reason. Maybe theres a bunch of steps. c gives you d, d and c give you e, d and e give you f, f and e give you g, g and f give you h, …. up until 'n' I'd guess.

 

c d e f g h i j k l m n -finally -> a or b

 

>VQC: Twelve steps are the key

 

c -n is 12 steps if we follow this pattern. Now obviously I have no idea what these steps are in specific, but if this is the case then vqc has given us up to f. This is 1/3 of the way already.

VA !!Nf9AmQNR7I ID: 341dcd Feb. 14, 2019, 7:39 p.m. No.8571   πŸ—„οΈ.is πŸ”—kun   >>8572

I just noticed a cool pattern.

Starting at the (na transform) element,

a[t] (e) - a[t] (-f) = an increasing pattern.

128-127=1

100-97=3

76-71=5

56-49=7 which is (n-1)

So the difference between a[t] (e,1) compared to a[t] (-f,1) moves upward in an ascending pattern until the first element in a given e column.

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

So this info greatly limits our search area.

Thinking out loud over here.

We limit the search using the (na transform) element and the a[1] element.

VA !!Nf9AmQNR7I ID: 341dcd Feb. 14, 2019, 8:42 p.m. No.8572   πŸ—„οΈ.is πŸ”—kun

>>8571

To be clear, I’m not suggesting iterating by x.

I’m suggesting using every available piece of info to limit the search area.

BigN is also a limiting piece of info

Along with 2(sqrt((f-1)/8)-1

Set boundaries, then factor

VA !!Nf9AmQNR7I ID: 341dcd Feb. 14, 2019, 9:06 p.m. No.8573   πŸ—„οΈ.is πŸ”—kun

More patterns.

This one has (-f,1) a[t]=c and its equivalent element in (e,1) is distance of (a) apart, 287 - 270= 7 = correct a value

Also, the a[t] = (bn) value = 328

in (e,1) 328 - 287 = 41 = b

so c, a, and b are available/calculable in a triangle pattern right next to each other in adjacent elements.

Here's a diagram.

VA !!Nf9AmQNR7I ID: 341dcd Feb. 14, 2019, 9:34 p.m. No.8574   πŸ—„οΈ.is πŸ”—kun   >>8575

Once you c it you can’t un-c it?? Why does a[t]=c appear in many of our cases?

 

for c287

2(sqrt(d))-1 = 7 = (n-1)

Must be a fluke, but still analyzing for patterns over here.

 

Lol, for c6107 2(sqrt(d))-1= 15 and correct (n-1)=35

Shared factor of 5. Interesting.

and 2(sqrt((f-1)/8)-1 = 7 which is another factor for 35

 

Hello PMA, Jan, and 5DAnon can I please request your assistance?

Could it be a characteristic of ONLY semiprime c values that we always have one a[t]=c element?

Makes sense that c always appears at a[t] since the prime factors have to show up again.

VA !!Nf9AmQNR7I ID: 341dcd Feb. 14, 2019, 9:37 p.m. No.8575   πŸ—„οΈ.is πŸ”—kun

>>8574

And if that element exists for a given semiprime c, we can solve the problem with the adjacent elements. (maybe!)

 

You guys know how I work, calculator in hand with a pencil at the ready. If the idea sucks, that's fine. Working from small examples over here.

AA ID: 344e49 Feb. 16, 2019, 4:34 p.m. No.8579   πŸ—„οΈ.is πŸ”—kun

>>8578

I wouldn't hold my breath if I were you. How many times has he said "hey guys let's go through the solution right now/tomorrow/next week" and then acts like he never said that? And if he actually goes through with it this time, keep in mind, there are people who have been following along with this board for months if not more than a year who don't even understand it so if he legitimately goes through the solution (or "part 1", whatever that means) nobody will even know what the fuck he's talking about aside from maybe 5 or 6 of us who have put significant effort in already.

AA ID: 344e49 Feb. 18, 2019, 4:33 p.m. No.8585   πŸ—„οΈ.is πŸ”—kun   >>8586

In columns where e is a square, c is the sum of two squares. We already know this. What hasn't been pointed out is that in columns where e is a trianglular number, c is the sum of a square and a triangle. This could potentially be useful given some of the things that Chris has been saying on /qresearch/ recently (speaking of which, Topol, you should post all of those screenshots here rather than just putting them on Discord).

VA !!Nf9AmQNR7I ID: b69141 Feb. 18, 2019, 10:40 p.m. No.8587   πŸ—„οΈ.is πŸ”—kun   >>8588

aan(n-1) rings a bell with all these new posts.

Supposedly this type of integer is easy to spot.

Although we haven't discovered the identification process.

Lol.

product of triangles and squares, i think.

Best idea I can think of is to find the next largest perfect square, which should be (aann)

Subtract the difference of (aann) and aan(n-1)

For c145 it's 625-500 = 125

Then, aan(n-1) / 125 = 500 / 125 = 4 = (n-1)

VA !!Nf9AmQNR7I ID: b69141 Feb. 18, 2019, 11:04 p.m. No.8588   πŸ—„οΈ.is πŸ”—kun

>>8587

Above is a small example, but here is another key idea we have yet to solve.

aan(n-1) is supposed to be easy to spot when comparing (-f,1) and (e,1) a[t] values.

We worked for a week on this, and we need to finish it, as Saga correctly pointed out yesterday.

We fucking work our asses off on an idea, then bail on it when VQC distracts us with a new idea. That distracting faggot.

He's doing it on purpose to slow us down.

 

Let’s make a list of unsolved crumbs.

And then work on them independently, while sharing our results. That would work well for us as a group. We are all very independent souls but love to check out each other's ideas.

And we share a common purpose, to do our part for Q / VQC etc. and our fellow humans worldwide.

Thoughts, fellow Anons and Math Fam?

VA !!Nf9AmQNR7I ID: b69141 Feb. 18, 2019, 11:26 p.m. No.8590   πŸ—„οΈ.is πŸ”—kun

The easiest method I can think of is this:

Get the (e na transform) and (-f na transform) elements.

Then create the list of a[t] values for both columns up to a[1].

Then simply move up subtracting e a[t] from -f a[t] and divide c by each result.

First whole integer result is prime a

It could be millions and millions of calcs, but it will run super fast, bc it’s so simple.

(e,1) gives us all factors.

Anonymous ID: 2344ac Feb. 19, 2019, 11:47 a.m. No.8593   πŸ—„οΈ.is πŸ”—kun

an interesting property, admitedly tested only on short integers

not directly connected to the problem presented, but nevertheless

 

sum of digits in base(b+1) is same as last digit in base(b)

 

e.g (base8):

0c1420=1+4+2+0=7(in base8) means its divisible by 8 and 7

0c3774=3+7+7+4=7(in base8) means its divisible by 7 and remainder when dividing by 8 is 4

0c7120=7+1+2+0=3(in base8) means its divisible by 8 and remainder when dividing by 7 is 3

0c6305=6+4+0+4=7(in base8) means its divisible by 7 and remainder when dividing by 8 is 5

0c6620=6+6+2+0=7(in base8) means its divisible by 8 and 7