>>7050

… one of you two people who solved it want to help us out?

So I was trying to think of something and I was visualizing our records as sums of consecutive odd numbers. This is very evident through d+n and x+n if you look at L numbers. Another thing that helps this is if you know that a square is a sum of consecutive odd numbers. Anyway here is the picture I did it in excel. lmk if unclear in any way. Basically I'm proposing a new pathway to study. Maybe these records are linked in another way that we can visualize through these pictures and this idea but we can see in the records in another way.

>>7147

Also if we did the gcd algorithm on the two numbers we would have solved it sooner

>>7153

I wasn't thinking about the circle with this. It would be an iterative approach. After work I'm gonna try to find relevant records and map it out more clearly.

>>7147

Relevant to this, I did the same type of thing with F from the inside.

Aside, I think that if you look at the e values they could be candidates for x in row n=1.

>>7147

>>7165

Realized these are the same two algorithms but the first was a little optimized (skipped every other). The goal from this perspective is to find a number K which is equal to (d+n)^2 - d^2 BUT this number must also be equal to e + m^2

(d+n)^2 - d^2 = e + m^2

dd + 2dn + nn - dd = e + m^2

n(2d+n) = e + m^2

for c=145

5(212+5) = 1 + m^2

5*(24+5) = 1 + m^2

5*(29) = 1 + m^2 =m = d [convenient because n=a]

From start (c=145):

(1,61,12,11,1,145)

61*(24+61) = e + m^2

13,31

[3, 2, 20, 7, 13, 31]

2(202+2) = 3 + m^2

2*(42) = 3 + m^2

84 = 3 + m^2

m = 9

84/4 = 21

This might be a path to a recursive solution of sorts but you need the solution to generate the numbers (as far as I know yet)

Also I've been thinking about this

>>7155

We need a way to use this info to our advantage. I think that we can use this to look at the different values in (e,1) and (-f,1). So we know that at the correct x in (e,1), A[t] = na. At the same x+1 in (-f,1), A[t]=(n-1)a. Also I've done some looking and I remember (pic related) that if A[t] is divisible by a, then A[t+a] is divisible by a. This also means that if A[t] is divisible by n, then A[t+n] is divisible n. Maybe we can multiply that stuff out. Here is my attempt it might be good.

n = 1

2na = xx + e

e%2==0:

X[t] = 2(t-1)

A[t] = (X[t]*X[t] + e)/2

= (2(t-1)*2(t-1) + e)/2

= (4(t-1)(t-1) + e)/2

= (4(tt-2t+1) + e)/2

= (4tt-8t+4+e)/2

= 2tt - 4t + 2 + e/2

e%2==1:

X[t] = 2t-1

A[t] = (X[t]*X[t] + e)/2

= ((2t-1)(2t-1) + e)/2

= (4tt - 4t + 1 + e)/2

= 2tt - 2t + (e+1)/2

diff even e:

eA[T]a - eA[t] = fA[T]a - fA[t]

[2TT - 4T + 2 + e/2]a - [2tt - 4t + 2 + e/2] = [2TT - 2T + (-f+1)/2]a - [2tt - 2t + (-f+1)/2]

2TTa - 4Ta + 2a + ea/2 - 2tt + 4t - 2 - e/2 = 2TTa - 2Ta + (-f+1)a/2 - 2tt + 2t - (-f+1)/2

• 4Ta + 2a + ea/2 + 2t - 2 - e/2 = - 2Ta + (-f+1)a/2 - (-f+1)/2

• 4Ta + 2a + ea/2 + 2t - 2 - e/2 = -2Ta + (-(2d+1-e)+1)a/2 - (-(2d+1-e)+1)/2

• 4Ta + 2a + ea/2 + 2t - 2 - e/2 = -2Ta + (-2d + e)a/2 - (-2d+e)/2

• 4Ta + 2a + ea/2 + 2t - 2 - e/2 = -2Ta + -da + ea/2 + d - e/2

• 4Ta + 2a + 2t - 2 + ea/2 - e/2 = -2Ta + -da + d + ea/2 - e/2

• 4Ta + 2a + 2t - 2 = -2Ta + -da + d

• 2Ta + 2a + 2t = -da + d

• 2a(T+1) + 2t = d - da

Since our e is even, our x value is even. This means 2(t-1)=x, so 2t-2=x, so 2t=x+2)

• 2a(T+1) + x + 2 = d - da

Now we have T to decode.

eA[T] = N (because we multiply by a outside the function)

N = 2TT - 4T + 2 + e/2

N - e/2 = 2TT - 4T + 2

0 = 2TT - 4T + (2 + e/2 - N)

quad

4 +/- sqrt(16 - 42(2 + e/2 - N))

/2*2

4 +/- sqrt(16 - 16 - 4e + 8N)

/4

1 +/- sqrt(8N - 4e)/4

So T = 1 +/- sqrt(8N - 4e)/4

So now we can calculate big T whenever we want.

• 2a(T+1) + x + 2 = d - da

• 2(d-x)(T+1) + x + 2 = d - d(d-x)

x + 2 - d = 2(d-x)(T+1) - d(d-x)

x + 2 - d = (d-x)(2(T+1) - d)

x + 2 - d = (d-x)(2T + 2 - d)

x + 2 - d = 2dT + 2d - dd - 2Tx - 2x + xd

2 - d - 2dT - 2d + dd = - 2Tx - 2x - x + xd

(2 - d - 2dT - 2d + dd)/(- 2T - 2 - 1 + d) = x

Someone check this??

>>7184

Second half I must have made a mistake

>>7151

>>7156

BIG post

We can at least begin to understand the complex plain.

Don't think of it as a + bi, think of it as Ae^(iθ)

e^(iθ) = cos(θ) + i * sin(θ)

[proof]

Probably the coolest proof of all time. Conceptually its out of this world.

Taylor Series [look up for e, sin, cos to validate]

e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + …

e^(iθ)= 1 + (iθ) + (iθ)^2/2! + (iθ)^3/3! + (iθ)^4/4! + …

e^((iθ) = 1 + (iθ) - θ^2/2! - iθ^3/3! + θ^4/4! - …

e^2(iθ) = (1 - θ^2/2! + θ^4/4! - θ^6/6! + …) + i(θ - θ^3/3! + θ^5/3 - …)

e^2(iθ) = cos(θ) + isin(θ)

Then, from this, we can get any angle from θ because these are all the points on a circle. At that point if we multiply by a constant, we 'shoot' or expand the line out in its current direction. Then numbers are Ae(iθ).

If we want to multiply two numbers, the way we can think about it is you multiply the LENGTHS together an that is your A value, then you add the θs and that is your angle. I made you people some pics to show you how it works. The vertical axis is +i, horizontal is real numbers

The first is all different numbers with the same radius and different thetas randomly made. It's made to be viewed each pic seperately. Basically you notice that both first points are on the same inner circle and no matter the angle they always end up on the circle of radius^2.

The second is to give you an idea of the angles and I made it like a clock with hours and minutes because why not.

The third and fourth are to help you imagine the complex conjugate numbers. These are of the form,

so they are on the same real axis and have opposite imaginary numbers.

This means vertically they differ (in the pics) but horizontally they are the same

(a+bi)(a-bi) = aa +abi - abi -bibi = aa - bb

So this will always land you in the real numbers.

Moreover, if the radius of the number and its conjugate is equal to the radius of others, then they will land on the same point pics related.

These look like the death star.

Lastly are the 4 pics for squaring a number.

These look the coolest. They always end up making a cardioid with other pics related, which looks like the rebel alliance

http://demonstrations.wolfram.com/

ModularMultiplicationOnACircle/

https://www.youtube.com/watc

h?v=qhbuKbxJsk8

>>7217

>>7222

Okay so here is what I did for that picture I wanted to see what squaring complex numbers would be like. And I know that complex numbers typically deal with circular stuff so I decided to keep them all the same length and just change their angle. Then I multiplied the number by itself. Then I drew a line from 0 to the point and one from the point to the square of the point. As I changed the number (went around counter clockwise from 3:00) I went from red to green.

Or do this:

iterations = 200 #arbitarary

radius = 4 #arbitrary

for i in range(iterations):

theta = 2.0pii/iterations

point = radius * (cos(theta) + i*sin(theta))

square = point * point

draw_line_series(0,point,square)

>>7225

It doesn't change angle where did you find that?

>>7227

>>7440

KEEP GOING

This shit is crazy.

The babylonians wrote in base 60 and they also had phi, the golden ratio, the one from fibonacci, etched into stones. The fact that this is the case is actually really cool. Also you always have those people posting the cyclical digit tesla math stuff which looks like 2, 4, 8, 6 with doubling.

>>7440

I'll help you look. Also remember to check this in other bases (binary)

>>7440

These minor patterns,

1,7,9,3 and 2,4,8,6

come from different multiplications

1,7,9,3, comes from multiplying repeatedly by 7

>>> 7*7

49

>>> 49*7

343

>>> 343*7

2401

>>> 2401*7

16807

>>> 16807*7

117649

>>> 117649*7

823543

If you do it by 3's then you get

3 -9 -> 27 -> 81 -> 243

3,9,7,1 which is the pattern in reverse

Then the 2,4,8,6 obviously comes doubling or multiplying by 2, because

22 = 4, 42 = 8, 82 = 16, 162 = 32, 32*2 = 64

2,4,8,6,2,4,…

Maybe we can use the fact that binary is doubling and maybe encode our number into a fibonacci number then do somethign from there.

12 step program

We admitted we were powerless over alcohol—that our lives had become unmanageable.

Came to believe that a power greater than ourselves could restore us to sanity.

Made a decision to turn our will and our lives over to the care of God as we understood Him.

Made a searching and fearless moral inventory of ourselves.

Admitted to God, to ourselves, and to another human being the exact nature of our wrongs.

Were entirely ready to have God remove all these defects of character.

Humbly asked Him to remove our shortcomings.

Made a list of all persons we had harmed, and became willing to make amends to them all.

Made direct amends to such people wherever possible, except when to do so would injure them or others.

Continued to take personal inventory, and when we were wrong, promptly admitted it.

Sought through prayer and meditation to improve our conscious contact with God as we understood Him, praying only for knowledge of His will for us and the power to carry that out.

>Having had a spiritual awakening as the result of these steps, we tried to carry this message to alcoholics, and to practice these principles in all our affairs.

>>7471

This started right around the time when I stopped drinking… No coincidences right?

//The VQC is a lookup table

//The question you answer is what do you do to find the lookup x for c that you gives you na or (n-1)

//On-c-e you c i-t you c[an]not un-c i[t]

Math in the last line:

on-c-e potentially c-e

c i-t

c[an] NOT (logical not)

UN-c (logical inverse of negative c)

i[t]

I think in the original vqc code

a = i - j

b = i + j

so i = (b+a)/2 = (d+n)

Values to maybe check

-c-e = -(dd+e) - e = -dd - 2e

i - t for t where i = i[t] = d+n

i[t] for various t

Inverse of c*an and -c

This last line wasn't an accident and there is something in it

>>7485

>>7494

>Could you quickly lookup or figure out any values that are a square multiples by two consecutive numbers like n(n-1)?

>Think about the values in 0,1 and 0,1.

Well interestingly enough in cell (0,1) the d values (also the a values in (0,0)) are these:

4, 12, 24, 40, 60, 84, 112, 144

Ignore 4 and 12 for now they're special.

24=2(34)

40=2(45)

60=2(56)

84=2(67)

112=2(78)

144=2(89)

Remember how also if you multiply any cell in column zero by a constant it is the same cell in another row

2 * (0,1,12,4,8,18) = (0,2,24,8,16,36)

So this would be a way to get any factor where the factor is divisible by 2.

So I thought about the values in (0,1)

If we think about these in 0,1 (binary) (double meaning) then we can just bit shift all these numbers over by one (or just divide by two for you newbies) and get every entry higher than 6.

So now we know that n(n-1)a*a MUST exist in column 0

This doesn't take care of the square multiple thing but I think its worth noting.

>>7503

Dude everything is in this grid

>>7485

http://mathworld.wolfram.com/EulersNumberTriangle.html

Here is some examples for various c

>>7536

I posted that second one. That is a program I wrote to view records with the colors he defined in the map. I think you're right though the colors must be linked.

>>7547

What happened last night? I posted like 4 posts with regards to the tetrahedral numbers and their locations in the grid. Also the numbers of the form t(t+h)/2 locations in the grid. I guess it was sort of a mass shitpost. Did BO delete it?

Also this post right here that I'm quoting has a different trip but I remember posting that yesterday. Is this a timeline shift or something? Am I the only one??

Also AA made a bunch of posts that got delete too. WTF

>>7559

>>6506

I'm stupid

>>7561

I did not know you were BO. Also trip change.

>>7563

Confirmed.

>>7552

>>7553

>>7554

So I took these patterns and I wrote some code to figure out where these were. Here is a summary of my findings.

For any odd number s, the series is defined by this function:

f(i) = i*(i+s) / 2

The first, for s=1, is the triangular numbers: 1, 3, 6, 10, 15, 21

The second for s=2, is 2, 5, 9, 14, 20, 27

Here is a table of the values horizontally.

What I found is that If your shift is s, an odd number, then

f(1) = 1*(1+s)/2

This value is found in the A[t] value where X[t] = 1 at (e,n)=(1+s,1).

If you increase to f(2) the new (e,n) would be (1+2s,1) with x=2.

Generally for any shift s:

for any x:

let f(x) = x*(x+s)/2

The f(x) is equal to the a value at the entry (1+x*s, 1, x)

Moreover, if you look at this sequence of records, the d values are equal to the sequence with the shifts + 2.

Example, if I'm looking at A[t] values equal to x(x+3)/2, then the D[t] values for those same records are gonna be equal to x(x+5)/2.

Code output related

>>7565

So if we want to find our c value as an A[t] value in row n=1, then we could find it in 3 places [that I know of for now], maybe 6 if we can use negative s too complex for this point in time.

We would need c = i*(i+s)/2 for some i and s

2c = ii + is

This is a little weird but let's do it with c=145.

2c = 290 = 2529

10 * 29 = i*(i+s) if i = 10 and s = 19

Then this would put it at (e,n,x) = (1+si, 1, i) = (1+1910, 1, 10) = (191,1,10)

[191, 1, 155, 10, 145, 167]

Or we could go to this:

2c = 290 = 2529

=558 = i(i+s) if i = 5 and s=53

(e,n,x) = (1+5*53, 1, 5) = (1+265, 1, 5)

[266, 1, 150, 5, 145, 157]

Or:

2c = 290 = 2529

= 2145 = i(i+s) if i=2 and s = 143

(e,n,x) = (1 + 2*143, 1, 2) = (287, 1, 2)

[287, 1, 147, 2, 145, 151]

Could we triangulate with these or something??

>>7566

But of course the only one we could access at first would be the third one (e,n,x) = (2c-3, 1, 2), so maybe it would be worthwhile to check out the negative value.

2c = i*(i+s)

i=c

i+s = 2 =c+s=2 => s=2-c (negative value)

Then of course this would appear at (e,n,x) = (1+s*i, 1, i)

= (1 + (2-c)c, 1, c) = (1 + 2c - cc, 1, c)

for c=145 it would be:

[-20734, 1, 290, 145, 145, 437]

>>7565

>>7566

So all of these equations are of the form f(x) = x(x+2h+1)/2 for any h.

Basically if x is even, then x+2h+1 is odd, but we can still divide it by two because of the x. If x is odd, then 2+2h+1 will be even, so that is divisible by two, so either way the product is divisible by two.

For the tetrahedral numbers, 1, 4, 10, 20, …, they have the formula:

f(x) = x(x+1)(x+2) / 6

This is similar to above, but instead of dividing by 2 we divide by 6.

Therefore we need one of x, x+1, and x+2 to be divisible by 2 and another (or the same) to be divisible by 3.

So basically if we want to do the same analysis for tetrahedral-type numbers, then the equation must be of the form:

x(x+6g+1)(x+6h+2)/6 for any g,h.

My lunch break is almost over so I'll try and find these in the grid after work.

>>7573

Nevermind. The 4,10,20 is NOT the tetrahedral numbers in (4,1), because the next number is 34 not 35. This kind of throws a whole wrench in my idea so I'm going to go back to looking at the other sequences

>>1762

I've updated some of this long overdue D-grid thread I made before. I will try to post more stuff. Maybe I'll get the A grid in there too.

>>7636

>Think col 0

>Think col 1

>Think -1

Column 0 and column 1 are specified here.

Think -1. He doesn't say column -1 so I'd say they mean row -1, which as all the negative squares in it. And if you notice from all these n=-1 entries there are little projections of them that map going from (-d^2, -1) to (-d^2 + 2, 0), (-d^2 + 4, 2), …. (-d^2 + 2i, i)

So maybe we could start from these -1 entries with the right d (or d and d+1 because our f record) and check something like in between these entries in the column. Obviously since one of d or d+1 will be odd and the other will be even, one will get cast into the col 0 and one will go into col -1. (I thought it was gonna be 1 so it would make more sense with this crumb but whatever) maybe this could be a path into using the center columns.

So if you draw a square and a smaller square inside of it, you can tilt the inner square a bit and then it makes four identical right triangles around the edge. One length of the triangle is the base of the big square, (d+n), then from the corner it goes into the middle corner of the tilted one like pic related. These are all the squares for products where the factors differ by 2. Then there is 4 and 10. Now I don't know if this is completely related, but it is a completely unique shape to every product. Also it looks like they start off moving then sort of set into place but approach a limit. Idk. You could also see that this is just two rectangles also and you could pull this out and fractal it up

>>7660

>>7658

So the dimensions of this boils down to a quadratic equation. If you have the inner square as x+n and the outer as d+n, then d+n would be the hypotenuse. You would also see that the little extra length (on the far right and far left) would be the same distance, lets call it w for no specific reason.

This function screenshot returns the length and angle

>>7676

check the grid patterns board. I thought this once too