/vqc/ - Virtual Quantum Computer

CodePost Guide

Because of the tendency for post links to disappear I will now be using Pastebin.

C#

BigInteger Square Root: pastebin.com/rz1SdACZ

Create Bitmap in: pastebin.com/hMTtJF6E

Generate Bitmap within original code: pastebin.com/JUdtehb4

Generate the large square for e and t: pastebin.com/nbjs2kz4

Original VQC code: pastebin.com/XFtcAcrz

Unity Script: pastebin.com/QgAXLQj3

Unity Script 2: pastebin.com/Y38nVWgT

Java

Create a Bitmap using the VQC Generator: pastebin.com/Dgu9aP1h

VQCGenerator: pastebin.com/VMRnkXFP

Traverse the VQC cells in real-time: anonfile.com/W44cofd6b6/VQCGUI.7z

NodeJS

BigInteger Library and Sqrt: pastebin.com/y8AXtFFr

Python

Create the VQC: pastebin.com/NZkjtnZL

3D VQC: pastebin.com/vdf8SpYt

3D VQC (v2): pastebin.com/wZM5Thzu

Generate genesis cell: pastebin.com/GKzcCpMF

''Generate positive AND negative genesis cells: pastebin.com/9ixjRyxt

Get A and B from C and N example: pastebin.com/s0SZ9BNF

VQC + t: pastebin.com/Lgufk0db

Rust

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

Create Bitmap using the VQC Generator: play.rust-lang.org/?gist=c2446efeec452fe14e1ddd0d237f4173&version=stable

Create Bitmap using the VQC Generator [V2]: pastebin.com/zGSusyz5

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

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

Previous Threads

RSA #0, or the VQC thread: archive.fo/XmD7P

RSA #1: archive.fo/RgVko

RSA #2: archive.fo/fyzAu

RSA #3: archive.fo/uEgOb

RSA #4 (not finished, but inactive): >>>/cbts/94250

RSA General (#5): >>7

Videos and Links

I've added some videos to show the exact details and equations that we would use to hack RSA once we find the solution (and some introductions to ECC), but we have to finish this before we can get to any of that.

Integer Factorization of any arbitrary integer — Part One

youtube.com/watch?v=9FeROMe0KBU

The RSA Encryption Algorithm (1 of 2: Computing an Example) (Very Simple)

youtube.com/watch?v=4zahvcJ9glg

Encryption and HUGE numbers - Numberphile

youtube.com/watch?v=M7kEpw1tn50

Public Key Cryptography: RSA Encryption Algorithm

youtube.com/watch?v=wXB-V_Keiu8

RSA-129 - Numberphile

youtube.com/watch?v=YQw124CtvO0

Elliptic Curve Cryptography Overview

youtube.com/watch?v=dCvB-mhkT0w

A (relatively easy to understand) primer on elliptic curve cryptography

arstechnica.com/information-technology/2013/10/a-relatively-easy-to-understand-primer-on-elliptic-curve-cryptography/2/

Elliptic Curve Cryptography: a gentle introduction

andrea.corbellini.name/2015/05/17/elliptic-curve-cryptography-a-gentle-introduction/

Elliptic Curve Point Addition

youtube.com/watch?v=XmygBPb7DPM

I stickied the thread. I thought it would be easier to navigate the board if the main objective was stickied. That might be a problem if we start focusing on multiple things, so if it seems like a bad idea anyone can feel free to whinge to me about it.

>>877

Continuing discussion from yesterday.

I've tried a number of combinations to link the various (0,n) records.

Following is the closest I've come to a workable path which seems to arrive at the desired a prime value.

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

The theory being, we can figure out how to walk backwards from the (1xcc) record, or really figure out the relationship between the various x values.

—————————

15=3x5

1 x c *=> (6,5,2) = {6:5:3:2:1:15} = 15;

a x b **=> (6,1,1) = {6:1:3:0:3:5} = 15;

c x c => (0,0,1) = {0:0:15:0:15:15} = 225;

aa x bb *=> (0,2,4) = {0:2:15:6:9:25} = 225;

b x ca => (0,10,6) = {0:10:15:10:5:45} = 225;

a x cb => (0,24,7) = {0:24:15:12:3:75} = 225;

1 x cc *=> (0,98,8) = {0:98:15:14:1:225} = 225;

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((14-6)/(14-12)) - 1 == 3 = 3; (a matches)

—————————

65=5x13

1 x c *=> (1,25,4) = {1:25:8:7:1:65} = 65;

a x b **=> (1,1,2) = {1:1:8:3:5:13} = 65;

c x c => (0,0,1) = {0:0:65:0:65:65} = 4225;

aa x bb *=> (0,32,21) = {0:32:65:40:25:169} = 4225;

b x ca => (0,104,27) = {0:104:65:52:13:325} = 4225;

a x cb => (0,360,31) = {0:360:65:60:5:845} = 4225;

1 x cc *=> (0,2048,33) = {0:2048:65:64:1:4225} = 4225;

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((64-40)/(64-60)) - 1 == 5 = 5; (a matches)

—————————

145=5x29

1 x c *=> (1,61,6) = {1:61:12:11:1:145} = 145;

a x b **=> (1,5,4) = {1:5:12:7:5:29} = 145;

c x c => (0,0,1) = {0:0:145:0:145:145} = 21025;

aa x bb *=> (0,288,61) = {0:288:145:120:25:841} = 21025;

b x ca => (0,232,59) = {0:232:145:116:29:725} = 21025;

a x cb => (0,1960,71) = {0:1960:145:140:5:4205} = 21025;

1 x cc *=> (0,10368,73) = {0:10368:145:144:1:21025} = 21025;

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((144-120)/(144-140)) - 1 == 5 = 5; (a matches)

—————————

785=5x157

1 x c *=> (1,365,14) = {1:365:28:27:1:785} = 785;

a x b **=> (1,53,12) = {1:53:28:23:5:157} = 785;

c x c => (0,0,1) = {0:0:785:0:785:785} = 616225;

aa x bb *=> (0,11552,381) = {0:11552:785:760:25:24649} = 616225;

b x ca => (0,1256,315) = {0:1256:785:628:157:3925} = 616225;

a x cb => (0,60840,391) = {0:60840:785:780:5:123245} = 616225;

1 x cc *=> (0,307328,393) = {0:307328:785:784:1:616225} = 616225;

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((784-760)/(784-780)) - 1 == 5 = 5; (a matches)

—————————

901=17x53

1 x c *=> (1,421,15) = {1:421:30:29:1:901} = 901;

a x b **=> (1,5,7) = {1:5:30:13:17:53} = 901;

c x c => (0,0,1) = {0:0:901:0:901:901} = 811801;

aa x bb *=> (0,648,307) = {0:648:901:612:289:2809} = 811801;

b x ca => (0,6784,425) = {0:6784:901:848:53:15317} = 811801;

a x cb => (0,22984,443) = {0:22984:901:884:17:47753} = 811801;

1 x cc *=> (0,405000,451) = {0:405000:901:900:1:811801} = 811801;

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((900-612)/(900-884)) - 1 == 17 = 17; (a matches)

—————————

6107=31x197

1 x c *=> (23,2976,39) = {23:2976:78:77:1:6107} = 6107;

a x b **=> (23,36,24) = {23:36:78:47:31:197} = 6107;

c x c => (0,0,1) = {0:0:6107:0:6107:6107} = 37295449;

aa x bb *=> (0,13778,2574) = {0:13778:6107:5146:961:38809} = 37295449;

b x ca => (0,88650,2956) = {0:88650:6107:5910:197:189317} = 37295449;

a x cb => (0,595448,3039) = {0:595448:6107:6076:31:1203079} = 37295449;

1 x cc *=> (0,18641618,3054) = {0:18641618:6107:6106:1:37295449} = 37295449;

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((6106-5146)/(6106-6076)) - 1 == 31 = 31; (a matches)

—————————

20413=137x149

1 x c *=> (249,10065,71) = {249:10065:142:141:1:20413} = 20413;

a x b **=> (249,1,3) = {249:1:142:5:137:149} = 20413;

c x c => (0,0,1) = {0:0:20413:0:20413:20413} = 416690569;

aa x bb *=> (0,72,823) = {0:72:20413:1644:18769:22201} = 416690569;

b x ca => (0,1377952,10133) = {0:1377952:20413:20264:149:2796581} = 416690569;

a x cb => (0,1500424,10139) = {0:1500424:20413:20276:137:3041537} = 416690569;

1 x cc *=> (0,208324872,10207) = {0:208324872:20413:20412:1:416690569} = 416690569;

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((20412-1644)/(20412-20276)) - 1 == 137 = 137; (a matches)

>>879

It gets even simpler.

The solution is here.

(1xcc).x - (aaxbb).x + 1 == (aaxbb).a

Difference in x between the c^2 record and the a^2, b^2 record + 1 == the prime solution a^2.

Who wants to figure out x ?!?!?!?

===

785=5x157

1 x c *=> (1,365,14) = {1:365:28:27:1:785} = 785;

a x b **=> (1,53,12) = {1:53:28:23:5:157} = 785;

c x c => (0,0,1) = {0:0:785:0:785:785} = 616225;

aa x bb *=> (0,11552,381) = {0:11552:785:760:25:24649} = 616225;

b x ca => (0,1256,315) = {0:1256:785:628:157:3925} = 616225;

a x cb => (0,60840,391) = {0:60840:785:780:5:123245} = 616225;

1 x cc *=> (0,307328,393) = {0:307328:785:784:1:616225} = 616225;

(1xcc).x - (aaxbb).x + 1 == (aaxbb).a

784 - 760 + 1 = 25.

(1xcc).x - (axcb).x + 1 == (axcb).a

784 - 780 + 1 = 5.

>>882

ok. appologies. this is exciting.

Simplified formula, and removed unnecessary calculations:

sqrt( (1xcc).x - (aaxbb).x + 1 ) == (axb).a

We just need the formula for the x value in (aa x bb).

—————————

65=5x13

1 x c *=> (1,25,4) = {1:25:8:7:1:65} = 65;

a x b **=> (1,1,2) = {1:1:8:3:5:13} = 65;

aa x bb *=> (0,32,21) = {0:32:65:40:25:169} = 4225;

1 x cc *=> (0,2048,33) = {0:2048:65:64:1:4225} = 4225;

sqrt( (1xcc).x - (aaxbb).x + 1 ) == (axb).a

sqrt( 64 - 40 + 1 ) = 5. (a matches!!!!)

—————————

145=5x29

1 x c *=> (1,61,6) = {1:61:12:11:1:145} = 145;

a x b **=> (1,5,4) = {1:5:12:7:5:29} = 145;

aa x bb *=> (0,288,61) = {0:288:145:120:25:841} = 21025;

1 x cc *=> (0,10368,73) = {0:10368:145:144:1:21025} = 21025;

sqrt( (1xcc).x - (aaxbb).x + 1 ) == (axb).a

sqrt( 144 - 120 + 1 ) = 5. (a matches!!!!)

—————————

785=5x157

1 x c *=> (1,365,14) = {1:365:28:27:1:785} = 785;

a x b **=> (1,53,12) = {1:53:28:23:5:157} = 785;

aa x bb *=> (0,11552,381) = {0:11552:785:760:25:24649} = 616225;

1 x cc *=> (0,307328,393) = {0:307328:785:784:1:616225} = 616225;

sqrt( (1xcc).x - (aaxbb).x + 1 ) == (axb).a

sqrt( 784 - 760 + 1 ) = 5. (a matches!!!!)

>>883

Doesn't d=c=ab=a^2-x for the record {0,n,c,x,a^2,b^2}

So x=a^2-ab

>>884

d=c yes.

solution can be simplified yet again to

sqrt( c - (aaxbb).x )

not following you on the formula for x.

Sorry

ab=a^2+x

So x=ab-a^2

>>886

if we knew a and b, I think the formula would be x = sqrt(ab - a^2).

We don't have the ab starting position.

The solution depends on figuring out the correct x value for (aa x bb) from only (1xcc).

>>887

I'm trying to understand now. Whats the latest equation to test?

sqrt( c - (aaxbb).x ) == (axb).a?

>>888

I fear you may be proving d-x=a?

>>885

For record {0,N,D=c=a*b,X,a^2,b^2}

ab=a^2+x so X=ab-a^2

a and b being from record

{e,n,d,x,a,b}

Where capital represent C^2 record and small letters represent the a*b=c record

>>889

anything is possible. but I'm using d, x, and a from completely different records.

>>890

sorry. I'm not seeing it.

>>891

For a and b primes

and a*b=c

there is a record for C =c^2

where A=a^2 and B=b^2

Using cap for C^2 record

{0,N,D=c=ab,X,A=a^2,B=b^2}

So D=c=a*b=A+X

a*b=a^2+X

X=a*b-a^2

>>892

don't we want to figure out x going the other way?

For a and b primes

a*b=c

There is a special record

a^2*b^2=c^2

where you can solve for X of c^2 record in terms of a and b

D=c=ab

D=A+X

D=c=ab=a^2 +X

X=ab-a^2

a and b being from the a*b=c record

>>893

the goal is to figure out a and b

We got rid of a veriable

You can do N in terms of a and b also.

X=ab-a^2=a(b-a)

X^2=2AN=a^2*2N=a^2*(b-a)^2

N=((b-a)^2)/2

just for the a^2*b^2=c^2 record

we got rid of X and N

so now

{0,N=((b-a)^2)/2,D=c,X=a*b-a^2,a^2,b^2}

So c^2 is fully described by a and b eliminating e,n,d,x variables

seems like progress

>>894

Interesting! So you can find the c^2 record and put that in some equations from the factorization we want?

It gets real hairy but you can factor the c^2 record entirely in a and b

>>902

Which a and b? I suggest using _ to denote which one it is. Such as a_1 and b_2

The a and b we are trying to solve.So I use caps for c^2 record. A of the c^2 record is a^2. B of c^2 record is b^2. a and b is the goal doesn't make sense to change that.

>>903

There are 5 factorizations of c^2 if a and b are prime. One is a^2*b^2

>>905

Good evening Anons! Re-reading VQC's crumbs and looking for patterns over here. Nothing to report yet, working on the factors of a in row one.

>>905

Baker, nice to see ya.

>>883

PMA, great work as always.

>>888

Checked!! Teach got the trip 7's last thread too. Hello Teach, nice digits!

Wow you guys seem to think I'm off my rocker! I just tested my formula for A and N for {0,2,35,10,25,49} based on 5*7=35 squared and for {0,8,21,12,9,49} based on 3*7=21 squared and for {0,98,51,42,9,289} based on 3*17=51 squared and it worked every time

>>910

Elaborate?

X = a*b-a^2

N = ((b-a)^2)/2

where a*b=c for record a^2*b^2=c^2

so for {0,8,21,12,9,49} based on 3*7=21 squared.

X=3*7-3*3=12

N=(7-3)*(7-3)/2=8

>>912

What's the next step?

>>913

By substitution for a^2*b^2=c^2

e is 0

d=c

N=((b-a)^2)/2

X= b*a - a^2

So you can describe C^2 using only the original a and b. It may be possible to substitute and solve for a or b but it kind of hairy. I'm still working with the equations. I'm just a GO player with little math training. I'd love some new eyes. This has been a blast watching and learning.

>>914

Is it easier to find a factor of c^2 than c, given c is semi-prime?

>>915

There are more factors of c^2 than c

but the a^2*b^2=c^2 factoring is obviously special

>>916

I wouldn't say special, it's to be expected. The problem is going from the A=1, B=C cell, to the factor cells. Dealing with the C^2 cell, this problem is ^2 as hard. We know the D=C, A=C,B=C cell, or the D=C, A=1, B=C^2 cell, but finding the factors is again, ^2 as hard.

>>917

It may be dead end. But eliminating 4 variables seems like progress and maybe there is more that can be done. Potentially even solving for a in terms of b

>>917

I disagree, given that a c^2 cell has many more factors.

>>919

It has more factors, but they're ^2 apart, with no way to find them.

>>920

Good point.

>>916

Have you looked into settings d^2 as c?

We can step through factors by X–, A++, but again this is useless when C is the size of an RSA number. We need a magic way to predict N, or predict X.

>>921

>>920

>>919

It was VCQ who suggested the a^2*b^2=c^2

>>888

>>907

Bit of an update here.

You can switch from an (e,n) record to (e,1) by setting n = 1 and a = na.

There is a shortcut to get to the (1xcc) a=na record for any c.

e = 0

n = 1

t = (c+1)/2.

Example below shows shortcut to (1 x cc(na)) for c =145.

The (0,1) space has known patterns.

Trying to figure out a jump to (aa x bb).

—————————

145=5x29

1 x c *=> (1,61,6) = {1:61:12:11:1:145} = 145;

a x b **=> (1,5,4) = {1:5:12:7:5:29} = 145;

aa x bb *=> (0,288,61) = {0:288:145:120:25:841} = 21025;

1 x cc *=> (0,10368,73) = {0:10368:145:144:1:21025} = 21025;

1 x cc(na) development where a=na at (0,1)

1 x cc(na) confirmed at (e,n,t) from t=(c+1)/2: (0,1,73)

1 x cc(na) *=> (0,1,73) = {0:1:10512:144:10368:10658} = 110502144;

a root = sqrt(a/2) = 72

d root = 2*(t-1)*(t) = 2*72*73

>>924

Creating an e,1 from e,n is huge.

>>925

not (e,1). (0,1).

From that, you can switch a=na again and get to the (1xcc) record. I think these are like pointers.

>>926

This means the reverse is true? You can create an E,N from E,1? THAT'S huge.

>>926

I found your notation hard to understand initially. * should always be used for multiplication.

>>924

You said

"You can switch from an (e,n) record to (e,1) by setting n = 1 and a = na."

Does it only work for e=0 still huge

>>927

I have some sample code somewhere and examples posted in these threads that showing jump from (e,n) to (e,1) and back. Will try and cleanup when we get to the end.

This is a shortcut directly to (0,1,t). Pretty powerful stuff.

>>928

my bad…

>>929

>huge

would be finding a way to get to (aa x bb)!!!

You can't predict X with a given E and N.

Hi fags, I'm the guy in previous threads concerned not about cool stuff, but the fundamental efficiency of reaching the correct cells. So with that in mind, I've been looking at {0:n:d:x:1:c^2}, because in a factoring situation, we don't have a, b, or n, but we do have c and c^2.

There seems to be an efficiently discoverable pattern in {0:n:d:x:1:c^2}. Let's take a == 3, b == 3, c == 9 simply because it appears on the downloadable spreadsheets and .png files for The End. c^2 == 81.

Now let's look through column 0 for "1:81}", and you'll find {0:32:9:8:1:81}. This can be quickly constructed via the general pattern {0:n:c:c - 1:1:c^2}. You can solve for n = ((a + b) / 2) - d. For this case:

n = ((81 + 1) / 2) - 9

n = (82 / 2) - 9

n = 41 - 9

n = 32

So, we can immediately construct a starting point, here {0:32:9:8:1:81}. Now search column 0 for ":81}". You will find:

{0:2:63:14:49:81}

{0:8:45:20:25:81}

{0:18:27:18:9:81}

{0:32:9:8:1:81}

[cont]

>>934

Now let's look at the following values of n in increasing order: 2, 8, 18, 32. Let's start by looking at the differences between elements:

8 - 2 = 6

18 - 8 = 10

32 - 18 = 14

Or, put differently,

6 = 2 + 4

10 = 2 + 4 + 4

14 = 2 + 4 + 4 + 4

Let's just look at the 4's:

4

44

444

This is called a Triangular number (https://en.wikipedia.org/wiki/Triangular_number). If these were all ones, the formula for summing them (where h is height of tree, here 3, and where s is the sum) is:

s = (h(h + 1)) / 2

So for:

1

11

111

… we would have:

s = (3(4)) / 2

s = 12 / 2

s = 6

Which you can easily verify.

[cont]

>>935

The formula for each element being 4 and not 1 would be simply 4 times the previous equation:

s = 4((h(h + 1)) / 2)

So for:

4

44

444

… we would have:

s = 4((3(4)) / 2)

s = 4(12 / 2)

s = 4(6)

s = 24

So let's go back to the sequence above:

6 = 2 + 4

10 = 2 + 4 + 4

14 = 2 + 4 + 4 + 4

Here, h = 3. We can sum up all of the 4's with the above equation. All we need to add in are the 2's, and there are exactly h 2's. So:

4((h(h + 1)) / 2) + 2h

Recall the original sequence 2, 8, 18, 32. Since it starts at 2 and not 0, we have to add 2 to the final result:

4((h(h + 1)) / 2) + 2h + 2

Let's now denote the first element 2 as element 0 and 32 as element 3, or h = 3.

s = 4((3(4)) / 2) + 2(3) + 2

s = 4(12 / 2) + 6 + 2

s = 4(6) + 8

s = 24 + 8

s = 32

>>936

Since the sum 's' is equal to the n field in {0:32:9:8:1:81}, n = 32, and in general, we can use n above where we used s. If we simplify:

n = 4((h(h + 1)) / 2) + 2h + 2

n = 2h^2 + 4h + 2

0 = 2h^2 + 4h + (2 - n)

We can apply the quadratic formula:

(-b [+-] sqrt(b^2 - 4ac)) / 2a

a = 2 (the constant in 2h^2)

b = 4 (the constant in 4h)

c = (2 - n) (the plain numbers)

(-4 [+-] sqrt(4^2 - 4(2)(2 - n))) / 2(2)

(-4 [+-] sqrt(16 - 8(2 - n))) / 4

To work the example above with n = 32, we can find "where we are" in the series in {0:n:d:x:1:c^2}. So for {0:32:9:8:1:81}, where n = 32:

(-4 [+-] sqrt(16 - 8(2 - 32))) / 4

(-4 [+-] sqrt(16 - 8(-30))) / 4

(-4 [+-] sqrt(16 - (-240))) / 4

(-4 [+-] sqrt(16 + 240)) / 4

(-4 [+-] sqrt(256)) / 4

(-4 [+-] 16)

(-4 + 16) / 4 [choosing +]

12 / 4

3

So at {0:32:9:8:1:81}, we know that we're at index 3 (zero based), or if you want to talk -th numbers, add 1, so we're at the 4th.

Interesting, right?

>>937

The interesting bit is the in the cells you listed, A is a perfect square, and N is a perfect square * 2. Maybe because you used c=81?

>>938

>A is a perfect square

This is part of the pattern

>N is a perfect square * 2

This is random luck and not part of a pattern

>>931

>>924

Ding Ding Ding! Nice work PMA! VQC hinted at this ability to move c into row 1. I'm following all your examples, reading your output and explanations, and I think I get it. Can you give us the quick rundown on the ideas and patterns you're seeing, and how they give us the ability to move from any (e,n) to (0,1)?

>>924

This looks pretty good.

Can you run through another example? I think I'm following, but i'm not sure.

>>940

currently I'm playing with the idea of moving any (e,n,t) to (0,1,t). Seeing what relationships are there.

>>924

in my example above, I don't think it's coincidence that the c^2 record is at (1,10368,73) and it's corresponding a=na record is at (0,1,73).

I'm trying to find a pattern that can bridge the gap between the (c^2).x and (aa x bb).x

Or even move directly to the (ab) solution.

It looks like all the heavy lifting has been done by the grid.

>>942

>in my example above, I don't think it's coincidence that the c^2 record is at (1,10368,73) and it's corresponding a=na record is at (0,1,73).

I agree. Very interesting.

From (0,1) we should easily be able to find a, right? Since if p is a factor of a at t, p will also be a factor at p+t, 2p+t, 3p+t, and p+1-t, 2p+1-t, 3p+1-t, etc.

>>944

Checked. For some reason I keep finding myself put in the position of being required to bake things. Over on /pol/ the mods repeatedly censored people who wanted to discuss the curious case of Sholom Rubashkin and why he deserved a commutation. Given they are constantly being censored, I need to make a thread in a non-comped board.

>>942

sqrt(a/2) in the (0,1, t for ccna) row is always equal to the 1xc row x+n.

>>946

confirmed.

my latest test case for c=145. Looking for a pattern to calculate the x at (aa x bb).

or any other suggestions!

—————————

145=5x29

1 x c *=> (1,61,6) = {1:61:12:11:1:145} = 145;

a x b **=> (1,5,4) = {1:5:12:7:5:29} = 145;

aa x bb *=> (0,288,61) = {0:288:145:120:25:841} = 21025;

1 x cc *=> (0,10368,73) = {0:10368:145:144:1:21025} = 21025;

(0,1) records to analyze

1 x c at (0,1,t) => (0,1,6) = {0:1:60:10:50:72} = 3600;

x=10

a root = sqrt(a/2) = 5 from 50

d root = 2*(t-1)*(t) = 2*5*6 = 60

a x b at (0,1,t) => (0,1,4) = {0:1:24:6:18:32} = 576;

x=6

a root = sqrt(a/2) = 3 from 18

d root = 2*(t-1)*(t) = 2*3*4 = 24

aa x bb at (0,1,t) => (0,1,61) = {0:1:7320:120:7200:7442} = 53582400;

x=120

a root = sqrt(a/2) = 60 from 7200

d root = 2*(t-1)*(t) = 2*60*61 = 7320

1 x cc at (0,1,t) => (0,1,73) = {0:1:10512:144:10368:10658} = 110502144;

x=144

a root = sqrt(a/2) = 72 from 10368

d root = 2*(t-1)*(t) = 2*72*73 = 10512

1 x cc (0,1,t) confirmed at (e,n,t) from t=(c+1)/2: (0,1,73)

1 x cc (0,1,t) sqrt(a/2) = 1 x c (x+n); 72 = 72

(1 x cc).x - (aa x bb).x = 144 - 120 = 24

sqrt( (1xcc).d - (aaxbb).x ) == (axb).a

sqrt( 145 - 120 ) = 5. (a matches!!!!)

>>947

The x at (aa x bb) is equal to c-a^2, though that's kind of the same result you already have.

You guys must be close look at bitcoin price!

Below $13,000

>>950

Interesting, but cracking RSA doesn't crack Bitcoin. It's based on ECC. However, cracking RSA would make them lose a lot of confidence in crypto and compromise millions of passwords-Not that I care.

So, I don't expect Bitcoin to die when we reach the solution, but it will be a shitstorm.

>>951

Not knowledgeable but QVC thinks cracking RSA problem for bitcoin.

>>952

Yeah, it's a major problem given all communications secured by RSA are compromised. But it doesn't hurt the mathematical foundation of Bitcoin itself.

>>951

I agree, bitcoin price reaction is unrelated.

I expect we're a month out from a true crash due to cracking ecc. If we're lucky.

I've been monitoring bitcoin for a while, there's always a dip at every major milestone, normally 25%, but this time I think people are getting scrared, its dropping closer to 33%.

>>950

I have been watching this very close and No one has mentioned the vqc,. zero zip nada. I doubt people would be selling like crazy on some conspiracy theorists on 8ch… They SHOULD but normies wont understand. Could be a super smart lurker or team.

Since nobody seems to pay attention to the meta thread, I have to bring this here. One particular tripfag keeps getting reported for being a "namefag shitposter" (it's not Topol). They aren't really posting anything that helps us, but we haven't discussed rules at all so I can't really use my own discretion. I don't want to be distracting, but if we don't discuss rules, I can't really delete anything without having the power to delete whatever I want. I don't want that and I know none of you do. If nobody wants to discuss rules, I'm not going to do anything about this particular person, so whoever keeps reporting them, suggest some rules and see what everyone thinks.

>>959

Sorry to post this here, but these are my comprehensive suggestions: >>962

Oooooh I think I figured out how to find a and b for (e, n) for any given n for t=1.

That's devious VQC

>>894

>>947

>X=ab-a^2 = c - a^2

You just used the other formula, anon mentioned earlier.

(aa x bb).x = c - a^2

a = sqrt(c - (aa x bb).x)

But I agree, we need to get to the (aa x bb) record, some way.

>>974

thanks

>>924

Is finding element t = (c^2+1)/2 a short cut? We can find n for any c for the c^2 record

{0, n, c, x, 1, c^2}

for e 0 x^2=2na

x=c^2-1

n=((c^2-1)^2)/2

Now create record in 0,1

using n=1 a =na

{0,1, sqrt(A*B), c^2-1, A=((c^2-1)^2)/2, B=c^2}

for any c=a*b prime

Now find the factors which have a pattern of repeating in this column

>>935

Thanks for show how to calculate a record

{0, n, t} for any t.

Even easier for {0, 1, t}

x=2t

d= triangular sum of h=t * 4

d = 4*(t(t+1)/2

Crazy stuff for record {0, 1, t}

If a number m has only prime factors

If t >=m then m is a factor of t{a} if t has those factors. t doesn't have to have duplicate factors. E.G. 225 for any t multiple of 15 and greater than 225 t{a} will be divisible by 225.

>>978

"A number only having prime factors" means it is semiprime. And what the hell is t{a}?

>>979

a of {0,1,t}

a number m will show up as a factor of a for {0,1,t}

if t >= m and has the same factors as m duplicate factors not necessary for t

>>980

I agree, Anon! This is a feature of the catalog of values for var a VQC has mentioned. My focus is solving the generation of a values in row 1. Key issue is, starting from only c, how do we derive a? In row 1 (0,0) a = 2. For odd e, at t=1 a=(e+1)/2. For even e, at t=1 a=e/2. This gives us the starting seeds for all of row one and also allows us to derive x, t, b, and hopefully n as well.

Another issue we're dealing with is column 0 and how to use the grid to find products of c and c^2. Finding these patterns in conjunction with knowing how to generate the seeds of var a will be a big step forward! Seems like PMA, Teach, other anons, and yourself are working on this and making progress. I also think your idea about new factors only appearing from that point forward in the chain of a values is cool. I'll check it out, and thanks for your work!

Lord Rothschild Vows To Destroy Bitcoin – Launches ‘FedCoin’

December 21, 2017

http:/ /yournewswire.com/rothschild-fedcoin-bitcoin/

Initial target identified!

>>984

Haha what? Is that a joke?

>>981

>>980

Typo, corrected. *In row 1 (0,1) a = 2*

>>985

Just relaying news article - little motivation.

Isn't even real yet. MSM is hyping it though.

http:/ /www.zerohedge.com/news/2017-12-20/washington-post-propagandizes-again-says-fedcoin-will-be-bigger-bitcoin

U.S. Government Awards $7 Billion to Develop New Fedcoin

http:/ /moneywise411.com/tbo-awards-7-billion-to-develop-new-fedcoin/

>>987

What is this Jewing and why does my government have the authority to create a new cryptocurrency?

>>989

(((Coinbase)))

(((Bitrex)))

It's like all the Jews these people mistakenly trusted unsurprisingly fucked them over.

>>990

Go figure.

Can we create a Blockchain Markets & News thread so we don't cuck this up for those doing serious maths?

Litecoin Founder Cashes Out, Sells Entire Stake After 9,300% Rally

Dec 20, 2017 8:28 PM

http:/ /www.zerohedge.com/news/2017-12-20/litecoin-founder-cashes-out-sells-entire-stake-after-9300-rally

Charlie Lee, the creator of the world’s fifth-biggest cryptocurrency, Litecoin, announced shortly after midnight that he was cashing in his profits after a torrid, 9,300% rally in the past 12 months. In a post on reddit, the San Francisco-based software engineer who founded litecoin in 2013, said that he sold and donated all of his holdings over the past few days.

"Litecoin has been very good for me financially, so I am well off enough that I no longer need to tie my financial success to Litecoin’s success. For the first time in 6+ years, I no longer own a single LTC that’s not stored in a physical Litecoin" Lee said in the post.

>>986

I was confused

I'm afraid converting to c^2 record to a (0,1) record is a dead end because the factors show up only when they are a factor of t. When there is a remainder it changes and if i understand QVC they show up twice as frequently.

>>993

In an (e,1) record

>>991

>>995

anyone having any luck? going in circles.

>>1012

Nope, I guess VQC needs to drop by again. I bet he feels like a teacher explaining maths to a special needs kid.

>>1013

I'll ask him. He's probably sent a crumb regarding where we should be going, he should just point to it.

>>1012

Nope, being confined to a crappy netbook doesn't help either. Seriously considering buying a new laptop tomorrow just to use it for xmas…

>>1015

Do you use Windows?

If you get Linux it will be many times faster and it will never get slower.

>>1012

>>1012

Hey PMA! Here's what I'm working on.

We are given c.

Find d and remainder e.

Using e to find starting value of a in (e,1).

In (0,1) e =2

For odd (e,1), t=1 we can calculate a var as (e+1)/2

Example: in column 3 (3,1,1) a=2

For even (e,1), t=1, a var = e/2

Example : in column 6 (6,1,1) a=3

So we can find a at t=1 starting from only c. I think so, please check my work, lads.

If true, we can now calculate x,t,b, and n for element one (t=1) in any column. Odd e and even e have different equations for t.

Knowing the pattern of how a and b drop and move as t increases in row 1, we can literally create a ladder to run t up as high as we need to to find a match for c???

VQC said all factors for a column exist in row one, the "One Row to rule them all," and if all factors exist there, then all potential c's exist there also.

So if we can ride the t elevator up to our c value, or maybe even multiple a and b parings that equal c, then maybe from that height we can look around the grid to find other locations where the answers are similar.

PMA, your method to move between (e,n) and row 1 would help us find other matching places in the grid record.

I know I've got more work to do on this, so please let me know if you find errors or have corrections. I've been feeling stuck a bit, but persistence and focus will win the day. Let's keep searching!

>>1018

I always tell people to stop using "Plebdows" because Linux does exactly what I want and never gets slow.

>>1019

Guess I have high standards. Only so much you can do with two 1.65GHz cores on any OS, but I imagine it would be unusable with Windows on it

>>1017

The problem with running up the ladder is its still too slow for RSA numbers even if we skip over a lot of steps.

My thinking is the big_n hint which should be "simple to calculate" could be a maximum bound, hopefully not too much larger than n, and we could go backwards from there instead? Anyone figured out what big_n is supposed to be?

>>1012

>>1017

Another way to say it: if c has a remainder e, we know it's var a value at t=1 in row one. All factors of a and b, as well as product c are also contained in that column row. So we start at t=1 and build the ladder up to c, filling in all the elements on the way up. Thoughts, Anons?

>>1021

Not sure, maybe it's too slow? Certainly eliminates a whole lot of the possibilities. Maybe we need a way to quickly jump to higher t values? I like your idea, and agree that if we can get closer to c, the calcs will be much faster. A maximum bound based on how long the c value is in digits? Thinking out loud here.

>>1023

I tested a few variants of this earlier on the RSA100 example (since we have the whole solution for it and can also create row one) and found the increases really useless after getting up to around a values of ~20 digits or so. This was with jumping to next factors only, not every t value. But I didn't verify my code properly so I may have fucked up

>>1024

I envy your dedication. I have some half-finished code to just brute force generate every possible formula ever into an abstract syntax tree and compile that to code, and some basic ideas for verification against the grid. Feels a bit like cheating TBH, but hey. Just print a list of formulas with > 99% coverage for each symbol after.

Also considered throwing genetic programming at it, but then even if we get a solution, deciphering that program would be even harder than what we're already doing.

>>1025

The code is a bit sloppy, but the ability to create various records is sound, and every record is verified in the constructor.

So I know the values that I’m looking at are accurate.

The relationships are difficult to understand. And the number of records to bring together is a bit much.

I’m sticking around for the solution - it would be terribly frustrating otherwise.

>>1026

I'm still here and working too.

I'm going back through Chris's posts and trying to re-read all the information to see if we're missing anything.

I'll post what I find.

>>1027

Sounds good. I need some new avenues to explore.

>>1026

I did check some of your generation code against the VQC code for some large grids with no problems. Haven't been able to work as much for the last few days just cause shitty netbook.

Mostly I just didn't verify my own implementations as much as I should cause now I doubt some of the tests I ran. Writing checks into the code is a good idea

The difficult relationships is why I'm at the stage of wanting to throw insane computing power at it and have it just tell me the good results :)

Side note, if you just bruteforced every possible formula, how would you check that against the grid, aside from matching every value in every record and counting results? I feel it needs some jump logic too based on the latest progress here.

Obviously we're sticking around till TheEnd :)

>>1028

One thing I'm obsessed with is the fact that x&n are 2 squares, as are (d+n) and (x+n).

There's some pattern either, x+n's or x's for 1xc and axb are always factors of each other.

There's more too it then that, but I think maybe the triangles made by the axb and the 1xc x+n squares may be similar.

>>1030

To continue that, VQC said:

>I thought from near the start in 2011 that it would feel good if "x" marked the spot.

I think factoring x, or 2x might be useful, so we recursively repeat that until row 1 or column 0.

>>1031

x^2 + e = 2an

and (d+n)^2 -(x+n)^2 I think are the algebraic keys to this theory.

x1 = x at 1 * c

xa = x at a * b

n1 = n at 1 * c

na = n at a * b

n1 - (x1^2 - xa^2) = aa * m

Where m is a odd number.

This implies that if we knew the multiple, we'd know a.

Examples (sorry about the lack of cell references):

1261 - (50^2 - 8^2)/2 = 43 * 1

13 - (4^2 - 0)/2 = 5 * 1

69 - (11^2 - 5^5)/2 = 7 * 3

1237 - (49^2 - 19^2)/2 = 31 * 7

369 - (27^2 - 15^2)/2 = 13 * 9

815 - (40^2 - 12^2)/2 = 29 * 3

>>1033

I'm working on the algebraic side now, since we know e=e, c=c and d=d at both rows, I'm setting the equations as equal to each other and solving.

>>1032

>>1034

Agreed, Teach. I've been thinking about (d+n) and (x+n) often. I'm working on the algebra side too. What are your thoughts on the "catalog f values" for a in row one? Any insight at this point?

>>1032

I’ve looked at x and n in (0,1) up the yin yang. x, x^2, 2x, x/2, +1, -1, multiples between records, relationships to a, b, c, small square, big square, square root.

Every time I thought I had a breakthrough I went to a different c value and it all broke down.

I know there’s a simple solution here.

In the (0,1) space the formulas get a bit easier as e = 0.

And we’re now dealing with perfect squares.

You’d think that the prime solution would be smack in the middle of the c and cc records. Or that the original e, d, or x would factor in somehow.

>>1035

*catalog of values* sorry lads, I'm out in the field so doing math challenge sneaking in when I can till I get home.

>>1039

I did some basics as well but kinda ran into a dead end using formulas I didn't understand properly and went back to other stuff. But the altitude of a right-angle triangle may be a useful variable.

And as posted above somewhere, triangular numbers are also relevant somehow (sorry I suck at this part)

>>1040

Well, like in the picture you can basically turn any 2 variables into a pair of triangles and change them around.

>>1040

For example the hypotenuse of na is sqrt(n^2+a^2)

>>1041

>>1042

Yeah that part I get, its pretty easy. But some of the more advanced math of what you can do with triangles I haven't figured out. I did manage to calculate the altitude but didn't study it enough to figure out why those formulas worked so I went back to another approach

>>1037

Please be warned that this might be a big rabbit hole as I haven't managed to find any useful correlations to the grid myself. I am not smart enough to prove it is a dead end yet

>>1037

This looks interesting, I'll take a look.

>>1036

Another quote by VQC:

>(e,n,d,x,a,b) in a grid (e,n)

>That is one way to organise the grid.

>Selecting two of the six variables (dimensions).

>There are SEVEN dimensions because each (e,n,d,x,a,b) element represent c.

>What are ways or grids or cubes are there to represent each elements?

>You could use another pair apart from (e,n)

>What would that produce?

>What other ways or shapes are there to view the six (seven) dimensions?

>What images result?

>Does this help find patterns?

>Does this help you c(see)?

Here's something else for you, assuming our 2 rows, 1xc and axb, labeling all my variables n1 or na for the n values corresponding to the cells etc.

2n1 = -(aa)^2 + 2aa(na + d) - (2d-1)

For those who dont want to click around for all VQC's posts, all the important stuff in 1 place:

https:/ /pastebin.com/eHJTEwWY

>>1046

>>1045

I’ll look into the formula later.

Re-reading vqc notes.

>>1046

Thanks Teach. Re-reading crumbs now.

I've been rendering in other dimensions, but haven't seen any useful patterns. Taking requests.

Can I get some eyes on something real quick please?

We made an assumption that the equations with t applied to more than n=1, but I'm not convinced they do.

If you assume that in every cell t always starts from 1, then, using the grid:

Choose your 2 primes, and find the cell 1xc.

Then *if* your n here is odd, the following will hopefully work (I haven't found a way to do it when n is even).

Move to the negative side by following F.

1xc for F will always be the first element in the cell.

Find the element in the cell when a=c, look at the t (assuming that always t starts from 1).

So far for me, t has always been a factor.

Am I crazy? Maybe? I dunno?

Help please guys!?

So far, whenever I

>>1051

Forget the last line, I dun goofed.

>>1051

So what you're saying is, ignoring our current formula for t - cells with odd n, moving to the negative side using f, f_t will always be a factor of f_x * f_c? Did I get this right? Rephrase it maybe?

>>1053

I think you understand correctly. But so far, I've only been using the grid to check, because if so our formulas for t are wrong.

I'm coding something now to test.

>>1053

actually, correction f_t will be a factor of c I'm finding.

This would explain the p being a factor of a at t then it also is one at p+t, 2p+t, 3p+t, etc, and p+1-t, 2p+1-t, 3p+1-t

>>1056

I'm sorry, this is wrong.

Seems to be just luck when its correct or not.

Very sorry all.

>>1057

No worries, posted worse mistakes already :)

But I have to second PMA's suggestion for notations, would be way easier if we agreed on something. My f_t thing just felt convenient for replying to you.

Maybe something for moving between records is needed? First idea (please improve it)

orig.e

orig.n

formula (f, whatever)

new.e

new.n

>>1058

Yeah, agreed. I'll use dot notation from now on.

>>1059

Not saying dot notation is the best and will be happy to use whatever as long as its consistent :)

Though it does suit us programmers

The formatting in my post is kinda verbose though, we could do p for prev and n for next (confusing?)

formula
p.e = 0 p.n = 0   >   n.e = 0 n.n = 0

But prev and next is still bad if we want to do more than one jump. Maybe

formula
e[0] = 0 n[0] = 0
e[1] = 0 n[1] = 0

Just suggestions for now

>>1061

Or we could keep the original notation and do something like this (yes I am to lazy to make real examples)

{0:0:0:0:0:0}
formula
{0:0:0:0:0:0}
{0:0:0:0:0:0}
{0:0:0:0:0:0}
formula
{0:0:0:0:0:0}
{0:0:0:0:0:0}

But we may want 7 dimensions not 6, so if we're going for this one we need to agree on a position for t

Well lads, we're currently stalled. Senpai, can you please help your Disciples? Just like Jesus and his disciples, we're having trouble understanding.

>>1064

Or one more:

* pythagorean triples

Thank you!

>>1064

Teach! Are we cool? Did I offend you with the Pepe panty post? If so, I apologize. No love from Teach no mo. :(

>>1066

Of course we're cool!

Sorry! Didn't mean to be rude, my head was just down in the numbers today.

>>1066

Also, what happened to Topol and CollegeAnon?

We're losing numbers!

I'm looking into Pythagorean triples and they appear everywhere in the grid, I wish I knew more math.

I am learning more math in this process though!

>>1068

Topol got booted by faggots who just want math and no crazy fun. Click on catalog to find Topol's ez bake oven. Boring. :(

CA? I have no clue.

>>1069

I'm sad Topol got booted. I like his posts, makes it feel alive here, and his pics are always great.

>>1070

Figured not voting would count as don't care/let him stay but guess not. Quiet here without him

>>1070

Oh, don't worry. If summoned, he will return with many amemes.

I'll go to ez bake and ask him to visit us.

>>1074

>>1070

I've summoned the Topol! Fuck banning people from the general. We need math and crazy ideas and fun! That's who our Senpai is!

>>1069

>>1070

Nobody has a problem with the occasional inspirational shitpost. What we have problems with is when it's 20% of total posts.

>>1077

I'm gonna post all the VQC maps in a bit. Then you guys can see who we're working with. VQC is the man, and he's way out there too. Who we gonna ban?

>>1047

f = 2d+1-e (f is what you add to c to make a square)

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

Research further…

>>1078

Nobody wants bans. But nobody wants every other post to be four huge pictures of deep dream wankery plus stoner memes. Especially when we're crunching the numbers and formulas in each others posts. One in twenty I think we can absorb.

So it's a matter of degree. Like, for example, bumps on .5 chan.

>>1063

Sounds like you guys are a bit down in the dumps on the math at the moment. I figured that for tonight's entertainment Instead of mildly dank memes Instead I will deliver something completely different. I have been working on this for a few days now.

As I look at this stuff I feel over my head so I figured I would play to my strengths. This is either a fan-fiction story OR it is a bit of a 1 hobo think tank document. Really it is just some ideas about what could happen if you fuckers get this thing to work. I kinda have a feeling someone would have crushed this in a moment if it was a known direction that has been explored mathematically. Basically if it was fake it would have been found out by now.

So I am leaning to this being the real deal and if that is the case then SOMEONE needs to start thinking about this stuff that I wrote about in this document. Im shit with programming and only slightly better with math (though this math is shockingly simple) but big picture stuff is fun for me. So Please enjoy this doc. I tried to make it funny and an easy read with a few little easter eggs so I hope you enjoy it! It is totally on topic as well so no grumpy anons angry because there is not a math problem to worry over in it.

https://anonfile.com/D2Fdp3d3b4/Ramifications.rtf

No. I'm tired of this. VQC is a spiritual and mathematical adventure. Some anons want to ban the spiritual. GTFO faggots. Who is contributing? Get out if you have no formulas or inspiration! Topol is more welcome than haters. Contribute or not.

>>1082

Forgot your name, VeritasAequitas? Why are you hopping around on multiple overlapping IDs, and posting with and without your "name"?

Nobody is saying ban the spiritual or ban even shitposting itself. The bans should be reserved for when there are high signal levels, such as when VQC is posting or when there is significant technical progress being made, and spiritposting or shitposting every other post is disruptive.

You guys, we don't need any more drops or crumbs from VQC. We already have enough.

I've already solved generating (e, n) for any given e and n. It's just a matter of studying the patterns in (0,n) and (1, n).

I've started to look into the controlling of c and it looks very promising.

>>1088

So why don't you share how you did it?

>>1088

What? We knew how to create an element from e and n since the first thread. But t makes it a lot more complicated.

I almost became blackpilled earlier. We need to finish this so clowns' hell-deserving activities get exposed.

>>1076

Pretty sure I'm the faggot who tipped over an anon or two. Still here every night though. Just posting in the general on cbts (never name/trip fagged there once), in Topol's EZ Bake, and doing my thing.

And the crypto news and bitcoin discounting in here was intended as motivation for you faggots. At least we have a thread for that now too.

I'm here because I've already seen this before in a vision, in a different state. Five years or so ago, I was literally SHOWN a GRID. It was 2D, but in a 3D perspective. All started black, then X-Y axis showed up, first ELEMENT dropped in with a BOOM (hard to describe when you're getting hit *hard* with a point by something beyond, but have had it a few times like that). Then the next, and next as the grid filled with thin colorful tiles, extending the rows and column heights. And then they each started coming to life a little and shimmering, while more dropped in filling beyond visual limit. And then, as it continued to increase in complexity… I went IN…

- that vision isn't why I have faith in this thing here. There were plenty of 'signs' in Chris's questions that he's legit. cbts anons couldn't always handle/understand his questions. We're all on the path at different points, they'll come around.

>>1081

Looking forward to the read Lord Hobo!

>>1085

Juicy juicy, just got started! And the look on the King's face is -> priceless! Trumps too, like "Yup, I think we have a deal coming." Faggot on the left is just smiling for the photo op, they *love* their group photo ops over there.

>>1088

Figured you had when I saw >>971 last night, 'devious' made it clear you weren't throwing BS.

The other reason I'm here is ANOTHER thing I was *shown* was about a new currency, EARTH DOLLARS. Was supposed to get that going over 5 years ago but I pussied out and became a coward once a family came into the picture. I was all set to get suicided for sake of doing my bit to save the world, had started lining things up, cleaning house, but then, when suddenly you've got a family and a life or two to raise in this world that you love more than anything, it gets DAMN hard! I think I was brought/meandered here for help with that mission, maybe its time has passed, who knows. I will say, it's all about seeds, roots, and trees -> 1=2, 2=4, 4=8…

Out of the 12,000 I'd pulled for the initial seeding, I have a mere 333 left, sitting here, waiting, and I haven't given up hope.

Might not sound like a lot, but there's a quote to the effect of "From a single seed, an Empire can be built".

Oh, and rainbows are important, frequencies, octaves. So important, perhaps this symbol was co-opted as well?? Just like the swastika symbol. Embrace your faggotry and all will be fine.

"Remember upon the conduct of each depends the fate of all". ~ Alexander Ptolemy III

Back to work! No more secrets!

K fags

for a*b=c

a and b prime

c^2= b^4/(b^4-b^2=1)

Is that useful?

c completely described by b

Minecraft how about a picture.

for a*b=c a and b prime

a^2*b^2=c^2

using caps for the squared expression

{0,N,c,X,a^2,b^2}

X=a^2*(b^2-1)

N=a^2*((b^2-1)^2)/2

see earlier posts

a^2+2X+2N=b^2

a^2+2a^2*(b^2-1)+a^2*(b^2-1)^2=b^2

a^2(1+b^2-1+b^4-2b^2+1)=b^2

a^2(b^4-b^2+1)=b^2

a^2=b^2/(b^4-b^2+1)

c^2=b^4/(b^4-b^2+1)

Square a value c, where c is the product of two different primes.

What are the values of n in column zero where these appear?

What is the value of d where these appear?

Enumerate the pattens of n in each cell of column zero.

The pattern GROWS in a certain way as n increases in column zero.

This pattern is related to the pattern across row one (e,1)

>>1094

Heya!

TERRIBLE way to relay messages.

Almost wasn't passed along. :D

Freaked her out and I'm not even sure I got the correct message and all I'm finding is some… deleted thread? or something?

>>1094

Aight, I believe I found it. Someone calling for banning namefags? And?

(Don't call that number again, plz. Yes, I know that those stupid background sites give bad info.)

>>1102

all good.

Hit me up on twitter lol

>>1103

Nah, nothing to do with me. People can shit on me and choose to ban me if they so choose (not that message was about that), and I don't give two shits who calls me what. I know what I know, I am what I am, and I'm all good.

Carry on and keep sending your love to these Math/Programmer fags!!!

>>1099

The d at the location where a = c*c appear all to occur in (0, 2) and will contain the factors of c and then some extra factors.

>>1107

And the extra factors appear to be the ones from the b.

>>1097

Please ignore made mistake

>>1098

More inspiration in the EZ bake if anyone wants some snacks. <;3=

>>1099

VQC, can you please take a quick look at my post >>934 where I show how to calculate any n for any column 0 record where c^2 appears?

>>1099

These are weird

For odd n, they appear to be the same as in row one (next.a = prev.b)

{0:3:12:6:6:24=144}
{0:3:36:12:24:54=1296}
{0:3:72:18:54:96=5184}

For even n, this makes no sense yet. Sometimes there are two interleaved patterns

{0:6:9:6:3:27=81} - pattern 1
{0:6:24:12:12:48=576} - pattern2
{0:6:45:18:27:75=2025} - pattern1
{0:6:72:24:48:108=5184} - pattern2

{0:12:18:12:6:54=324} - p1
{0:12:48:24:24:96=2304} - p2
{0:12:90:36:54:150=8100} - p1
{0:12:144:48:96:216=20736} - p2

Sometimes more

{0:8:5:4:1:25=25}
{0:8:12:8:4:36=144} - p1
{0:8:21:12:9:49=441} - p2
{0:8:32:16:16:64=1024} - p3
{0:8:45:20:25:81=2025} - p4
{0:8:60:24:36:100=3600} - p1
{0:8:77:28:49:121=5929} - p2
{0:8:96:32:64:144=9216} - p3
{0:8:117:36:81:169=13689} - p4

tested results thanks QVC

for a*b=c a and b prime

a^2*b^2=c^2

using capitals to designate squared record

{0, N, D=c=a*b, X, a^2, b^2}

X=ab-a^2 from X=D-A

N=((ab-a^2)^2)/(2*a^2) from 2AN=X^2

N=a^2/2-ab+b^2/2 square and factor above

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

D+N = ab+a^2/2-ab+b^2/2=b^2/2+a^2/2

D+N large side = b^2/2 +a^2/2

X+N = ab-a^2+b^2/2-ab+a^2/2=b^2/2-a^2/2

X+N small side equals b^2/2-a^2/2

>>1115

Didn't realize that what you were doing. Got some good ideas anyway. Thanks

>>1089

Take a look at (1, 1). The t = 2 has a = 5. Then go t + 5 steps, a = 85.

(1, 5) starts with a = 5/5 and b = 85/5. This appears to hold true for all (e, n).

Again, we want to find t = 1 for (1, 13). We find 13 at t = 3. We "save" this a (13) and then add 13 to t (3 + 13). We should get a = 481.

13/13, 481/13 = 1, 37. Also the first t = 1 for (1, 13).

>>1121

For clarity, the (1, 13) example we find 13 at t = 3 in (1, 1) and we find 481 at t = 16 in (1, 1).

>>1122

This only seems to work for generating n that appear as a's in (e, 1).

For instance take (1, 37), 37 doesn't appear as any a or b in (1, 1), but it appear in (1, 13). I still haven't figured it out entirely yet.

>>1123

However, there might be more to it.

In (1, 1) you'll find 481 (13 * 37). If you use this as a starting point you'll get (1, 37) starting with 13 (which is correct).

I think, in order to enumerate the "possible" n's you'll have to "jump" between n's.

Since we know that a=1, b=37 appears in (1, 13) you'll either have to jump from (1, 1) to (1, 13) in order to find it, or we might have to employ recursive factorization to "know" that 37 is a possible 'n' by factorizing 481 when we encounter it in (1, 1). But I don't think the recursive method is the way to do it.

>>1099

I'd love a thumbs up if I'm moving in the proper direction

Found a link between our entry c value and the cc record in (0,1).

(0,1).d == ( ( c - 1 ) / 2 ) * ( c + 1 );

Still looking for elusive d formula from x…

—————————

145=5x29

a x b *=> (1,5,4) = {1:5:12:7:5:29} = 145;

1 x c => (1,61,6) = {1:61:12:11:1:145} = 145;

aa x bb => (0,288,61) = {0:288:145:120:25:841} = 21025;

1 x cc => (0,10368,73) = {0:10368:145:144:1:21025} = 21025;

(0,1) perfect squares to analyze

aa x bb (0,1) => (0,1,61) = {0:1:7320:120:7200:7442} = 53582400;

1 x cc (0,1) => (0,1,73) = {0:1:10512:144:10368:10658} = 110502144;

ccna.d = 10512 compared to 10512 = 0

((ter.c - 1) / 2) * (ter.c + 1) = ((145 - 1) / 2) * (145 + 1)

(ccna.d record match!)

>>1120

Thanks. Yeah I should have been more clear. You calculate the 1 * c^2 value in column 0, then convert that n to "index", aka "height". This is zero-based, so it goes index 0, index 1, index 2, index 3…

My example had n = 32 and I calculated a index or height of 3.

The point I should have better stated is that from that you can calculate the n values of index 2, 1, 0 in constant time (which are 18, 8, and 2 respectively).

This is interesting because any 0, n cell you generate this way has perfect squares for a and b. Work through it once or twice on paper; it's fucking magic.

>>1119

>>1121

>>1125

>>1128

Wow, great work Anons! Thanks for getting some forward momentum going again.

>>1099

>>1117

Also analyzing the (0,n) space.

There is a path to the correct record. Below are 2 examples for 3x5 and 3x7.

Substituting the a-> b values at e = 0.

For the 3x5 example, I tried simply to use the new.d = (prev.d / 2). Not quite right.

Looks like we need to figure out this movement to solve.

15=3x5

a x b *=> (6,1,1) = {6:1:3:0:3:5} = 15;

1 x c => (6,5,2) = {6:5:3:2:1:15} = 15;

aa x bb => (0,2,4) = {0:2:15:6:9:25} = 225;

1 x cc => (0,98,8) = {0:98:15:14:1:225} = 225;

(0,1) perfect squares to analyze

aa x bb (0,1) => (0,1,4) = {0:1:24:6:18:32} = 576;

1 x cc (0,1) => (0,1,8) = {0:1:112:14:98:128} = 12544;

Path subtituting a -> b

(0,1,8) = {0:1:112:14:98:128} = 12544;

(0,9,13) = {0:9:56:24:32:98} = 3136;

(0,1,4) = {0:1:24:6:18:32} = 576;

21=3x7

a x b *=> (5,1,1) = {5:1:4:1:3:7} = 21;

1 x c => (5,7,2) = {5:7:4:3:1:21} = 21;

aa x bb => (0,8,7) = {0:8:21:12:9:49} = 441;

1 x cc => (0,200,11) = {0:200:21:20:1:441} = 441;

(0,1) perfect squares to analyze

aa x bb (0,1) => (0,1,7) = {0:1:84:12:72:98} = 7056;

1 x cc (0,1) => (0,1,11) = {0:1:220:20:200:242} = 48400;

Path subtituting a -> b

(0,1,11) = {0:1:220:20:200:242} = 48400;

(0,9,22) = {0:9:140:42:98:200} = 19600;

(0,1,7) = {0:1:84:12:72:98} = 7056;

>>1132

Thanks for the reply. I really am glad you liked it. I will keep it up. This has been a very inspiring experience and I am glad others have liked my thoughts and ideas.

If I am off on something PLEASE feel free to call me out. I have very thick skin and am always open to other ideas or information. Much of what I stated in this doc is "kluged" together. It will have insight but at the same time will be messed up on the details. I have always been a big picture guy so that is what I like to do. It leaves me with some holes sometimes.

Im glad you liked Sneakers. I feel Like I am living that movie sometimes. Well, I will post the next episode of this document now renamed "Unforeseen Consequences"

I think that is a good name (shamelessly stolen from my favorite video game). I am not sure I Have seen "Go for it" but I want to check it out.

Thanks for your support and I hope to provide some guidance and direction from the 40 foot view. I do not claim to KNOW what that is but I love to explore it.

>>1135

Kek. Dicsword.

What journal?

>>1139

The Martin Luther "What you say?!" copy-pasta meme made me kek.

The rest of it… I feel like my story was at least somewhat touching and then… Absurdly unnecessary hatred laden with appeals to emotion followed by a slight at the God Emperor.

-le sigh-

>>1140

I mean, you could have at least called him a Zionist fuckstain. Or a low level Qabal member or something.

There is a difference… Hell, he mighta been in Kosher Nostra (not joking) or that sycophantic "follow the messiah rebbi" hassidic mafia shit.

So like…

At least complete the thought.

>>1140

Read it yourself. Watch the documentaries on him. Sounds nice enough eh?

Until you read the FBI reports on the bank fraud, meth lab and animal abuse.

No, you'll know why I felt that way if you take the time to actually read into it.

>>1142

I feel like you penned that before my last comment loaded… lol

>>1143

I did. But really, you will have a very hard time reconciling why this guy is going to be free if you read all sides of his life story.

>>1142

Maybe it was the pics that made feel it was overly generalized. Makes sense to me.

It's not like I was arguing with ya about the piece of shit or about shekel grubbing kikes, but I generally leave your "non-MoT obsessed" Jew out of it at least.

Hell, I haven't attended a regular service in a synagogue since I was like… 8.

>>1144

Free but unprotected?

Maybe "free soon" happens after "swamp draining" and golly… he's not available to be freed anymore.

Positive thinking.

Positive thinking.

>>1145

And why not since I was 8?

Because my parents were sick of "jew society" and their bullshit. And we never looked back.

>>1147

>>1146

Yup. Every time I fall for the "regular Jews are okay" I meet one and they treat me like a disgusting rat. If that is really how it's going to go I'd be so happy.. Because this guy does not deserve to be free. I didn't even mention how he treated the immigrants or how they completely destroy any discussion of this now with muh holocaust and muh nazis.

I'm not crazy or a maniac. I just needed to get that out. It's basically how my week went.

What the fuck you two?? Seriously, get your own fucking board or start another thread and stop shitting this up. What the fuck??

And !!UrvFpU0has, enough with your recruitment and shilling for collaborators on some fucking occult mission. This shit is inside everyone, hardwired, stop being an egotistical fuck, and take it to the Bake, you've got your space in which to spew. Lord have mercy!

>>1157

I was trying to motivate again since it was so important I did so last time that folks tracked me down. I saw a lull and tried related a story. Then OP came in all down in the dumps and I tried to cheer OP up and got him over to the EZ Bake as I could. Aaaaand it went there.

I'm trying to behave, damn it.

>>1157

And this isn't an occult mission.

This is a mission of vision, so that all may see.

Literally the opposite of that thing you said.

>>1134

I’m here, Baker! Still working on the row (e,1) patterns, and also looking for patterns of c^2 in column zero. Any chance you could post a link to the original .csv output? I’m on the road, and my small printed copy isn’t cutting it for the column one research.

>>1158

Thanks Topol for the inspiration! The thread needs some encouragement.

>>1094

Cool vision, Mr E!

>>1133

Thanks Hobo! I haven’t had a chance to read it yet, but appreciate the memes a lot.

>>1160

Unfortunately I can't. I am with family now.

>>1162

Me too, lol. I’ll get back to work tomorrow. Merry Christmas!

>>1157

And yes. This is inside everyone.

But not everyone is focusing there at the moment.

Some of them are focusing on code.

And I'm here to support them.

<;3=

It's like gay sex.

Some other guy has it covered and I can experience vicariously through him to understand the idea.

Savvy?

>>1165

Literally you might as well say that the Math and the capability to figure this out is within ME, as well. Which it is. But we're all at different points on the same… whaaaaatever… and the collective consciousness is the delivery system.

Don't forget… all of this work is actually already accomplished… this is how it's disseminated to the world.

Once it's understood here by the few, it can be passed around through the channel(s) to the many.

If VQC wanted credit for this, like the whole Lucifer serving Man story, he'd have just handed it to you. But… it wouldn't be as thoroughly understood and… how in fuck would you have found THIS TEAM of ABSOLUTELY CAPABLE MINDS if you were just handing out something that seemed so monstrously absurd to begin with? When would you have even heard about it? Teach seems to be on the academic up and up but how many filters would it have gone through before TEACH ever saw it, let alone us academic and corporate outsiders?

I'm not even a fucking mathnerd and I understand that.

So… quit with the discouragement.

You want to piss and moan about thread sliding in a timeless environment, you can do it elsewhere.

I was brought back into this space for a purpose… Anons needed -headpats-. I did my job, and now I'm fuckering back off until I either pop back in, or my attention is brought back here.

>>1166

Topol, keep it up. Words and insights are very helpful.

>>1168

Teach - thanks for adding clarity!

>>1131

Teach - in case you missed it yesterday, I was able to piece together a path through various (0,n) records starting from (1xcc) and getting to (aaxbb).

I can see the path clearly, and there most likely is some elegant math to jump directly to it.

In the meantime, I've gone back to the "drawing board", as VQC suggested and think I've figured out the pattern to walking up and down the tree at (0,n).

Rather than flood with examples, following are the formulas:

to walk up the tree at (0,n)

n is odd: new.d = d + 2*(x+n) + 2n

n is even: new.d = d + (x+n) + n/2

to walk down the tree at (0,n)

n is odd: new.d = d - 2*(x+n) + 2n

n is even: new.d = d - (x+n) + n/2

I have verified this with data from (0,1) through (0,6). Both up and down.

>>1169

Thanks PMA, I just re-read.

I don't fully get it. Do you think you could explain the part from:

>(0,1) perfect squares to analyze

on?

And what exactly do you mean by substituting a and b?

Sorry for the questions, I want to understand!

I feel like we may be close again…

>>1170

Records in the (0,1) space are very well defined. We can break them into their component parts very easily to get an answer.

For example at (0,1,4) we know that the value of a = 18 can be represented as 2*(3^2). And the prime answer is "3".

Squaring the c, puts us into the (0,n) space. Substituting a = na, puts us in the (0,1) space.

And creating the record directly by (0,1) and t = (c+1)/2, is a shortcut to getting to (1xcc) for any c.

Now there, the trick is finding that aa*bb record.

Subsequent records in a column are connected by a -> b movements.

In reviewing some data, I found a way to get to the b value of the (aa*bb) record, by going through 1 jump.

Start with the (cc) record, move it's a to the next record b (figure out the d), and then one more jump to the aa*bb record.

I've verified this path twice, but calculating the d value presented a problem.

Hence my previous post.

>>1171

Ok, I think I gotcha. I'm going to need to sit with this a bit.

I'm going to run my own tests to re-prove your tests, and see how they correlate to my path. I think we might be coming full circle to the same point!

Do you think you could test with one of these pairs next?

3*13, 3*17, 3*19, 7*17

>>1171

>move it's a to the next record b (figure out the d)

So I'm with you until here. You move b to a. you know e=0, and you need 1 more variable to find a specific cell, and you're saying you can calculate d?

I'm stuck PMA.

>>1168

>>1169

Y'all legit made me tear up

>>1163

Merry Christmas to you too :)

Writing this with 1% battery.

>>1173

I posted earlier in this thread a way to calculate the d value for the (1xcc) record directly from c.

So I figured to focus a bit on navigating records by changing d values.

Path subtituting a -> b

(0,1,8) = {0:1:112:14:98:128} = 12544;

(0,9,13) = {0:9:56:24:32:98} = 3136;

(0,1,4) = {0:1:24:6:18:32} = 576;

First record is the (1xcc) record. Last is (aaxbb).

Notice the movement of a to b in each step. e is always 0, n is 1 in the first and last record. We have the a value in the first record. So we need one more variable to move to the second step, and then we will have the a value to use in the last calculation.

This may not be the end solution, but this is the first time I've seen a direct connection between an (axb) and (1xc) record.

>>1179

Ok, I'm with you.

Look at your example. 2 things to notice (if you haven't already):

1) taking your a from (0,1,8), how many times in the 0 column will it appear as b? It is always going to be 1/2 of a perfect square. And as such it'll appear in the 0 column at every n where n is a perfect square (in your example n was 9). Because this value (98) appears many time, the pattern of the value that it appears next to (the a to its b) is all possible values of 1 half perfect squares, that have an e=0.

2) There's a row directly below that one that, (0,9,14), where a = 18, and b = 128, and n will equal the n of (0,9,13).n.

I'm not sure what this means, I'm having trouble visualizing it now.

Were you already aware of these patterns?

>>1169

well, shit. turns out the d movement formulas work for all cases of (e,n).

Walking up tree 5 iterations from a=5, b=41.

(9,9,5) = {9:9:14:9:5:41} = 205

(9,9,14) = {9:9:68:27:41:113} = 4633

(9,9,23) = {9:9:158:45:113:221} = 24973

(9,9,32) = {9:9:284:63:221:365} = 80665

(9,9,41) = {9:9:446:81:365:545} = 198925

(9,9,50) = {9:9:644:99:545:761} = 414745

Walking down tree 5 iterations from a=157, b=273.

(12,8,26) = {12:8:207:50:157:273} = 42861

(12,8,22) = {12:8:153:42:111:211} = 23421

(12,8,18) = {12:8:107:34:73:157} = 11461

(12,8,14) = {12:8:69:26:43:111} = 4773

(12,8,10) = {12:8:39:18:21:73} = 1533

(12,8,6) = {12:8:17:10:7:43} = 301

>>1180

Thanks Teach. I will look into this soon.

>>1181

fucking nice.

>>1181

Can you show me with this example?

{59:5:61:19:42:90}

From the grid, the cell values are:

{59:5:61:19:42:90}

{59:5:71:21:50:102}

{59:5:119:29:90:158}

{59:5:133:31:102:174}

{59:5:197:39:158:246}

{59:5:215:41:174:266}

>>1169 Amazing body of work. You seem to have all the pieces!!

I'm having trouble verifying your new.d formula. Could you give an example?

>>1168

>>1169

>>1171

>>1179

Hey guys! I'm tripping out on how d values increase by 1 every other column! Along with a and b too. So for odd e, d,a,b increase evenly as columns increase. Same for even columns. PMA your example in >>1181 jumps up by 9t every time. Awesome pattern.

Chris is gone from twitter. So is Julian Assange on same day. Remember there are no coincidences!

>>1169

Hello PMA! So your formula needs x and n to walk the records. Only problem is that we have to derive them from c. What are your thoughts on getting x? We need a first, correct? I could be wrong.

>>1187 You start with b=1 a=cc record so you have n and x

>>1189

I'm sure q has some good pedo intel on @Jack.That would be awesome to see him perp walk!

>>1179

So how much for you first RSA cracking program!

>>1183

>>1184

Teach - good eye.

Following are what I have for all grid entries at (59,5)

(59:5:10) = {59:5:61:19:42:90} = 3780; (x+n)=24x24=576; (d+n)=66x66=4356

(59:5:11) = {59:5:71:21:50:102} = 5100; (x+n)=26x26=676; (d+n)=76x76=5776

(59:5:15) = {59:5:119:29:90:158} = 14220; (x+n)=34x34=1156; (d+n)=124x124=15376

(59:5:16) = {59:5:133:31:102:174} = 17748; (x+n)=36x36=1296; (d+n)=138x138=19044

(59:5:20) = {59:5:197:39:158:246} = 38868; (x+n)=44x44=1936; (d+n)=202x202=40804

(59:5:21) = {59:5:215:41:174:266} = 46284; (x+n)=46x46=2116; (d+n)=220x220=48400

(59:5:25) = {59:5:295:49:246:354} = 87084; (x+n)=54x54=2916; (d+n)=300x300=90000

(59:5:26) = {59:5:317:51:266:378} = 100548; (x+n)=56x56=3136; (d+n)=322x322=103684

(59:5:30) = {59:5:413:59:354:482} = 170628; (x+n)=64x64=4096; (d+n)=418x418=174724

(59:5:31) = {59:5:439:61:378:510} = 192780; (x+n)=66x66=4356; (d+n)=444x444=197136

And here is my walking up formula for a=42, b=90 and a=50, b=102.

Walking up tree 5 iterations from a=42, b=90.

(59,5,10) = {59:5:61:19:42:90} = 3780

(59,5,15) = {59:5:119:29:90:158} = 14220

(59,5,20) = {59:5:197:39:158:246} = 38868

(59,5,25) = {59:5:295:49:246:354} = 87084

(59,5,30) = {59:5:413:59:354:482} = 170628

(59,5,35) = {59:5:551:69:482:630} = 303660

Walking up tree 5 iterations from a=50, b=102.

(59,5,11) = {59:5:71:21:50:102} = 5100

(59,5,16) = {59:5:133:31:102:174} = 17748

(59,5,21) = {59:5:215:41:174:266} = 46284

(59,5,26) = {59:5:317:51:266:378} = 100548

(59,5,31) = {59:5:439:61:378:510} = 192780

(59,5,36) = {59:5:581:71:510:662} = 337620

My walking tree formula seems to iterate only the movements from a -> b. Not sure if this is a bug or a feature.

>>1171

Getting to the aa bb record is ballgame! I'm working on understanding each of your steps.

>>1193

Ok, I'm with you now.

So does it always work for walking in steps of 5? or n?

I also still don't get how this helps us get to aa*bb, but i'm trying!

>>1196

Merry Christmas to you VA. And PMA, and Topol, and MA, and Hobo, and AA, and CA (if he's still alive), and Baker and all the other anons!

Merry Christmas VQC, I hope the plan plays out smoothly over the coming days, and I hope we can help you.

I enjoy being here with you all too!

>>1195

>>1197

Happy holidays to everyone.

I adjusted my test case to include the latest walking of the tree using calculated values of d.

Review the TestIterateMovementToAABB line below.

It may look like I'm simply subtracting 1 from t each time, but the reality is that the previous d value is being calculated from the current record using the formula:

return ter.d - 2 * (ter.x + ter.n) + 2 * ter.n;

where ter is the current record.

I think that if we can figure out now how to skip multiple records using a factor, we have a workable solution.

—————————

145=5x29

a x b *=> (1,5,4) = {1:5:12:7:5:29} = 145;

1 x c => (1,61,6) = {1:61:12:11:1:145} = 145;

aa x bb => (0,288,61) = {0:288:145:120:25:841} = 21025;

1 x cc => (0,10368,73) = {0:10368:145:144:1:21025} = 21025;

(0,1) perfect squares to analyze

aa x bb (0,1) => (0,1,61) = {0:1:7320:120:7200:7442} = 53582400;

1 x cc (0,1) => (0,1,73) = {0:1:10512:144:10368:10658} = 110502144;

TestDCalculationFromC

———————

ccna.d = 10512 compared to 10512 = 0

((ter.c - 1) / 2) * (ter.c + 1) = ((145 - 1) / 2) * (145 + 1)

(ccna.d record match!)

TestIterateMovementToAABB

1) (0,1,72) = {0:1:10224:142:10082:10368} = 104530176

2) (0,1,71) = {0:1:9940:140:9800:10082} = 98803600

3) (0,1,70) = {0:1:9660:138:9522:9800} = 93315600

4) (0,1,69) = {0:1:9384:136:9248:9522} = 88059456

5) (0,1,68) = {0:1:9112:134:8978:9248} = 83028544

6) (0,1,67) = {0:1:8844:132:8712:8978} = 78216336

7) (0,1,66) = {0:1:8580:130:8450:8712} = 73616400

8) (0,1,65) = {0:1:8320:128:8192:8450} = 69222400

9) (0,1,64) = {0:1:8064:126:7938:8192} = 65028096

10) (0,1,63) = {0:1:7812:124:7688:7938} = 61027344

11) (0,1,62) = {0:1:7564:122:7442:7688} = 57214096

12) (0,1,61) = {0:1:7320:120:7200:7442} = 53582400

Matching aabbna record found!

(1 x cc).x - (aa x bb).x = 144 - 120 = 24

sqrt( (1xcc).d - (aaxbb).x ) == (axb).a

sqrt( 145 - 120 ) = 5. (a matches!!!!)

>>1198

Good job PMA.

I am worried about the t– though…

I like the process, but so far we're just decrementing x by 2 each time, which is essentially a search for "a".

But I agree that this might unlock the ability to algorithmically move through it.

For this example the n = 61 from 1xc the t == 61 from 0,1 at the end…

Is that a coincidence?

>>1199

>Is that a coincidence?

yes.

There is a formula here.

Goodnight fags.

This is epic!

Merry Christmas!

>>1198

>>1199 Checked!!

All primes, With multiples of 3. Just like column x in row one.

Ugh, this thread sucks. I mean, where is that fag Topol on Christmas Eve? I would love to see ponies, math lsd trips, memes, whatever. Do I really have to go over to ez bake, Topol?

>>1198

How long does it take to iterate? It’s mostly multiplication and subtraction and a computer is quite good at those. But 100s of digits?

>>1205

Patriots are winning, Baker! 10 days?

>>1205

Do you know whats happening?

I'm concerned for Chris.

>>1204

+1 for math lsd pics

and +1 for topol summoning generally

>>1211

Lads, think about the challenge we've undertaken at VQC's request. WHEN we solve this, we can make a major contribution to POTUS and our world. Using math. Let's go, Faggots!! Spirits Up!! Christmas is here, let's solve this. We can help with a gift for Christmas.

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———-Freedom.

>>1210

Thanks Teach! Merry Christmas! "When the infinite became the finite." Season 1:God Makes Mary Pregnant, via energy beams using incredibly complex math. Stay tuned. Lol, I totally have faith guys. But it's good to laugh at yourself too, Jesus is the man. Merry Christmas, again.

>>1218

Yeah, they do. Jesus and his teachings are beautiful. BC they love evil and serve it, they hate him. Our spiritual and biological and mental evolution has outgrown them. Jesus is a symphony of love and truth. Religion, different story. I always go to the source document.

Yah yah, source documents, etc. just saying, it's Christmas. Love this powerful story. Just like Tolkien.

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Freedom.

>>1207

>>1198

It won't be as fast as Julian oops Chris says so there is work to do. However the numbers in those cells grow fast. PMA may have already done enough to endanger RSA for those with ill intent.

>>1196

b = 1 was typo. Its the a =1 b = cc record which is solvable for x and n because e is 0. PMA has put together all the steps you need to move around grid to solve problem. Now it needs speeding up.