I have a theory on possible values of n.

for every a and b combination in (e,n) a valid n record exists at (e,a) and (e,b).

examples:

(1,1,1) = {1:1:2:1:1:5} = 5; (x+n)=2x2=4; (d+n)=3x3=9

(1,1,2) = {1:1:8:3:5:13} = 65; (x+n)=4x4=16; (d+n)=9x9=81

(1,1,3) = {1:1:18:5:13:25} = 325; (x+n)=6x6=36; (d+n)=19x19=361

you will find n records at (1,1), (1,5), (1,13), (1,25) etc.

for (7,1)

(7,1,4) = {7:1:35:7:28:44} = 1232; (x+n)=8x8=64; (d+n)=36x36=1296

you will find valid n for both (7,28) and (7,44).

for (6,3)

(6,3,25) = {6:3:433:48:385:487} = 187495; (x+n)=51x51=2601; (d+n)=436x436=190096

I bet you can find valid records at (6,385) and (6,487).

So I believe we can effectively follow a and b to different levels.

Now we need to understand the small square relationship at n 1 and I believe we can "walk" the tree.

Please verify.

>>294

sorry. maybe it didn't process the first time.

what if (9,3) comes from negative space?

>>300

There are quite a number of e records where (e,2) and (e,3) exist, but aren't in the root (e,1) key.

(20,1,1) = {20:1:10:0:10:12} = 120; (x+n)=1x1=1; (d+n)=11x11=121

valid n's include (20,2), (20,3), (20,4), (20,5), (20,6), (20,7), (20,9).

Perhaps valid a and b combinations are only part of the story.

>>306

I see where you are going here.

If you recall, there is a way to create small and large squares from only e and t.

Odd values of e:

small square (x+n)^2 = (2t)^2

large square (d+n)^2 = (2 * ( (t+1)*(t+1) - 2 * (t + 1) + 2 ) + ( ( e / 2 ) - 1 ) )^2

Even values of e:

small square (x+n)^2 = (2t-1)^2

large square (d+n)^2 = (2 * ( (t*t) - t + 1 ) + ( ( e / 2 ) - 1 ) )^2

I've verified these formulas for all entries in the grid.

Do we know the relationship between a and (x+n)^2?

>>325

can you clarify?

we can get to the large square where a = 1.

Finding the correct n that represents a factor of c is still a problem.

Looks like a relationship between the small and/or large squares. I believe the small squares as they appear easier to predict.

>>326

appreciate the kind words.

>>327

yes. I believe it works everywhere. I wasn't able to take it further, as knowing the value of t in different levels of n is a problem.

>>333

in (e,1) t follows a defined pattern. Next records are incremented by t = t + 1.

we haven't identified the pattern of t for (e,n) yet.

>>334

Not sure if they add any value. Maybe Teach can make some sense of them. It could just be another way of stating the same thing that doesn't lead to a solution.

>>339

Noticed a pattern for a between e and e+1 records.

(e+1).a = orig.a + orig.x + 1

Example:

(16,1,11) = {16:1:228:20:208:250} = 52000; (x+n)=21x21=441; (d+n)=229x229=52441

a = 208

x = 20

(17,1,11).a = 208 + 20 + 1

(17,1,11) = {17:1:250:21:229:273} = 62517; (x+n)=22x22=484; (d+n)=251x251=63001

>>699

>>702

Starting to look at column zero..

Following is a test case for c = 149.

TestColumnZeroForAB a=5, b=29

P(e,n) =(1,61,6) = {1:61:12:11:1:145} = 145; (x+n)^2=72x72=5184; (d+n)^2=73x73=5329

P(e,1) na =(1,1,6) = {1:1:72:11:61:85} = 5185; (x+n)^2=12x12=144; (d+n)^2=73x73=5329

P(0,n) for c^2 =(0,10368,73) = {0:10368:145:144:1:21025} = 21025; (x+n)^2=10512x10512=110502144; (d+n)^2=10513x10513=110523169

P(0,1) na for c^2 =(0,1,73) = {0:1:10512:144:10368:10658} = 110502144; (x+n)^2=145x145=21025; (d+n)^2=10513x10513=110523169

p(e,n) a*b =(1,5,4) = {1:5:12:7:5:29} = 145; (x+n)^2=12x12=144; (d+n)^2=17x17=289

>How do you USE the Row and Column to find what you need?

processing….

>>717

nothing yet…

>>717

>>715

>for c = 149

c = 145.

Thinking out loud.

Interesting relationship here:

(0,1,3) = {0:1:12:4:8:18} = 144; (x+n)^2=5x5=25; (d+n)^2=13x13=169

(1,5,4) = {1:5:12:7:5:29} = 145; (x+n)^2=12x12=144; (d+n)^2=17x17=289

same d, e = e + 1, a = x + 1, c = c + 1.

From >>31

>Note that all the values of d for (1,1) are identical to the values of a for (0,1)

>Notice that all the values of a at (1,1) are the values of d in (0,1) with one added.

>Notice that all the values of a in this cell are also each twice the value of a perfect square.

>1+1 = 2

>4+4 = 8

>9+9 = 18

>Notice that all the values of d for this cell also follow a pattern:

>2x(1x2) = 4

>2x(2x3) = 12

>2x(3x4) = 24

Can we create a (0,n) record with the details from our entry point?

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

New record values: d = 12, c = 144, e=0, and we have patterns that define both d and a per VQC.

If this (0,n) record is created, is the relationship to the prime answer simply a = x + 1?

>>717

>>720

Teach

solve 2 x ( ( t - 1 ) x t ) = d for t? We always have d!

>>721

>2 x ( ( t - 1 ) x t ) = d

thanks wolframalpha

t = (sqrt(2d + 1) + 1) / 2

>>723

>>724

int t = ( (int)Math.Sqrt( 2 * d + 1 ) + 1 ) / 2;

solves for t given d in (0,1)

>>746

not sure of the value. The factors for small numbers are easily calculated.

Currently looking into formulas for (0,n).

>>749

sure.

there are 2 different ways to iterate records once you have a factor p.

Method 1: newT = depth * p + orig.t;

Method 2: newT = depth * p + 1 - orig.t;

With the newT, you can create a record using e,n,t values.

I showed an example and also found a case that didn't work properly.

>>623

>>624

Not sure I fully understand this.

>>751

>Does your method 1 apply to even e, and method 2 apply to odd e

With a factor of 5, I was able to use both methods and get different results on (e,1).

I don't think it matters odd/even. Though more testing is required.

>>699

>What did Einstein say about co-In(c)iden(c)e???

A coincidence is a small miracle when God chooses to remain anonymous.

>>757

yes.

>>760

ok, fine.

My mind is a bit blown away. So i'll let you guys ponder.

Following are tests for c^2 with records labeled as follows:

"P(0,n) for c^2" = an entry record where a = 1, c = c^2.

"p(e,n) a*b" = the prime solution.

"P(0,n) a^2, b^2" = a test where a=a^2 and b=b^2.

From the results below, you can see that the value c for "P(0,n) for c^2" equals the value c of "P(0,n) a^2, b^2".

So if we start from c^2, then figure out the movement down to "P(0,1) a^2, b^2", the prime solution turns out to be the square root of a, square root of b.

I'm currently working on the movement.

The difference in x values between the c^2 record and the a^2, b^2 record appears to a factor of 12.

TestABSquared a=5, b=13

P(0,n) for c^2 =(0,2048,33) = {0:2048:65:64:1:4225} = 4225; (x+n)^2=2112x2112=4460544; (d+n)^2=2113x2113=4464769

P(0,n) a^2, b^2 =(0,32,21) = {0:32:65:40:25:169} = 4225; (x+n)^2=72x72=5184; (d+n)^2=97x97=9409

n division =2048/32 = 64;

x diff =64-40 = 24;

xdiffBy24 =24/24 = 1;

p(e,n) a*b =(1,1,2) = {1:1:8:3:5:13} = 65; (x+n)^2=4x4=16; (d+n)^2=9x9=81

TestABSquared a=5, b=29

P(0,n) for c^2 =(0,10368,73) = {0:10368:145:144:1:21025} = 21025; (x+n)^2=10512x10512=110502144; (d+n)^2=10513x10513=110523169

P(0,n) a^2, b^2 =(0,288,61) = {0:288:145:120:25:841} = 21025; (x+n)^2=408x408=166464; (d+n)^2=433x433=187489

n division =10368/288 = 36;

x diff =144-120 = 24;

xdiffBy24 =24/24 = 1;

p(e,n) a*b =(1,5,4) = {1:5:12:7:5:29} = 145; (x+n)^2=12x12=144; (d+n)^2=17x17=289

TestABSquared a=5, b=157

P(0,n) for c^2 =(0,307328,393) = {0:307328:785:784:1:616225} = 616225; (x+n)^2=308112x308112=443724032; (d+n)^2=308113x308113=444340257

P(0,n) a^2, b^2 =(0,11552,381) = {0:11552:785:760:25:24649} = 616225; (x+n)^2=12312x12312=151585344; (d+n)^2=12337x12337=152201569

n division =307328/11552 = 26;

x diff =784-760 = 24;

xdiffBy24 =24/24 = 1;

p(e,n) a*b =(1,53,12) = {1:53:28:23:5:157} = 785; (x+n)^2=76x76=5776; (d+n)^2=81x81=6561

TestABSquared a=17, b=53

P(0,n) for c^2 =(0,405000,451) = {0:405000:901:900:1:811801} = 811801; (x+n)^2=405900x405900=1546052752; (d+n)^2=405901x405901=1546864553

P(0,n) a^2, b^2 =(0,648,307) = {0:648:901:612:289:2809} = 811801; (x+n)^2=1260x1260=1587600; (d+n)^2=1549x1549=2399401

n division =405000/648 = 625;

x diff =900-612 = 288;

xdiffBy24 =288/24 = 12;

p(e,n) a*b =(1,5,7) = {1:5:30:13:17:53} = 901; (x+n)^2=18x18=324; (d+n)^2=35x35=1225

>>877

please see >>879