>>5515

Hey Topol,

Here are the grid records for your spreadsheets. c, prime solution, P, and p.

c = 155147

(698,77181,197) = {698:77181:393:392:1:155147} = 155147; f=89; (x+n)=77573; u=38786; (d+n)=77574

(698,1281,174) = {698:1281:393:346:47:3301} = 155147; f=89; (x+n)=1627; u=813; (d+n)=1674

(52,1594,29) = {52:1594:57:56:1:3301} = 3301; f=63; (x+n)=1650; (d+n)=1651

(11,18,3) = {11:18:6:5:1:47} = 47; f=2; (x+n)=23; u=11; (d+n)=24

c = 7430551

(4926,3712551,1363) = {4926:3712551:2725:2724:1:7430551} = 7430551; f=525; (x+n)=3715275; u=1857637; (d+n)=3715276

(4926,51,238) = {4926:51:2725:474:2251:3301} = 7430551; f=525; (x+n)=525; u=262; (d+n)=2776

(52,1594,29) = {52:1594:57:56:1:3301} = 3301; f=63; (x+n)=1650; (d+n)=1651

(42,1079,24) = {42:1079:47:46:1:2251} = 2251; f=53; (x+n)=1125; u=562; (d+n)=1126

AND, I just noticed something quite remarkable.

Take the various (x+n) values from the c=7430551 as an example.

a=2251 (x+n)=1125

b=3301 (x+n)=1650

(x+n) diff = 1650 - 1125 = 525

(x+n) from the prime solution record = 525!!!!!!

Holy cow:

here's c = 6107

(23,2976,39) = {23:2976:78:77:1:6107} = 6107; f=134; (x+n)=3053; u=1526; (d+n)=3054

(23,36,24) = {23:36:78:47:31:197} = 6107; f=134; (x+n)=83; u=41; (d+n)=114

(1,85,7) = {1:85:14:13:1:197} = 197; f=28; (x+n)=98; (d+n)=99

(6,11,3) = {6:11:5:4:1:31} = 31; f=5; (x+n)=15; u=7; (d+n)=16

a=31 (x+n) = 15

b=197 (x+n) = 98

(x+n) diff = 98 - 15 = 83

(x+n) from the prime solution record = 83!!!!!!

>>5491

>The 'heart' of the problem is that breaking the problem down involves two sets of triangle solving (otherwise it is prime), with one set of triangles (say half of the eight) working on a triangular problem that is one unit longer than the other set.

Is this the 2 triangle iteration?!?!?!?!?!?!

So one of the prime numbers has an odd (x+n), the other prime number has an even (x+n), but an odd (d+n). We can iterative search on both records at the same time. And the difference between the 2 small squares is our prime solution!

>>5518

Sure. But usable in searching?

>>5491

Trying something different with the split triangles.

Attached pics are for c=6107 and c=7463 and show an attempt to improve the iterative search process based on VQC's split triangle hint.

Previous number of iterations for these examples were 23 and 83 for c=6107 and c=7463 respectively.

As you can see, these spreadsheets match the prime solution XPN in just 2 calculations.

The revised formula for testing is:

1 + 4T(u1) + 4T(u2) + 2*mod - (rm 2d(n-1))

Where u1 and u2 are each based on different factors of (f-2) div 40 chunks.

The efficiency comes in from the "step factor" and "u2 factor offset" variables. I arrived at these by trial and error. But it appears that the "step factor" variable may be somehow related to the factor tree d value.

u1 = step * "step factor" * (f-2 div 40 chunks).

u2 = u1 - "u2 factor offset"

I haven't gotten this to work for small values of n or yet figured out larger test cases.

But it is certainly interesting enough to share at this point.

>>5521

One more example for c=14904371.

Previous iterative search solved this in 2166 steps.

For this example, the u2 formula needed to be adjusted slightly to:

u2 = (u1 factor - "u2 factor offset") * (f-2 div 40) - 1

previous examples used the formula:

u2 = (u1 factor - "u2 factor offset") * (f-2 div 40)

The "step factor" column is pretty irrelevant and just indicates the total number of iterations required to get to the correct solution. So in this example, the total number of iterations would be 60 instead of 2166. Still a pretty good performance improvement.

The main issues with this approach are:

1) determining where that "u2 factor offset" value comes from.

2) when the u2 needs the -1 additional offset.

3) How to make this approach work where the small n values are lower than the first f-2 div 40 triangle.

>>5522

Continuing with work on a revised iterative approach from 2 triangles.

Pics attached are for c=6107 and c=209785063.

These examples show multiple pathways to a solution using different combinations of factors and offsets.

The two triangles are calculated as follows:

u1 = "u1 factor" * (f-2 div 40)

u2 = ("u1 factor" - "u2 factor offset") * (f-2 div 40) - "u2 offset"

I'm not quite sure yet how to pick appropriate "u2 factor offset" or "u2 offset" values. These tests just run through multiple combinations and show only prime solution matches. (u2 factor offsets from 1 to 100, and u2 offsets from 0 to 5)

The "u1 factor" column + 1 is the number of iterations required for the "u2 factor offset" and "u2 offset" values.

For c=209785063, my current best case scenario is 289 iterations, compared to 3705 in the previous iterative search solution.

For c=6107, there are a number of cases where even negative u2 values find the correct solution. This may equate to the -x records we previously explored in the grid. Best scenario here is 1 calculation. (cool)

Haven't gotten this to work for all examples yet.

>>5562

Pic attached is a summary of relevant odd x+n squares showing parities and values. Additional odd x+n records from e=1 to 24, n=1 to 10, and t=1 to 10 can be downloaded via pastebin.com/XJagiXg0.

Interesting that n-1 parity always matches e parity.

Also, noticed some weird behaviour in d parities for e values at e mod 8 - 1. (i.e. e=7, 15, 23, etc.)

Sample of this odd x+n data is posted at pastebin.com/AvSzBeFQ for e=7, n=1 to 50, and t=1 to 50.

Specifically, where n mod 4 = 0, the odd and even parities cycle, whereas other n values they look to always be odd.

>>5582

Excellent approach. Looking into this further.

Initial observation is that perhaps there is an upper bound on n values for each possible x+n combination, which would enable us to make much larger jumps during iteration.

Is there a formula to calculate that maximum (n-1) from an (x+n)?