I may have found an alternate way to iterate. Not sure how effective it will be, but it's interesting nonetheless.

Attached pics are for c=209325931.

Using the current linear approach of incrementing f-2 div 40 chunks, it takes 51699 iterations to solve. The solution is found in the 12th increment of the T(est u) triangle, and at the 4886th iteration in remainder portion T(rm u).

We can calculate the "triangle base" for the remainder portion by T(rm u) - T(est u).

Here's the interesting part.

If you divide this "triangle base" by (f-2 div 40), you get a u that is exactly half way between the rm u and est u values. And the "triangle base" mod (f-2 div 40) is 1/2 the starting (f-2 div 40).

If you repeat that calculation, you basically find a way to subdivide the triangle base between T(rm u) and T(est u) by valid chunks of (f-2 div 40), but using triangle values instead of increments.

For example, at iteration 12:

(f-2) div 40 = 650

(T(8450) - T(7800)) / 650 = 8125 with remainder 325

The next iterations:

(T(8125) - T(7800)) / 325 = 7963

(T(7963) - T(7800)) / 162 = 7930

and so on.

The rm_u_tree pic attached is the partial result of a parse tree that starts from the mid point between est u and rm u, and splits u values in both the positive and negative direction. So the triangle base is divided into 1/2, 1/4, 1/8, 1/16, 1/32, etc in a tree structure until the leaf nodes are just 1 u apart.

There are only 255 entries for this parse tree, so we're down to 3060 total calculations to solve. But most likely a few less than that. Notice from the picture that T(8196) is only 5 levels down in this tree, and is only 1 away from the prime solution of 8197.

Triangles, Triangles, and more Triangles

>>5239 (f-2) div 40 iterations

>>5251 rm 2d(n-1) fine tuning

>>5257 overview

>>5260 woah another triangle?

>>5272 more than 1 triangle??

This post is an attempt to further explain the iteration process. How the numbers "click" into place, that everything can be represented as triangles, a possible connection to the factor tree that needs more research, and perhaps even a connection to the grid using x+n values.

First off, thanks to Topol for posting the attached pic (topol_triangles.png) in the EZ Bake >>5304. This pic shows how I believe the "remainder" portions come together as separate triangles to form the solution square.

I have created a new "triangle" prime solution output for all my odd x+n test cases, which you can find at pastebin.com/qanuvycg (again if anyone is interested).

From that list, I've chosen c=208932019 for the walkthough, but the same approach applies to all numbers. The full explanation is in the pic (c208932019_formula_breakdown.png).

The final XPN value we are searching for is comprised of:

XPN = 1 + 8T(u) + mod + 8 * (remainder) + mod - (rm 2d(n-1))

XPN = 1 + 8T(7875) + 4 + 8 * 2440500 + 4 - 1128

The (remainder) portion can be broken down further into a series of 0 or more triangle numbers. By removing the largest possible triangle, then the next, and so on until the remainder is zero.

2440500 = T(2208) + T(58) + T(9) + T(3) + T(1) + T(1)

The 2 mod values can be added together, and then subtract the 2d(n-1) remainder. This ALWAYS leaves you with a number that is divisible by 8. I'm referring to this as the "balancing total" and is the "bal total" column in the sample output.

balancing total = (4 + 4) - 1128 == -1120

The balancing total can be positive or negative. To visualize this, see the "dark green" remainder blocks in the triangle iteration mapping picture at >>5260. It is the adjustment from this "balancing total" that enables the estimated XPN from the iteration process to "click" into the exact final XPN value. This number can also be converted into triangle representation.

balancing total = -( T(16) + T(2) + T(1) )

Therefore, we can represent the ENTIRE final XPN formula as a series of triangular numbers.

final XPN = 1 + 8 * ( T(7875) + T(2208) + T(58) + T(9) + T(3) + T(1) + T(1) - T(16) - T(2) - T(1) )

See the pic (c208932019_triangle_breakdown.png) for a further breakdown of how these triangles fit together to form the (x+n)(x+n) solution square. It also shows how the (x+n) = 2u + 1 relationship can be used to solve for (x+n)(x+n). The color coding for the respective triangles matches pictures at >>5260. So the u and (x+n) formulas are really interchangeable.

Also notice how the triangle numbers grow to form the solution. There are most likely many combinations of triangle numbers that can be pieced together to arrive at a solution. My methodolgy is just an attempt to illustrate the point.

I believe that this growth in u values is what the factor tree is meant to provide. And might very well be the last remaining key to an extremely fast solution.

>>5323

>>5327

>this board is the best place to get work done, AND if people want to supplement that with chatting on Discord, then great.

AA - I think VA summarized this pretty well.