VQC - welcome back. What an enlightening journey this has been. Indeed. Thank you again.
MM - nothing doing. Just responding to CA. >>5770
Attached spreadsheet explodes the pyramid numbers into left and right side polite numbers from 1 to 10 (and 73 just for fun).
XPN columns follow the formula 1+4T(right polite number) + 4T(left polite number).
Some interesting numbers are showing up.
Based on VA's new square layout, and the pyramid numbers posted earlier, attached is another test case for c6107 that attempts to bring these ideas together.
The solution takes a few more iterations than posted in >>5728, and includes the following changes:
1) u1 and u2 are now incremented by pyramid staircase values offset by 4 (meaning each iteration is descending the pyramid).
2) There are 4 squares made up of 2 triangles each that are themselves 1 staircase unit apart.
3) The estimated XPN formula has been adjusted to:
1 + 4*(T(u1) + T(u2) + (2d + 1)) - (2 * f mod 8)
4) the 2d+1 portion represents the border around the 2 adjoining triangles and fills the remainder portion. The d value in each iteration is calculated as (u1+u2)/2.
5) Subtracting the (2 * f mod 8) resulted in each iteration returning a perfect square.
This method hasn't been tested on any other cases and is highly depended on identifying the correct starting u1, u2, and the "polite" way to descend the pyramid.
Pics attached show two possible layouts for the first iteration of the c6107 example previously posted.
Left square is the f-1 starting position.
Right squares represent the T(u1), T(u2), 2d+1 and f mod 8 portions of the formula.
>(x+n) is odd.
>f is divided amonst the 8 triangles (some have a bit more than others).
May have an idea of how to proceed.
Pics attached are the (x+n)=83 small squares for c6107 and c27707 arranged according to the nn + 2d(n-1) + f - 1 formula.
nn around the edge, then 2d(n-1), and finally the f-1 portion in the middle including the center square.
c27707 is a new test case, so for reference, here are the relevant details from the grid representing the prime solutions.
{23:36:78:47:31:197} = 6107; f=134; (x+n)=83; u=41; (d+n)=114
{151:20:166:63:103:269} = 27707; f=182; (x+n)=83; u=41; (d+n)=186;
Notice the f-1 distribution for both cases is uneven with respect to the 8 triangles.
(1/2)
And here is (I think) the find.
Per VQC,
>f is within each triangle but OUTSIDE of (n-1), where the (n-1) square should go in the middle.
By taking one of the (n-1) portions out of the 2d(n-1) piece of the nn + 2d(n-1) + f - 1 equation, we are left with an equation that looks like:
nn + (2d-1)(n-1) + (n-1) + f - 1
This new (n-1) + f - 1 piece turns an "out of balance" f portion around the middle into something that is equally distributed among the 8 triangles.
The revised full square pics attached for the same c6107 and c27707 tests show how this new layout comes together. Also attached are close ups of the (n-1) + f - 1 in the centers.
The green squares in the middle represent the single (n-1) portion that has been removed from 2d(n-1).
The yellow squares are the portions of (f-1) that have been "displaced" from their original positions.
The red square is a single unit taken from (2d-1)(n-1).
And this is most likely why (n-1) is so important as a factor to all of the transforms that we worked on previously. (d[t]-d)/(n-1), a(n-1) na transform into (e,1).
(n-1) balances out the f center square.
(2/2)