>>5690
Have been focusing on the following hints from VQC's recent post.
>For any idea, ensure that it scales.
>Using an arbitrary divisor for f, then each of the eight triangles will have one OR one of two … configurations in each triangle.
>The difference between those two configurations of a portion of f are that they are both staircase numbers where the base of one is a unit longer than the other.
The quickest iterative search so far has been using 2 different sized triangles. (see >>5522, >>5522, >>5525 for some background)
The problem here has been determining the different u values to use, and perhaps we now have a bit of a clue.
First pic attached is a spreadsheet calculation for c=14904371 showing the solution in 18 steps using the dual triangle approach. Previous best result using f-2 iterative search and valid n jumps was 1083 iterations. So a pretty good performance bump.
In this test case, the u1 and u2 values of 271 and 84 were chosen by trial and error. But there may be a way to get to those values - which is the reason for the post.
The second pic is the square layout for f-1. It is broken into 8 triangles of 368, 4 remainder squares, and 1 square in the middle. 1 + 8*368 + 4 = 2949 == f - 1.
One of the triangles has been subdivided into f-2 div 40 chunks of 73, with 2 separate remainder areas of 12 and 64, to try and figure out where these two configurations come from.
For the working solution:
u1 = 271 == (1+2) * 73 + 52 ???
u2 = 84 == (0+1) * 73 + 11 ???
The staircase numbers may be the factors that are applied to the arbitrary divisor of f (40 in this case). (0+1) and (1+2) in my guestimate.
But then where do the differences of 52 and 11 come from?
And why is the solution found with only 4 chunks of 73, when there are 5 available in the 368 triangle?
This may all be coincidental, of course, as these large u values would prevent this solution from working with small values of n.