I've been looking more at the tetrahedral numbers (4,10,20), and I may have found something. Look at these grids. Basically the white columns are the seeds, and each value is equal to the sum of the one above and one to the left of it. By default, the first row is a 1. Then the top left box is just regular numbers seeded by a 1. The highlighted column in that box are the tetrahedral numbers. The highlighted column in the next is square numbers. Then the next is pentagonal numbers, then hexagonal, heptagonal, etc. If you look at the highlighted row, then you can see that there is a simple pattern to get from grid to grid. (ie, from 1 to 2 you subtract 1, from 2 to 3 you subtract 2, from 3 to 4 you subtract 3) so you can easily navigate through these grids. Also for each box n, the first column is all the numbers where mod n is the seed. So maybe we could start with a zero seed (cuz then its divisible) or something like that. Or we could use the identity n*n = T(n) + T(n-1) where T(n) is the nth triangular number. Get the number in the square grid and translate it to a different grid and do stuff. Also these grids can be extended into the negative (I'll get on that).

Basically the idea is to get our number. E would be the seed and D would be the start column. Then we can use some of these identities to navigate around to the grid where we are in the A column with a 0 seed.

>>4871

Refer to the seed as the 1 row

>>4872

For this, notice that the top left grid, is just 4* the 1 and 1 grid. Also if you start at the top left cell within a grid, that is the sum of the seed and the column. Then if you go down and to the right, you have that number x3, x10, x35, x126, and this series is the same as the numbers going diagonally down from 3 of the one column one seed grid from my last post.

>>4873

Because of this multiplicity, if we start with a number with the h column seed, k row seed, then your diagnoal entries will be (h+k), (h+k)3, (h+k)10, (h+k)*35, etc. So if we start in G(h,k) (this will mean grid columnseed h, rowseed k), the diagonal entries for this will be the same as G(i,j) where i+j=h+k.

Going with our 145 = 12*12 + 1, we could start in the G(12,1) cuz, then we would have multiples of 1 mod 12, which would include 145. Then if we could navigate to the diagonal (probably by going to the right, because if you go up you would be at the 13 right away) we would have some value and we could then navigate to any other grid G(m,d) where m+d = 13. Then we could do something once there, this is where I'm lost.

>>4874

Here are some potential pathways to the solution using this grid

>>4875

Also for some D's I reduced until it was odd (divide by 2 repeatedly). I don't think you can do that for E though (maybe I could set it to E%D though)

>>4876

Also stuff is fucked up in the upper part of the third pic but not the stuff that matters

>>4875

Also if you try this, be careful, the positive (bottom right) part stays the same, but if you switch the center cell (e or d) then the negative parts change.