>So back to trying to integrate the factor tree.
>Breaking the problem down.
>First by roots.
>Then by triangular numbers.
I have been working through an idea of how the factor tree could potentially integrate with the iterative solution.
Pics attached are for c=6107 and show the full factor tree (c6107_full_factor_tree.png) and a trimmed down version with squares (c6107_factor_tree_and_squares.png) that I will explain further.
The large number test results in >>5428 confirmed that the iterative formula works for any odd x+n. And it is EXTREMELY fast for small n values.
If there was some way we could turn every operation into a small n value search, then I think we would have our solution. (Reminder that in RSA #8 >>3264, an anon showed that there are multiple grid entries for c with prime a and b at various n values. A few samples for c=115 posted at >>3270.)
First thing to note is that the parsing of the e values has been removed. Could be wrong here, but I don't see the point of further factoring remainders.
There are three groups of nodes to the c=6107 parse tree. The first group corresponds to how the iterative search is currently working. The c, e, d, and f values get us to two iterative results where rm 2d(n-1)=0. First for the prime solution, and second for the starting c value.
The second group is for the 39 (d/2) node and is a result of 78/2, it's d and e values. And the third group is for the 3 (d/2) node, from 6/2 with it's d and e values.
By not further factoring the e values, we get a linear path of d values to the root of the tree. (First by root?)
So here's the working theory:
At the bottom of the tree, there is a way to iteratively search for two n0 values that satisfy an rm 2d(n-1)=0 constraint. (This is the "Then by triangular numbers" piece.)
These n0 values won't be the n values we are looking for. But from that operation (should be really quick), we will have something that can be subsequently used.
One of the n0 values will correspond to a path back up the tree in the "Prime Solution" column, while the other will lead us up the "Starting C" column.
This process will repeat at each level until we have the "final" result back at the top of the tree.
Still need to work out what the calculations look like at each level in terms of inputs and outputs. And what the relationship between the squares at different levels could be.