After testing several examples I noticed a pattern for a lot of c's that N or N-1 is a multiple of n or n-1.

An example of what I mean, is 145. Its N is 61, and its n is 5. (61-1) = 5 * 12

So, for c's that adhere to this pattern (not all of them do, I haven't figured out what determines whether they do or not), having the factors of N or N-1 (whichever one is a multiple of n or n-1) would give the n for c.

Now, even though factorization is itself the problem we are trying to solve, there have already been presented some shortcuts that would factor N or N-1 easily. Since N and N-1 appear in e,1 (one as the a[t] value and the other as the d[t]-d value), those hints about factor appearances in e,1 would be the shortcut that we need to make an efficient path from c -ab for the semiprimes that adhere to this pattern. Some other c's this pattern works for are 95,123,287,6107 and 73891