Remember all those times you took the difference of some value or the remainder of some value from c and a solution value surfaced, but it didn't work for other example c values?
Well, inspired by those lists of values I developed that happen to make x and x+n show up somewhere, I don't think those paths are useless coincedences and should be forgotten, I think we should make a list of them.
I believe that making a list of paths to solution values and numbers they work for and don't work for will lead to a deeper understanding of why those paths worked and what types of semiprimes there are.
And the solution itself.
Just be sure to keep it neat and organized, because searching for solution paths can get out of hand easily (as in organization collapses and you end up with too many numbers to make sense of).
α = T(c-2) + ((c-e)/2)
γ = T(c-1) + ((c-f-1)/2)
δ = T(c-1) + (c-e-1/2)
Observed with c34117 (data in the main thread):
(-f, 1, (d+N))[b] - α = x+1
γ - α - c = x
γ - δ = x+2
Observed with c6107 (data in the main thread):
(e, 1, (d+N)[b] - (e,1,(d+N)[a] - 2α = x-1
This occurs because
(e, 1, (d+N)[b] - (e,1,(d+N)[a] - 2α = 2e
and 2* e of 6107 = x-1
These variables are rearranged forms of formulae that generate the a[t] and b[t] values in the place where Chris hinted the lookup would be in. (t) and (i) -set t = -i and x=-c
so I looked back at these and realize that sometimes this works even for massive multiples of 5
when gcd(d,e) != 1 it is equal to a
"there is an entire family of numbers that can be factored in 1 step"
which is when x appears in the tree
The elements that d is between in row one relatively reliably give the correct i value for c (nontrivial). Previously, it took 4 element calculations to check these, but with parity matching, it can be found in 2 calculations.
With the addition of the elements that c is between in i[t], the solution i is also given by j[t] relatively reliably, and it works on examples that fail the original SeeIt calculation.