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Also, [result] has an index that happens to be i=a-1 or a-m. Already said that, but the important bit is if there is a method to derive [result]s factors, or at least the upper bound on the number of factors or the largest factor, then the only other step is deriving an identity for converting from [result] to the relevant index, which happens to be a factor p, +/- m.
And of course most of the factors of [result] are trivial by themselves.
Using something other than v=3 for (i*v) may also have implications for the length of the factor list of [result], and the largest factor of it, though I haven't looked into it.
Math is hard.
When you wanted to know deep learning immediately
:)) finally a good answer
Hang with me! This is *not* a math shitpost, I repeat, it is NOT a math shitpost, not entirely anyway.
It appears there is for products of two non-trivial factors, a real number n (well a rational number anyway) such that p/n = i (some number in the set of integers), whos factor chain is apparently no greater than floor(log(log(p))**2)-2, and whos largest factor is never greater than p^(1/4).
And that this number is at least derivable, laboriously with the following:
where p=a*b
https://pastebin.com/Z4thebha
And assuming you have the factors of p/z = jkl..
then instead of doing
p/(jkl..) = z
you can do
p-(jkl) to get the value of [result] whos index is a-1
Getting the actual factor tree of p/z is another matter, but its a start.
Edit: you have to provide your own product.
Preferably import Decimal first.
random
notatotalshitpost
prime factorization
math