At 1 and 0 you get the identies 1 and 0. 0 can't be divided at all and 1 has only one possible factor including itself and itself. Hehe we'll just say "period" to make more sense.
At 2 total factors there's only primes. (works both ways: A>B and B>A)
The solutions for 2 factors are a kind of 'key' to the higherfactorcount sets in the 3D tree since obviously primes are the simplest factors possible.
If there's an infinite number of primes, is there an infinite number of the 3factor results? 4factor? All? See 2nd post to see why there must be.
Yes, there are infinite primes but no telling how long you'll have to wait to find the next one! I left this in the spoiler so those who want to can do the work themselves to find and understand the proof.
