Quote:
Originally Posted by nesio
S  the number of all divisors of m (excluding 1 and m).

Ok, now we understand that one.
Quote:
Originally Posted by nesio
A rough model of m:
Find the maximum of (S/(m^1/3)) when m < n^1/3,
where S  the number of all divisors of m (excluding 1 and m).
Here are S and m^1/3 have equal weights in maximize function.

I'ld say this is not a model of m, but a procedure to choose the best m. A model would give a kind of score to any m. (at least)
Quote:
Originally Posted by nesio
What will be?

A simple weighting function giving a "plus" for smooth m, a "minus" for big m.
This is no rocket science. We tested dozens of ways to arrange k's according to such principles, and many other constellations...