I had previously shown by construction that every number of form (2^k)*(odd-number) has the same number of sums for all k - that number only depends on (odd-number).
I have gone further with my numerical experiments, and I have found that the number of sums depends on the factorization of that odd number. It depends on the powers of distinct primes in that factorization, but not the primes themselves. Powers of single primes give that power, two distinct primes give 3, etc. After some further experimentation, I found that adding 1's gives
( (something) * 2^(number of 1's) ) - 1
For that (something), I found something similar for other powers: 3^(number of 2's), 4^(number of 3's), etc.
So I came up with this formula. For distinct odd-prime powers m repeated k(m) times, I find
( product over m of (m+1)^(k(m)) ) - 1
That subtraction of 1 is for a trivial case: sum of one number, itself. Including that case gives us
product over m of (m+1)^(k(m))
This is a conjecture, not a proof, I concede.
