@martin I've figured out how to get the coefficients for arbitrary exponent sets, not sure it's very practical, but it has no practical limit on number of terms/sizes of exponents, perhaps might serve to spur other thoughts: consider the internal and external powers as defining a multiset, where there are external power of internal power entries, e.g. for (a^2+b^2)^3 (a^3+b^3)^2, the multiset w/b {2,2,2,3,3}.
Now, take some exponent set, say {10,2}. Find the distinct ways 10 can be formed from elements selected from the multiset w/o replacement. For each of those, find the same for 2. In this trivial case, we'd have {3,3,2,2} and {2}. Now, starting with the first, we took 2 of the 2 threes, and two of the 3 twos. For the second, we took 1 of the remaining 1 twos.
Binomial[2,2]*Binomial[3,2]*Binomial[1,1]=3, the coefficient of {10,2}. When there is more than the one way shown here, enumerate the distinct ones, sum the results.
@martin It's a interesting transformation, have not spent much time thinking about how to do it efficiently... but going through my computer algebra books & combinatorics books has led to nowhere so far for alternatives.
@martin Whoa! That's a lotta code ;-) I'll be lounging later, will ponder this, my gut tells me there's an efficient way of doing this, though I'm not sure so efficient that getting all for all valid sets will be nearly as fast as the earlier simpler case. I'm also playing with "truncating" results - that is, doing one poly multiplication, pulling distinct terms, multiplying by next, pulling distinct... rinse and repeat
. If I can get it to work (right now most coeff. are fine, some are < than actual from truncation, so need to figure out correction) this c/b quite fast.
