Re: [math-fun] partitions into distinct part: spell this one out

joerg> Let P(x) = prod(n=1, oo, (1+x^n)).

This is the generating function for partitions into distinct parts.

We have P(x^2)^8 + 16*x*P(x^2)^16*P(x)^8 = P(x)^16

So the number of partitions of n into distinct parts of 16 kinds (rhs.) equals the number of partitions of n into even parts of 8 kinds (1st term lhs.) plus *WHAT*?

P.S.: the equality really is k^2+k'^2=1 massaged to render a combinatorial relation.

How about: The number of partitions of n into distinct integers using 16 different fonts equals the number of partitions of n into distinct even parts using 8 fonts, plus 16 times the number of partitions of n-1 into distinct even parts using 24 fonts and distinct odd parts using 8 fonts? --rwg