Let $D(x)= (x+1)(x^2 +1 ) (x^3 +1 ).... $ and let F(x) be inverse of $D(x)$
I know, that $ D$ is the number of ways to write n as a sum of positive integers without repeated summands. Sums only differing by the order of the summands are counted only once. But I don't see what is inverse for that. Help me.
Using the result of this question, we have:
$$D(x)=\prod_{n=1}^\infty (1+x^n) = \prod_{n=1}^{\infty} \frac{1}{1-x^{2n-1}}$$
Then:
$$\frac{1}{D(x)} = \prod_{n=1}^{\infty} \left(1-x^{2n-1}\right)$$
So this is the generating function for $a_n=(-1)^np_n$ where $p_n$ is the number of ways to partition $n$ into distinct odd numbers.
Aside: Note that $\frac{1}{D(x)} = \frac{D(-x)}{D(x^2)}$.