What is the generating function for the number of partitions of an integer in which each part is used an even number of times?

Question

What is the generating function for the number of partitions of an integer in which each part is used an even number of times? I'm trying to prove the number of partitions of an integer in which each part is used an even number of times equals the number of partitions of an integer in which each part is even using generating function. I know the generating function for the number of partitions of an integer in which each part is even would be $\prod_{i=1}^{\infty} \frac{1}{1-q^{i*2}}$, but I'm really stuck on how to find the generating function for the other case.

Answer 1

If each part appears even number of times, then the number $n$ is even, and this partition gives a partition of $n/2$. On the other hand for every partition of $n$ we can get a partition of $2n$ with every part repeating even number of times. Thus your sequence $q_n$ is such that $q_{2k+1}=0$ and $q_{2k}=p_k$, the number of partitions of $k$. This gives the generation function which is easy to compute given that $$\sum_{n=1}^\infty p_nx^n=\prod_{k=1}^\infty \frac{1}{1-x^k}.$$ So $$\sum_{n=1}^\infty q_nx^n=\sum p_n x^{2n}=\prod_{k=1}^\infty \frac{1}{1-x^{2k}}.$$