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How I would write the generating function for a partition of a positive integer n with an even number of odd parts?

Any hints or suggestions will be greatly appreciated.

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Let $P_k^{\,\mathrm{odd}}(n)$ be the set of partitions $\lambda\vdash n$ with $k$ odd parts, and $P_k(n)$ the set of partitions $\mu\vdash n$ with $k$ generic parts. Then we construct a bijection $f:P_{2k}^{\,\mathrm{odd}}(2n)\to P_{2k}(n+k)$ as follows: for each part of the partition $\lambda\in P_{2k}^{\,\mathrm{odd}}(2n)$, add $1$ and then divide by $2$. Hence $$\# P_{\mathrm{even}}^{\,\mathrm{odd}}(n)=\begin{cases}\sum_{k=1}^{n/2}\#P_{2k}\left‌​( \frac{n}{2} +k\right) & \text{if }n\text{ is even,} \\ 0& \text{ if }n\text{ is odd}.\end{cases}$$ Just thought I'd put that out there. – anon Jan 29 '12 at 9:01

If $n$ is odd, then no partition of $n$ has an even number of odd parts. If $n$ is even, then every partition of $n$ has an even number of odd parts. So the generating function is just $Q(x)=\sum_0^{\infty}p(2k)x^{2k}$, where $p$ is the ordinary partition function. If we write $P(x)=\sum_0^{\infty}p(n)x^n$, then $Q(x)=(1/2)(P(x)+P(-x))$. You can play around with this to see if you get anything nicer.

EDIT: One way to write this generating function is $$\prod_{k=1}^{\infty}(1-x^{4k-2})^{-2}(1-x^{4k})^{-1}$$

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