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I denote by $e(n)$ is the number of partitions of $n$ with an even number of even parts, $o(n)$ the number of those with an odd number of even parts, and $k(n)$ the number of those that are self-conjugate.

The claim is that $e(n)-o(n)=k(n)$.

By the answer in Aryabhata's comment, I know that $$ \sum_{n\geq 0}(e(n)-o(n))x^n=\prod_{i\geq 1}(1+(-1)^ix^i)^{-1}. $$

I also know identity $$ \prod_{i\geq 1}(1+x^i)=\prod_{i\geq 1}(1-x^{2i-1})^{-1}, $$ so plugging in $-x$ yields and inverting gives $$ \prod_{i\geq 1}(1+(-1)^ix^i)^{-1}=\prod_{i\geq 1}(1+x^{2i-1}). $$ Does it somehow follow that this last expression above equals $\sum_{n\geq 0}k(n)x^n$ to get the result? Thank you.

(P.S. Is there a combinatorial argument for this equality? I'd like to see one if there is, but I'd be perfectly satisfied to just see the explanation of the sketch above. Either or.)

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For $1^{st}$ question, see… – Aryabhata Jan 25 '12 at 7:44
Thanks @Aryabhata, seems like someone else had the same issue as me. – sprtof Jan 25 '12 at 7:57
This question is identical to this one, except that the formula given there has an extra factor $(-1)^n$ tacked onto one of the members of the equation. The cases $n=1$, $n=3$ show that such a factor is indeed necessary. The answers there contain a combinatorial argument I gave. – Marc van Leeuwen Jan 25 '12 at 8:27
@MarcvanLeeuwen Maybe I'm mistaken, but in that question $p_E(n)$ is the number of partitions with an even number of parts, but here $e(n)$ is the number of partitions with an even number of even parts, and the same for $p_O(n)$ and $o(n)$. – sprtof Jan 25 '12 at 8:31
@sprtof: Sorry about that, I overlooked, the notation looked so similar. Still the combinatorial proof I gave could inspire one for this case. – Marc van Leeuwen Jan 25 '12 at 8:36
up vote 3 down vote accepted

In fact, this is virtually the same question as this one, even though this question considers only the parity of the number of even parts while that question does so for the parity of all parts (and throws in a factor $(-1)^n$). The connection is due to the fact that for any partition of $n$, the parity of its number of odd parts is necessarily the same as the parity of $n$ and therefore unalterable; therefore changing the parity of the number of parts always also changes the parity of the number of even parts. As a consequence one can prove this using the same bijection to cancel out non-contributing partitions. For your convenience I've edited my answer so as to apply directly to the current question, without needing translation.

Firstly, the Young diagram of a self-conjugate partition can be decomposed it into hooks with a diagonal square as corner, and can be reconstructed from the sequence of the (distinct odd) sizes of the hooks. As a consequence $k(n)$ equals the number of partitions of $n$ into distinct odd parts (and the generating series for the $k(n)$ is indeed $\prod_{i\geq 1}(1+x^{2i-1})$ as you guessed).

To equate this number to $e(n)-o(n)$, consider the following involution to match any partition of $n$ in the subset $S=p(n)\setminus p_{DO}(n)$ of partitions that are not into distinct odd parts, with another element of $S$ with the opposite parity of its number of its even parts. For $\lambda\in S$, test the odd numbers $m=1,3,5,\ldots$ in order, considering the set of parts of $\lambda$ of the form $2^km$ with $k\in\mathbf N$. If the set is either empty or consists of a single part $m$, move on to the next odd number. Since $\lambda\in S$, there is some $m$ that is not skipped; stop at the smallest such $m$. Now take the maximal $k$ such that $\lambda$ has a part of size $2^km$. If this part is unique, then $k\neq0$ by the choice of $m$; break the part into two parts of size $2^{k-1}m$. Otherwise $\lambda$ has at least two such parts; glue them together to form a part of size $2^{k+1}m$. One readily checks that the partition $\mu$ so produced is in $S$, that this procedure is an involution on $S$, and that $\lambda,\mu$ always have opposite parities for the number of their even parts (any pair of equal-size parts does not contribute to this parity, regardless of their size).

So the elements of $S$ together produce a null contribution to $e(n)-o(n)$, while each element of $p_{DO}(n)$ is in $e(n)$ (as it has no even parts) and therefore contributes $1$ to this difference.

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Thanks for translating the old answer! – sprtof Jan 25 '12 at 18:36

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