Let $n$ be a nonnegative integer. Prove that the total number of symmetric partitions of $n$ is equal to $[x^n]\prod_{j=1}^{\infty}(1+x^{2j-1})$.
(Here, $[x^n]f$ means the coefficient of $x^n$ in a formal power series $f$. A partition is said to be symmetric if it is self-conjugate, i.e., if it equals its own conjugate.)
My professor discussed this question in class, but my blanked out for some reason and I didn't make a note of it.
She used something along the lines of a hook length to find a weight preserving bijection.
Unfortunately, that is all I can recall and understand. Can someone please help me with the proof.
Lets look at a symmetric partition of $n=24$,
If we bend the hooks of the symmetric partition like so,
we see that $f$ is a function that turns symmetric integer partitions into partitions with distinct odd parts. Similarly we can turn any partition with distinct odd parts into a symmetric partition by applying $f^{-1}$. Since $f$ and $f^{-1}$ yields unique results going both ways (and they apply to all symmetric and distinct odd partitions), this establishes that $f$ is a bijection between all symmetric partitions and all distinct odd partitions.
This means that the generating function of partitions with distinct odd parts, $\mathcal{O}_d(x)$, will be exactly the generating function of symmetric integer partitions, $\mathcal{S}(x)$. So we have
$$\begin{align*} \mathcal{S}(x) &=\mathcal{O}_d(x) \\ &= \prod_{n=1}^\infty\left(1+x^{2n-1} \right ). \end{align*}$$