Let us call a partition *odd* if all its parts are odd, and let $Odd(n)$ be the set of all odd partitions of $n$, e.g. $Odd(6)=\{(5\,1),(3\, 3),(3\,1^3),(1^6)\}$.

Let $H(n)$ denote the set of all *hook* partitions of $n$. I have made the following surprising observation (to me, at least): For every $\mu\in H(m)$,
$$\frac{1}{2^{\ell(\nu)}}\sum_{\lambda\in H(n+m)} \chi_{\lambda\backslash\mu}(\nu)=\begin{cases} 0, \text{ if }\nu\notin Odd(n)\\
1, \text{ if }\nu\in Odd(n),\end{cases}$$ where $\chi_{\lambda\backslash\mu}$ are skew characters of the symmetric group.

I think a proof would come from the combinatorial description of skew characters in terms of tableaux (although I didn't really develop this all the way through). My question is: is this result already in the literature somewhere?

EDIT: Upon a moment of reflection it becomes clear that my sum is indeed independent of $\mu$. So we could use any $\mu$ to compute it, even the trivial one $\mu=(0)\vdash0=m$.

However, user61318 mentioned a paper in his comment which proves precisely that $$ \frac{1}{2^{\ell(\nu)-1}}\sum_{\lambda\in H(n)} \chi_{\lambda}(\nu)=\begin{cases} 0, \text{ if }\nu\notin Odd(n)\\ 1, \text{ if }\nu\in Odd(n),\end{cases}$$ which differs from the $m=0$ case of my sum by a factor of $2$. I have checked that both formulas are right as they stand, so this difference is a puzzle.

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