Sum of skew characters over hooks and "odd" partitions 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.  
 A: I know you were asking for a reference, and there may be better approaches, but just to offer one proof of your statement based on the Murnaghan–Nakayama formula. Assume $m>0$ then any skew tableaux $\lambda/\mu$ has shape $(a,1^b)$ with $a+b=n$. This means $a$ boxes in the first row and $b$ boxes in the first column but the row and column are disconnected. We want to evaluate the sum
$$
\Lambda_n = \sum_{a=0}^n \chi_{(a,1^{n-a})}
$$
If $1 \leqslant k \leqslant n$ then the skew partition $(a,1^b)$ has at most two $k$-hooks, one of leg length 0 if $k \leqslant a$ and one of leg length $k-1$ if $k \leqslant b$. Now assume $\nu\vdash n$ is a partition with part equal to $k$ and let $\hat{\nu} \vdash n-k$ be the partition obtained by removing this part. By the Murnaghan–Nakayama formula for skew partitions we have
\begin{align*}
\Lambda_n(\nu) &= \sum_{a=0}^{k-1}\chi_{(a,1^{n-a})}(\nu) + \sum_{a=k}^{n-k}\chi_{(a,1^{n-a})}(\nu) + \sum_{a=n-k+1}^n\chi_{(a,1^{n-a})}(\nu)\\
&= \sum_{a=0}^{k-1}(-1)^{k-1}\chi_{(a,1^{n-a-k})}(\hat{\nu}) + \sum_{a=k}^{n-k}(\chi_{(a-k,1^{n-a})}(\hat{\nu}) + (-1)^{k-1}\chi_{(a,1^{n-a-k})}(\hat{\nu})) + \sum_{a=n-k+1}^n\chi_{(a-k,1^{n-a})}(\hat{\nu})\\
&= \sum_{a=0}^{n-k}\chi_{(a,1^{n-k-a})}(\hat{\nu}) + (-1)^{k-1}\sum_{a=0}^{n-k}\chi_{(a,1^{n-k-a})}(\hat{\nu})\\
&= (1+(-1)^{k-1})\Lambda_{n-k}(\hat{\nu}).
\end{align*}
Hence if $k$ is even then $\Lambda_n(\nu) = 0$ and an easy induction shows that $\Lambda_n(\nu) = 2^{\ell(\nu)}$ if $\nu \in \mathrm{Odd}(n)$.
Maybe I would just add that the same argument essentially works when $m=0$. For this case let $\tilde{\chi}_{(a,1^{n-a})}$ be the irreducible character labelled by a hook partition and let $\tilde{\Lambda}_n = \sum_{a=1}^n \tilde{\chi}_{(a,1^{n-a})}$. The calculation above goes through in this case except when $\nu = (n)$ in which case we have
\begin{equation*}
\tilde{\Lambda}_n(\nu) = \sum_{a=1}^n \chi_{(a,1^{n-a})}(\nu) = \sum_{a=1}^n (-1)^{n-a}
\end{equation*}
which is 0 if $n$ is even and 1 if $n$ is odd. This explains why one gets $\tilde{\Lambda}_n(\nu) = 2^{\ell(\nu)-1}$ when $\nu \in \mathrm{Odd}(n)$.
EDIT: It should also be possible to show that $\Lambda_n = 2\tilde{\Lambda}_n$ directly using the fact that $\chi_{\lambda/\mu} = \sum_{\tau \vdash n} c^{\lambda}_{\mu\tau}\tilde{\chi}_{\tau}$, where $c^{\lambda}_{\mu\tau}$ is the Littlewood–Richardson coefficient.
