Recently, during the research, I came across a sum, denoted by $H(n,L)$, involving irreducible characters of the symmetric group,
\begin{equation}
H(n,L)\colon=\sum\_{Y\_{i,j,w}} \frac{\chi^{Y\_{i,j,w}([2^n])} \chi^{Y\_{i,j,w}}(\tau)}{\chi^{Y\_{i,j,w}}([1^{2n}])} s\_{Y\_{i,j,w}}(1,\ldots,1).
\end{equation}
Notations: 


Let $S\_{2n}$ denote the symmetric group of all permutations over $2n$ objects. Let $\chi^{Y}$ denote the irreducible character of $S\_{2n}$ labeled by Ferrers diagram (or shape) $Y$. 


Let $Y\_{i,j,w}$ denote a special type of Ferrers diagram having one row of $j+w+1$ boxes, one row of $j+1$ boxes, $i-1$ rows of two boxes and $w$ rows of one boxes, where $i\geq 1$, $j\geq 1$ and $i+j+w=n$. 


Let $[2^n]$ and $[1^{2n}]$ denote the conjugacy classes of $S\_{2n}$ having the cycle types $[2^n]$ and $[1^{2n}]$, respectively. Set $\tau = (1,\ldots,n) \circ (n+1,\ldots,2n)$, a permutation of cylce type $[n^2]$. 


Let $s\_{Y\_{i,j,w}}(x\_1,\ldots,x\_L)$ denote the Schur-polynomial of $Y\_{i,j,w}$ over $L\geq 2n$ indeterminants.


The sum $H(n,L)$ runs over all the Ferrers diagrams $Y\_{i,j,w}$ such that $i\geq 1$, $j\geq 1$ and $i+j+w=n$.


Using the Murnaghan-Nakayama Rule, we obtain $\chi^{Y\_{i,j,w}}(\tau)=2 (-1)^w$. The Schur-polynomial can be experssed as $s\_{Y\_{i,j,w}}(1,\ldots,1)= \prod\_{(p,q)\in Y\_{i,j,w}}\frac{(L-p+q)}{h\_{pq}} $, where $h\_{pq}$ denote the hook length at position $(p,q)$ of $ Y\_{i,j,w}$. Utilizing the hook length formula, we have $\chi^{Y\_{i,j,w}}([1^{2n}]) =\frac{(2n)!}{\prod\_{(p,q)\in Y\_{i,j,w}}h\_{pq} }$. Therefore, we can simplify the sum $H(n,L)$
\begin{equation}
H(n,L)=\frac{2}{(2n)!} \sum\_{Y\_{i,j,w}} (-1)^w \chi^{Y\_{i,j,w}}([2^n])
 \prod\_{(p,q)\in Y\_{i,j,w}} (L-p+q)
\end{equation}
We can also use the Murnaghan-Nakayama Rule to compute $\chi^{Y\_{i,j,w}}([2^n])$. But I can't find a formula! We can also show that the character values $\chi^{Y\_{i,j,w}}([2^n])$ have the property: when $n$ is odd, $\chi^{Y\_{i,j,w}}([2^n])=-\chi^{Y\_{j,i,w}}([2^n])$, and when $n$ is even, $\chi^{Y\_{i,j,w}}([2^n])=\chi^{Y\_{j,i,w}}([2^n])$. For specific $n$, the sum $H(n,L)$ is a polynomial of $L$. A few examples are listed below
\begin{equation}
H(2,L) = \frac{1}{6} L^2 (L^2-1) 
\end{equation}
\begin{equation}
H(3,L) = \frac{1}{20} L^3 -\frac{1}{20} L^5
\end{equation}
\begin{equation}
H(4,L) = \frac{1}{140} L^2 -\frac{131}{5040} L^4+\frac{47}{2520}L^6+\frac{1}{5040} L^8
\end{equation}


Questions:
I am wondering if there exists a formula for these character values $\chi^{Y\_{i,j,w}}([2^n])$. Are there any other ways to compute this sum $H(n,L)$ smartly? Is there a formula for the sum $H(n,L)$? Any things about these character values $\chi^{Y\_{i,j,w}}([2^n])$ and sums $H(n,L)$ would also be appreciated.



A sum involving irreducible characters of the symmetric group 
For any irreducible character $\chi^\lambda$ of $S\_{2n}$, the value
$\chi^\lambda([2^n])$ can be computed as follows. If $\lambda$ has a
nonempty 2-core, then $\chi^\lambda([2^n])=0$. Otherwise $\lambda$ has
a 2-quotient $(\mu,\nu)$, where $\mu$ and $\nu$ are partitions
satisfying $|\mu|+|\nu|=n$. Say $\mu$ is a partition of $k$, so $\nu$
is a partition of $n-k$. Then $\chi^\lambda([2^n])=\pm{n\choose k}f^\mu
f^\nu$, where $f^\rho$ denotes the dimension of the irrep of $S\_n$
indexed by $\rho$ (given explicitly by the hook-length formula). The
sign is $(-1)^{m/2}$, where $\lambda$ has $m$ odd parts. This result
follow from the theory of cores and quotients, e.g., Section 2.7 of
James and Kerber, *The Representation Theory of the Symmetric Group*.


