A sum involving irreducible characters of the symmetric group  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: Hi Thomas,
This is very similar to things I have been working on or thinking about. I would be very interested to know the source of this problem. My own source for similar questions was in Random Matrix Theory. What do you want to know about those sums? 
First a remark. In your definition of H(n,L), you could see this as a sum over all partitions of $2n$. The term $ \chi^{Y_{i,j,w}}(\tau)$ ensures only the diagrams you describe give non-zero contribution.
You might know this, but the $-p+q$ is the content of the cell $(p,q)$. 
You can compute $\chi^{Y_{i,j,w}}([2^n])$ via the Murnaghan-Nakayama rule indeed: it is counting all the different ways to express $Y_{i,j,w}$ as an incremental union of 2-ribbons (with signs!). Another way uses the so-called rim hook lattice. If the 2-core $Y_{i,j,w}$ is non-empty, this character value is 0. If it is empty, take the two 2-quotients of $Y_{i,j,w}$ and compute the dimension of each. Combining this with a binomial sum, you can obtain the character value you are looking for. [*]
Finally, I would send you to Stanley's recent paper (look at the version on his homepage http://math.mit.edu/~rstan/papers.html#hooks ) called 
''Some combinatorial properties of hook lengths, contents, and parts of partitions''
There he has sums over partitions of functions of the contents (like you do), but weighted by the Plancherel measure. I think you can make that measure aappear out of your sum. His result would typically take the form of a polynomiality in $n$ of the $L^i$ coefficient of $H(n,L)$. 
[Edit], concerning part [*]: This is essentially what's done in the accepted answer to this question: Statistics of irreps of S_n that can be read off the Young diagram, and consequences of Kerov-Vershik  (I was describing the Fomin-Lulov method, while that answer to the MO question describes their result).
A: 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.
