Identity involving dimensions and contents of partitions Let $d_\lambda$ with $\lambda\vdash n$ be the dimensions of irreducible representions of the permutation group $S_n$. Let $C^{\nu}_{\lambda,\mu}$ be the standard Littlewood-Richardson coefficients.
I have been led to define the quantity
$$ F_n=\frac{1}{n!(n+1)!}\sum_{\lambda\vdash n}\sum_{\nu\vdash n+1}C^{\nu}_{\lambda,(1)} d_\lambda d_\nu M_\nu^2,$$
where $M_\nu$ is the product of all non-vanishing contents in the partition $\nu$, i.e.
$$ M_\nu=\prod_{(i,j)\in\nu, i\neq j}(j-i).$$
By comparing two different ways of computing the same result, I was expecting that $F_n$ should be equal to 1 for all $n$. Instead, when I actually calculated the values, I got $F_0=F_1=F_2=1$, fine, but $F_3=\frac{37}{36}$, which is curious.
Does anyone know how to evaluate $F_n$ in general, or something very similar to it (maybe by some mistake I ended up with a slightly wrong definition of $F_n$) that could shed some light into this situation?
 A: This is a partial observation:
The Littlewood-Richardson coefficient in your case is simply 1, 
if and only if $\nu/\lambda$ is a skew shape, and 0 otherwise.
Hence, (by using more standard notation for dimensions)
$$
F_n = \frac{1}{n!(n+1)!}\sum_{\lambda \vdash n} \sum_{\substack{\nu\vdash n+1 \\ \nu \supset \lambda}} (f^\lambda) (f^\nu) (M_\nu)^2.
$$
Moreover, you also have that for $\nu \vdash n+1$,
$$
f^\nu = \sum_{\substack{\mu\vdash n \\ \nu \supset \mu}} f^\mu, 
$$
since every SYT of shape $\nu$ can be constructed by adding a box to a SYT of shape $\mu$.
Rearranging a bit,
$$
F_n = \frac{1}{n!(n+1)!}\sum_{\lambda\vdash n}
f^\lambda 
 \sum_{\substack{\nu\vdash n+1 \\ \nu \supset \lambda}} 
 \left(
\sum_{\substack{\mu\vdash n \\ \nu \supset \mu}} f^\mu\right)
 (M_\nu)^2.
$$
Further rearranging,
$$
F_n = \frac{1}{n!(n+1)!}\sum_{\lambda,\mu\vdash n}
f^\lambda f^\mu
 \sum_{\substack{\nu\vdash n+1 \\ \nu \supset \lambda,\\ \nu \supset \mu}} 
 (M_\nu)^2.
$$
Now it might be a good idea to consider the separate cases $\lambda=\mu$, and $\lambda \neq \mu$.
This might help a bit.
A: I realize now that, in the original quantity
$$ F_n(\mu)=\frac{1}{n!(n+m)!}\sum_{\lambda\vdash n}\sum_{\nu\vdash n+m}C^{\nu}_{\lambda,\mu} d_\lambda d_\nu M_\nu^2 \delta_{D(\nu),D(\mu)},$$
I should have imposed the condition that the Durfee square of $\nu$ is equal to the Durfee square of $\mu$.
In the particular case $\mu=(1)$, mentioned in the question, this implies that $\nu$ is a hook, which of course simplifies things enormously. Once this condition is imposed, we indeed get $F_n((1))=1$, as I had expected. 
