It is remarkable that the symplectic content equals the hook length (up to sign) if and only if the product of the symplectic contents is non-zero. It's sign being opposite to the sign of the (ordinary) content.

A fun fact is that $e_{(l-k)}(\lambda)$ equals the coefficient of $t^k$ in $\prod (\lambda + t )$ where $l$ is the length of $\lambda$ ; Example: $e_{2}({2,2,1}) = 8$ and the coefficient of $t$ in $(2+t)(2+t)(1+t)$ is $8$.

maybe this arxiv.org/pdf/1203.4419.pdf could be of use?

I expected to see the product (and not the sum) over the inner arguments in the Nekrasov-Okounkov hook length formula, see hal.archives-ouvertes.fr/hal-00395682/document Theorem 1.2.

maybe of interest, maybe also obvious: expand the square into a sum of Schur functions (using LR) and look at the sum in representation space: replace $s_\lambda$ by $\chi_\lambda$ in $S_{n(n+1)}$ and check that the non-zero classes are indexed by partitions $\mu$ of n(n+1) with odd parts <= 2n-1.

It appears that the count of skew SYT of shape $\lambda / \alpha$ equals the count of SYT of all partitions generated in the decomposition of $s_{\lambda / \alpha}$ including LR-multiplicities. Can't prove it though.

Nice! It never occurred to me that the first column of the character table of Sn (class {n}) equals 0 for irreps whose partitions are not hook shaped and (-1)^l for hook-shaped partitions (k,1^l). It's like a sister to the well-known fact that the characters of class {1^n} equal the tableaux-count.

@PietroMajer: indeed! I checked the Mathematica implementation, and that at least is correct. Well spotted. Apologies.

@GerryMyerson: function f takes a partition as argument; f(n=2) is intended to mean the sum of f over all partitions of n. Btw: conjecture tested overnight to n=52. Computationally, summing f over the partitions is much faster than iterating the differential equation. So, as 'counting' goes, the conjecture is not much good.

@GerryMyerson: yes, I tried to be more explicit : f(n) at value n=2. Overkill becoming murky.

@NateEldredge: extra dollar signs left as an excercise to the poster?

@GerryMyerson: should look better now?