I want to evaluate the following sum:


$\sum\_{\lambda}e\_k(\lambda)$ where the sum is over all partitions $\lambda$ that fit in an $m$ by $n$ grid and where $e\_k(\lambda):=e\_k(\lambda\_1,\lambda\_2,\dots,\lambda\_m)$ for $\lambda$ possibly padded with zeros if $\lambda$ does not have $m$ parts. Also $k\leq m$.



elementary symmetric function identity
I do not expect a nice closed formula for arbitrary $k$, in general. However, we could use the following recursive relation to extract some evaluations. This recurrence, we're alluding to, is already immediate from that of the elementary symmetric functions. Denote the given sum in the problem by
$$F\_k(m,n)=\sum\_{\lambda}e\_k(\lambda\_1,\dots,\lambda\_m)$$
where $\lambda\subset m\times n$. Then, we have
$$F\_k(m,n+1)=\sum\_{a=0}^{n+1}aF\_{k-1}(m-1,a)+\sum\_{a=0}^{n+1}F\_k(m-1,a),$$
together with $F\_k(k,n)=S(n+k,n)$ as noted above.


For example, we know $F\_0(m,n)=\binom{n+m}m$. Therefore, starting with $F\_1(1,a)=S(a+1,a)=\binom{a+1}2$ one can build $F\_1(m,n)=\frac{nm}2\binom{n+m}m$, successively.


Another example is, based on $F\_1(m,n)$, the evaluation
$$F\_2(m,n)=\binom{n+m}m\frac{n(3n+1)m(m-1)}{24}.$$


