denote $p=mn$, then the mean $\bar{k}$ you seek equals $\bar{k}=\frac{p!}{(p-j)!}f_j(p)$, with
$$f_j(p)=\sum _{k=1}^{\infty } \frac{k  \mathcal{S}_k^{(j)} }{p^{k}}.$$
Mathematica will evaluate this in closed form as a function of $p$ for given $j$, for example
<sub>
$$\{f_1(p),f_2(p),\ldots f_5(p)\}=\left\{\frac{p}{(p-1)^2},\frac{p (2 p-3)}{\left(p^2-3 p+2\right)^2},\frac{p (3 (p-4) p+11)}{\left(p^3-6 p^2+11 p-6\right)^2},\frac{2 p (2 p-5) ((p-5) p+5)}{(p-4)^2 (p-3)^2 (p-2)^2 (p-1)^2},\frac{p (5 (p-6) p ((p-6) p+15)+274)}{(p-5)^2 (p-4)^2 (p-3)^2 (p-2)^2 (p-1)^2}\right\}.$$
</sub>