This is a follow-up question on the question computing length of integer partitions and it is loosely related with the paper "On a partition identity".
Let $\lambda$ be a partition of $n$, in the multiplicity representation $\lambda=(a_1,a_2,a_3,\dots)$ meaning that
$$n=\underbrace{1+1+\dots}_{a_1}+\underbrace{2+2+\dots}_{a_2}+\underbrace{3+3+\dots}_{a_3}+\dots$$
I can express $\lambda$ by drawing $a_k$ squares in the $k$-th row of a diagram (this is not the usual Ferrers or Young diagram), e.g. \begin{array}{cccc} a_1&\leftarrow&\square&\square&\square\\ a_2&\leftarrow\\ a_3&\leftarrow&\square&\square&\square&\square\\ a_4&\leftarrow&\square&\square\\ \vdots\\ a_n&\leftarrow&\\ \ &\ &\downarrow&\downarrow&\downarrow&\downarrow&\ &\downarrow\\ \ &\ &\mu_1&\mu_2&\mu_3&\mu_4&\cdots&\mu_n \end{array}
The $\mu_k$ numbers indicate how many squares there are in each column, so one can write the total number of squares $S$ (i.e. the length of the partition $\lambda$) in two ways: $S=\sum_ka_k=\sum_k\mu_k$. It is then very easy to show that the following holds:
$$ G[\lambda]:=\prod_{k=1}^n a_k!=\prod_{k=1}^n k^{\mu_k} $$
Now, there can be many partitions with the same length $S$, which one obtains by rearranging the squares in the diagram above whilst maintaining the number $n=\sum_k ka_k$ constant. So we can divide the set $\Lambda$ of all the partitions of $n$ into non-overlapping subsets $\Lambda_S$ of partitions of fixed length $S$, i.e. $\Lambda=\bigcup_S\Lambda_S$.
I would really like to be able to compute $$F(S)=\sum_{\lambda_i\in\Lambda_S}\frac{1}{G[\lambda_i]}$$
without resorting to the computation of all the partitions of $n$. Is there a way of doing this? Or, if it were not possible, is there a way of obtaining at least the list of the numbers $G[\lambda_i]$ with $\lambda_i\in\Lambda_S$?