I'm looking for something making tractable the sum, over all partitions into k terms of an integer n, of the product of the factorials of all the terms. Thanks,
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1$\begingroup$ Are 2+3 and 3+2 different partitions? $\endgroup$– Fedor PetrovCommented May 4, 2016 at 19:27
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1$\begingroup$ Could you provide an example? $\endgroup$– WojowuCommented May 4, 2016 at 19:34
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2$\begingroup$ The numbers for the sum over all partitions of $n$ are tabulated at oeis.org/A077365 with a link that goes up to $n=300$. Some references, formulas, and codes are given there. See also oeis.org/A134134 $\endgroup$– Gerry MyersonCommented May 4, 2016 at 22:54
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$\begingroup$ I am very interested to see that somebody has already come across this problem. Personally I am trying to find any information about the partition of a given integer n as a sum of positive integers, for which the product of factorials is minimal: so for $n\in \mathbb{N}$ $(n_1,n_2, \ldots,n_p) \in (\mathbb{N}^*)^p / \sum_{i=1}^pn_i = n ~\wedge \prod_{i=1}^p (n_i!) $ is minimal $\endgroup$– QuantumCommented Mar 28, 2017 at 19:53
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2 Answers
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Is equals the coefficient of $x^ny^k$ in $$\left(\prod_{i=1}^{\infty} (1-i!x^iy)\right)^{-1}.$$
answered May 5, 2016 at 10:59
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I don't know if this is any use, but I think that the sum is bounded above by $n!p(n)$, where $p(n)$ is the number of partitions of $n$, and this upper bound may not be too far from the truth:
The reason is that $\sum_{\pi \in S_{n}} |C_{S_{n}}(\pi)| = p(n)n!$, since $S_{n}$ has $p(n)$ conjugacy classes. Now if a permutation $\pi$ has distinct cycle sizes $n_{1},n_{2},\ldots,n_{k}$ with respective multiplicities $a_{1},\ldots a_{k}$, then $|C_{S_{n}}(\pi)| = \prod_{j=1}^{k} (n_{i}!)^{a_{i}} a_{i}!$.
answered Jun 6, 2016 at 14:03
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1$\begingroup$ An even better upper bound would be $(1+n-k)!p_k(n)$, where we count the number of partitions of $n$ with $k$ pieces by $p_k(n)$. The lower bound has a multiplicative term close to $(n/k)!^k$, and I imagine the harmonic mean of the two terms is a good approximating multiplicative term for $p_k(n)$ for the desired sum. Gerhard "Or Maybe The Geometric Mean" Paseman, 2016.06.06. $\endgroup$ Commented Jun 6, 2016 at 22:49