Cardinality of a certain subset of group $S_k$ defined in connection with cycle decompositions

Question

I am interested in a subset of the permutation group of $k$ elements, $\Sigma_k$.

Any element in $\Sigma_k$ can be decomposed into disjoint cycles in a unique way. Conversely, if we take a partition of $k$, say $(r_1,\dots, r_m)$, and $\alpha_i=\#\left\{j:r_j=i\right\}$, one can find an element $\sigma\in\Sigma_k$ such that it can be decomposed into $\alpha_i$ pair-wise disjoint cycles of length $i$.

Example: the permutation $\sigma=(1\ 2\ 6\ 3)(4\ 5)(8\ 9)(7)(10)$ has cycle shape $(4,2,2,1,1)$, where $\alpha_4=1$, $\alpha_3=0$, $\alpha_2=2$ and $\alpha_1=2$.

My question: I consider the subset $$ S_{M} := \big\{\sigma\in\Sigma_k: k-\small{\sum} \alpha_i>M\big\}.$$ How many elements does $S_M$ have in terms of $M$ and $k$?

Answer 1

The (unsigned) Stirling cycle number (aka Stirling number of the first kind) $[{n\atop k}]$ counts the number of permutations of an $n$-element set with exactly $k$ cycles. They form a triangle similar to Pascal's triangle. There is no simple closed formula, but there are recurrence formulas etc., check the Wikipedia article.

Your set $S_M$ contains all permutations of a $k$-element set with less than $k-M$ cycles. Hence $$ \#S_M = \sum_{i=0}^{k-M-1} \left [{k\atop i}\right]. $$ I don't think that this sum can be simplified.

Addition We have $[{n\atop 0}] = \delta_{n,0}$, for $n \geq 1$ we have $[{n\atop 1}] = (n-1)!$ and for $n \geq 2$ we have $[{n\atop 2}] = (n-1)! H_{n-1}$, where $H_n = \sum_{i=1}^n \frac{1}{i}$ is the $n$-th Harmonic Number. For the harmonic numbers, no simple closed form is known.

Hence in the special case $k - M = 3$ and $k \geq 2$ we have $$\#S_M = (k-1)! (H_{k-1} + 1),$$ and I don't think there is a simpler closed formula.