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I was given the following identities (note: $s_{n,k},S_{n,k}$ are the Stirling numbers of first and second kind respectively):

(1) $s_{n+1,k+1}=\sum_{i=k}^{n}\binom{i}{k}s_{n,i}$

(2) $S_{n+1,k+1}=\sum_{i=o}^{n}\binom{n}{i}S_{i,k}$

(1): The left hand side counts the number of permutations of $\{1,...,n+1\}$ that have exactly $k+1$ cycles. The right hand side counts the number of permutations of $\{1,...,n\}$ that have any number of cycles $\ge k$ where $k$ of them are designated in some way. My idea is to give a bijection between both sides by inserting the element $n+1$ at a certain place and combining the $i-k$ non-designated cycles to one cycle. But I couldn't find a clever way to make that invertible.

(2): Here, the left hand side counts the number of ways to partition a $(n+1)$-set into $k+1$ disjoint, nonempty sets. The right hand side counts the number of ways we can choose $i$ out of $n$ elements, partition them into $k$ blocks, and put the left over $n-i$ elements in the $(k+1)$-th block. I don't see the bijection here at all.

I appreciate any help.

blst
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1 Answers1

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The bijection in (2) can be established by noting that the $(k + 1)$-th block also contains the initially omitted $(n + 1)$-th element. So when you want to determine which $k$ out of $k + 1$ sets were those that spawned from $S_{i,k}$ you just pick those that do not contain the $(n + 1)$-th element.

I'm also struggling with (1) right now.

Update: Stirling number of the first kind: Proof of Recursion formula has what you seek for (1).

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Marcin Kaczmarek
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