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I have a recursive definition of the $q$-analogue of the Stirling numbers of the second kind. They are given by the recurrence $S_{q}(n,k) = [k]_{q}S_{q}(n-1,k) + q^{k-1}S_{q}(n-1,k-1)$, where $[k]_q=1+q+q^2+\cdots+q^{k-1}$ and the initial conditions are $S_q(n, 0)=S_q(0, n)=\delta_{0,n}$.

Is there some set that this class of numbers counts? I need to prove some identities, and it may be helpful for me to have a combinatorial interpretation.

Nishant
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  • There seems to be one in [Michelle Wachs, Dennis White, p,q-Stirling numbers and set partition statistics, Journal of Combinatorial Theory, Series A, Volume 56, Issue 1, January 1991, Pages 27-46] – Mariano Suárez-Álvarez Feb 02 '15 at 04:28

1 Answers1

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Here’s an interpretation involving partitions. It’s my modification of an interpretation in Chapter $3$ of Johann Cigler’s lecture notes that applies to a slightly different $q$-analogue of Stirling numbers of the second kind.

Let $\{P_1^\pi,\ldots,P_k^\pi\}$ be a partition of $[n]$. For $i\in[k]$ let $a_i^\pi=\min P_i^\pi$; we will always assume that $a_1<\ldots<a_k$. An $i$-inversion of $\pi$ is a $b\in\bigcup_{j<i}P_j^\pi$ such that $b>a_i$, and $I(\pi,i)$ is the number of $i$-inversions of $\pi$. Let $w(\pi,i)=I(\pi,i)+i-1$; this is the smallest possible cardinality of $\bigcup_{j<i}P_j^\pi$, given $i$ and $I(\pi,i)$. Let $w(\pi)=\sum_{i\in[k]}w(\pi,i)$, and define the weight of $\pi$ to be $q^{w(\pi)}$. Let $\mathscr{P}(n,k)$ be the set of $k$-partitions of $[n]$, and define

$$p(n,k)=\sum_{\pi\in\mathscr{P}(n,k)}q^{w(\pi)}\;;$$

I claim that $p$ satisfies the recurrence

$$p(n,k)=q^{k-1}p(n-1,k-1)+[k]_qp(n-1,k)\;.\tag{1}$$

Proof: Let $\pi\in\mathscr{P}(n,k)$. If $P_k^\pi=\{n\}$, let $\tau=\{P_1^\pi,\ldots,P_{k-1}^\pi\}\in\mathscr{P}(n-1,k-1)$; it’s easy to check that $w(\pi)=w(\tau)+k-1$. Each $(k-1)$-partition of $[n-1]$ gives rise to such a $k$-partition of $[n]$, and the correspondence is bijective, so the members of $\mathscr{P}(n,k)$ having a part $\{n\}$ (which is necessarily the last part) contribute $q^{k-1}p(n-1,k-1)$ to $p(n,k)$.

Otherwise there is an $i\in[k]$ such that $n\in P_i^\pi\ne\{n\}$. Let $\tau$ be the partition of $[n-1]$ obtained from $\pi$ by removing $n$ from $P_i^\pi$; clearly $\tau\in\mathscr{P}(n-1,k)$. $I(\pi,j)=I(\tau,j)$ if $j\le i$, and $I(\pi,j)=I(\tau,j)+1$ if $j>i$, so $w(\pi)=w(\tau)+k-i$. Conversely, if $\tau\in\mathscr{P}(n-1,k)$, we can add $n$ to any part of $\tau$ to get an element of $\mathscr{P}(n,k)$, and the weight of that element is $q^{k-i}q^{w(\tau)}$ if we add $n$ to $P_i^\tau$. Thus, the total weight of the $k$ partitions of $[n]$ generated in this way from $\tau$ is $[k]_qq^{w(\tau)}$, and hence the members of $\mathscr{P}(n,k)$ not having $\{n\}$ as a part contribute $[k]_qp(n-1,k)$ to $p(n,k)$. This establishes the recurrence $(1)$. $\dashv$

The initial conditions $p(n,0)=p(0,n)=\delta_{n,0}$ are exactly what’s needed in order to make $(1)$ yield the correct values of $p(1,k)$ and $p(n,1)$, so $p(n,k)=S_q(n,k)$.


Stephen C. Milne, ‘Restricted Growth Functions, Rank Row Matchings of Partition Lattices, and $q$-Stirling Numbers’, Advances in Mathematics $\mathbf{43}$, $173$-$196$ ($1982$), has a different interpretation in terms of partitions. For $\pi\in\mathscr{P}(n,k)$ and $i\in[n]$ let $J(\pi,i)$ be the number of parts $P_j^\pi$ such that $j<i$ and $P_j^\pi$ contains a number smaller than $i$, and let $J(\pi)=\sum_{i\in[n]}J(\pi,i)$; then $$S_q(n,k)=\sum_{\pi\in\mathscr{P}(n,k)}q^{J(\pi)}\;.$$


A.M. Garsia and J.B. Remmel, ‘Q-Counting Rook Configurations and a Formula of Frobenius’, Journal of Combinatorial Theory, Series A $\mathbf{41}$, $246$-$276$ ($1986$) give $S_q(n,k)$ an interpretation involving the number of configurations of $n-k$ non-taking rooks on the staircase Ferrers board.


Michelle Wachs and Dennis White, ‘$p,q$-Stirling Numbers and Set Partitions’, Journal of Combinatorial Theory, Series A $\mathbf{56}$, $27$-$46$ ($1991$), have nice generalizations of the interpretations of Milne and of Garsia & Remmel.

Brian M. Scott
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  • Thanks! This is just what I was looking for. – Nishant Feb 03 '15 at 14:48
  • @Nishant: You’re welcome; glad to help! – Brian M. Scott Feb 03 '15 at 19:28
  • @Nishant: I don’t know, I’m afraid; I’m only moderately familiar with this area, and I’ve not had time to give it serious thought. I’ll try to give it some thought over the next couple of days, but as I say, it’s not really my field. – Brian M. Scott Feb 08 '15 at 01:30
  • Oh sorry about that, I don't think there's an obvious correspondence, but soon after I posted that comment, I was able to prove that my weighting formula also satisfies the recursion in basically the same way you did, which is why I deleted the comment after. – Nishant Feb 08 '15 at 06:57
  • @Nishant: Oh, good; then I'll not worry about it! By the way, if I'm remembering it correctly, I believe that the sum of the weights of all compositions of $n$ is OEIS A$001788$, in case that's of any use to you. – Brian M. Scott Feb 08 '15 at 07:58