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The stirling numbers of the second kind satisfy the formula $x^n=\sum_k\left\{n\atop k\right\}(x)_k$, where $(x)_k$ is the falling factorial.

Consider the $q$-analog recursive definition of the stirling numbers, given by $$ \left\{n\atop k\right\}_q=(k)_q\left\{n-1\atop k\right\}_q+q^{k-1}\left\{n-1\atop k-1\right\}_q. $$

Why do they satisfy an analog to the standard formula, $$ ((r)_q)^n=\sum_k\left\{n\atop k\right\}_q(r)_q(r-1)_q\cdots(r-k+1)_q? $$ Thank you.

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  • $\begingroup$ Hi, Is there any special type of formula that you're looking for? I'm trying to find analogs such as ${(x)_q}^n$ and ${(x^n)}_q$, and I might try looking for some others, but I'm not sure if you want to restrict the possible formulas in any way. $\endgroup$
    – Matt Groff
    Feb 15, 2012 at 21:10
  • $\begingroup$ @MattGroff Hi, Matt, I'm specifically looking for an analog for $((r)_q)^n$ to be in the same spirit of $x^n$. $\endgroup$
    – Carmen So
    Feb 15, 2012 at 21:44

2 Answers 2

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The $q$-Stirling numbers defined by your recurrence satisfy the operator equation

$$(xD_q)^n=\sum_k\left\{n\atop k\right\}_qx^kD_q^k\;,\tag{1}$$

where $D_q$ is the $q$-derivative operator defined by $$D_qf(x)=\frac{f(qx)-f(x)}{qx-x}$$ and satisfying $$D_qx^n=\frac{(qx)^n-x^n}{(q-1)x}=\frac{q^n-1}{q-1}x^{n-1}=(n)_qx^{n-1}$$ and $$D_qx=qxD_q\;;$$ for a proof see this answer.

Now $xD_qx^r=x(r)_qx^{r-1}=(r)_qx^r$, so an easy induction shows that $(xD_q)^nx^r=\big((r)_q\big)^nx^r$. Thus, applying $(1)$ to $x^r$ yields

$$\begin{align*} \Big((r)_q\Big)^nx^r&=(xD_q)^nx^r\\ &=\sum_k\left\{n\atop k\right\}_qx^kD_q^kx^r\\ &=\sum_k\left\{n\atop k\right\}_qx^k(r)_q(r-1)_q\cdots(r-k+1)_qx^{r-k}\\ &=\sum_k\left\{n\atop k\right\}_q(r)_q(r-1)_q\cdots(r-k+1)_qx^r\;. \end{align*}$$

Now just substitute $x=1$.

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If you can read German you find an answer in my lecture notes Elementare q-Identitaeten, Chapter 3. (http://homepage.univie.ac.at/johann.cigler/skripten.html).

Your $ \left\{n\atop k\right\}_q$ is the same as my $ q^{\binom{k}{2}} S[n,k]$

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