\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{amscd} \usepackage{graphicx} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics} \usepackage{latexsym} \usepackage{epsf} \usepackage{breakurl} \usepackage{mathrsfs} \usepackage{mathtools} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{https://oeis.org/#1}{\underline{#1}}} \newcommand{\NN}{\mathbb{N}} \newcommand{\DD}{\mathfrak{D}} \newcommand{\Der}{\mathscr{D}} \newcommand{\falling}[2]{{#1}^{\underline{#2}}} \newcommand{\rising}[2]{{#1}^{\overline{#2}}} \newcommand{\Mchoose}[2]{{\left(\!\!\!{ #1 \choose #2 }\!\!\!\right)}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf $q$-Derangement Identities\\ \vskip 1cm} \large Emanuele Munarini \\ Dipartimento di Matematica \\ Politecnico di Milano \\ Piazza Leonardo da Vinci 32 \\ Milano \\ Italy \\ \href{mailto:emanuele.munarini@polimi.it}{\tt emanuele.munarini@polimi.it}\\ \end{center} \vskip .2 in \begin{abstract} In this paper, we derive several combinatorial identities involving the $q$-derangement numbers (for the major index) and many other $q$-numbers and $q$-polynomials of combinatorial interest, such as the $q$-binomial coefficients, the $q$-Stirling numbers, the $q$-Bell numbers, the $q$-Pochhammer symbol, the Gaussian polynomials, the Rogers-Szeg\H{o} polynomials and the Galois numbers, and the Al-Salam-Carlitz polynomials. We also obtain two determinantal identities expressing the $q$-derangement numbers as tridiagonal determinants and as Hessenberg determinants. \end{abstract} \section{Introduction} The \emph{derangement number} $ d_n $ (sequence \seqnum{A000166} in the {\it On-Line Encyclopedia of Integer Sequences}) counts the derangements (i.e., permutation with no fixed points) of an $n$-set. By a simple application of the principle of inclusion-exclusion, we have the formula \begin{equation}\label{def-derN} d_n = \sum_{k=0}^n { n \choose k } (n-k)!\, (-1)^k \, . \end{equation} Moreover, these numbers satisfy the recurrences \begin{align} & d_{n+1} = (n+1)\, d_n + (-1)^{n+1} \label{rec-derN01} \\ & d_{n+2} = (n+1)\, d_{n+1} + (n+1)\, d_n \label{rec-derN02} \end{align} with initial conditions $ d_0 = 1 $ and $ d_1 = 0 $. The \emph{$q$-derangement numbers} \cite{Wachs,ChenRota} are defined by the formula \begin{equation}\label{def-qderN} d_n(q) = \sum_{k=0}^n { n \choose k }_{\!\!q} [n-k]_q!\, (-1)^k q^{{ k \choose 2 }} \, , \end{equation} where $ [n]_q = 1+q+q^2+\cdots+q^{n-1} $, and satisfy the recurrences \begin{align} & d_{n+1}(q) = [n+1]_q\, d_n(q) + (-1)^{n+1} q^{{ n+1 \choose 2 }} \label{rec-qderN01} \\ & d_{n+2}(q) = [n+1]_q\, d_{n+1}(q) + [n+1]_q\, q^{n+1} d_n(q) \label{rec-qderN02} \end{align} with initial conditions $ d_0(q) = 1 $ and $ d_1(q) = 0 $. The \emph{major index} $ \mathrm{maj}(\sigma) $ of a permutation $ \sigma $ of $ \{1,2,\ldots,n\} $ is the sum of all positions $ k $ for which $ \sigma(k) > \sigma(k+1) $. The $q$-numbers $ d_n(q) $ arise from the $q$-counting of derangements by major index and have several interesting combinatorial properties. Indeed, if $ \Der_n $ is the set of all derangements of $ \{1,2,\ldots,n\} $, then we have \cite{Wachs} $$ d_n(q) = \sum_{\sigma\in\Der_n} q^{\mathrm{maj}(\sigma)} \, . $$ Moreover, considered as a polynomial in $ q $, the $q$-derangement number $ d_n(q) $ has non-negative integer coefficients forming a unimodal sequence \cite{ChenRota}. These coefficients have a spiral property \cite{Zhang}, which implies their unimodality and also the fact that the maximum coefficient of $ d_n(q) $ appears exactly in the middle of the polynomial, i.e., is the coefficient of $ q^{\lfloor n(n-1)/4 \rfloor} $ (as conjectured by Chen and Rota \cite{ChenRota}). Furthermore, they have the ratio monotonicity property (for $ n \geq 6 $) \cite{ChenXia} which implies log-concavity and the spiral property. For the ordinary derangement numbers $ d_n $ (and their generalizations \cite{CFMunariniZ,FerrariMunarini,MunariniA,MunariniC,MunariniO}) there are a lot of combinatorial identities. In this paper, we derive several $q$-analogues of these identities. They involve the $q$-derangement numbers $ d_n(q) $ and many other $q$-numbers or $q$-polynomials, such as the $q$-binomial coefficients, the $q$-Stirling numbers and the $q$-Bell numbers, the $q$-Pochhammer symbol, the Gaussian polynomials, the Rogers-Szeg\H{o} polynomials and the Galois numbers, and the Al-Salam-Carlitz polynomials. Finally, we obtain two determinantal identities expressing the $q$-derangement numbers as tridiagonal determinants and as Hessenberg determinants. \section{$q$-binomial identities}\label{sec-qBin} We start by recalling some basic definitions of $q$-number theory. For every $ n \in \NN $, we have the \emph{$q$-natural number} $ [n]_q = 1+q+q^2+\cdots+q^{n-1} $ and the \emph{$q$-factorial number} $ [n]_q! = [n]_q[n-1]_q\cdots[2]_q[1]_q $. Then, for every $ n, k \in \NN $, we have the \emph{$q$-binomial coefficients} (or \emph{Gaussian coefficients} \cite{Gauss}) defined by $$ { n \choose k }_{\!\!q} = \begin{cases} \displaystyle \frac{[n]_q!}{[k]_q![n-k]_q!}, & \mbox{if } k \leq n; \\[2mm] 0, & \mbox{otherwise,} \end{cases} $$ and satisfying the recurrence \begin{equation}\label{RecqBin} { n+1 \choose k+1 }_{\!\!q} = { n \choose k }_{\!\!q} + q^{n+1} { n \choose k+1 }_{\!\!q} \end{equation} with initial conditions $ { n \choose 0 }_{\!\!q} = 1 $ and $ { 0 \choose k }_{\!\!q} = \delta_{k,0} $. Moreover, we have the relations \begin{equation}\label{qBinRel} [n]_{q^{-1}} = \frac{1}{q^{n-1}}\,[n]_q \, , \qquad [n]_{q^{-1}}! = [n]_q!\; q^{-{n \choose 2 }} \, , \qquad { n \choose k }_{\!\!q^{-1}} = { n \choose k }_{\!\!q} q^{-k(n-k)} \, . \end{equation} \medskip In this first section, we derive some $q$-binomial identities using an elementary approach \cite{MunariniSP} which exploits the properties of the $q$-binomial coefficients and the recurrences of the $q$-derangement numbers. In Section \ref{sec-qES}, we derive some other $q$-binomial identities by using the more advanced technique of the $q$-exponential series. Our first result is the following: \begin{theorem} For every $ n \in \NN $, we have the identity \begin{equation}\label{IdqBinqDer} 1 + \sum_{k=1}^n { n \choose k }_{\!\!q} \frac{d_{k+1}(q)}{[k]_q} = \sum_{k=0}^n { n+1 \choose k+1 }_{\!\!q} d_k(q) \, . \end{equation} \end{theorem} \begin{proof} By recurrence (\ref{rec-qderN02}), we have $$ \frac{d_{k+2}(q)}{[k+1]_q} = d_{k+1}(q) + q^{k+1} d_k(q)\, . $$ Consequently, we have $$ \sum_{k=0}^{n-1} { n \choose k+1 }_{\!\!q} \frac{d_{k+2}(q)}{[k+1]_q} = \sum_{k=0}^{n-1} { n \choose k+1 }_{\!\!q} d_{k+1}(q) + \sum_{k=0}^{n-1} { n \choose k+1 }_{\!\!q} q^{k+1} d_k(q) $$ or $$ \sum_{k=1}^n { n \choose k }_{\!\!q} \frac{d_{k+1}(q)}{[k]_q} = \sum_{k=1}^n { n \choose k }_{\!\!q} d_k(q) + \sum_{k=0}^n { n \choose k+1 }_{\!\!q} q^{k+1} d_k(q) $$ or $$ \sum_{k=1}^n { n \choose k }_{\!\!q} \frac{d_{k+1}(q)}{[k]_q} = \sum_{k=0}^n \left( { n \choose k }_{\!\!q} + q^{k+1} { n \choose k+1 }_{\!\!q} \right) d_k(q) - d_0(q) \, . $$ By recurrence (\ref{RecqBin}) and the initial condition $ d_0(q) = 1 $, we have identity (\ref{IdqBinqDer}). \end{proof} Similarly, we have the following formula. \begin{theorem} For every $ n \in \NN $, we have the identity \begin{equation}\label{IdqBinqDer2} \sum_{k=0}^n { n \choose k }_{\!\!q} d_{k+1}(q) = \sum_{k=0}^n { n+1 \choose k+1 }_{\!\!q} [k]_q d_k(q) + \sum_{k=1}^n { n \choose k }_{\!\!q} q^{2k-1} d_{k-1}(q) \, . \end{equation} \end{theorem} \begin{proof} By recurrence (\ref{rec-qderN02}), we have $$ \sum_{k=0}^{n-1} { n \choose k+1 }_{\!\!q} d_{k+2}(q) = \sum_{k=0}^{n-1} { n \choose k+1 }_{\!\!q} [k+1]_q d_{k+1}(q) + \sum_{k=0}^{n-1} { n \choose k+1 }_{\!\!q} [k+1]_q q^{k+1} d_k(q) $$ or $$ \sum_{k=1}^n { n \choose k }_{\!\!q} d_{k+1}(q) = \sum_{k=1}^n { n \choose k }_{\!\!q} [k]_q d_k(q) + \sum_{k=0}^n { n \choose k+1 }_{\!\!q} [k+1]_q q^{k+1} d_k(q) $$ or $$ \sum_{k=0}^n { n \choose k }_{\!\!q} d_{k+1}(q) - d_1(q) = \sum_{k=0}^n { n \choose k }_{\!\!q} [k]_q d_k(q) + \sum_{k=0}^n { n \choose k+1 }_{\!\!q} ([k]_q+q^k) q^{k+1} d_k(q) $$ or $$ \sum_{k=0}^n { n \choose k }_{\!\!q} d_{k+1}(q) - d_1(q) = \sum_{k=0}^n \left( { n \choose k }_{\!\!q} + q^{k+1} { n \choose k+1 }_{\!\!q} \right) [k]_q d_k(q) + \sum_{k=0}^n { n \choose k+1 }_{\!\!q} q^{2k+1} d_k(q) \, . $$ By recurrence (\ref{RecqBin}) and the initial condition $ d_1(q) = 0 $, we have identity (\ref{IdqBinqDer2}). \end{proof} We also have the following formula. \begin{theorem} For every $ m,n \in \NN $, we have the identity \begin{equation}\label{IdBin11} \sum_{k=0}^n { m+n+1 \choose m+k+1 }_{\!\!q} \frac{[n]_q!}{[k]_q!}\, (-1)^k q^{{ m+k+1 \choose 2 }} d_k(q) = \sum_{k=0}^n { m+n \choose m+k }_{\!\!q} \frac{[n]_q!}{[k]_q!}\, (-1)^k q^{{ m+k+1 \choose 2 }+{ k \choose 2 }} \, . \end{equation} In particular, for $ m = 0 $, we have the identity \begin{equation}\label{IdBin12} \sum_{k=0}^n { n+1 \choose k+1 }_{\!\!q} \frac{[n]_q!}{[k]_q!}\, (-1)^k q^{{ k+1 \choose 2 }} d_k(q) = \sum_{k=0}^n { n \choose k }_{\!\!q} \frac{[n]_q!}{[k]_q!}\, (-1)^k q^{k^2}. \end{equation} \end{theorem} \begin{proof} From recurrence (\ref{rec-qderN01}), we have $$ \frac{d_{k+1}(q)}{[k+1]_q!} = \frac{d_{k}(q)}{[k]_q!} + (-1)^{k+1} \frac{q^{{ k+1 \choose 2 }}}{[k+1]_q!}\, . $$ and consequently \begin{align*} & \sum_{k=0}^{n-1} { m+n \choose m+k+1 }_{\!\!q} (-1)^{k+1} [n]_q! q^{{ m+k+2 \choose 2 }} \frac{d_{k+1}(q)}{[k+1]_q!} \\ & = \sum_{k=0}^{n-1} { m+n \choose m+k+1 }_{\!\!q} (-1)^{k+1} [n]_q! q^{{ m+k+2 \choose 2 }} \frac{d_{k}(q)}{[k]_q!} + \sum_{k=0}^{n-1} { m+n \choose m+k+1 }_{\!\!q} [n]_q! q^{{ m+k+2 \choose 2 }} \frac{q^{{ k+1 \choose 2 }}}{[k+1]_q!}\, ; \end{align*} that is, \begin{align*} & \sum_{k=1}^n { m+n \choose m+k }_{\!\!q} \frac{[n]_q!}{[k]_q!}\, (-1)^k q^{{ m+k+1 \choose 2 }} d_k(q) + \sum_{k=0}^{n-1} { m+n \choose m+k+1 }_{\!\!q} \frac{[n]_q!}{[k]_q!}\, (-1)^k q^{{ m+k+1 \choose 2 }} q^{m+k+1} d_k(q) \\ & \qquad = \sum_{k=1}^n { m+n \choose m+k }_{\!\!q} \frac{[n]_q!}{[k]_q!} q^{{ m+k+1 \choose 2 }} q^{{ k \choose 2 }}\, , \end{align*} or \begin{align*} & \sum_{k=0}^n \left( { m+n \choose m+k }_{\!\!q} + q^{m+k+1} { m+n \choose m+k+1 }_{\!\!q} \right) \frac{[n]_q!}{[k]_q!}\, (-1)^k q^{{ m+k+1 \choose 2 }} d_k(q) \\ & \qquad = { m+n \choose m }_{\!\!q} [n]_q! q^{{ m+1 \choose 2 }} + \sum_{k=1}^n { m+n \choose m+k }_{\!\!q} \frac{[n]_q!}{[k]_q!} q^{{ m+k+1 \choose 2 }+{ k \choose 2 }}\, . \end{align*} Hence, by formula (\ref{RecqBin}), we have identity (\ref{IdBin11}). \end{proof} Using the same approach, we also have the following result. \begin{theorem} We have the identity \begin{equation}\label{IdU} q \sum_{k=0}^n [k]_q d_k(q) + \sum_{k=0}^n (-1)^k q^{{ k \choose 2 }} = [n+1]_q d_n(q) \, . \end{equation} \end{theorem} \begin{proof} Since $ [k+1]_q = 1+q[k]_q $, recurrence (\ref{rec-qderN01}) can be rewritten as $$ d_{k+1}(q) = (1+q[k]_q)\, d_k(q) + (-1)^{k+1} q^{{ k+1 \choose 2 }} $$ or $$ d_{k+1}(q)-d_k(q) = q[k]_q\, d_k(q) + (-1)^{k+1} q^{{ k+1 \choose 2 }}\, . $$ Hence, we have $$ \sum_{k=0}^n d_{k+1}(q) - \sum_{k=0}^n d_k(q) = q \sum_{k=0}^n [k]_q\, d_k(q) + \sum_{k=0}^n (-1)^{k+1} q^{{ k+1 \choose 2 }} $$ or $$ \sum_{k=1}^{n+1} d_k(q) - \sum_{k=0}^n d_k(q) = q \sum_{k=0}^n [k]_q\, d_k(q) + \sum_{k=1}^{n+1} (-1)^k q^{{ k \choose 2 }} $$ or $$ d_{n+1}(q) - d_0(q) = q \sum_{k=0}^n [k]_q\, d_k(q) + \sum_{k=0}^{n+1} (-1)^k q^{{ k \choose 2 }} - 1 \, . $$ Hence, by the recurrence (\ref{rec-qderN01}) once again, we have the identity $$ [n+1]_q d_n(q) + (-1)^{n+1} q^{{ n+1 \choose 2 }} = q \sum_{k=0}^n [k]_q\, d_k(q) + \sum_{k=0}^n (-1)^k q^{{ k \choose 2 }} + (-1)^{n+1} q^{{ n+1 \choose 2 }} $$ which simplifies to identity (\ref{IdU}). \end{proof} Similarly, we also have the following property. \begin{theorem} We have the identity \begin{equation}\label{IdUU} \sum_{k=0}^n { n+1 \choose k+1 }_{\!\!q} (-1)^k q^{{ k+1 \choose 2 }} d_k(q) = q \sum_{k=1}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k+1 \choose 2 }} [k-1]_q d_{k-1}(q) + \sum_{k=0}^n { n \choose k }_{\!\!q} q^{k^2}. \end{equation} \end{theorem} \begin{proof} Once again, we start by the recurrence (\ref{rec-qderN01}) written as $$ d_{k+1}(q)-d_k(q) = q[k]_q\, d_k(q) + (-1)^{k+1} q^{{ k+1 \choose 2 }}\, . $$ Then we have $$ \begin{aligned} \MoveEqLeft \sum_{k=0}^{n-1} { n \choose k+1 }_{\!\!q} (-1)^{k+1} q^{{ k+2 \choose 2 }} d_{k+1}(q) - \sum_{k=0}^{n-1} { n \choose k+1 }_{\!\!q} (-1)^{k+1} q^{{ k+2 \choose 2 }} d_k(q) = \\ & q \sum_{k=0}^{n-1} { n \choose k+1 }_{\!\!q} (-1)^{k+1} q^{{ k+2 \choose 2 }} [k]_q\, d_k(q) + \sum_{k=0}^{n-1} { n \choose k+1 }_{\!\!q} q^{{ k+2 \choose 2 }} q^{{ k+1 \choose 2 }}, \end{aligned} $$ which is $$ \begin{aligned} \MoveEqLeft \sum_{k=1}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k+1 \choose 2 }} d_k(q) + \sum_{k=0}^n { n \choose k+1 }_{\!\!q} (-1)^k q^{{ k+1 \choose 2 }} q^{k+1} d_k(q) = \\ &\qquad q \sum_{k=1}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k+1 \choose 2 }} [k-1]_q\, d_{k-1}(q) + \sum_{k=1}^n { n \choose k }_{\!\!q} q^{{ k+1 \choose 2 }} q^{{ k \choose 2 }} \end{aligned} $$ or $$ \begin{aligned} \MoveEqLeft \sum_{k=0}^n \left( { n \choose k }_{\!\!q} + q^{k+1} { n \choose k+1 }_{\!\!q} \right) (-1)^k q^{{ k+1 \choose 2 }} d_k(q) - 1 \\ & = q \sum_{k=1}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k+1 \choose 2 }} [k-1]_q\, d_{k-1}(q) + \sum_{k=0}^n { n \choose k }_{\!\!q} q^{{ k+1 \choose 2 }} q^{{ k \choose 2 }} - 1\, . \end{aligned} $$ By recurrence (\ref{RecqBin}), this last identity simplifies to identity (\ref{IdUU}). \end{proof} \section{$q$-Stirling identities}\label{sec-qSti} The \emph{$q$-Stirling numbers of the second kind} are defined as the connection constants \cite{MunariniCC,MunariniSP} between the ordinary powers $ x^n $ and the $q$-falling factorials $ \falling{x}{n}_q = x(x-[1]_q)(x-[2]_q)\cdots(x-[n-1]_q) $, that is, as the coefficients $ { n \brace k }_{\!\!q} $ for which $$ x^n = \sum_{k=0}^n { n \brace k }_{\!\!q} \falling{x}{k}_q \, . $$ Equivalently, they are the numbers defined by the recurrence \begin{equation}\label{Rec-StirlingS2} { n+1 \brace k+1 }_{\!\!q} = { n \brace k }_{\!\!q} + [k+1]_q { n \brace k+1 }_{\!\!q} \end{equation} with initial values $ { n \brace 0 }_{\!\!q} = \delta_{n,0} $ and $ { 0 \brace k }_{\!\!q} = \delta_{k,0} $. Similarly, the \emph{$q$-Stirling numbers of the first kind} are defined as the connection constants \cite{MunariniCC,MunariniSP} between the $q$-rising factorials $ \rising{x}{n}_q = x(x+[1]_q)(x+[2]_q)\cdots(x+[n-1]_q) $ and the ordinary powers $ x^n $, that is, as the coefficients $ { n \brack k }_{\!q} $ for which $$ \rising{x}{n}_q = \sum_{k=0}^n { n \brack k }_{\!q} x^k \, . $$ Equivalently, they are the numbers defined by the recurrence \begin{equation}\label{Rec-StirlingS1} { n+1 \brack k+1 }_{\!q} = { n \brack k }_{\!q} + [n]_q { n \brack k+1 }_{\!q} \end{equation} with initial values $ { n \brack 0 }_{\!q} = \delta_{n,0} $ and $ { 0 \brack k }_{\!q} = \delta_{k,0} $. For the $q$-Stirling numbers, we have the inverse relations \begin{equation}\label{Stirling-inversion} f_n = \sum_{k=0}^n { n \brace k }_{\!\!q} g_k \qquad\iff\qquad g_n = \sum_{k=0}^n { n \brack k }_{\!q} (-1)^{n-k} f_k \end{equation} and \begin{equation}\label{Stirling-inversion1} f_n = \sum_{k=0}^n { n+1 \brace k+1 }_{\!\!q} g_k \qquad\iff\qquad g_n = \sum_{k=0}^n { n+1 \brack k+1 }_{\!q} (-1)^{n-k} f_k\, . \end{equation} Consider the \emph{$q$-Bell numbers} defined by \begin{equation}\label{qBellN} b_n(q) = \sum_{k=0}^n { n \brace k }_{\!\!q} q^{{ k \choose 2 }} \, . \end{equation} Although they are not the cumulative constants of the $q$-Stirling numbers considered above, we have the following formulas relating the $q$-derangement numbers and the $q$-Bell numbers, \begin{theorem} We have the identities \begin{align} & \sum_{k=0}^n { n+1 \brace k+1 }_{\!\!q} (-1)^k d_k(q) = b_n(q) \label{Id-qDer-qBell} \\ & \sum_{k=0}^n { n+1 \brack k+1 }_{\!q} (-1)^k b_k(q) = d_n(q) \label{Id-qBell-qDer} \, . \end{align} \end{theorem} \begin{proof} By recurrence (\ref{rec-qderN01}), we have $$ d_{k+1}(q) - [k+1]_q d_k(q) = (-1)^{k+1} q^{{ k+1 \choose 2 }}\, . $$ Hence, we have the identity $$ \sum_{k=0}^{n-1} { n \brace k+1 }_{\!\!q} (-1)^{k+1} d_{k+1}(q) - \sum_{k=0}^{n-1} { n \brace k+1 }_{\!\!q} (-1)^{k+1} [k+1]_q d_k(q) = \sum_{k=0}^{n-1} { n \brace k+1 }_{\!\!q} q^{{ k+1 \choose 2 }} $$ or $$ \sum_{k=1}^n { n \brace k }_{\!\!q} (-1)^k d_k(q) + \sum_{k=0}^{n-1} { n \brace k+1 }_{\!\!q} (-1)^k [k+1]_q d_k(q) = \sum_{k=1}^n { n \brace k }_{\!\!q} q^{{ k \choose 2 }} $$ or $$ \sum_{k=0}^n \left( { n \brace k }_{\!\!q} + [k+1]_q { n \brace k+1 }_{\!\!q} \right) (-1)^k d_k(q) = \sum_{k=0}^n { n \brace k }_{\!\!q} q^{{ k \choose 2 }} $$ By recurrence (\ref{Rec-StirlingS2}) and definition (\ref{qBellN}), we have identity (\ref{Id-qDer-qBell}). Then, by this identity, we get identity (\ref{Id-qBell-qDer}) at once as its inverse relation (by property (\ref{Stirling-inversion1})). \end{proof} To prove the next theorem, we need the following result. \begin{lemma} For every $ n,k \in \NN $, we have the identity \begin{equation}\label{IdqStirling} { n+1 \brace k+1 }_{\!\!q} = \sum_{i=k}^n { n \choose i } { i \brace k }_{\!\!q} q^{i-k} \, . \end{equation} \end{lemma} \begin{proof} Since \begin{align*} x^{n+1} = \sum_{k=1}^{n+1} { n+1 \brace k }_{\!\!q} \falling{x}{k}_q = \sum_{k=0}^n { n+1 \brace k+1 }_{\!\!q} \falling{x}{k+1}_q = \sum_{k=0}^n { n+1 \brace k+1 }_{\!\!q} x(x-[1]_q)\cdots(x-[k]_q)\, , \end{align*} we have \begin{align*} x^n &= \sum_{k=0}^n { n+1 \brace k+1 }_{\!\!q} (x-[1]_q)(x-[2]_q)\cdots(x-[k]_q) \\ &= \sum_{k=0}^n { n+1 \brace k+1 }_{\!\!q} (x-1)(x-1-q[1]_q)\cdots(x-1-q[k-1]_q) \end{align*} or $$ (qx+1)^n = \sum_{k=0}^n { n+1 \brace k+1 }_{\!\!q} q^k x(x-[1]_q)\cdots(x-[k-1]_q) = \sum_{k=0}^n { n+1 \brace k+1 }_{\!\!q} q^k \falling{x}{k}_q \, . $$ Then, from this relation, we have $$ \sum_{k=0}^n { n+1 \brace k+1 }_{\!\!q} q^k \falling{x}{k}_q = \sum_{i=0}^n { n \choose i } q^i x^i = \sum_{i=0}^n { n \choose i } q^i \sum_{k=0}^i { i \brace k }_{\!\!q} \falling{x}{k}_q = \sum_{k=0}^i \left( \sum_{i=k}^n { n \choose i } { i \brace k }_{\!\!q} q^i \right) \falling{x}{k}_q\, . $$ By equating the coefficients of $ \falling{x}{k}_q $, we obtain identity (\ref{IdqStirling}). \end{proof} Now we can prove the following result. \begin{theorem} We have the identity \begin{equation}\label{IdSSB1} \sum_{k=0}^n { n \brace k }_{\!\!q} (-1)^k q^{n-k} d_k(q) = \sum_{k=0}^n { n \choose k } (-1)^{n-k} b_k(q)\, . \end{equation} \end{theorem} \begin{proof} By identities (\ref{Id-qDer-qBell}) and (\ref{IdqStirling}), we have $$ b_n(q) = \sum_{k=0}^n \left(\sum_{i=0}^n { n \choose i } { i \brace k }_{\!\!q} q^{i-k}\right) (-1)^k d_k(q) = \sum_{i=0}^n { n \choose i } \sum_{k=0}^i { i \brace k }_{\!\!q} (-1)^k q^{i-k} d_k(q)\, . $$ Thus, if we set $$ z_n(q) = \sum_{k=0}^n { n \brace k }_{\!\!q} (-1)^k q^{n-k} d_k(q) \, , $$ then we have the identity $$ b_n(q) = \sum_{i=0}^n { n \choose i } z_i(q), $$ whose inverse is $$ z_n(q) = \sum_{i=0}^n { n \choose i } (-1)^{n-i} b_i(q)\, . $$ This is identity (\ref{IdSSB1}). \end{proof} We also have the following result. \begin{theorem} We have the identity \begin{equation}\label{IdSS11} \sum_{k=0}^n { n+1 \brace k+1 }_{\!\!q} (-1)^k d_{k+1}(q) = \sum_{k=0}^n { n \brace k }_{\!\!q} (-1)^k q^k [k]_q d_{k-1}(q)\, . \end{equation} \end{theorem} \begin{proof} By recurrence (\ref{rec-qderN02}), we have $$ d_{k+2}(q) - [k+1]_q d_{k+1}(q) = [k+1]_q q^{k+1} d_k(q)\, . $$ Then we have $$ \begin{aligned} \MoveEqLeft \sum_{k=0}^{n-1} { n \brace k+1 }_{\!\!q} (-1)^{k+1} d_{k+2}(q) - \sum_{k=0}^{n-1} { n \brace k+1 }_{\!\!q} (-1)^{k+1} [k+1]_q d_{k+1}(q) = \\ & = \sum_{k=0}^{n-1} { n \brace k+1 }_{\!\!q} (-1)^{k+1} [k+1]_q q^{k+1} d_k(q), \end{aligned} $$ or $$ \sum_{k=1}^n { n \brace k }_{\!\!q} (-1)^k d_{k+1}(q) + \sum_{k=0}^n { n \brace k+1 }_{\!\!q} (-1)^k [k+1]_q d_{k+1}(q) = \sum_{k=1}^n { n \brace k }_{\!\!q} (-1)^k [k]_q q^k d_{k-1}(q), $$ or $$ \sum_{k=1}^n \left( { n \brace k }_{\!\!q} + [k+1]_q { n \brace k+1 }_{\!\!q} \right) (-1)^k d_{k+1}(q) = \sum_{k=1}^n { n \brace k }_{\!\!q} (-1)^k [k]_q q^k d_{k-1}(q)\, . $$ By recurrence (\ref{Rec-StirlingS2}), we have identity (\ref{IdSS11}). \end{proof} Similarly, we also have the following formula. \begin{theorem} We have the identity \begin{equation}\label{IdSS01} \sum _{k=0}^n { n \brace k }_{\!\!q} (-1)^k q^{-{k+1 \choose 2 }} d_k(q)^2 = \sum _{k=0}^n { n+1 \brace k+1 }_{\!\!q} (-1)^k q^{-{ k+1 \choose 2 }} d_k(q) d_{k+1}(q) \, . \end{equation} \end{theorem} \begin{proof} By recurrence (\ref{rec-qderN02}), we have $$ d_{k+1}(q) d_{k+2}(q) = [k+1]_q d_{k+1}(q)^2 + [k+1]_q q^{k+1} d_k(q) d_{k+1}(q), $$ or $$ [k+1]_q d_{k+1}(q)^2 = d_{k+1}(q) d_{k+2}(q) - [k+1]_q q^{k+1} d_k(q) d_{k+1}(q)\, . $$ Hence, we have \begin{align*} & \sum_{k=0}^{n-1} { n \brace k+1 }_{\!\!q} (-1)^{k+1} q^{-{ k+2 \choose 2 }} [k+1]_q d_{k+1}(q)^2 \\ & = \sum_{k=0}^{n-1} { n \brace k+1 }_{\!\!q} (-1)^{k+1} q^{-{ k+2 \choose 2 }} d_{k+1}(q) d_{k+2}(q) \\ & \qquad - \sum_{k=0}^{n-1} { n \brace k+1 }_{\!\!q} (-1)^{k+1} q^{-{ k+2 \choose 2 }} [k+1]_q q^{k+1} d_k(q) d_{k+1}(q) \end{align*} or \begin{align*} & \sum_{k=1}^n { n \brace k }_{\!\!q} (-1)^k q^{-{ k+1 \choose 2 }} [k]_q d_k(q)^2 \\ & = \sum_{k=1}^n { n \brace k }_{\!\!q} (-1)^k q^{-{ k+1 \choose 2 }} d_k(q) d_{k+1}(q) \\ & \qquad + \sum_{k=0}^n { n \brace k+1 }_{\!\!q} [k+1]_q (-1)^k q^{-{ k+1 \choose 2 }} d_k(q) d_{k+1}(q)\, . \end{align*} Since $ [0]_q = 0 $ and $ d_1(q) = 0 $, we have \begin{align*} & \sum_{k=0}^n { n \brace k }_{\!\!q} (-1)^k q^{-{ k+1 \choose 2 }} [k]_q d_k(q)^2 \\ & \qquad = \sum_{k=0}^n \left( { n \brace k }_{\!\!q} + [k+1]_q { n \brace k+1 }_{\!\!q} \right) (-1)^k q^{-{ k+1 \choose 2 }} d_k(q) d_{k+1}(q) \, . \end{align*} Finally, by the recurrence (\ref{Rec-StirlingS2}), this identity simplifies to identity (\ref{IdSS01}). \end{proof} Now consider the \emph{$q$-Bell numbers} $ B_n(q) $ defined by the recurrence \begin{equation}\label{rec-qBellD} B_{n+1}(q) = \sum_{k=0}^n { n \choose k }_{\!\!q} q^{n(n-k)} B_k(q) \end{equation} with initial value $ B_0(q) = 1 $. Notice that the $q$-numbers $ b_n(q) $ and $ B_n(q) $ are different $q$-analogues of the ordinary Bell numbers (\seqnum{A000110}). For these $q$-Bell numbers, we have the following result. \begin{theorem} We have the identity \begin{equation}\label{IdDerBellD} \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^{n-k} q^{-{ k+1 \choose 2 }} B_k(q)\, d_{n-k}(q) = \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^{n-k} q^{-{ k+1 \choose 2 }} B_{k+1}(q)\, [n-k]_q! \end{equation} \end{theorem} \begin{proof} Let $ \sigma_n(q) $ be the sum on the right-hand side of identity (\ref{IdDerBellD}). Then, by the recurrence (\ref{rec-qBellD}), we have \begin{align*} \sigma_n(q) &= \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^{n-k} q^{-{ k+1 \choose 2 }} [n-k]_q!\, B_{k+1}(q) \\ &= \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^{n-k} q^{-{ k+1 \choose 2 }} [n-k]_q! \sum_{i=0}^k { k \choose i }_{\!\!q} q^{k(k-i)} B_i(q) \\ &= \sum_{i=0}^n \left( \sum_{k=i}^n { n \choose k }_{\!\!q} { k \choose i }_{\!\!q} [n-k]_q! (-1)^{n-k} q^{-{ k+1 \choose 2 }} q^{k(k-i)} \right) B_i(q) \\ &= \sum_{i=0}^n { n \choose i }_{\!\!q} \left( \sum_{k=i}^n { n-i \choose k-i }_{\!\!q} [n-k]_q! (-1)^{n-k} q^{-{ k+1 \choose 2 }} q^{k(k-i)} \right) B_i(q) \\ &= \sum_{i=0}^n { n \choose i }_{\!\!q} \left( \sum_{k=0}^{n-i} { n-i \choose k }_{\!\!q} [n-k-i]_q! (-1)^{n-k-i} q^{-{ k+i+1 \choose 2 }} q^{(k+i)k} \right) B_i(q) \\ &= \sum_{i=0}^n { n \choose i }_{\!\!q} (-1)^{n-i} \left( \sum_{k=0}^{n-i} { n-i \choose k }_{\!\!q} [n-k-i]_q! (-1)^k q^{-{ i+1 \choose 2 }-{ k+1 \choose 2 }-ik} q^{(k+i)k} \right) B_i(q) \\ &= \sum_{i=0}^n { n \choose i }_{\!\!q} (-1)^{n-i} q^{-{ i+1 \choose 2 }} \left( \sum_{k=0}^{n-i} { n-i \choose k }_{\!\!q} [n-k-i]_q! (-1)^k q^{{ k \choose 2 }} \right) B_i(q) \, . \end{align*} Finally, by formula (\ref{def-qderN}), we have $$ \sigma_n(q) = \sum_{i=0}^n { n \choose i }_{\!\!q} (-1)^{n-i} q^{-{ i+1 \choose 2 }} d_{n-i}(q)\, B_i(q), $$ and this is the claimed identity. \end{proof} \section{Elementary identities}\label{sec-Eul} Several combinatorial identities can be derived from the following property of linear recurrences of the first order: the general solution of the recurrence $$ y_{n+1} = a_{n +1}y_n + b_{n+1} $$ is given by \begin{equation}\label{LinRecGenSol} y_n = a_n^* y_0 + \sum_{k=1}^n \frac{a_n^*}{a_k^*}\, b_k, \end{equation} where $ a_n^* = a_1 a_2 \cdots a_n $, provided that $ a_n \ne 0 $ for all $ n \in \NN $. First of all, we have the following simple result. \begin{theorem} For every $ m, n \in \NN $, we have the identity \begin{equation}\label{IdE000} \begin{aligned} \MoveEqLeft d_{m+n+2}(q) = { m+n+1 \choose m }_{\!\!q} [n+1]_q! d_{m+1}(q) \\ &\hspace{1.5cm} + [m+n+1]_q \sum_{k=0}^n { m+n \choose m+k}_{\!\!q} [n-k]_q! q^{k+m+1} d_{m+k}(q) \end{aligned} \end{equation} In particular, for $ m = 0 $, we have the identity \begin{equation}\label{IdE001} d_{n+2}(q) = [n+1]_q \sum_{k=0}^n {n \choose k }_{\!\!q} [n-k]_q! q^{k+1} d_k(q)\, . \end{equation} \end{theorem} \begin{proof} Let $ y_n(q) = d_{m+n+1}(q) $. By recurrence (\ref{rec-qderN02}), we have $$ y_{n+1}(q) = d_{m+n+2}(q) = [m+n+1]_q\, d_{m+n+1}(q) + [m+n+1]_q\, q^{m+n+1} d_{m+n}(q), $$ or $$ y_{n+1}(q) = [m+n+1]_q\, y_n(q) + [m+n+1]_q\, q^{m+n+1} d_{m+n}(q)\, . $$ This is a linear recurrence of the first order with coefficients $ a_n = [m+n]_q $ and $\: b_n = [m+n]_q\, q^{m+n} d_{m+n-1}(q) $. Since $$ a_n^* = [m+n]_q\cdots[m]_q = \frac{[m+n]_q}{[m]_q} = { m+n \choose m }_{\!\!q} [n]_q! \, , $$ then the solution, being $ y_0(q) = d_{m+1}(q) $, is \begin{align*} y_n(q) &= { m+n \choose m }_{\!\!q} [n]_q! d_{m+1}(q) + \sum_{k=1}^n \frac{[m+n]_q}{[m]_q} \frac{[m]_q}{[m+k]_q} [m+k]_q\, q^{m+k} d_{m+k-1}(q) \\ &= { m+n \choose m }_{\!\!q} [n]_q! d_{m+1}(q) + \sum_{k=1}^n \frac{[m+n]_q}{[m+k]_q}\; [m+k]_q\, q^{m+k} d_{m+k-1}(q) \\ &= { m+n \choose m }_{\!\!q} [n]_q! d_{m+1}(q) + \sum_{k=1}^n { m+n \choose m+k }_{\!\!q} [m+k]_q [n-k]_q!\, q^{m+k} d_{m+k-1}(q) \\ &= { m+n \choose m }_{\!\!q} [n]_q! d_{m+1}(q) + [m+n]_q \sum_{k=1}^n { m+n-1 \choose m+k-1 }_{\!\!q} [n-k]_q!\, q^{m+k} d_{m+k-1}(q) \\ &= { m+n \choose m }_{\!\!q} [n]_q! d_{m+1}(q) + [m+n]_q \sum_{k=0}^{n-1} { m+n-1 \choose m+k }_{\!\!q} [n-k-1]_q!\, q^{m+k+1} d_{m+k}(q) \, . \end{align*} Finally, by replacing $ n $ by $ n+1 $, we obtain formula (\ref{IdE000}). \end{proof} \begin{theorem} For every $ m, n \in \NN $, we have the identity \begin{equation}\label{IdE01} \frac{d_{m+n+2}(q)}{[m+n+1]_q!} = \frac{d_{m+1}(q)}{[m]_q!} + \sum _{k=0}^n q^{m+k+1} \frac{d_{m+k}(q)}{[m+k]_q!} \, . \end{equation} In particular, for $ m = 0 $, we have the identity \begin{equation}\label{IdE02} \frac{d_{n+2}(q)}{[n+1]_q!} = \sum _{k=0}^n q^{k+1} \frac{d_k(q)}{[k]_q!} \, . \end{equation} \end{theorem} \begin{proof} Let $ y_n(q) = \frac{d_{m+n+1}(q)}{[m+n]_q!} $. By recurrence (\ref{rec-qderN02}), we have \begin{align*} y_{n+1}(q) &= \frac{d_{m+n+2}(q)}{[m+n+1]_q!} = \frac{[m+n+1]_q(d_{m+n+1}(q) + q^{m+n+1} d_{m+n}(q))}{[m+n+1]_q!} \\ &= \frac{d_{m+n+1}(q)}{[m+n]_q!} + q^{m+n+1} \frac{d_{m+n}(q)}{[m+n]_q!}, \end{align*} or $$ y_{n+1}(q) = y_n(q) + q^{m+n+1} \frac{d_{m+n}(q)}{[m+n]_q!} \, . $$ This is a linear recurrence of the first order with $ a_n = 1 $ and $ b_n = q^{m+n} \frac{d_{m+n-1}(q)}{[m+n-1]_q!} $ for $ n \geq 1 $. So, by formula (\ref{LinRecGenSol}), we have the solution $$ y_n(q) = y_0(q) + \sum_{k=1}^n q^{m+k} \frac{d_{m+k-1}(q)}{[m+k-1]_q!} = y_0(q) + \sum_{k=0}^{n-1} q^{m+k+1} \frac{d_{m+k}(q)}{[m+k]_q!}, $$ or $$ \frac{d_{m+n+1}(q)}{[m+n]_q!} = \frac{d_{m+1}(q)}{[m]_q!} + \sum_{k=0}^{n-1} q^{m+k+1} \frac{d_{m+k}(q)}{[m+k]_q!}\, . $$ Now by replacing $ n $ by $ n+1 $, we obtain identity (\ref{IdE01}). \end{proof} \begin{theorem} For every $ m, n \in \NN $, we have the identity \begin{equation}\label{IdE03} \frac{d_{m+n+1}(q) d_{m+n+2}(q)}{q^{\binom{n+2}{2}} [m+n+1]_q!} = q^{m(n+1)}\,\frac{d_m(q) d_{m+1}(q)}{[m]_q!} + \sum _{k=0}^n q^{m (n-k)} \frac{d_{m+k+1}(q)^2}{q^{\binom{k+2}{2}} [m+k]_q!} \end{equation} In particular, for $ m = 0 $, we have the identity \begin{equation}\label{IdE04} \frac{d_{n+1}(q) d_{n+2}(q)}{q^{\binom{n+2}{2}} [n+1]_q!} = \sum _{k=0}^n \frac{d_{k+1}(q)^2}{q^{{ k+2 \choose 2 }} [k]_q!} \end{equation} \end{theorem} \begin{proof} Let $ y_n(q) = \frac{d_{m+n}(q)d_{m+n+1}(q)}{[m+n]_q!} $. By recurrence (\ref{rec-qderN02}), we have \begin{align*} y_{n+1}(q) &= \frac{d_{m+n+1}(q)d_{m+n+2}(q)}{[m+n+1]_q!} = \frac{d_{m+n+1}(q)\,(d_{m+n+1}(q) + q^{m+n+1} d_{m+n}(q))}{[m+n]_q!} \\ &= \frac{d_{m+n+1}(q)^2}{[m+n]_q!} + q^{m+n+1} \frac{d_{m+n}(q)d_{m+n+1}(q)}{[m+n]_q!}, \end{align*} or $$ y_{n+1}(q) = q^{m+n+1} y_n(q) + \frac{d_{m+n+1}(q)^2}{[m+n]_q!} \, . $$ This is a linear recurrence of the first order with $ a_n = q^{m+n} $ and $ b_n = \frac{d_{m+n}(q)^2}{[m+n-1]_q!} $ for $ n \geq 1 $. Since $ a_n^* = q^{mn+{ n+1 \choose 2 }} $, by formula (\ref{LinRecGenSol}), we have the solution \begin{align*} y_n(q) &= q^{mn+{ n+1 \choose 2 }} y_0(q) + \sum_{k=1}^n \frac{q^{mn+{ n+1 \choose 2 }}}{q^{mk+{ k+1 \choose 2 }}} \frac{d_{m+k}(q)^2}{[m+k-1]_q!} \\ &= q^{mn+{ n+1 \choose 2 }} y_0(q) + q^{{ n+1 \choose 2 }} \sum_{k=0}^{n-1} q^{m(n-k-1)} \frac{d_{m+k+1}(q)^2}{q^{{ k+2 \choose 2 }}[m+k]_q!}, \end{align*} or $$ \frac{d_{m+n}(q)d_{m+n+1}(q)}{q^{{ n+1 \choose 2 }}[m+n]_q!} = q^{mn} \frac{d_{m}(q)d_{m+1}(q)}{[m]_q!} + \sum_{k=0}^{n-1} q^{m(n-k-1)} \frac{d_{m+k+1}(q)^2}{q^{{ k+2 \choose 2 }}[m+k]_q!}\, . $$ Now by replacing $ n $ by $ n+1 $, we obtain identity (\ref{IdE03}). \end{proof} The next formula can be obtained with the same elementary approach used in Section \ref{sec-qBin}. \begin{theorem} We have the identity \begin{equation}\label{IdE06} \begin{aligned} \MoveEqLeft \frac{d_{m+n+1}(q)^2-q^{2{ m+n+1 \choose 2 }}}{[m+n+1]_q!^2} = \frac{d_m(q)^2-q^{2{ m \choose 2 }}}{[m]_q!^2} + \\ & + 2 \sum_{k=0}^n (-1)^{m+k+1} q^{{ m+k+1 \choose 2 }} \frac{d_{m+k}(q)}{[m+k]_q![k+m+1]_q!} + \sum_{k=0}^n \frac{q^{2{ m+k \choose 2 }}}{[m+k]_q!^2} \, . \end{aligned} \end{equation} In particular, for $ m = 0 $, we have the identity \begin{equation}\label{IdE07} \frac{d_{n+1}(q)^2-q^{n(n+1)}}{[n+1]_q!{}^2} = 2 \sum_{k=0}^n (-1)^{k+1} q^{\binom{k+1}{2}} \frac{d_k(q)}{[k]_q! [k+1]_q!} + \sum_{k=0}^n \frac{q^{k (k-1)}}{[k]_q!{}^2} \, . \end{equation} \end{theorem} \begin{proof} By recurrences (\ref{rec-qderN01}), we have $$ \begin{aligned} \MoveEqLeft d_{m+k+1}(q)^2 = \Big( [m+k+1]_q d_{m+k}(q) + (-1)^{m+k+1} q^{{ m+k+1 \choose 2 }} \Big)^{\!2} \\ &= [m+k+1]_q^2 d_{m+k}(q)^2 + 2 (-1)^{m+k+1} q^{{ m+k+1 \choose 2 }} [m+k+1]_q d_{m+k}(q) + q^{2{ m+k+1 \choose 2 }} \, . \end{aligned} $$ Hence, we can write $$ \frac{d_{m+k+1}(q)^2}{[m+k+1]_q!^2} = \frac{d_{m+k}(q)^2}{[m+k]_q!^2} + 2 (-1)^{m+k+1} q^{{ m+k+1 \choose 2 }} \frac{d_{m+k}(q)}{[m+k]_q![m+k+1]_q!} + \frac{q^{2{ m+k+1 \choose 2 }}}{[m+k+1]_q!^2} $$ and, consequently, we have $$ \begin{aligned} \MoveEqLeft \sum_{k=0}^n \frac{d_{m+k+1}(q)^2}{[m+k+1]_q!^2} = \sum_{k=0}^n \frac{d_{m+k}(q)^2}{[m+k]_q!^2} \\ & + 2 \sum_{k=0}^n (-1)^{m+k+1} q^{{ m+k+1 \choose 2 }} \frac{d_{m+k}(q)}{[m+k]_q![m+k+1]_q!} + \sum_{k=0}^n \frac{q^{2{ m+k+1 \choose 2 }}}{[m+k+1]_q!^2}, \end{aligned} $$ or $$ \begin{aligned} \MoveEqLeft \sum_{k=1}^{n+1} \frac{d_{m+k}(q)^2}{[m+k]_q!^2} = \sum_{k=0}^n \frac{d_{m+k}(q)^2}{[m+k]_q!^2} \\ & + 2 \sum_{k=0}^n (-1)^{m+k+1} q^{{ m+k+1 \choose 2 }} \frac{d_{m+k}(q)}{[m+k]_q![m+k+1]_q!} + \sum_{k=1}^{n+1} \frac{q^{2{ m+k \choose 2 }}}{[m+k]_q!^2}\, . \end{aligned} $$ By simplifying, we get the identity $$ \begin{aligned} \MoveEqLeft \frac{d_{m+n+1}(q)^2}{[m+n+1]_q!^2} = \frac{d_{m}(q)^2}{[m]_q!^2} + 2 \sum_{k=0}^n (-1)^{m+k+1} q^{{ m+k+1 \choose 2 }} \frac{d_{m+k}(q)}{[m+k]_q![m+k+1]_q!} \\ &\hspace{2.4cm} + \sum_{k=0}^n \frac{q^{2{ m+k \choose 2 }}}{[m+k]_q!^2} + \frac{q^{2{ m+n+1 \choose 2 }}}{[m+n+1]_q!^2} - \frac{q^{2{ m \choose 2 }}}{[m]_q!^2} \end{aligned} $$ which yields identity (\ref{IdE06}) at once. \end{proof} \section{$q$-exponential series}\label{sec-qES} Many identities can be obtained by using the $q$-exponential generating series. Recall that the product of two $q$-exponential series $ f(t) = \sum_{n\geq0} f_n \frac{t^n}{[n]_q!} $ and $ g(t) = \sum_{n\geq0} g_n \frac{t^n}{[n]_q!} $ is given by $$ f(t)\cdot g(t) = \sum_{n\geq0} \left(\sum_{k=0}^n { n \choose k }_{\!\!q} f_k g_{n-k}\right) \frac{t^n}{[n]_q!}, $$ and that the \emph{$q$-derivative} (\emph{Jackson's derivative}) $ \DD_q $ of a $q$-exponential generating series $ f(t) = \sum_{n\geq0} f_n \frac{t^n}{[n]_q!} $ is defined \cite{Jackson,Jackson1908,Jackson1910} by the formula $$ \DD_q f(t) = \frac{f(qt)-f(t)}{(q-1)t} = \sum_{n\geq0} f_{n+1} \frac{t^n}{[n]_q!} \, . $$ The \emph{$q$-exponential series} (\emph{Jackson's $q$-exponential}) \cite{Jackson} \begin{equation}\label{series-E} E_q(t) = \sum_{n\geq0} \frac{t^n}{[n]_q!} = \prod_{k\geq0} \frac{1}{1+(q-1)q^kt} \end{equation} is the eigenfunction of the $q$-derivative, that is, $$ \DD_q E_q(\lambda t) = \lambda E_q(t) \, . $$ In particular, since $ \DD_q E_q(t) = E_q(t) $, we have the relation \begin{equation}\label{EEqq} E_q(qt) = (1-(1-q)t)\,E_q(t) \, . \end{equation} Consequently, considering the \emph{$q$-Pochhammer symbol} $ (x;q)_m = (1-x)(1-qx)\cdots(1-q^{m-1}x) $, we have, for every $ m \in \NN $, the identity \begin{equation}\label{mEEqq} E_q(q^mt) = \prod_{k=0}^{m-1} (1-(1-q)q^kt)\cdot E_q(t) = ((1-q)t;q)_m\, E_q(t)\, . \end{equation} Moreover, the inverse of the $q$-exponential series is \begin{equation}\label{series-Einv} E_q(t)^{-1} = \sum_{n\geq0} (-1)^n q^{{ n \choose 2 }} \frac{t^n}{[n]_q!} \end{equation} and we have the identities \cite{Ubriaco} \begin{align} E_q(-t)\, E_{q^{-1}}(t) &= 1 \label{qExp01} \\ E_q(t)\, E_q(-t) &= E_{q^2}\Big(\frac{1-q}{1+q}\,t^2\Big)\, . \label{qExp02} \end{align} By definition (\ref{def-qderN}) and series (\ref{series-Einv}), we have at once that the $q$-exponential generating series of the $q$-derangement numbers is \begin{equation}\label{series-derN} D_q(t) = \sum_{n\geq0} d_n(q) \frac{t^n}{[n]_q!} = \frac{E_q(t)^{-1}}{1-t} \, . \end{equation} We consider the following $q$-polynomials: \begin{itemize} \item the \emph{$q$-Pochhammer symbol} \begin{equation}\label{qPochhammer} (x;q)_n = (1-x)(1-qx)\cdots(1-q^{n-1}x) = \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} x^k\, , \end{equation} \item the \emph{Gaussian polynomials} \cite{GoldmanRota,GoldmanRota70,MunariniCC} $$ g_n(q;x) = (x-1)(x-q)\cdots(x-q^{n-1}) = \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^{n-k} q^{{ n-k \choose 2 }} x^k\, , $$ \item the \emph{$q$-Hermite polynomials} (or \emph{Rogers-Szeg\H{o} polynomials}) (\cite{Szego,Carlitz,AlSalamCarlitz,GoldmanRota}, \cite[p.\ 180]{Roman}) $$ H_n(q;x) = \sum_{k=0}^n { n \choose k }_{\!\!q} x^k $$ and the \emph{Galois numbers} \cite{GoldmanRota,NijenhuisSolowWilf} \begin{equation}\label{GaloisN} G_n(q) = \sum_{k=0}^n { n \choose k }_{\!\!q}\, , \end{equation} \item the \emph{$q$-Carlitz polynomials} (or \emph{Al-Salam-Carlitz polynomials}) (\cite{AlSalamCarlitzU}, \cite[p.\ 195]{Chihara}, \cite{Ismail,ChenSaadSun,Kim}) $$ U_n^{(\alpha)}(q;x) = \sum_{k=0}^n { n \choose k }_{\!\!q} (-\alpha)^{n-k} g_k(x)\, , $$ \end{itemize} having $q$-exponential generating series \begin{align} P_q(x,t) &= \sum_{n\geq0} (x;q)_n \frac{t^n}{[n]_q!} = \frac{E_q(t)}{E_q(xt)} = E_q(t)\, E_q(xt)^{-1} \label{series-P} \\ g_q(x,t) &= \sum_{n\geq0} g_n(q;x) \frac{t^n}{[n]_q!} = \frac{E_q(xt)}{E_q(t)} = E_q(t)^{-1} E_q(xt) \label{series-g} \\ H_q(x,t) &= \sum_{n\geq0} H_n(q;x) \frac{t^n}{[n]_q!} = E_q(t)\, E_q(xt) \label{series-G} \\ G_q(t) &= \sum_{n\geq0} G_n(q) \frac{t^n}{[n]_q!} = E_q(t)^2 \label{series-Galois} \\ U_q(x,t) &= \sum_{n\geq0} U_n^{(\alpha)}(q;x) \frac{t^n}{[n]_q!} = \frac{E_q(xt)}{E_q(t)E_q(\alpha t)} = E_q(\alpha t)^{-1} g_q(x,t) \, . \label{series-U} \end{align} Using the properties of the $q$-exponential series, we have at once the following results. \begin{theorem} We have the identities \begin{align} \sum_{k=0}^n { n \choose k }_{\!\!q} d_{n-k}(q)\, (x;q)_k &= \sum_{k=0}^n { n \choose k }_{\!\!q} [n-k]_q!\, (-1)^k q^{{ k \choose 2 }} x^k \label{IdqP00} \\ \sum_{k=0}^n { n \choose k }_{\!\!q} d_{n-k}(q)\, x^k &= \sum_{k=0}^n { n \choose k }_{\!\!q} [n-k]_q!\, g_k(q;x) \label{IdqP01} \\ \sum_{k=0}^n { n \choose k }_{\!\!q} d_{n-k}(q)\, G_k(q,x) &= \sum_{k=0}^n { n \choose k }_{\!\!q} [n-k]_q!\, x^k \label{IdqP02} \\ \sum_{k=0}^n { n \choose k }_{\!\!q} \alpha^{n-k} d_{n-k}(q)\, g_k(x) &= \sum_{k=0}^n { n \choose k }_{\!\!q} \alpha^{n-k} [n-k]_q!\, U^{(\alpha)}_k(q;x) \label{IdqP03} \, . \end{align} \end{theorem} \begin{proof} By series (\ref{series-derN}), (\ref{series-P}), (\ref{series-g}), (\ref{series-G}), (\ref{series-U}) and (\ref{series-Einv}), we have the identities \begin{align*} & D_q(t)\, P_q(x,t) = \frac{E_q(xt)^{-1}}{1-t} \\ & D_q(t)\, E_q(x,t) = \frac{g_q(x,t)}{1-t} \\ & D_q(t)\, H_q(x,t) = \frac{E_q(xt)}{1-t} \\ & D_q(\alpha t)\, g_q(x,t) = \frac{U_q^{(\alpha)}(xt)}{1-\alpha t} \end{align*} which are equivalent to identities (\ref{IdqP00}), (\ref{IdqP01}), (\ref{IdqP02}) and (\ref{IdqP03}), respectively. \end{proof} Moreover, we also have the next result. \begin{theorem} We have the identity \begin{equation}\label{IdqQ1} \sum_{k=0}^n (-1)^k \frac{d_k(q)}{[k]_q!}\,\frac{d_{n-k}(q^{-1})}{[n-k]_{q^{-1}}!} = \frac{1+(-1)^n}{2} \end{equation} or, equivalently, \begin{equation}\label{IdqQ2} \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k d_k(q)\, d_{n-k}(q^{-1}) = \frac{1+(-1)^n}{2}\,[n]_q!\, . \end{equation} \end{theorem} \begin{proof} By identity (\ref{qExp01}), we have $$ D_q(-t)\, D_{q^{-1}}(t) = \frac{E_q(-t)\, E_{q^{-1}}(t)}{(1-t)(1+t)} = \frac{1}{1-t^2}\, . $$ Now we have $$ D_q(-t)\, D_{q^{-1}}(t) = \sum_{i\geq0} (-1)^i d_i(q) \frac{t^i}{[i]_q!} \sum_{j\geq0} d_j(q^{-1}) \frac{t^j}{[j]_{q^{-1}}!} = \sum_{i,j\geq0} (-1)^i \frac{d_i(q)}{[i]_q!}\,\frac{d_j(q^{-1})}{[j]_{q^{-1}}!}\;t^{i+j} \, . $$ Setting $ i+j=n $ and replacing $i $ by $ k $, we have $$ D_q(-t)\, D_{q^{-1}}(t) = \sum_{n\geq0} \left( \sum_{k=0}^n (-1)^k \frac{d_k(q)}{[k]_q!}\,\frac{d_{n-k}(q^{-1})}{[n-k]_{q^{-1}}!}\right) t^n \, . $$ Hence, we have the identity $$ \sum_{n\geq0} \left( \sum_{k=0}^n (-1)^k \frac{d_k(q)}{[k]_q!}\,\frac{d_{n-k}(q^{-1})}{[n-k]_{q^{-1}}!}\right) t^n = \sum_{n\geq0} \frac{1+(-1)^n}{2}\; t^n $$ and this yields identity (\ref{IdqQ1}). This identity and $ [n]_{q^{-1}}! = [n]_q!\, q^{-{ n \choose 2 }} $, immediately yield identity (\ref{IdqQ2}). \end{proof} \begin{theorem} We have the identity \begin{equation}\label{IdqConv} \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k d_k(q)\, d_{n-k}(q) = \frac{1+(-1)^n}{2} \sum_{k=0}^{n/2} (-1)^k q^{k^2-k} \left(\frac{1-q}{1+q}\right)^{\!\!k} \frac{[n]_q!}{[k]_{q^2}!}\, . \end{equation} \end{theorem} \begin{proof} By identity (\ref{qExp02}), we have \begin{align*} D_q(t) D_q(-t) &= \frac{E_q(t)^{-1}E_q(-t)^{-1}}{(1-t)(1+t)} \\ &= \frac{1}{1-t^2}\; E_{q^2}\Big(\frac{1-q}{1+q}\,t^2\Big)^{\!\!-1} \\ &= \sum_{i\geq0} t^{2i}\cdot \sum_{k\geq0} (-1)^k q^{2{ k \choose 2 }} \left(\frac{1-q}{1+q}\right)^{\!\!k} \frac{t^{2k}}{[k]_{q^2}!} \\ &= \sum_{i,k\geq0} (-1)^k q^{k^2-k} \left(\frac{1-q}{1+q}\right)^{\!\!k} \frac{[2i+2k]_q!}{[k]_{q^2}!} \frac{t^{2i+2k}}{[2i+2k]_q!} \\ &= \sum_{n\geq0} \left(\sum_{k=0}^n (-1)^k q^{k^2-k} \left(\frac{1-q}{1+q}\right)^{\!\!k} \frac{[2n]_q!}{[k]_{q^2}!}\right) \frac{t^{2n}}{[2n]_q!} \\ &= \sum_{n\geq0} \left(\frac{1+(-1)^n}{2}\sum_{k=0}^{n/2} (-1)^k q^{k^2-k} \left(\frac{1-q}{1+q}\right)^{\!\!k} \frac{[n]_q!}{[k]_{q^2}!}\right) \frac{t^n}{[n]_q!} \, . \end{align*} Taking the coefficients of $ \frac{t^n}{[n]_q!} $ in the first and in the last series, we have identity (\ref{IdqConv}). \end{proof} Recall that for the $q$-binomial coefficients we have the $q$-series \begin{equation}\label{series-qBin} \sum_{n\geq0} { m+n \choose m }_{\!\!q} t^n = \frac{1}{(1-t)(1-qt)(1-q^2t)\cdots(1-q^mt)} = \frac{1}{(t;q)_{m+1}}\, . \end{equation} Then we have the following result. \begin{theorem} For every $ m, n \in \NN $, we have the identities \begin{align} \sum_{k=0}^n { n \choose k }_{\!\!q} { m+k \choose m }_{\!\!q} [k]_q! q^k d_{n-k}(q) &= \sum_{k=0}^n { n \choose k }_{\!\!q} { m+k+1 \choose m+1 }_{\!\!q} (-1)^{n-k} [k]_q! q^{{ n-k \choose 2 }} \label{IdBB01} \\ \sum_{k=0}^n { n \choose k }_{\!\!q} { \alpha+k \choose k } [k]_q! d_{n-k}(q) &= \sum_{k=0}^n { n \choose k }_{\!\!q} { \alpha+k+1 \choose k } (-1)^{n-k} [k]_q! q^{{ n-k \choose 2 }} \label{IdBB02} \, . \end{align} \end{theorem} \begin{proof} From the $q$-series (\ref{series-qBin}), we have the $q$-exponential series \begin{align*} \frac{1}{(1-t)(1-qt)\cdots(1-q^mt)} &= \sum_{n\geq0} { m+n \choose m }_{\!\!q} [n]_q! \frac{t^n}{[n]_q!} \\ \frac{1}{(1-qt)(1-q^2t)\cdots(1-q^{m+1}t)} &= \sum_{n\geq0} { m+n \choose m }_{\!\!q} [n]_q! q^n \frac{t^n}{[n]_q!} \, . \end{align*} Then, by formula (\ref{series-derN}), we have the identity $$ \frac{D_q(t)}{(1-qt)(1-q^2t)\cdots(1-q^{m+1}t)} = \frac{E_q(t)^{-1}}{(1-t)(1-qt)(1-q^2t)\cdots(1-q^{m+1}t)} $$ which is equivalent to identity (\ref{IdBB01}). Similarly, we have the identity $$ \frac{D_q(t)}{(1-t)^{\alpha+1}} = \frac{E_q(t)^{-1}}{(1-t)^{\alpha+2}} $$ which is equivalent to identity (\ref{IdBB02}). \end{proof} To prove the next theorem, we need the following result. \begin{lemma} We have the $q$-exponential series \begin{equation}\label{seriesE-2} E_q(t)^{-2} = \sum_{n\geq0} (-1)^n q^{{ n \choose 2 }}\, G_n(q^{-1})\,\frac{t^n}{[n]_q!} \, . \end{equation} \end{lemma} \begin{proof} By formula (\ref{series-Einv}) and relations (\ref{qBinRel}), the coefficient of $ \frac{t^n}{[n]_q!} $ in the $q$-exponential series $ E_q(t)^{-2} $ is $$ \sum_{k=0}^n { n \choose k }_{\!\!q} (-1)^k q^{{ k \choose 2 }} (-1)^{n-k} q^{{ n-k \choose 2 }} = (-1)^n q^{{ n \choose 2 }} \sum_{k=0}^n { n \choose k }_{\!\!q} q^{-k(n-k)} = (-1)^n q^{{ n \choose 2 }} \sum_{k=0}^n { n \choose k }_{\!\!q^{-1}}. $$ By the definition (\ref{GaloisN}) of the Galois numbers, this implies identity (\ref{seriesE-2}). \end{proof} Recall that the \emph{$q$-multiset coefficients} are defined by $$ \Mchoose{n}{k}_{\!\!q} = \begin{cases} \displaystyle { n+k-1 \choose k }_{\!\!q}, & \mbox{if } k \geq 1; \\ 1, & \mbox{if } k = 0 \end{cases} $$ and that they have $q$-generating series \begin{equation}\label{series-qMulti} \sum_{k\geq0} \Mchoose{n}{k}_{\!\!q} t^k = \frac{1}{(1-t)(1-qt)(1-q^2t)\cdots(1-q^{n-1}t)} = \frac{1}{(t;q)_n}\, . \end{equation} \begin{theorem} For every $ m, n \in \NN $, we have the identity \begin{equation}\label{IdW} \sum_{k=0}^n { n \choose k }_{\!\!q} q^{mk}\, d_k(q) = [n]_q! \sum_{k=0}^n \Mchoose{m}{k}_{\!\!q} (1-q)^k q^{m(n-k)} \, . \end{equation} \end{theorem} \begin{proof} We have the $q$-exponential series $$ L(q;t) = \sum_{n\geq0} \left( \sum_{k=0}^n { n \choose k }_{\!\!q} q^{mk}\, d_k(q) \right) \frac{t^n}{[n]_q!} = E_q(t) D_q(q^mt) = \frac{E_q(t) E_q(q^mt)^{-1}}{1-q^mt}\, . $$ By identity (\ref{mEEqq}), we have \begin{align*} L(q;t) &= \frac{E_q(t) E_q(t)^{-1}}{(1-q^mt)((1-q)t;q)_m} = \frac{1}{1-q^mt}\cdot \frac{1}{((1-q)t;q)_m} \\ &= \sum_{n\geq0} \left( [n]_q!\sum_{k=0}^n \Mchoose{m}{k}_{\!\!q} (1-q)^k q^{m(n-k)} \right) \frac{t^n}{[n]_q!} \end{align*} from which we have at once identity (\ref{IdW}). \end{proof} We conclude this section proving the following elementary identity involving the $q$-Pochhammer symbol. \begin{theorem} We have the identity \begin{equation}\label{IdZ} \frac{d_{n+2}(q)}{[n+1]_q!}\, \frac{x^{n+1}}{(qx;q)_{n+1}} + \sum_{k=0}^n \frac{d_{k+1}(q)+d_k(q)}{[k]_q!}\, \frac{x^k}{(qx;q)_k} = \sum_{k=0}^n \frac{ d_{k+1}(q)\,x^{k+1}+d_k(q)\,x^k}{[k]_q!(qx;q)_{k+1}} \, . \end{equation} \end{theorem} \begin{proof} By recurrence (\ref{rec-qderN02}), we have \begin{align*} d_{k+2}(q)\, x &= [k+1]_q\, d_{k+1}(q)\, x + [k+1]_q\, q^{k+1} x \, d_k(q) \\ &= [k+1]_q\, d_{k+1}(q)\, x + [k+1]_q \, d_k(q) - [k+1]_q\, (1-q^{k+1} x) \, d_k(q), \end{align*} and consequently $$ \frac{d_{k+2}(q)}{[k+1]_q!}\, \frac{x^{k+1}}{(qx;q)_{k+1}} = \frac{ d_{k+1}(q)}{[k]_q!}\, \frac{x^{k+1}}{(qx;q)_{k+1}} + \frac{d_k(q)}{[k]_q!}\, \frac{x^k}{(qx;q)_{k+1}} - \frac{d_k(q)}{[k]_q!}\, \frac{x^k}{(qx;q)_k} \, . $$ Then we have $$ \sum_{k=0}^n \frac{d_{k+2}(q)}{[k+1]_q!}\, \frac{x^{k+1}}{(qx;q)_{k+1}} = \sum_{k=0}^n \frac{ d_{k+1}(q)}{[k]_q!}\, \frac{x^{k+1}}{(qx;q)_{k+1}} + \sum_{k=0}^n \frac{d_k(q)}{[k]_q!}\, \frac{x^k}{(qx;q)_{k+1}} - \sum_{k=0}^n \frac{d_k(q)}{[k]_q!}\, \frac{x^k}{(qx;q)_k} \, , $$ which is $$ \sum_{k=1}^{n+1} \frac{d_{k+1}(q)}{[k]_q!}\, \frac{x^k}{(qx;q)_k} + \sum_{k=0}^n \frac{d_k(q)}{[k]_q!}\, \frac{x^k}{(qx;q)_k} = \sum_{k=0}^n \frac{ d_{k+1}(q)}{[k]_q!}\, \frac{x^{k+1}}{(qx;q)_{k+1}} + \sum_{k=0}^n \frac{d_k(q)}{[k]_q!}\, \frac{x^k}{(qx;q)_{k+1}} \, , $$ or $$ \frac{d_{n+2}(q)}{[n+1]_q!}\, \frac{x^{n+1}}{(qx;q)_{n+1}} + \sum_{k=0}^n \frac{d_{k+1}(q)+d_k(q)}{[k]_q!}\, \frac{x^k}{(qx;q)_k} = \sum_{k=0}^n \frac{ d_{k+1}(q)\,x^{k+1}+d_k(q)\,x^k}{[k]_q!(qx;q)_{k+1}} \, . $$ This is the claimed identity. \end{proof} \section{Determinantal identities}\label{sec-det} Since the $q$-derangement numbers satisfy a three-term recurrence, they can be represented in terms of \emph{tridiagonal determinants} (or \emph{continuants} (\cite[pp.\ 516--525]{Muir}, \cite{VeinDale})). \begin{theorem} We have the identity \begin{equation}\label{d-tridiagonal} d_n(q) = \begin{vmatrix} [0]_q & -q \\ [1]_q & [1]_q & -q^2 \\ & [2]_q & [2]_q & -q^3 \\ & & \ddots & \ddots & \ddots \\ & & & [n-2]_q & [n-2]_q & -q^{n-1} \\ & & & & [n-1]_q & [n-1]_q \end{vmatrix}_{n\times n}\hspace{-5mm}. \end{equation} \end{theorem} \begin{proof} The tridiagonal determinants in formula (\ref{d-tridiagonal}) satisfy recurrence (\ref{rec-qderN02}) with the appropriate initial values. This implies at once the claimed identity. \end{proof} The $q$-derangement numbers can also be represented in terms of \emph{Hessenberg determinants} \cite[p.\ 90]{VeinDale}, as follows. \begin{theorem} Consider the $ n \times n $ lower Hessenberg matrix $$ A_n(q) = \begin{bmatrix} a_{00}(q) & -1 & 0 & 0 & \cdots & 0 \\ a_{10}(q) & a_{11}(q) & -1 & 0 & \cdots & 0 \\ a_{20}(q) & a_{21}(q) & a_{22}(q) & -1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{n-2,1}(q) & a_{n-2,2}(q) & a_{n-2,3}(q) & a_{n-2,4}(q) & \cdots & -1 \\ a_{n-1,1}(q) & a_{n-1,2}(q) & a_{n-1,3}(q) & a_{n-1,4}(q) & \cdots & a_{n-1,n-1}(q) \end{bmatrix} $$ where $$ a_{i,j}(q) = \begin{cases} \displaystyle { i \choose j }_{\!\!q} a_{i-j}(q), & \mbox{if } i \geq j; \\ -1, & \mbox{if } i = j-1; \\ 0, & \mbox{otherwise} \end{cases} $$ where \begin{equation}\label{coeff-a} a_k(q) = \frac{q}{2q-1}\;(q^n-(1-q)^n)\,[n]_q!\, . \end{equation} Then we have the identity \begin{equation}\label{formula-Fx-det} d_n(q) = \det A_n(q) \, . \end{equation} \end{theorem} \begin{proof} Let $ b_n(q) = \det A_n(q) $. By expanding the determinant along the last column, we get the recurrence $$ b_{n+1}(q) = \sum_{k=0}^n { n \choose k }_{\!\!q} a_k(q) b_{n-k}(q) $$ with initial value $ b_0(q) = 1 $. Hence, considering the $q$-exponential generating series $$ a(q;t) = \sum_{n\geq0} a_n(q) \frac{t^n}{[n]_q!} \qquad\text{and}\qquad b(q;t) = \sum_{n\geq0} b_n(q) \frac{t^n}{[n]_q!} \, , $$ we have the $q$-differential equation $$ \DD_q b(q;t) = a(q;t)\, b(q;t) \, . $$ If $ b(q;t) = D_q(t) $, then $ b_0(q) = d_0(q) = 1 $, as requested, and $$ a(q;t) = \frac{\DD_q b(q;t)}{b(q;t)} = \frac{\DD_q D_q(t)}{D_q(t)} \, . $$ By series (\ref{series-derN}) and relation (\ref{EEqq}), we have \begin{align*} \DD_q D_q(t) &= \frac{D_q(qt)-D_q(t)}{(q-1)t} \\ &= \frac{1}{(q-1)t}\left( \frac{E_q(qt)^{-1}}{1-qt} - \frac{E_q(t)^{-1}}{1-t} \right) \\ &= \frac{1}{(q-1)t}\left( \frac{E_q(t)^{-1}}{(1-qt)(1+(q-1)t)} - \frac{E_q(t)^{-1}}{1-t} \right) \\ &= \frac{qt\,E_q(t)^{-1}}{(1-t)(1-qt)(1+(q-1)t)}, \end{align*} which is $$ \DD_q D_q(t) = \frac{qt}{(1-qt)(1+(q-1)t)}\;D_q(t) \, . $$ Therefore, we have $$ a(q;t) = \frac{qt}{(1-qt)(1+(q-1)t)} = \frac{q}{2q-1}\,\frac{1}{1-qt} - \frac{q}{2q-1}\,\frac{1}{1-(1-q)t}\, . $$ This decomposition yields identity (\ref{coeff-a}), and, consequently, this proves identity (\ref{formula-Fx-det}). \end{proof} \section{Final remarks} In the literature, there are also other $q$-analogues for the derangement numbers. For instance, we have the $q$-derangement numbers \cite{GarsiaRemmel} \begin{equation}\label{def-qderNN} D_n(q) = \sum_{k=0}^n { n \choose k }_{\!\!q} [n-k]_q!\, (-1)^k \end{equation} satisfying the recurrences \begin{align} & D_{n+1}(q) = [n+1]_q\, D_n(q) + (-1)^{n+1} \label{rec-qderNN01} \\ & D_{n+2}(q) = q\, [n+1]_q\, D_{n+1}(q) + [n+1]_q\, D_n(q) \label{rec-qderNN02} \end{align} with initial conditions $ D_0(q) = 1 $ and $ D_1(q) = 0 $. Moreover, they have $q$-exponential generating series $$ \sum_{n\geq0} D_n(q) \frac{t^n}{[n]_q!} = \frac{E_q(-t)}{1-t} \, . $$ The $q$-numbers $ d_n(q) $ and $ D_n(q) $ are not independent, as shown in the next theorem. \begin{theorem} For every $ n \in \NN $, we have the relation \begin{equation}\label{dnqDnq} d_n(q^{-1}) = q^{-{ n \choose 2 }} D_n(q) \, . \end{equation} Moreover, we have the formulas \begin{align} & d_n(q) = \sum_{k=0}^n (-1)^{n-k}q^{{ n-k \choose 2 }}\frac{(q^{n-k+1};q)_k}{(1-q)^k}\label{qDqPoch}\\ & D_n(q) = \sum_{k=0}^n (-1)^{n-k}\frac{(q^{n-k+1};q)_k}{(1-q)^k}\label{qDDqPoch}\, . \end{align} \end{theorem} \begin{proof} By formula (\ref{def-qderN}) and relations (\ref{qBinRel}), we have \begin{align*} d_n(q^{-1}) &= \sum_{k=0}^n { n \choose k }_{\!\!q^{-1}} [n-k]_{q^{-1}}!\, (-1)^k q^{-{ k \choose 2 }} \\ &= \sum_{k=0}^n { n \choose k }_{\!\!q} q^{-k(n-k)} [n-k]_q! q^{-{ n-k \choose 2 }}\, (-1)^k q^{-{ k \choose 2 }} \\ &= q^{-{ n \choose 2 }} \sum_{k=0}^n { n \choose k }_{\!\!q} [n-k]_q!\, (-1)^k \, . \end{align*} By formula (\ref{def-qderNN}), this is relation (\ref{dnqDnq}). Since $ [n]_q! = \frac{(q;q)_n}{(1-q)^n} $, from definition (\ref{def-qderN}) we have $$ d_n(q) = \sum_{k=0}^n \frac{[n]_q!}{[k]_q!}\, (-1)^k q^{{ k \choose 2 }} = \sum_{k=0}^n \frac{(-1)^k q^{{ k \choose 2 }}}{(1-q)^{n-k}}\,\frac{(q;q)_n}{(q;q)_k} = \sum_{k=0}^n \frac{(-1)^k q^{{ k \choose 2 }}}{(1-q)^{n-k}}\,(q^{k+1};q)_{n-k} \, . $$ This is equivalent to identity (\ref{qDqPoch}). Similarly, formula (\ref{def-qderNN}) can be rewritten as formula (\ref{qDDqPoch}). \end{proof} We also have the following result. \begin{theorem} We have the ordinary generating series \begin{align} & d(q;t) = \sum_{n\geq0} d_n(q)\,t^n = \sum_{k\geq0} \frac{(-1)^k q^{{ k+1 \choose 2 }}}{(1-q)^k}\, \frac{t^kQ(q;q^kt)}{\big(\frac{t}{1-q};q\big)_{k+1}} \label{qseriesD} \\ & D(q;t) = \sum_{n\geq0} D_n(q)\,t^n = \sum_{k\geq0} \frac{(-1)^k q^{{ k+1 \choose 2 }}}{(1-q)^k}\, \frac{t^k}{(1+q^kt)\big(\frac{t}{1-q};q\big)_{k+1}} \label{qseriesDD} \end{align} where $$ Q(q;t) = \sum_{n\geq0} (-1)^n q^{{ n \choose 2 }}\, t^n \, . $$ \end{theorem} \begin{proof} From recurrence (\ref{rec-qderN01}), we have $$ \sum_{n\geq0} d_{n+1}(q) \, t^n = \sum_{n\geq0} \frac{1-q^{n+1}}{1-q}\, d_n(q) \, t^n + \sum_{n\geq0} (-1)^{n+1} q^{{ n+1 \choose 2 }} \, t^n, $$ which is $$ \frac{d(q;t)-d_0(q)}{t} = \frac{1}{1-q} ( d(q;t ) - q d(q;qt) ) + \frac{Q(q;t)-1}{t}, $$ or $$ d(q;t) - 1 = \frac{t}{1-q} d(q;t) - \frac{qt}{1-q}\, d(q;qt) + Q(q;t) - 1, $$ or $$ \left( 1 - \frac{t}{1-q} \right) d(q;t) = Q(q;t) - \frac{qt}{1-q}\, d(q;qt) ,$$ which is $$ d(q;t) = \frac{Q(q;t)}{1-\frac{t}{1-q}} - \frac{qt}{(1-q)\big(1-\frac{t}{1-q}\big)}\, d(q;qt)\, . $$ By repeatedly applying this formula, we get \begin{align*} d(q;t) &= \sum_{k=0}^n \frac{(-1)^kq^{{ k+1 \choose 2 }}}{(1-q)^k} \frac{t^k\,Q(q;q^kt)}{\big(1-\frac{t}{1-q}\big)(1-\frac{qt}{1-q}\big)\cdots(1-\frac{q^kt}{1-q}\big)} \\ &\qquad\qquad + \frac{(-1)^{n+1}q^{{ n+2 \choose 2 }}}{(1-q)^{n+1}} \frac{t^{n+1}} {\big(1-\frac{t}{1-q}\big)(1-\frac{qt}{1-q}\big)\cdots(1-\frac{q^nt}{1-q}\big)}\, d(q;q^{n+1}t) \\ &= \sum_{k=0}^n \frac{(-1)^kq^{{ k+1 \choose 2 }}}{(1-q)^k}\frac{t^k\,Q(q;q^kt)}{\big(\frac{t}{1-q};q\big)_{k+1}} + \frac{(-1)^{n+1}q^{{ n+2 \choose 2 }}}{(1-q)^{n+1}}\frac{t^{n+1}}{\big(\frac{t}{1-q};q\big)_{n+1}}\, d(q;q^{n+1}t) \, . \end{align*} Taking the limit for $ n $ tending to $ +\infty $, we obtain series (\ref{qseriesD}). Similarly, from recurrence (\ref{rec-qderNN01}), we have $$ \sum_{n\geq0} D_{n+1}(q)\, t^n = \sum_{n\geq0} \frac{1-q^{n+1}}{1-q}\, D_n(q)\, t^n + \sum_{n\geq0} (-1)^{n+1} t^n, $$ which is $$ \frac{D(q;t)-D_0(q)}{t} = \frac{1}{1-q} ( D(q;t ) - q D(q;qt) ) - \frac{1}{1+t}, $$ or $$ \left(1-\frac{t}{1-q}\right) D(q;t) = \frac{1}{1+t} - \frac{qt}{1-q} D(q;qt) , $$ or $$ D(q;t) = \frac{1}{(1+t)\big(1-\frac{t}{1-q}\big) } - \frac{qt}{(1-q)\big(1-\frac{t}{1-q}\big) } D(q;qt). $$ By repeatedly applying this formula, we get \begin{align*} D(q;t) &= \sum_{k=0}^n \frac{(-1)^kq^{{ k+1 \choose 2 }}}{(1-q)^k} \frac{t^k} {(1+q^kt)\big(1-\frac{t}{1-q}\big)(1-\frac{qt}{1-q}\big)\cdots(1-\frac{q^kt}{1-q}\big)} \\ &\qquad\qquad + \frac{(-1)^{n+1}q^{{ n+2 \choose 2 }}}{(1-q)^{n+1}}\frac{t^{n+1}} {\big(1-\frac{t}{1-q}\big)(1-\frac{qt}{1-q}\big)\cdots(1-\frac{q^nt}{1-q}\big)}\, D(q;q^{n+1}t) \\ &= \sum_{k=0}^n \frac{(-1)^kq^{{ k+1 \choose 2 }}}{(1-q)^k}\frac{t^k}{(1+q^kt)\big(\frac{t}{1-q};q\big)_{k+1}} \\ &\qquad\qquad + \frac{(-1)^{n+1}q^{{ n+2 \choose 2 }}}{(1-q)^{n+1}}\frac{t^{n+1}}{\big(\frac{t}{1-q};q\big)_{n+1}}\, D(q;q^{n+1}t) \, . \end{align*} Taking the limit for $ n $ tending to $ +\infty $, we obtain series (\ref{qseriesDD}). \end{proof} \begin{thebibliography}{99} \bibitem{AlSalamCarlitz} W. A. Al-Salam and L. Carlitz, A $q$-analog of a formula of Toscano, \emph{Boll. Un. Mat. Ital.} (3) \textbf{12} (1957), 414--417. \bibitem{AlSalamCarlitzU} W. A. Al-Salam and L. Carlitz, Some orthogonal $q$-polynomials, \emph{Math. Nachr.} \textbf{30} (1965), 47-61. \bibitem{CFMunariniZ} S. Capparelli, M. M. Ferrari, E. Munarini, and N. Zagaglia Salvi, A generalization of the ``\emph{Probl\`eme des Rencontres}'', \emph{J. Integer Sequences} \textbf{21} (2018), \href{https://cs.uwaterloo.ca/journals/JIS/VOL21/Munarini/muna8.html}{Article 18.2.8}. \bibitem{Carlitz} L. Carlitz, Some polynomials related to the theta functions, \emph{Ann. Mat. Pura Appl.} \textbf{41} (1955), 359--373. \bibitem{ChenRota} W. Y. C. Chen and G.-C. Rota, $q$-analogs of the inclusion-exclusion principle and permutations with restricted position, \emph{Discrete Math.} \textbf{104} (1992), 7--22. \bibitem{ChenSaadSun} W. Y. C. Chen, H. L. Saad, and L. H. Sun, An operator approach to the Al-Salam–Carlitz polynomials, \emph{J. Math. Phys.} \textbf{51} (2010), Paper 043502. \bibitem{ChenXia} W. Y. C. Chen and E. X. W. Xia, The ratio monotonicity of the $q$-derangement numbers, \emph{Discrete Math.} \textbf{311} (2011), 393--397. \bibitem{Chihara} T. S. Chihara, \emph{An Introduction to Orthogonal Polynomials}, Gordon and Breach, 1978. \bibitem{MunariniCC} O. M. D'Antona and E. Munarini, A combinatorial interpretation of the connection constants for persistent sequences of polynomials, \emph{European J. Combin.} \textbf{26} (2005), 1105--1118. \bibitem{FerrariMunarini} M. M. Ferrari and E. Munarini, Decomposition of some Hankel matrices generated by the generalized rencontres polynomials, \emph{Linear Algebra Appl.} \textbf{567} (2019), 180-–201. \bibitem{GarsiaRemmel} A. M. Garsia and J. Remmel, A combinatorial interpretation of $q$-derangement and $q$-Laguerre numbers, \emph{Europ. J. Combin.} \textbf{1} (1980), 47--5. \bibitem{Gauss} C. F. Gauss, Summatio quarumdam serierum singularium, \emph{Comm. soc. reg. sci. Gottingensis rec.} \textbf{1} (1808), 147--186. Reprinted in Gauss's {\it Werke} \textbf{2} (1863), 9--45. \bibitem{GoldmanRota} J. Goldman and G.-C. Rota, The number of subspaces of a vector space, \emph{Recent Progress in Combinatorics} (Proc. Third Waterloo Conf. on Combinatorics, 1968) Academic Press, 1969, pp.~75--83. \bibitem{GoldmanRota70} J. Goldman and G.-C. Rota, On the foundations of combinatorial theory. IV. Finite vector spaces and Eulerian generating functions, \emph{Studies in Appl. Math.} \textbf{49} (1970), 239--258. \bibitem{Ismail} M. E. H. Ismail, \emph{Classical and Quantum Orthogonal Polynomials in One Variable}, Cambridge University Press, 2005. \bibitem{Jackson} F. H. Jackson, A basic-sine and cosine with symbolical solutions of certain differential equations, \emph{Proc. Edinburgh Math. Soc.} \textbf{22} (1904), 28--39. \bibitem{Jackson1908} F. H. Jackson, On $q$-functions and a certain difference operator, \emph{Trans. Roy. Soc. Edin.} \textbf{46} (1908), 253--281. \bibitem{Jackson1910} F. H. Jackson, $q$-Difference equations, \emph{Am. J. Math.} \textbf{32} (1910), 305--314. \bibitem{Kim} D. Kim, On combinatorics of Al-Salam Carlitz polynomials, \emph{European J. Combin.} \textbf{18} (1997), 295--302. \bibitem{Muir} T. Muir, \emph{The Theory of Determinants in the Historical Order of Developement}, Dover, 1960. \bibitem{MunariniSP} E. Munarini, Shifting property for Riordan, Sheffer and connection constants matrices, \emph{J. Integer Sequencesf} \textbf{20} (2017), \href{https://cs.uwaterloo.ca/journals/JIS/VOL20/Munarini/muna4.html}{Article 17.8.2}. \bibitem{Munarini-qFiboSP} E. Munarini, Some $q$-binomial identities involving the generalized $q$-Fibonacci numbers, \emph{Integers} \textbf{18} (2018), Paper No.~A76. \bibitem{MunariniA} E. Munarini, Combinatorial identities for Appell polynomials, \emph{Appl. Anal. Discrete Math.} \textbf{12} (2018), 362--388. \bibitem{MunariniC} E. Munarini, Callan-like identities, \emph{Online J. Anal. Comb.} \textbf{14} (2019), Article \#06. \bibitem{MunariniO} E. Munarini, An operational approach to the generalized rencontres polynomials, \emph{Kragujevac J. Math.} \textbf{46} (2022), 461--469. \bibitem{NijenhuisSolowWilf} A. Nijenhuis, A. E. Solow, and H. S. Wilf, Bijective methods in the theory of finite vector spaces, \emph{J. Combin. Theory Ser. A} \textbf{37} (1984), 80--84. \bibitem{Roman} S. Roman, \emph{The Umbral Calculus}, Academic Press, 1984. \bibitem{Sloane} N. J. A. Sloane et al., \emph{The On-Line Encyclopedia of Integer Sequences}, 2020. Available at \url{https://oeis.org/}. \bibitem{Szego} G. Szeg\H{o}, \emph{Orthogonal Polynomials}, Amer. Math. Soc. Colloquium Publication \textbf{23}, Amer. Math. Soc., 1975. \bibitem{Ubriaco} M. R. Ubriaco, Time evolution in quantum mechanics on the quantum line, \emph{Phys. Lett.} A \textbf{163} (1992), 1--4. \bibitem{VeinDale} R. Vein and P. Dale, \emph{Determinants and Their Applications in Mathematical Physics}, Springer-Verlag, 1999. \bibitem{Wachs} M. L. Wachs, On $q$-derangement numbers, \emph{Proc. Amer. Math. Soc.} \textbf{106} (1989), 273--278. \bibitem{Zhang} X.-D. Zhang, On the spiral property of the $q$-derangement numbers, \emph{Discrete Math.} \textbf{159} (1996), 295--298. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 05A30; Secondary 11B65, 05A19, 11B73, 33B10, 39A13. \noindent \emph{Keywords: } combinatorial sum, $q$-binomial sum, $q$-derangement number, $q$-Stirling number, $q$-Bell number, connection constant, $q$-exponential generating series. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A000110} and \seqnum{A000166}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received October 27 2019; revised version received February 18 2020; March 11 2020. Published in {\it Journal of Integer Sequences}, March 17 2020. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}