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\begin{document}

\title{Factorial Euler--Seidel matrix: Refinement and $q$-analogue}

\newbox\Adr
\setbox\Adr\vbox{
\centerline{\sc
Dimbiniaina Ratovilebamboavison$^\ast$, Mamy Laingo Nomenjanahary}
\centerline{\sc Rakotoarison$^\star$, and Fanja Rakotondrajao$^\diamond$}
\vskip18pt
\centerline{\footnotesize
$^\ast$ISTRALMA Ambatondrazaka, Universit\'e de Toamasina, Toamasina 503, Madagascar}
\centerline{\footnotesize Email: \tt ratovilebamboavison@gmail.com}
\centerline{\footnotesize $^\star$D\'epartement de Math\'ematiques et Informatiques, Universit\'e d'Antananarivo, Antananarivo 101, Madagascar}
\centerline{\footnotesize Email: \tt rmlnomenjanahary@gmail.com}
\centerline{\footnotesize $^\diamond$D\'epartement de Math\'ematiques et Informatiques, Universit\'e d'Antananarivo, Antananarivo 101, Madagascar}
\centerline{\footnotesize Email: \tt frakoton@yahoo.fr}
}

\author[D. Ratovilebamboavison, M. L. N. Rakotoarison and F. Rakotondrajao]
{\box\Adr}

\begin{abstract}
We determine the Riordan matrix associated to  Eulerian numbers and establish a formula defining Eulerian polynomials as well. We refine the factorial Euler--Seidel matrix considering the excedance distribution over permutations and derangements. We introduce the so called $(n,k)$-permutations. Combining with the number of cycles, we establish exponential generating functions and their $q$-analogues as well.
\end{abstract}


\keywords{Excedance distribution; Euler--Seidel matrix; Eulerian polynomial; Riordan matrix; $q$-analogue; generating function}

\subjclass[2020]{05A05; 05A15}

\maketitle

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\markright{\its S\'eminaire Lotharingien de
Combinatoire \bfs 93 \rms (2025), Article~B93b\hfill}
\def\thepage{}

\section{Reminder and introduction}

Clarke~et~al.~\cite{bib2}, Dumont and Randrianarivony~\cite{bib4}, and Rakotondrajao~\cite{bib9} studied  Euler's difference table $(d_n^k)_{n,k \geq 0}$, also called difference factorial numbers, which are given by
\begin{equation}
\begin{cases} 		d_n^n = n!,\\
  d_n^k = d_n^{k+1} - d_{n-1}^k,&
  \text{for }1 \leq k \leq n-1.
\end{cases}
\label{equ1}
\end{equation} 
Their matrix is presented in Table~\ref{tab1}.
\begin{table}[h]
	\begin{center}
		\caption{Euler's difference matrix $(d_n^k)_{n,k \geq 0}$} \label{tab1}
		\begin{tabular}{@{}l|cccccl@{}}
			\hline
			\multicolumn{7}{@{}c@{}}{$d_n^k$}\\
			\hline
			$n\backslash k$ & 0 & 1 & 2 & 3 & 4 & $\cdots$ \\
			\hline
			0 & 0!& & & & & \\
			1 & 0 &1!& & & & \\
			2 & 1 & 1&2!& & & \\
			3 & 2 & 3& 4&3!& & \\
			4 & 9 &11&14&18&4!& \\
			$\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$& $\ddots $ \\
			\hline
		\end{tabular}
	\end{center}
\end{table}

\noindent
Dumont~\cite{bib5} reintroduced the Euler--Seidel matrix associated to a given sequence $(a_n)_{n \geq 0}$. It is an infinite matrix and is determined recursively by the formula
\begin{equation}
\begin{cases} 		a_n^0 = a_n,&\text{for }n\geq 0,\\
  a_n^k = a_n^{k-1} +a_{n+1}^{k-1}, &
  \text{for } n\geq 0,\ k\geq 1.
\end{cases}
  \label{equ2}
\end{equation}   
The sequence $\left( a_{n}^{0}\right)_{n \geq 0}$ is called the \textit{initial sequence}, and $\left( a_{0}^{n}\right)_{n \geq 0}$ is called the \textit{final sequence}.

%First page headline in AmS-LaTeX for S\'eminaire Lotharingien de Combinatoire                
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\markboth{\SMALL D. RATOVILEBAMBOAVISON, M. L. N. RAKOTOARISON AND
  F. RAKOTONDRAJAO}
{\SMALL FACTORIAL EULER--SEIDEL MATRIX: REFINEMENT AND $q$-ANALOGUE}

\begin{theorem}[\sc Dumont~\cite{bib5}] \label{theo1}
	The relation between any entry and the initial sequence is determined by
	\begin{equation}
	  a_n^k=\sum_{i=0}^{k} \binom{k}{i} a_{n+i}^0,\quad
          \text{for }n\geq0,\ k\geq0.\label{equ3}
	\end{equation}
\end{theorem}

\noindent
Firengiz and Dil~\cite{bib6} reconsidered the Euler--Seidel matrix, by introducing the parameters $p$ and $q$ as follows:
\begin{align}
a_n^k&=a_n^k(p,q)\nonumber\\
&=p\ a_n^{k-1} +q\ a_{n+1}^{k-1},\quad \text{for }n\geq0,\ k\geq1. \label{equ4}
\end{align}

\begin{theorem}[\sc Firengiz and Dil~\cite{bib6}] \label{theo2}
The relation between any entry and the initial sequence is determined by
\begin{equation}
	a_n^k=\sum_{i=0}^{k}\binom{k}{i} p^{k-i} q^i a_{n+i}^0.\label{equ5}
\end{equation}
\end{theorem}

\begin{theorem}[\sc Firengiz and Dil~\cite{bib6}] \label{theo3}
The exponential generating function of $\left( a_n^k \right)_{k,n \geq 0}$ has the closed form
\begin{equation} 
	\sum_{n\geq 0}\sum_{k\geq 0} a_n^k \frac{u^k}{k!}\ \dfrac{t^n}{n!}=\exp(p u) \, A(t+q u), \text{ where } A(t)=\sum_{n \geq 0} a_n^0 \frac{t^n}{n!}. \label{equ6}
\end{equation}
\end{theorem}

\noindent
We study the factorial Euler--Seidel matrix deduced from Euler's difference table as follows. The initial sequence is $(n!)_{n \geq 0}$, and the infinite matrix is determined recursively by
\begin{equation}
\begin{cases} 
	a_n^0 = n!,\\
	a_n^k = -a_n^{k-1} + a_{n+1}^{k-1}, \quad \text{for } n \geq
        0,\ k \geq 1.
\end{cases}
\label{equ7}
\end{equation}

\begin{definition}[\sc Shapiro~\cite{bib12}, Sprugnoli~\cite{bib14}]\label{def1}
Consider an infinite matrix \mbox{$R=\left(r_{n,k}\right)_{n,k\geq0}$} with complex coefficients and two formal power series \mbox{$g(x)=\sum_{k=0}^\infty g_kx^k$}, \mbox{$f(x)=\sum_{k=1}^\infty f_kx^k$} with \mbox{$g_0=1,\ f_1\neq 0$}. Let \mbox{$R_k(x)=\sum_{n=0}^\infty r_{n,k}x^n$} be the formal power series of the $k$-th column of~$R$. Then $R$ is called a Riordan matrix if
\begin{equation} 
	R_k(x)=g(x)\big[f(x)\big]^k.\label{equ8}
\end{equation}
$R$ is an infinite lower-triangular matrix which we denote by a pair $R=\Big(g(x),f(x)\Big)$. Note that the identity $I=(1,x)$ is a Riordan matrix.
\end{definition}

\begin{remark}
If we have $g(x)=\sum_{k=0}^\infty g_k\frac{x^k}{k!}$ and 
$f(x)=\sum_{k=1}^\infty f_k\frac{x^k}{k!}$, with $g_0=1$ and $f_1\neq
0$,
then $R_k(x)=g(x)\frac{\big[f(x)\big]^k}{k!}$. 
We denote the matrix by $R=\Big[g(x),f(x)\Big]$.
\end{remark}

\begin{theorem}[\sc Shapiro~\cite{bib12}, Shapiro et al.~\cite{bib13}] \label{theo4}
Let $\left( a_{n}\right)_{n \geq 0}$ and $\left( b_{n}\right)_{n \geq
  0}$ be two sequences. Set $a(x)=\sum_{k=0}^\infty a_kx^k$ and $b(x)=\sum_{k=0}^\infty b_kx^k$. Then we have
\begin{equation}
	\Big( g(x),f(x)\Big)\ 
	\begin{bmatrix}
		a_0\\
		a_1\\
		a_2\\
		\vdots
	\end{bmatrix}=
	\begin{bmatrix}
		b_0\\
		b_1\\
		b_2\\
		\vdots
	\end{bmatrix} \iff g(x) a\big(f(x)\big)=b(x). \label{equ9}
\end{equation}
\end{theorem}

\begin{theorem}[\sc Shapiro~\cite{bib12}, Shapiro et al.~\cite{bib13}] \label{theo5}
For any Riordan matrix $\Big( g(x),f(x)\Big)$, its inverse $R^{-1}$ is given by
\begin{align}
R^{-1}&=\Big( g(x),f(x)\Big)^{-1}\nonumber \\
	&=\left( \frac{1}{g\Big(\bar{f}(x)\Big)},\bar{f}(x)\right),\label{equ10}
\end{align}
where $\bar{f}\Big(f(x)\Big)=f\Big(\bar{f}(x)\Big)=x$.
\end{theorem}

\noindent
Let us write $[n]=\{1,2,\dots,n\}$. We say that an integer $i\in [n]$ is a fixed point for a permutation~$\sigma$ if $\sigma (i)=i$. We denote by $\FIX(\sigma)$ the set of fixed points of $\sigma$. A derangement is a permutation without fixed points, that is, a permutation $\sigma$ such that $\FIX(\sigma)=\emptyset$. We denote by $D_n$ the set of all derangements of the symmetric group $S_n$. We say that a permutation $\sigma$ is an $(n,k)$-permutation if the length of $\sigma$ is equal to $(n+k)$ and $\FIX(\sigma)\subset [n]$.

This class of $(n,k)$-permutations should not be confused with the "$n$-fixed-points-permutations" introduced by Rakotondrajao~\cite{bib10}.
An $n$-fixed-points-permutation is an \mbox{$(n,k)$-permutation}. For example, \mbox{$(1)(2,3)(4)(5,6)$} is a \mbox{$(4,2)$-permutation}, but it is not a $4$-fixed-points-permutation.

We denote by $\mathcal{E}_n^k$ the set of $(n,k)$-permutations. Note
that the coefficient of the $n$-th column and $k$-th row of the
classical Euler--Seidel matrix is the cardinality of
$\mathcal{E}_n^k$. Since $n+1$ is either a fixed point or not for a
permutation $\sigma \in \mathcal{E}_{n+1}^k$: if $n+1$ is, then
$\sigma \in \mathcal{E}_{n}^k$; otherwise $\sigma \in
\mathcal{E}_{n}^{k+1}$.

We say that an integer $i$ is an excedance for $\sigma$ if
$\sigma(i)>i$. We denote the number of excedances by $e(\sigma)$ and
the number of cycles of $\sigma$ by $c(\sigma)$.

\noindent
The numbers $E_{n,k}$ count the number of permutations of size $n$ having $k$ excedances. They are given by the formula
\begin{equation}
	E_{n,k}=\begin{cases} 
		1, &\text{if } n=0,\ k=0 ,\\
		(n-k)E_{n-1,k-1} + (k+1)E_{n-1,k}, & \text{if } 0< k\le n ,\\
		0, &otherwise.
\end{cases}
\label{equ11}
\end{equation}
The Eulerian numbers $E_{n,k}$ are classical numbers in
mathematics. They can be found in many places in the literature (Brenti~\cite{bib1}, Comtet~\cite{bib3}, Foata and Sch\"{u}tzenberger~\cite{bib7}, Roselle~\cite{bib11}, Stanley~\cite{bib15}, Worpitzky~\cite{bib16}).

We are interested in refining the factorial Euler--Seidel matrix using the excedance distribution over permutations. The Eulerian polynomials \mbox{$E_n(x) = \sum_{i=0}^{n} E_{n,i} x^i = \sum_{i\geq 0}E_{n,i} x^i$} become the initial sequence.
We have
\begin{equation}
\begin{cases} 
		e_n^0(x) = E_n(x), &{\text{for }n\geq 0 ,}\\
		e_n^k(x) = -e_n^{k-1}(x)+e_{n+1}^{k-1}(x), &
                {\text{for } n\geq 0,\ k\geq 1,}
\end{cases} \label{equ12}
\end{equation}
and their matrix is presented in Table~\ref{tab2}.
\begin{table}[h]
\begin{center}
	\caption{Refined factorial Euler--Seidel matrix}\label{tab2}
	\begin{tabular}{@{}l|ccccl@{}}
		\hline
		\multicolumn{6}{@{}c@{}}{$e_n^k(x)$}\\
		\hline
		$k\backslash n $& 0 & 1 & 2 & 3 & $\cdots $\\
		\hline
		0 & 1 & 1 & $1+x$ & $1+4x+x^2 $ & $\cdots$ \\
		1 & 0 & $x$ & $3x+x^2$ & $\cdots$ & $\cdots$ \\
		2 & $x$ & $2x+x^2$ & $\cdots$ & $\cdots$ & $\cdots$ \\
		3 & $x+x^2$ & $\cdots$ & $\cdots$ &$\cdots$ & $\cdots$ \\
		$\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\ddots$\\  
		\hline
	\end{tabular}
\end{center}
\end{table}

\noindent
The Hurwitz generating function of the classical Eulerian
polynomials~\cite{bib7} is
\begin{align} 
E(x,t) &= \sum_{n \geq 0}\sum_{k \geq 0}E_{n,k}x^k\frac{t^n}{n!}\nonumber \\ &= \frac{x-1}{x-\exp\left(t(x-1)\right)}.\label{equ13}
\end{align}
We define the $q$-analogue of the refined factorial Euler--Seidel
matrix combining the excedance distribution and the number of cycles.
We write
\begin{align}
e_n^k(x)&=\sum_{\sigma \in \mathcal{E}_{n}^{k}}x^{e(\sigma)},\label{equ14}\\
e_n^k(x,q)&=\sum_{\sigma \in \mathcal{E}_{n}^{k}}x^{e(\sigma)}\ q^{c(\sigma)}.\label{equ15}
\end{align}
Mantaci and Rakotondrajao~\cite{bib8} extensively studied the excedance distribution over derangements. The number $a_{n,k}$ counts derangements over $[n]$ having $k$ excedances,
\begin{equation}
	a_{n,k}=\lvert\{\delta\in D_n \mid \delta \text{ has } k \text{ excedances}\}\rvert.\label{equ16}
\end{equation}
Let us write $A_k(x)=\sum_{i=0}^{k} a_{k,i} x^i$, and let $A_k(x,q)$ be its $q$-analogue.

\begin{remark}
	Note that a derangement of $D_n$ is a $(0,n)$-permutation, and a permutation of $S_n$ is a $(n,0)$-permutation.
\end{remark}

\noindent
Let us denote by $E_n(x,q)$ the $q$-analogue of the Eulerian polynomials \mbox{$E_n(x)=\sum_{i=0}^{n} E_{n,i} x^i$} given by
\begin{equation} 
E_n(x,q) = \sum_{\sigma \in S_n}x^{e(\sigma)}q^{c(\sigma)}.\label{equ17}
\end{equation}

\begin{theorem}[\sc Brenti~\cite{bib1}] \label{theo6}
The polynomials $E_n(x,q)$ satisfy the
recursion relation
\begin{equation}
	E_n(x,q) = \left(x(n-1)+q\right)E_{n-1}(x,q)+x(1-x)\frac{d}{dx}E_{n-1}(x,q) \label{equ18}
\end{equation}
with initial condition $E_0(x,q) = 1$.
\end{theorem}

\begin{theorem}[\sc Brenti~\cite{bib1}] \label{theo7}
The exponential generating function of $E_n(x,q)$ is given by
\begin{equation}
	E(x,t,q) = \sum_{n \geq 0} \frac{E_n(x,q)}{n!}t^n = \left(\frac{x-1}{x-\exp\left(t(x-1)\right)}\right)^q,\ E_0(x,q) = 1. \label{equ19}
\end{equation}
\end{theorem}

\section{Riordan matrix representation of Worpitzky identity}
For a fixed integer $n$ with $n\geq0$, let us consider the infinite matrix \mbox{$R_n=\left(r_{k,\ell}^{(n)}\right)_{k,\ell \geq 0}$} given by
\begin{equation}
R_n=\left(
\begin{matrix}
	1 & 0 & 0 & 0 & 0 & \cdots\\
	2^n & 1 & 0 & 0 & 0 & \cdots\\
	3^n & 2^n & 1 & 0 & 0 & \cdots\\
	4^n & 3^n & 2^n & 1 & 0 & \cdots\\
	5^n	& 4^n & 3^n & 2^n & 1 & \cdots \\
	\vdots	& \vdots & \vdots& \vdots & \vdots & \ddots 
\end{matrix}
\right). \label{equ20}
\end{equation}
\begin{theorem}\label{theo8}
$R_n$ is a Riordan matrix and
\begin{equation}
	R_n=\left( \frac{E_n(x)}{(1-x)^{n+1}}, x \right). \label{equ21}
\end{equation}
Its inverse $R_n^{-1}$ is given by 
\begin{equation}
	R_n^{-1}=\left( \frac{(1-x)^{n+1}}{E_n(x)}, x \right). \label{equ22}
\end{equation}
\end{theorem}

\begin{proof}
We have
\begin{equation*}
	\frac{E_n(x)}{(1-x)^{n+1}}= \sum_{k \geq 0} (k+1)^n x^k.
\end{equation*}
Let $U=R_n$. If $U=\left(g(x),f(x)\right)$, from Definition~\ref{def1}, we have Equation~(\ref{equ21}), that is
\begin{equation*}
	\left\{\begin{array}{l}
		U_0(x)=g(x)=\frac{E_n(x)}{(1-x)^{n+1}},\\
		U_1(x)=g(x)f(x)=x\frac{E_n(x)}{(1-x)^{n+1}},\\
		\vdots \\
		U_i(x)=g(x)\left(f(x)\right)^i=x^i\frac{E_n(x)}{(1-x)^{n+1}},\\
		\vdots
	\end{array}
	\right.
\end{equation*}
Then $U^{-1}=\left( \frac{(1-x)^{n+1}}{E_n(x)}, x \right)=R_n^{-1}$
(cf.\ Theorem~\ref{theo5}).
\end{proof}

\noindent
For a fixed integer $n$ with $n\geq0$, let us consider the Eulerian
number vector~$E_n$ and the signed binomial coefficient vector~$B_n$ given by
\begin{equation*}
	E_n=\begin{bmatrix}
		E_{n,0}\\
		E_{n,1}\\
		\vdots\\
		E_{n,n+1}
	\end{bmatrix} \text{and} \quad
	B_n =
	\begin{bmatrix}
		1\\
		-\binom{n+1}{1}\\
		\vdots\\
		(-1)^i\binom{n+1}{i}\\
		\vdots\\
		(-1)^{n+1}\binom{n+1}{n+1}
	\end{bmatrix}.
\end{equation*}

\begin{theorem}\label{theo9}
For a fixed integer $n$ with $n\geq0$, we have
\begin{equation}
	\left[
	\begin{array}{l}
		E_{n,0}\\
		E_{n,1}\\
		E_{n,2}\\
		E_{n,3}\\
		E_{n,4}\\
		\quad\vdots
	\end{array}
	\right] = \left(
	\begin{matrix}
		1 & 0 & 0 & 0 & 0 & \cdots\\
		2^n & 1 & 0 & 0 & 0 & \cdots\\
		3^n & 2^n & 1 & 0 & 0 & \cdots\\
		4^n & 3^n & 2^n & 1 & 0 & \cdots\\
		5^n	& 4^n & 3^n & 2^n & 1 & \cdots \\
		\vdots	& \vdots & \vdots& \vdots & \vdots & \ddots 
	\end{matrix}
	\right)\ \left[
	\begin{array}{r}
		1\ \ \ \\
		-\binom{n+1}{1}\\
		\binom{n+1}{2}\\
		-\binom{n+1}{3}\\
		\binom{n+1}{4}\\
		\vdots\quad
	\end{array}
	\right]. \label{equ23}
\end{equation}
\end{theorem}

\begin{proof}
Equation~(\ref{equ23}) is a matrix representation of the Worpitzky~\cite{bib16} formula
\begin{equation}
	E_{n,k} = \sum_{i=0}^{k}(-1)^i \, (k+1-i)^n\binom{n+1}{i}. \label{equ24}
\qedhere\end{equation}
\end{proof}

\begin{theorem}\label{theo10}
Let us consider the infinite matrix  $R=\left(r_{k,\ell}\right)_{k,\ell \geq 0}$ given by
\begin{equation*}
	r_{k,\ell}=\begin{cases} 
		(k-\ell+1)^p, & \text{if }  k \geq \ell \text{ and } \frac{2+(5+p)p}{2} \geq k \geq \frac{(3+p)p}{2}, \\
		0, & \text{otherwise,} 
	\end{cases}\quad 
\text{ with } p=\left\lfloor\frac{-3+\sqrt{9+8\ell}}{2}\right\rfloor.
\end{equation*}
We have
\begin{equation}
	\left(
	\begin{matrix}
		E_0&&&&\\
		&E_1&&\textup{\Large 0}&\\
		&&E_2&& \\
		&\textup{\Large 0}&&E_3&\\
		&&&&&\ddots
	\end{matrix}
	\right) = \raise-5pt\hbox{\huge $R$}\ \left(
	\begin{matrix}
		B_0&&&&\\
		&B_1&&\textup{\Large 0}&\\
		&&B_2&& \\
		&\textup{\Large 0}&&B_3&\\
		&&&&&\ddots
	\end{matrix}
	\right).\label{equ25}
\end{equation}
\end{theorem}

\begin{proof}
For a fixed integer $n$ with $n\geq0$, let us consider the matrix $\bar{R}_n=\left(\bar{r}_{k,\ell}^{(n)}\right)_{k,\ell \geq 0}$ of the first $(n+2)$ rows and $(n+2)$ columns extracted from $R_n$ in Equation~(\ref{equ20}), that is,
\begin{equation*}
	\bar{R}_n=\left(
	\begin{matrix}
		1 & 0 & 0 & \cdots & 0\\
		2^n & 1 & 0 & \cdots & 0\\
		3^n & 2^n & 1 & \cdots & 0\\
		\vdots	& \vdots & \vdots & \ddots & \vdots\\
		(n+2)^n & (n+1)^n & n^n & \cdots & 1
	\end{matrix}
	\right).
\end{equation*}
From Theorem~\ref{theo9}, we have $E_n = \bar{R}_n\ B_n $.
Hence, we have
\begin{equation*}
	\left(
	\begin{matrix}
		E_0&&&&\\
		&E_1&&\textup{\Large 0}&\\
		&&E_2&& \\
		&\textup{\Large 0}&&E_3&\\
		&&&&&\ddots
	\end{matrix}
	\right) = \left(
	\begin{matrix}
		\bar{R}_0&&&&\\
		&\bar{R}_1&&\textup{\Large 0}&\\
		&&\bar{R}_2&& \\
		&\textup{\Large 0}&&\bar{R}_3&\\
		&&&&&\ddots
	\end{matrix}
	\right)\ \left(
	\begin{matrix}
		B_0&&&&\\
		&B_1&&\textup{\Large 0}&\\
		&&B_2&& \\
		&\textup{\Large 0}&&B_3&\\
		&&&&&\ddots
	\end{matrix}
	\right).
\end{equation*}
That is,
\begin{equation*}
	\begin{pmatrix}
		\boxed{\begin{matrix}
				E_{0,0} \\ 0
		\end{matrix}} & 0 &  \cdots \\
		0 & \boxed{\begin{matrix}
				E_{1,0} \\ E_{1,1} \\ 0
		\end{matrix}} & \cdots \\
		\vdots & \vdots & \ddots
	\end{pmatrix}
	= 
	\begin{pmatrix}
		\boxed{\begin{matrix}
				1 & 0 \\
				1 & 1
		\end{matrix}} & 0 & \cdots \\
		0 & \boxed{\begin{matrix}
				1 & 0 & 0 \\
				2 & 1 & 0 \\
				3 & 2 & 1
		\end{matrix}} & \cdots \\
		\vdots & \vdots & \vdots & \ddots
	\end{pmatrix}\ 
	\begin{pmatrix}
		\boxed{\begin{matrix}
				\binom{1}{0} \\ -\binom{1}{1}\ \ \
		\end{matrix}} & 0 & \cdots \\
		0 & \boxed{\begin{matrix}
				\binom{2}{0} \\ -\binom{2}{1}\ \ \ \\ \binom{2}{2}
		\end{matrix}} & \cdots \\
		\vdots & \vdots & \vdots & \ddots
	\end{pmatrix}.
\end{equation*}

Identifying the coefficients, we get 
\begin{equation*}
	r_{k,\ell}=\begin{cases} 
		(k-\ell+1)^p, & \text{if }  k \geq \ell \text{ and } \frac{2+(5+p)p}{2} \geq k \geq \frac{(3+p)p}{2}, \\
		0, & \text{otherwise,} 
   \end{cases}
        \quad\text{with }p=\left\lfloor\frac{-3+\sqrt{9+8\ell}}{2}\right\rfloor.
\qedhere\end{equation*}
\end{proof}

\section{Refined factorial Euler--Seidel matrix and $q$-analogue}
Let us consider the refined factorial Euler--Seidel matrix coefficient $e_n^k(x)$.

\begin{theorem}\label{theo11}
For any integers $n\geq0$ and $k\geq0$, we have
\begin{equation}
	e_n^k(x)=\sum_{i=0}^{k} (-1)^{k-i} \binom{k}{i} E_{n+i}(x). \label{equ26}
\end{equation}
\end{theorem}

\begin{proof}
Setting $p=-1$ and $q=1$ in Theorem~\ref{theo2}, we get
\begin{align*}
	e_n^k(x)&=\sum_{i=0}^{k} (-1)^{k-i} \binom{k}{i} e_{n+i}^0(x)\\
	&=\sum_{i=0}^{k} (-1)^{k-i} \binom{k}{i} E_{n+i}(x).
\qedhere\end{align*}
\end{proof}

\begin{corollary}\label{theo12}
For any integer $k\geq0$, we have
\begin{equation}
	e_0^k(x)=\sum_{i=0}^{k} (-1)^{k-i} \binom{k}{i} E_i(x).\label{equ27}
\end{equation}
\end{corollary}

\begin{proof}
Set $n=0$ in Theorem~\ref{theo11}.
\end{proof}

\begin{remark} We have
$e_0^k(x)=A_k(x)=\sum_{i=0}^{k} a_{k,i} x^i$.
\end{remark}

\begin{theorem}\label{theo13}
The polynomial $e_n^k(x)$ is the polynomial of the excedance  distribution over $\mathcal{E}_n^k$.
\end{theorem}

\begin{proof}
From the construction of set $\mathcal{E}_{n+1}^{k-1}$, we have
\begin{equation*}
	e_{n+1}^{k-1}(x)=e_n^k(x)+e_n^{k-1}(x).
\end{equation*}
This follows from the interpretation of $e_n^k(x)$ as the excedance distribution over $\mathcal{E}_{n+1}^{k-1}$ where $n+1$ is not a fixed point, and $e_n^{k-1}(x)$ as the excedance distribution over $\mathcal{E}_{n+1}^{k-1}$ where $n+1$ is a fixed point (Equation~(\ref{equ14})).
\end{proof}

\begin{theorem} \label{theo14}
The exponential generating function of $e_n^k(x)$ has the closed form
\begin{equation} 
	F(x,u,t)=\sum_{n\geq 0} \sum_{k\geq 0} e_n^k(x) \frac{u^k}{k!}\ \dfrac{t^n}{n!}=\frac{(x-1)\exp(-u)}{x-\exp\left((t+u)(x-1)\right)}.\label{equ28}
\end{equation}
\end{theorem}

\begin{proof}
From Equation~(\ref{equ12}) and Theorem~\ref{theo3}, we have
\begin{equation*} 
	\sum_{n\geq 0}\sum_{k\geq 0} e_n^k(x) \frac{u^k}{k!} \dfrac{t^n}{n!}=exp(-u) \sum_{n \geq 0}E_n(x) \frac{(t+u)^n}{n!}.
\end{equation*}
From Equation~(\ref{equ13}), we have
\begin{equation*} 
	\sum_{n \geq 0}E_n(x) \frac{t^n}{n!}=\frac{x-1}{x-\exp\left(t(x-1)\right)}.
\end{equation*}
Hence,
\begin{equation*} 
	\sum_{n\geq 0} \sum_{k\geq 0} e_n^k(x) \frac{u^k}{k!} \dfrac{t^n}{n!}=\frac{(x-1)\exp(-u)}{x-\exp\left((t+u)(x-1)\right)}.
\qedhere\end{equation*}
\end{proof}

\begin{corollary}\label{theo15}
The exponential generating function of $A_k(x)$ has the closed form
\begin{equation} 
	A(x,u)=F(x,u,0) =
	\frac{(x-1)\exp(-u)}{x-\exp\left(u(x-1)\right)}.\label{equ29}
\end{equation}
\end{corollary}

\noindent
Moreover, we can express the initial sequence
$\left(e_n^0(x)\right)_{n \geq 0}$ as an infinite matrix. The
coefficient $[x^\ell] e_n^0(x)$ is the classical Eulerian number
$E_{n,\ell}$, and Equation~(\ref{equ12}) becomes
$$[x^\ell] e_n^k(x) = -[x^\ell] e_n^{k-1}(x)+[x^\ell]
e_{n+1}^{k-1}(x).$$
That is,
\begin{equation}
\begin{cases} 
	[x^\ell] e_n^0(x)  = E_{n,\ell}, &{\text{for }n\geq 0,}\\
	[x^\ell] e_n^k(x)  = -[x^\ell] e_n^{k-1}(x)+[x^\ell]
        e_{n+1}^{k-1}(x), &\text{for }n\geq 0,\ k\geq 1.	
\end{cases}\label{equ30}
\end{equation}

\begin{theorem}\label{theo16}
For fixed integers $k$ and $\ell$, the coefficient of $x^\ell$ in  $e_0^k(x)$ is
\begin{equation}
	[x^\ell] e_0^k(x) = a_{k,\ell} = \sum_{i=0}^{k}
        (-1)^{k-i}\binom{k}{i} E_{i,\ell},\quad \text{for } k\geq0 \text{ and } l\geq0. \label{equ31}
\end{equation}
\end{theorem}

\begin{proof}
We have
\begin{align*}
	a_{k,\ell}&=[x^\ell] e_0^k(x)\\ 
	&=\sum_{i=0}^{k}\binom{k}{i} (-1)^{k-i} \left([x^\ell] e_i^0(x)\right),\\
	&=\sum_{i=0}^{k}\binom{k}{i} (-1)^{k-i} E(i,\ell).
\end{align*}
Here, we obtained the second line by
setting $p=-1$ and $q=1$ in Equation~(\ref{equ5}).
\end{proof}

\begin{theorem}\label{theo17}
We have
\begin{equation}
	\left(
	\begin{matrix}
		1& 0 & 0 & 0 &0 &\cdots\\
		0& 0 & 0 & 0 & 0& \cdots \\
		0& 1 & 0 & 0 &0 & \cdots\\
		0& 1 &1 & 0 & 0& \cdots \\
		0& 1 &7 & 1 & 0 & \cdots \\
		\vdots	& \vdots & \vdots & \vdots & \vdots & \ddots 
	\end{matrix}
	\right) = \left(
	\begin{array}{rrrrrr}
		1& 0 & 0 & 0 &\ \ 0 &\cdots\\
		-1& 1 & 0 & 0 & 0& \cdots \\
		1 & -2 & 1 & 0 &0 & \cdots\\
		-1& 3 &-3 & 1 & 0& \cdots \\
		1& -4 & 6 & -4 & 1 & \cdots \\
		\vdots	& \vdots & \vdots & \vdots & \vdots & \ddots 
	\end{array}
	\right)\ \left(
	\begin{matrix}
		1 & 0 & 0 & 0  & 0&\cdots\\
		1 & 0  & 0 & 0 & 0 &\cdots\\
		1 &1 &0 &0  & 0 & \cdots\\
		1 &4 &1 &0 & 0 & \cdots\\
		1 & 11& 11 & 1 & 0& \cdots\\
		\vdots &\vdots & \vdots & \vdots & \vdots & \ddots
	\end{matrix}
	\right).\label{equ32}
\end{equation}
That is,
\begin{equation}
	\left(a_{k,\ell}\right)_{k,\ell \geq 0} = \Big[\exp(-t), t \Big]\, \left(E_{n,\ell}\right)_{n,\ell \geq 0},\label{equ33}	
\end{equation}
with $S=\Big[\exp(-t), t \Big]$ the Riordan matrix associated to the refined factorial Euler--Seidel matrix.
\end{theorem}

\begin{theorem} \label{theo18}
For all integers $k\geq0$ and $l\geq0$, we have
\begin{equation}
	a_{k,\ell}=\sum_{i=0}^{k}\sum_{j=0}^{\ell}
        (-1)^{k-i+j}\binom{k}{i}\binom{i+1}{j} (\ell+1-j)^i, \quad
        \text{for }k\geq0,\ \ell\geq0. \label{equ34}
\end{equation}
\end{theorem}

\begin{proof}
Using the Worpitzky identity in Equation~(\ref{equ24})
with $n=i$ and $k=\ell$
and Theorem~\ref{theo16}, we have
\begin{align*}
	a_{k,\ell}&=\sum_{i=0}^{k}\binom{k}{i}(-1)^{k-i}\sum_{j=0}^{\ell} \left((-1)^j (\ell+1-j)^i\right) \binom{i+1}{j} \nonumber \\
	&=\sum_{i=0}^{k}\sum_{j=0}^{\ell} (-1)^{k-i+j}\binom{k}{i}\binom{i+1}{j} (\ell+1-j)^i.
\qedhere\end{align*}
\end{proof}

\begin{theorem} \label{theo19}
We have
\begin{equation}
	\left(
	\begin{matrix}
		1 & 0 & 0 & 0  & 0&\cdots\\
		1 & 0  & 0 & 0 & 0 &\cdots\\
		1 &1 &0 &0  & 0 & \cdots\\
		1 &4 &1 &0 & 0 & \cdots\\
		1 & 11& 11 & 1 & 0& \cdots\\
		\vdots &\vdots & \vdots & \vdots & \vdots & \ddots
	\end{matrix}
	\right) = \left(
	\begin{matrix}
		1 & 0 & 0 & 0 & 0 & \cdots\\
		1 & 1 & 0 & 0 & 0 & \cdots \\
		1 & 2 & 1 & 0 & 0 & \cdots\\
		1 & 3 & 3 & 1 & 0 & \cdots \\
		1 & 4 & 6 & 4 & 1 & \cdots \\
		\vdots	& \vdots & \vdots & \vdots & \vdots & \ddots 
	\end{matrix}
	\right)\ \left(
	\begin{matrix}
		1& 0 & 0 & 0 &0 &\cdots\\
		0& 0 & 0 & 0 & 0& \cdots \\
		0 & 1 & 0 & 0 &0 & \cdots\\
		0& 1 &1 & 0 & 0& \cdots \\
		0& 1 &7 & 1 & 0 & \cdots \\
		\vdots	& \vdots & \vdots & \vdots & \vdots & \ddots 
	\end{matrix}
	\right). \label{equ35}
\end{equation}
That is,
\begin{equation}
	\left(E_{k,\ell}\right)_{k,\ell \geq 0} = \Big[\exp(t), t \Big]\, \left(a_{n,\ell}\right)_{n,\ell \geq 0}, \text{ with }\Big[\exp(t), t \Big]=S^{-1}. \label{equ36}
\end{equation}
\end{theorem}

\begin{proof}
This is obtained from Equation~(\ref{equ33}) and Theorem~\ref{theo5}.
\end{proof}

\begin{corollary} \label{theo20} 
We have
\begin{equation}
	\left(E_{k,\ell}\right)_{k,\ell \geq 0}=\left(\binom{k}{n}\right)_{k,n \geq 0}\ \left(a_{n,\ell}\right)_{n,\ell \geq 0}. \label{equ37}
\end{equation}
\end{corollary}

\begin{corollary} \label{theo21}
For a fixed integer $\ell \geq 0$, the sequence
$\left(E(n,\ell)\right)_{n\geq 0}$ of a row of the Eulerian number matrix is given by
\begin{equation}
	\begin{bmatrix}
		E_{0,\ell}\\
		E_{1,\ell}\\
		E_{2,\ell}\\
		E_{3,\ell}\\
		E_{4,\ell}\\
		\vdots
	\end{bmatrix}
	=\Big[\exp(t), t \Big]\ 
	\begin{bmatrix}
		a_{0,\ell}\\
		a_{1,\ell}\\
		a_{2,\ell}\\
		a_{3,\ell}\\
		a_{4,\ell}\\
		\vdots
	\end{bmatrix}.\label{equ38}
\end{equation}
\end{corollary}

\begin{theorem} \label{theo22}
The Euler--Seidel matrix preserves the excedance distribution.
\end{theorem}

\begin{proof}
This process is derived from the construction of the $(n+1,k-1)$-permutation. Consider first the case where $n+1$ is a
fixed point. By removing $n+1$ from the permutation and reordering the sequence, that is, decreasing by~1 all integers greater than $n+1$ in the remaining permutation, we obtain an $(n,k-1)$-permutation. Conversly, if $n+1$ is not a fixed point, then the \mbox{$(n+1,k-1)$}-permutation is, by definition, an $(n,k)$-permutation. This construction leads directly to the recurrence relation
\begin{equation*}
e_n^k(x) = -e_n^{k-1}(x)+e_{n+1}^{k-1}(x).
\qedhere
\end{equation*}
\end{proof}

\begin{theorem} \label{theo23}
For a fixed integer $\ell \geq 0$, the exponential generating function of\/ $[x^\ell] e_n^k(x)$, that is, the exponential generating function of all $(n,k)$-permutations with $\ell$ excedances, is given by
\begin{align}
	e_\ell(t,u) &
		= \sum_{n\geq 0} \sum_{k\geq 0} [x^\ell] e_n^k(x) \frac{u^k}{k!} \dfrac{t^n}{n!}\nonumber\\
&= \sum_{i=0}^\ell\frac{(-1)^i}{i!}\Big((\ell-i+1)(t+u)\Big)^{i-1} \Big((\ell-i+1)(t+u)+i\Big)
		\exp\Big(t+(\ell-i)(t+u)\Big).
	\label{equ39}
\end{align}
\end{theorem}

\begin{proof}
For fixed integers $\ell$ and $n$, from the Worpitzky identity in Equation~(\ref{equ24}), we consider the initial term of the sequence
\begin{equation*}
	E_{n,\ell}=\sum_{i=0}^{\ell}(-1)^i(\ell+1-i)^n \binom{n+1}{i}.
\end{equation*}
Hence,
\begin{align}	
	\sum_{n \geq 0} E_{n,\ell} \frac{t^n}{n!}&=\sum_{n \geq i-1} \sum_{i=0}^{\ell}(-1)^i(\ell+1-i)^n \frac{(n+1) t^n}{(n-i+1)!\, i!}\nonumber \\
	&=\sum_{i=0}^{\ell} \bigg[\sum_{n \geq i}\frac{(-1)^i}{i!} \frac{\left((\ell+1-i)\ t\right)^{n-i+1}}{(n-i)!}\nonumber \\
	&\qquad \quad +\sum_{n \geq i-1}\frac{(-1)^i\, i}{i!} \frac{\left((\ell+1-i)\ t\right)^{n-i+1}}{(n-i+1)!}\bigg]\Big(t (\ell+1-i)\Big)^{i-1}\nonumber \\
	&=\sum_{i=0}^{\ell} \frac{(-1)^i}{i!}\Big(t (\ell+1-i) \Big)^{i-1}\Big(t\ (\ell+1-i)+i\Big)\exp\Big(t (\ell+1-i)\Big)\nonumber \\
	&=\sum_{i=0}^{\ell} \frac{(-1)^i}{i!}\Big(t (\ell+1-i) \Big)^{i-1}\Big(t\ (\ell+1)+
 i (1-t)\Big)\exp\Big(t\ (\ell+1-i)\Big).\label{equ40}
\end{align}
From Equations~(\ref{equ6})~and~(\ref{equ30}), setting $p=-1$ and $q=1$, we obtain
\begin{equation*}
	\sum_{n\geq 0} \sum_{k\geq 0} [x^\ell] e_n^k(x) \frac{u^k}{k!} \dfrac{t^n}{n!}=\exp(-u)\sum_{n \geq 0} E_{n,\ell} \frac{(t+u)^n}{n!}.
\end{equation*}
By Equation~(\ref{equ40}), we have
\begin{align*}
	\sum_{n\geq 0} \sum_{k\geq 0} [x^\ell] e_n^k(x) \frac{u^k}{k!}
        \dfrac{t^n}{n!}&=\exp(-u)\sum_{i=0}^{\ell}\frac{(-1)^i}{i!}\Big((\ell+1-i)(t+u)\Big)^{i-1}\\
 &\kern2cm\cdot \Big((t+u)(\ell+1)
 +i (1-t-u)\Big) \exp\Big((\ell+1-i)(t+u)\Big)\\
	&=\sum_{i=0}^{\ell}\frac{(-1)^i}{i!}\Big((\ell+1-i)(t+u)\Big)^{i-1}\Big((t+u)(\ell+1)+i (1-t-u)\Big)\\
 &\kern2cm\cdot \exp\Big(t(\ell+1-i)+u (\ell-i)\Big).
\qedhere\end{align*}
\end{proof}

\begin{corollary}\label{theo24}
For a fixed integer \mbox{$\ell \geq 0$}, the exponential generating function \mbox{$a_\ell(u) = \sum_{k\geq 0}a_{k,\ell} \frac{u^k}{k!}$} of derangements with $\ell$ excedances  has the closed form
\begin{equation}
	a_\ell(u)=\sum_{i=0}^\ell\frac{(-1)^i}{i!}\Big(u(\ell-i+1)\Big)^{i-1}\Big(u(\ell-i+1)+i\Big)\exp\Big(u(\ell-i)\Big).\label{equ41}
\end{equation}
\end{corollary}

\begin{proof}
Setting $t=0$ in Equation~(\ref{equ39}),
we get $a_\ell(u)=e_\ell(0,u)$.
\end{proof}

\begin{corollary}\label{theo25}
For a fixed integer \mbox{$\ell \geq 0$}, the exponential generating function \mbox{$e_\ell(t) = \sum_{n\geq 0}E_{n,\ell} \frac{t^n}{n!}$} of permutations with $\ell$ excedances has the closed form
\begin{equation}
	e_\ell(t) = \sum_{i=0}^\ell \frac{(-1)^i}{i!}\Big(t(\ell-i+1)\Big)^{i-1}\Big(t(\ell-i+1)+i\Big)\exp\Big(t(\ell-i+1)\Big).\label{equ42}
\end{equation}
\end{corollary}

\begin{proof}
Setting $u=0$ in Equation~(\ref{equ39}),
we get $e_\ell(t)=e_\ell(t,0)$.
\end{proof}

\begin{theorem} \label{theo26}
The $q$-analogue of the refined factorial Euler--Seidel matrix $\left(e_n^k(x)\right)_{k, n \geq 0}$ is determined by the formula
\begin{equation}
	e_n^k(x,q) = -q\ e_n^{k-1}(x,q) + e_{n+1}^{k-1}(x,q). \label{equ43}
\end{equation}
\end{theorem}

\begin{proof}
From the construction of $(n+1,k+1)$-permutations, we distinguish
whether the integer $(n+1)$ is a fixed point or not. By doing the
corresponding case analysis, we get the result.
\end{proof}

\begin{theorem} \label{theo27}
We have
\begin{equation}
	e_n^k(x,q) = \sum_{i=0}^k (-q)^{k-i}\binom{k}{i} E_{n+i}(x,q). \label{equ44}
\end{equation}
\end{theorem}

\begin{proof}
From Equation~(\ref{equ43}) and Theorem~\ref{theo2}, we get the result.
\end{proof}

\begin{theorem}\label{theo28}
We have
\begin{equation}
	\begin{bmatrix}
		e_n^0(x,q)\\
		e_n^1(x,q)\\
		e_n^2(x,q)\\
		e_n^3(x,q)\\
		e_n^4(x,q)\\
		\vdots
	\end{bmatrix}
	= \left(
	\begin{array}{rrrrrr}
		1& 0 & 0 & 0 &\ \ 0 &\cdots\\
		-q& 1 & 0 & 0 & 0& \cdots \\
		q^2 & -2q & 1 & 0 &0 & \cdots\\
		-q^3& 3q^2 &-3q & 1 & 0& \cdots \\
		q^4& -4q^3 & 6q^2 & -4q & 1 & \cdots \\
		\vdots	& \vdots & \vdots & \vdots & \vdots & \ddots 
	\end{array}
	\right)\ 
	\begin{bmatrix}
		E_{n}(x,q)\\
		E_{n+1}(x,q)\\
		E_{n+2}(x,q)\\
		E_{n+3}(x,q)\\
		E_{n+4}(x,q)\\
		\vdots
	\end{bmatrix}
	\label{equ45}.
\end{equation}
That is,
\begin{equation}
	\left[ e_n^k(x,q) \right]_{k \geq 0} = \left[ \exp(-qt), t \right]\ \left[E_{n+k}(x,q)\right]_{k \geq 0}
	\label{equ46}.
\end{equation}
\end{theorem}

\begin{proof}
Equation~(\ref{equ45}) follows from Equation~(\ref{equ44}).
\end{proof}

\begin{theorem}\label{theo29}
The matrix $T$ given by
\begin{equation}
	T
	=\left[ \exp(-qt), t \right] = \left(
	\begin{array}{rrrrrr}
		1& 0 & 0 & 0 &\ \ 0 &\cdots\\
		-q& 1 & 0 & 0 & 0& \cdots \\
		q^2 & -2q & 1 & 0 &0 & \cdots\\
		-q^3& 3q^2 &-3q & 1 & 0& \cdots \\
		q^4& -4q^3 & 6q^2 & -4q & 1 & \cdots \\
		\vdots	& \vdots & \vdots & \vdots & \vdots & \ddots 
	\end{array}
	\right)
	\label{equ47}
\end{equation}
is the $q$-analogue of $\left[ \exp(-t), t \right]$.
\end{theorem}

\begin{proof}
We have
\begin{equation*}
	\sum_{\ell \geq 0} (-q)^\ell \frac{t^\ell}{\ell!} = \exp(-qt).
\end{equation*}
If $T=\left[g(t),f(t)\right]$, from Definition~\ref{def1}, we have Equation~(\ref{equ47}), that is,
\begin{equation*}
	\left\{\begin{array}{l}
		T_0(t)=g(t)=\exp(-qt),\\
		T_1(t)=g(t)f(t)=tg(t),\\
		\vdots \\
		T_i(t)=g(t)\left(f(t)\right)^i=t^ig(t),\\
		\vdots
	\end{array}
	\right.
\qedhere\end{equation*}
\end{proof}

\begin{remark}
$T^{-1}=\left[ \exp(qt), t \right]$ is the $q$-analogue of the inverse
Riordan matrix of $T$.
\end{remark}

\begin{theorem}\label{theo30}
We have
\begin{equation}
	\begin{bmatrix}
		e_0^k(x,q)\\
		e_1^k(x,q)\\
		e_2^k(x,q)\\
		e_3^k(x,q)\\
		e_4^k(x,q)\\
		\vdots
	\end{bmatrix}
	= \left(
	\begin{array}{rrrrrr}
		1& 0 & 0 & 0 &\ \ 0 &\cdots\\
		q& 1 & 0 & 0 & 0& \cdots \\
		q^2 & 2q & 1 & 0 &0 & \cdots\\
		q^3& 3q^2 &3q & 1 & 0& \cdots \\
		q^4& 4q^3 & 6q^2 & 4q & 1 & \cdots \\
		\vdots	& \vdots & \vdots & \vdots & \vdots & \ddots 
	\end{array}
	\right)\ 
	\begin{bmatrix}
		A_{k}(x,q)\\
		A_{k+1}(x,q)\\
		A_{k+2}(x,q)\\
		A_{k+3}(x,q)\\
		A_{k+4}(x,q)\\
		\vdots
	\end{bmatrix}. \label{equ48}
\end{equation}
That is,
\begin{equation}
	\left[ e_n^k(x,q) \right]_{n \geq 0} = \left[ \exp(qt), t \right]\ \left[A_{k+n}(x,q)\right]_{n \geq 0}
	,\text{ with } \left[ \exp(qt), t \right] = T^{-1}. \label{equ49}
\end{equation}
\end{theorem}

\begin{proof}
From Equation~(\ref{equ43}), we get
\begin{equation*}
	e_{n+1}^{k-1}(x,q) = q\ e_n^{k-1}(x,q) + e_n^k(x,q).
\end{equation*}
Let us consider $b_k^n(x,q) = e_{n+1}^{k-1}(x,q)$. Then
\begin{equation*}
	b_k^n(x,q) = q\ b_k^{n-1}(x,q) + b_{k+1}^{n-1}(x,q) \text{ and } b_k^0=A_k(x,q).
\end{equation*}
Combining this with Theorem~\ref{theo2}, we get the result.
\end{proof}

\begin{theorem}\label{theo31}
We have
\begin{equation}
	\begin{bmatrix}
		E_0(x,q)\\
		E_1(x,q)\\
		E_2(x,q)\\
		E_3(x,q)\\
		E_4(x,q)\\
		\vdots
	\end{bmatrix}
	= \left(
	\begin{array}{rrrrrr}
		1& 0 & 0 & 0 &\ \ 0 &\cdots\\
		q& 1 & 0 & 0 & 0& \cdots \\
		q^2 & 2q & 1 & 0 &0 & \cdots\\
		q^3& 3q^2 &3q & 1 & 0& \cdots \\
		q^4& 4q^3 & 6q^2 & 4q & 1 & \cdots \\
		\vdots	& \vdots & \vdots & \vdots & \vdots & \ddots 
	\end{array}
	\right)\ 
	\begin{bmatrix}
		A_0(x,q)\\
		A_1(x,q)\\
		A_2(x,q)\\
		A_3(x,q)\\
		A_4(x,q)\\
		\vdots
	\end{bmatrix}. \label{equ50}
\end{equation}
That is,
\begin{equation}
	\left[E_n(x,q)\right]_n = \left[ \exp(qt), t \right]\ \left[A_n(x,q)\right]_n. \label{equ51}
\end{equation}
\end{theorem}

\begin{proof}
Setting $k=0$ in Equation~(\ref{equ48}), we get the result.
\end{proof}

\begin{theorem} \label{theo32}
For a fixed integer $n$ with $n \geq 0$, the exponential generating function $A(x,t,q) = \sum_{n \geq 0} A_{n}(x,q) \frac{t^n}{n!}$ has the closed form
\begin{equation}
	A(x,t,q) = \exp(-qt)
	\left(\frac{x-1}{x-\exp\left(t(x-1)\right)}\right)^q. \label{equ52}
\end{equation}
\end{theorem}

\begin{proof}
Setting $n=0$ in Equation~(\ref{equ46}), we get
\begin{equation*}
	\left[ e_0^k(x,q) \right]_{k \geq 0} = \left[ \exp(-qt), t \right]\, \left[E_k(x,q)\right]_{k \geq 0}.
\end{equation*}
Hence 
\begin{equation*}
	\sum_{k \geq 0} \frac{E_k(x,q)}{k!}t^k =
        \left(\frac{x-1}{x-\exp\left(t(x-1)\right)}\right)^q
        \text{(cf.\ Brenti~\cite{bib1})},
\end{equation*}
and from Theorem~\ref{theo4} we get the result.
\end{proof}

\begin{corollary}\label{theo33}
We have
\begin{equation}
	A(x,t,q) = \left(A(x,t)\right)^q.
	\label{equ53}
\end{equation}
\end{corollary}

\begin{proof}
We have
\begin{align*}
	A(x,t) &= \frac{(1-x)\exp(-xt)}{1-x\exp\left((1-x)t\right)}
        \text{ (cf.\ Mantaci and Rakotondrajao~\cite{bib8}) }\\
	&= \frac{(x-1)\exp(-t)}{x-\exp\left(t(x-1)\right)}.
\end{align*}
Hence
\begin{align*}
	A(x,t,q) &= \exp(-qt)
	\left(\frac{x-1}{x-\exp\left(t(x-1)\right)}\right)^q\\
	&= 
	\left(\frac{(x-1)\exp(-t)}{x-\exp\left(t(x-1)\right)}\right)^q.
\end{align*}
It follows that $A(x,t,q) = \left(A(x,t)\right)^q$.
\end{proof}

\begin{theorem} \label{theo34}
For a fixed integer $n$ with $n \geq 0$, the exponential generating function $F_n(x,t,q) = \sum_{k \geq 0} e_n^k(x,q) \frac{t^k}{k!}$ of $(n,k)$-permutations has the closed form
\begin{equation}
	F_n(x,t,q) = \exp(-qt)
	\frac{d^n}{dt^n} \left(\frac{x-1}{x-\exp\left(t(x-1)\right)}\right)^q. \label{equ54}
\end{equation}
\end{theorem}

\begin{proof}
From Equation~(\ref{equ46}) and Theorem~\ref{theo4}, we get
\begin{align*}
	F_n(x,t,q) &= \exp(-qt)\ \sum_{k \geq 0} E_{n+k}(x,q) \frac{t^k}{k!}\\
	&= \exp(-qt) \frac{d^n}{dt^n} \left(\sum_{k \geq 0} E_k(x,q) \frac{t^k}{k!}\right) \\
	&= \exp(-qt) 
	\frac{d^n}{dt^n} \left(\frac{x-1}{x-\exp\left(t(x-1)\right)}\right)^q.
\qedhere\end{align*}
\end{proof}

\begin{corollary}\label{theo35}
We have
\begin{equation}
	F_n(x,t,q) = \exp(-qt)\ \frac{d^n}{dt^n} E(x,t,q).\label{equ55}
\end{equation}
\end{corollary}

\begin{theorem} \label{theo36}
For a fixed integer $k$ with $k \geq 0$, the exponential generating function $F_k(x,t,q) = \sum_{n \geq 0} e_n^k(x,q) \frac{t^n}{n!}$ of $(n,k)$-permutations has the closed form
\begin{equation}
	F_k(x,t,q) = \exp(qt) 
	\frac{d^k}{dt^k} \left(\frac{(x-1)\exp(-t)}{x-\exp\left(t(x-1)\right)}\right)^q. \label{equ56}
\end{equation}
\end{theorem}

\begin{proof}
From Equation~(\ref{equ49}) and Theorem~\ref{theo4}, we get
\begin{align*}
	F_k(x,t,q) &= \exp(qt)\ \sum_{k \geq 0} A_{k+n}(x,q) \frac{t^k}{k!}\\
	&= \exp(qt) \frac{d^k}{dt^k} \left(\sum_{n \geq 0} A_n(x,q) \frac{t^n}{n!}\right) \\
	&= \exp(qt) 
	\frac{d^k}{dt^k} \left(\frac{(x-1)\exp(-t)}{x-\exp\left(t(x-1)\right)}\right)^q.
\qedhere\end{align*}
\end{proof}

\begin{corollary}\label{theo37}
We have
\begin{equation}
	F_k(x,t,q) = \exp(qt)\ \frac{d^k}{dt^k} A(x,t,q). \label{equ57}
\end{equation}
\end{corollary}

\begin{theorem} \label{theo38}
The exponential generating function
$$F(x,t,u,q) = \sum_{n \geq 0}\sum_{k \geq 0} e_n^k(x,q)
\frac{u^k}{k!}\frac{t^n}{n!}$$
of $(n,k)$-permutations has the closed form
\begin{equation}
	F(x,t,u,q) = \exp(-qu)\left(\frac{x-1}{x-\exp\left((x-1)(u+t)\right)}\right)^q. \label{equ58}
\end{equation}
\end{theorem}

\begin{proof}
From Equation~(\ref{equ43}) and Theorem~\ref{theo3}, we have
\begin{align*}
	F(x,t,u,q) &= \sum_{n \geq 0}\sum_{k \geq 0} e_n^k(x,q) \frac{u^k}{k!}\frac{t^n}{n!}\\
	&= \exp(-qu)E(x,t+u,q)\\
	&= \exp(-qu)\left(\frac{x-1}{x-\exp\left((x-1)(u+t)\right)}\right)^q.
\end{align*}
Here, the last line follows from Equation~(\ref{equ19}).
\end{proof}

\begin{corollary}\label{theo39}
We have
\begin{equation}
	F(x,t,u,q) = \left(F(x,t,u)\right)^q. \label{equ59}
\end{equation}
\end{corollary}

\begin{proof}
We have
\begin{align*}
	\left(F(x,t,u)\right)^q
	&= \left(\frac{(x-1)\exp(-u)}{x-\exp\left((x-1)(u+t)\right)}\right)^q\\
	&= \exp(-qu)\left(\frac{x-1}{x-\exp\left((x-1)(u+t)\right)}\right)^q.
\qedhere\end{align*}
\end{proof}

%-------------------------------------------------------------------
\section*{Acknowledgements}
One of the authors, Ratovilebamboavison Dimbiniaina, would like to thank the European Union for the financial support through the project ``{Appui \`a la Formation, \`a la Recherche et \`a l’Innovation pour le D\'eveloppement Intra-Afrique (AFRIDI)}''.

\begin{thebibliography}{20}
	
\bibitem{bib1}
F. Brenti, \emph{A class of $q$-symmetric functions arising from plethysm}, Journal of Combinatorial Theory, Series~A, 91  (2000) 137--170.

\bibitem{bib2}
R. J. Clarke, G.-N. Han and J. Zeng, \emph{A combinatorial interpretation of the Seidel generation of $q$-derangement numbers}, Annals of Combinatorics, 4 (1997) 313--327.

\bibitem{bib3}
L. Comtet, \emph{Analyse combinatoire}, tome second, PUF, 1970.

\bibitem{bib4}
D. Dumont, A. Randrianarivony, \emph{D\'erangements et nombres de Genocchi}, Discrete Mathematics, 132 (1994) 37--49.

\bibitem{bib5}
D. Dumont, {\em Matrices d'Euler--Seidel}, S\'eminaire Lotharingien de
Combinatoire, 5 (1981) Art.~B05c, 16~pp.,
formely Publ. I.R.M.A. Strasbourg, 182 (1981) 59--76.

\bibitem{bib6}
M. C. Firengiz and Q. Dil, \emph{Generalized Euler--Seidel method for second order recurrence relations}, Notes on Number Theory and Discrete Mathematics, ISSN 1310-5132, 20(4) (2014) 21--32.

\bibitem{bib7}
D. Foata and M.-P. Sch\"{u}tzenberger, \emph{Th\'{e}orie g\'{e}om\'{e}trique des polyn\^{o}mes eul\'{e}riens}, Berlin, Heidelberg, New York, Springer-Verlag, 1970.

\bibitem{bib8}
R. Mantaci and F. Rakotondrajao, \emph{Exceedingly deranging!}, Advances in Mathematics, 30 (2003) 177--188.

\bibitem{bib9}
  F. Rakotondrajao, \emph{On Euler's difference table}, Proc. FPSAC'07, Nankai University, Tianjin, China, 2007; available at
  {\tt http://igm.univ-mlv.fr/\~{}fpsac/FPSAC07/SITE07/PDF-Proceedings/Posters/22.pdf}.

\bibitem{bib10}
F. Rakotondrajao, \emph{$k$-Fixed-points-permutations},
Pure Mathematics and Applications, 17 (2006) 165--173.

\bibitem{bib11}
D. P. Roselle, \emph{Permutations by number of rises and successions}, Proceedings of the American Mathematical Society, 19 (1968) 8--16.

\bibitem{bib12}
L. W. Shapiro, \emph{Bijections and the Riordan group}, Theoretical Computer Science, 307 (2003) 403--413.

\bibitem{bib13}
L. W. Shapiro, S. Getu, W. J. Woan and L.C. Woodson, \emph{The Riordan group}, Discrete Applied Mathematics, 34 (1991) 229--239.

\bibitem{bib14}
R. Sprugnoli, \emph{Riordan arrays and combinatorial sums}, Discrete Mathematics, 132 (1994) 267--290.

\bibitem{bib15}
R. P. Stanley, \emph{Enumerative Combinatorics, Volume~I}, Cambridge University Press, Cambridge, UK, 1997.

\bibitem{bib16}
J. Worpitzky, \emph{Studien \"{u}ber die Bernoullischen und Eulerschen Zahlen}, Crelle's Journal, 94 (1883) 203--232.
\end{thebibliography}

\end{document}



