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\begin{center} \vskip 1cm{\LARGE\bf 
Associated Stirling Eulerian Numbers \\
\vskip .1in and Their $q$-Analogues }
\vskip 1cm 
Imboa-Navalona Manjaka Rakotoson and Fanja Rakotondrajao\\
Department of  Mathematics and Computer Science\\
Antananarivo University \\
 101 Antananarivo\\
Madagascar\\ 
\href{mailto:2knaval@gmail.com}{\tt 2knaval@gmail.com} \\ \href{mailto:2knaval@gmail.com}{\tt frakoton@yahoo.fr}\\
\end{center}

\begin{abstract}
In this paper, we study the excedance distribution over permutations while considering the parameters of cycle length and the number of cycles. We refer to the number of such permutations as the \emph{associated Stirling Eulerian number}. Moreover, if we consider the permutations in which the first $s$ integers are in different cycles, we denote their count as the \emph{associated $s$-Stirling Eulerian number}. We provide a formula defining these numbers, along with their generating functions. We establish $q$-analogues and offer extensions of the results.

\end{abstract}

\section{Introduction}

The Stirling numbers of the first kind, denoted $S_{n,k}$, count the permutations with $k$ cycles over $n$ objects. These are classical numbers. Appell \cite{Appell}, Carlitz \cite{Carlitz1,Carlitz2}, Comtet \cite{Comtet}, Foata \cite{Foata}, Riordan \cite{Riordan}, Tricomi \cite{Tricomi} have defined the \textit{associated Stirling numbers of the first kind}, denoted $d(n,k)$, which count the numbers of permutations over $n$ objects having exactly $k$ cycles, where each cycle has a length $2$ or greater. These numbers are defined as follows: 

\begin{equation}\label{dnk}
d(n+1,k)=n\big(d(n,k)+d(n-1,k-1)\big),
\end{equation}
with $d(n,1)=(n-1)!\ $ and $d(n,k)=0$ for $n\leq 2k-1.$

The generating function $D(q,u)=\sum_k\sum_nd(n,k)q^k \frac{u^n}{n!}$ has the closed form \begin{equation}\label{qegfdn}
D(q,u)=(\exp (-u)/(1-u))^{q}.
\end{equation}

Belbachir and Bousbaa \cite{Belb}, as well as Broder \cite{Broder} have conducted in-depth studies of these numbers. 


\begin{definition}
An $r$-\textit{cycle} is a cycle in a permutation of length $r$. A \textit{circular permutation} over $[n]$ is a permutation consisting of an $n$-cycle. An $r$-\textit{cycle permutation} is a permutation in which each cycle has length $r$. 
\end{definition}


An $r^-$-\textit{cycle} (respectively, $r^+$-\textit{cycle}) \textit{permutation} is a permutation where each cycle length is at most $r$ (respectively, greater than $r$). An \textit{involution} is a $2^-$-cycle permutation. Following Broder's definitions \cite{Broder}, we define the number of $r^-$-cycle (respectively, $r^+$-cycle) permutations as the $r^-$-\textit{associated} (respectively, $r^+$-\textit{associated}) \textit{Stirling number of the first kind}.

An $(r^-,k)$-\emph{cycle} (respectively, $(r^+,k)$-\emph{cycle}) \emph{permutation} is an $r^-$-cycle (respectively, $r^+$-cycle) permutation consisting of $k$ cycles. If each cycle has length $r$, we refer to it as an $((r),k)$-\textit{cycle permutation}. Furthermore, if the first $s$ integers lie in distinct cycles, we define an $(r^-,k,s)$-\emph{cycle} (respectively, $(r^+,k,s)$-\emph{cycle}) \emph{permutation} as an $(r^-,k)$-cycle (respectively, $(r^+,k)$-cycle) permutation. The $r^-$-\textit{associated} $s$-\textit{Stirling number of the first kind} (respectively, $r^+$-\textit{associated} $s$-\textit{Stirling number of the first kind}) counts the number of such permutations. If the length of each cycle is equal to $r$, we call it an $((r),k,s)$-\emph{cycle permutation}.

An $(r^-,\star,s)$-cycle permutation (respectively, $(r^+,\star,s)$-cycle permutation) is an $r^-$-cycle (respectively, $r^+$-cycle) permutation where the first $s$ integers are in distinct cycles.

We can generalize this by considering an $(r_1)\cdots(r_\ell)$-\textit{cycle permutation} consisting of cycles of lengths $1 < r_1 < \cdots < r_\ell$, and $r_1 \cdots r_s$-\textit{cycle permutations}, where the first $s$ integers are in distinct cycles of lengths $r_i > 1$ for $i = 1, \ldots, s$.

An $((r_1)\cdots(r_\ell),k)$-\textit{cycle permutation} is an $(r_1)\cdots(r_\ell)$-cycle permutation with $k$ cycles, and an $((r_1)\cdots(r_\ell),k,s)$-\textit{cycle permutation} has the first $s$ integers in distinct cycles. An $r_1 \cdots r_s (r_{s+1}) \cdots (r_\ell)$-\textit{cycle permutation} is a permutation where the first $s$ integers belong to different cycles, each of length $r_i > 1$ for $i = 1, \ldots, s$, and the remaining cycles have lengths $1 < r_{s+1} < \cdots < r_\ell$.

An $r_1^\star \cdots r_\ell^\star$-\textit{cycle permutation} is a permutation that does not contain any cycles of length $r_i$ for $1 \leq i \leq \ell$.

\begin{definition}
An integer $i$ is a \textit{fixed point} of a permutation $\sigma$ if $\sigma(i)= i$.
\end{definition}


A \textit{derangement} is a permutation without fixed points. We can also define a derangement as a $1^+$-cycle permutation. 

\begin{definition}[Rakotondrajao \cite{Fanja1}] An $s$-\textit{fixed-points-permutation} is a permutation whose set of fixed points is a subset of the set $[s]$, and the first $s$ integers lie in different cycles.
\end{definition}

For example, the permutation $(1\ 4)(2)(3)(5\ 7)(6)$ is a $(0^+,5,3)$-cycle permutation and not a $3$-fixed-points-permutation.

\begin{definition}
An integer $i$ is an \textit{excedance} for a permutation $\sigma$ if $\sigma(i) >i$.
\end{definition}

The number of permutations of $n$ objects with exactly $m$ excedances is the classical \emph{Eulerian number} $A_{n,m}$. They satisfy the following relation:
\begin{equation}\label{eulerian}
A_{n,m}=(n-m)A_{n-1,m-1}+(m+1)A_{n-1,m},
\end{equation}
with $A_{0,0}=1$ and $A_{0,1}=0$. The egf of the Eulerian polynomial \begin{equation}\label{polE}
    A_n(x)=\sum_{\sigma \in S_n}x^{e(\sigma)}=\sum_m A_{n,m}x^m
\end{equation}
has the closed form 
\begin{equation}\label{egfE}
A(x,u)=\sum_n A_n(x)\dfrac{u^n}{n!}=\dfrac{x-1}{x-\exp{((x-1)u)}}.
\end{equation}

Many works on Eulerian numbers can be easily found in the literature, including those by Foata and Sch\"utzenberger \cite{Foata}, Mantaci and Rakotondrajao \cite{Mantaci-Fanja}, Riordan \cite{Riordan}, Stanley \cite{Stanley}. The function $\mathcal{P}(x,u)$ defined by
\begin{equation}\label{primitive1}
\mathcal{P}(x,u)=\dfrac{\ln(A(x,u))+u(x-1)}{x},
\end{equation}
is the primitive of $A(x,u)$ and is computed combinatorially in Sec. \ref{r-per}. 

Note that a permutation is a $0^+$-cycle permutation, and a derangement a $1^+$-cycle permutation.  The length of an $r$-cycle permutation is a multiple of $r$ and the length of an $((r),k)$-cycle permutation is equal to $kr$. We are interested in the study of associated Stirling Eulerian numbers. More precisely, we consider the parameter of the number of cycles with the constraint on the length of cycles in the study of excedance distribution over permutations. We also study the same distribution by adding the placement of the first $s$ integers. Recurrence relations, formulas, exponential generating functions, and their $q$-analogues are provided. We extend Brenti's results \cite[Sec. 7]{Brenti} on his study of a $q$-analogue of the Eulerian polynomials, as well as general identities. A general identity extends the result on derangement numbers established by Ksavrelof and Zeng \cite{Ksaverlof}, along with new identities. We provide generalizations of our results by specifying the lengths of cycles. Note that the associated Stirling numbers of the first kind $d(n,k)$ in Equation \eqref{dnk} enumerate the $(1^+,k)$-cycle permutations, while $D(q,x)$ in Equation \eqref{qegfdn} is a $q$-analogue of the exponential generating function (egf) of the derangement numbers. We denote by $e(\sigma)$ the number of excedances of the permutation $\sigma$ and by $c(\sigma)$ the number of cycles of $\sigma$. The parameter $q$ associated with the number of cycles defines the $q$-analogue in our study. We use the expression \textit{Stirling number} to refer to the Stirling number of the first kind and introduce the notation in each section throughout this paper.
  

\vphantom{a}
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\section{Circular permutations}\label{circper}

\begin{longtable}[htb]{cll}
\\\noalign{\smallskip}\hline\noalign{\smallskip} $\mathcal{C}_n$& {\footnotesize the set of circular permutations over $[n]$}& \\
 $C_{n,m}$& $|\{\sigma\in \mathcal{C}_n: e(\sigma)=m\}|$ &{\footnotesize Th.~\ref{t2.1} Eq.~\eqref{CnAn-1}}\\$C_n(x)$&$\sum_m C_{n,m}x^{m}=\sum_{\sigma\in \mathcal{C}_n}x^{e(\sigma)}$& {\footnotesize Th.~\ref{t2.1} Eq.~\eqref{Cnx}} \\\smallskip\\$C(x,u)$&$\sum_n\sum_{\sigma\in\mathcal{C}_n}x^{e(\sigma)}\dfrac{u^n}{n!}=\sum_n C_n(x) \dfrac{u^n}{n!}$ &{\footnotesize Th.~\ref{circulartheorem} Eq.~\eqref{circular}}\\
 $\mathcal{C}_{r^-}$& {\footnotesize the set of $r^-$-cycle circular permutations}&\\
 $C^{r^-}(x,u)$& $\sum_{n=0}^{r}\sum_{\sigma\in\mathcal{C}_n}x^{e(\sigma)}\dfrac{u^n}{n!}=\sum_{n=0}^{r}C_{n}(x)\dfrac{u^n}{n!}$&{\footnotesize Th.~\ref{t.egfCr-} Eq.~\eqref{egfCr-}}\\
  & &{\footnotesize Cor.~\ref{c.egfCr-2} Eq.~\eqref{egfCr-2}}\\
  $\mathcal{C}_{r^+}$& {\footnotesize the set of $r^+$-cycle circular permutations}&\\
  $C^{(r)}(x,u)$& $\sum_{\sigma\in\mathcal{C}_r}x^{e(\sigma)}\dfrac{u^r}{r!}=C_{r}(x)\dfrac{u^r}{r!}$&{\footnotesize Th.~\ref{t.egfcr} Eq.~\eqref{egfcr}}\\
 $C^{r^+}(x,u)$& $\sum_{n\ge r+1}\sum_{\sigma\in\mathcal{C}_n}x^{e(\sigma)}\dfrac{u^n}{n!}=\sum_{n\ge r+1}C_{n}(x)\dfrac{u^n}{n!}$ &{\footnotesize Th.~\ref{t.egfcr+} Eq.~\eqref{egfcr+}}\\
 &&{\footnotesize Cor.~\ref{c.egfcr+2} Eq.~\eqref{egfcr+2}}
\\\noalign{\smallskip}\hline\noalign{\smallskip}\caption{Notation.}
\end{longtable}

\begin{theorem}\label{t2.1}
For all integers $n\geq 2$ and $1\leq m\leq n$, we have
\begin{equation}\label{CnAn-1}
C_{n,m}=A_{n-1,m-1} \mbox{\ with\ } C_{1,0}=1
\end{equation}
and \begin{equation}\label{Cnx}
C_{n}(x)=xA_{n-1}(x).
\end{equation}
\end{theorem}

\begin{proof}
Let $n\ge 2 $ and $m\ge1$. Let $\pi \in \mathcal{C}_n$ be a permutation such that $e(\pi)=m$. It is common to put the smallest letter at the beginning of the cycle. The integer $1$ is always an excedance for a circular permutation. The remaining permutation after this integer, which is an ordinary permutation over the set $\{2,\ldots, n\}$, has $m-1$ excedances.
Multiplying by $x^m$ and summing over $m$ in Equation \eqref{CnAn-1}, we get
\begin{equation*}
C_{n}(x)=xA_{n-1}(x).
\end{equation*}
\end{proof}
\begin{theorem}
\label{circulartheorem}
The exponential generating function of the excedance distribution over circular permutations has the closed form
\begin{equation}\label{circular}
C(x,u)=\sum_{n\geq 0}\sum_{\pi \in \mathcal{C}_n}x^{e(\pi)}\dfrac{u^n}{n!}=\ln{A(x,u)}.
\end{equation}
\end{theorem}

\begin{proof}
Since 
$C_{n}(x)=xA_{n-1}(x)$, we get 
\begin{align*}
\sum_{n\geq 2}C_{n}(x)\dfrac{u^n}{n!}&=\displaystyle{\sum_{n\geq 2}}xA_{n-1}(x)\dfrac{u^n}{n!}, \\&=x(\mathcal{P}(x,u))-xu.
\end{align*} 
Then, we obtain
\begin{equation*}
u+\sum_{n\geq 2}C_{n}(x)\dfrac{u^n}{n!}=u+x\dfrac{\ln(A(x,u))+u(x-1)}{x}-xu. 
\end{equation*}
Therefore, we have
\begin{equation*}
C(x,u)=\ln(A(x,u)).
\end{equation*}
\end{proof}

\begin{theorem}\label{t.egfCr-}
For a fixed integer $r\ge 1$, the exponential generating function of the excedance distribution over circular $r^-$-cycle permutations has the closed form  
\begin{equation}\label{egfCr-}
C^{r^{-}}(x,u)=\sum_{n= 1}^{r} C_n(x) \dfrac{u^n}{n!}=u+x\sum_{n=1}^{r-1}A_{n}(x)\dfrac{u^{n+1}}{(n+1)!\ }.
\end{equation}
\end{theorem}

\begin{proof}
From Equation~\eqref{Cnx}, we have
\begin{align*}
\sum_{n= 2}^{r}C_{n}(x)\dfrac{u^{n}}{n!}&=x\sum_{n= 2}^{r}A_{n-1}(x)\dfrac{u^{n}}{n!}. \\
\sum_{n= 1}^{r}C_{n}(x)\dfrac{u^{n}}{n!}&=u+x\sum_{n= 1}^{r-1}A_{n}(x)\dfrac{u^{n+1}}{(n+1)!\ }.
\end{align*}
\end{proof}

\begin{theorem}\label{t.egfcr}
For a fixed integer $r\geq 2$, the exponential generating function of the excedance distribution over circular $r$-cycle  permutations has the closed form  
\begin{equation}\label{egfcr}
C^{(r)}(x,u)=xA_{r-1}(x)\dfrac{u^r}{r!}.
\end{equation}
\end{theorem}

\begin{theorem}\label{t.egfcr+}
For a fixed integer $r\geq 2$, the exponential generating function of the excedance distribution over circular $r^+$-cycle permutations has the closed form  
\begin{equation}\label{egfcr+}
C^{r^+}(x,u)=\sum_{n\geq r+1} C_n(x) \dfrac{u^n}{n!}=\ln \Biggl( A(x,u)\exp\Bigl(-u-x\sum_{i=1}^{r-1}A_{i}(x)\dfrac{u^{i+1}}{(i+1)!\ }\Bigr)\Biggr).
\end{equation}
\end{theorem}

\begin{proof} We have 
\begin{align*}
C^{r^+}(x,u)&=C(x,u)-C^{r^-}(x,u),\\
&=\ln{A(x,u)}-u-x\sum_{i=1}^{r-1}A_{i}(x)\dfrac{u^{i+1}}{(i+1)!\ },\\&=\ln\Biggl(A(x,u)\exp\Bigl(-u-x\sum_{i=1}^{r-1}A_{i}(x)\dfrac{u^{i+1}}{(i+1)!\ }\Bigr)\Biggr).
\end{align*}
\end{proof}

\section{$r^-$-cycle permutations}\label{r-per}

\begin{longtable}[htb]{cll}\\ \noalign{\smallskip}\hline\noalign{\smallskip}
$(n_1,\ldots, n_k)!$&$\dfrac{n_1+\cdots+n_k}{n_1!\cdots n_k!}$&\\
$\mathcal{S}_n^{r^-}$& {\footnotesize the set of $r^-$-cycle permutations}&\\
 $A_{n,m}^{r^-}$& $|\{\sigma\in \mathcal{S}_n^{r^-}: e(\sigma)=m\}|$ & {\footnotesize Th.~\ref{t2.2} Eq.~\eqref{recsmallerR}}\\&{\footnotesize \emph{ $r^-$-associated Stirling Eulerian number }} &\\
  $A_{n}^{r^-}(x)$& $\sum_m  A_{n,m}^{r^-}x^m=\sum_{\sigma\in \mathcal{S}_n^{r^-}}x^{e(\sigma)}$&\\
 $A^{r^-}(x,u)$& $\sum_{n}\sum_{\sigma\in \mathcal{S}_n^{r^-}}x^{e(\sigma)}\dfrac{u^n}{n!}=\sum_{n}A^{r^-}_{n}(x)\dfrac{u^n}{n!}$ &{\footnotesize Th.~\ref{t.egfCycleSmallerOrEqualToR} Eq.~\eqref{egfCycleSmallerOrEqualToR}}\\
 $\mathcal{P}(x,u)$ &{\footnotesize the primitive of $A(x,u)$ with respect to $u$}&{\footnotesize Cor.~\ref{t.primitive} Eq.~\eqref{primitive} } \\
 $\mathcal{S}_n^{r^-,k}$& {\footnotesize the set of $(r^-,k)$-cycle permutations}&\\
$A^{r^-}_{n,m,k}$ &  $|\{\sigma\in \mathcal{S}_n^{r^-,k}: e(\sigma)=m\}|$&\\
$A^{r^-}_{n,k}(x)$& $\sum_m A^{r^-}_{n,m,k}x^m=\sum_{\sigma \in \mathcal{S}_n^{r^-,k}}x^{e(\sigma)}$ & \\
$A^{r^-}_k(x,u)$&$\sum_n\sum_{\sigma \in \mathcal{S}_n^{r^-,k}}x^{e(\sigma)}\dfrac{u^n}{n!}=\sum_nA^{r^-}_{n,k}(x)\dfrac{u^n}{n!}$
&{\footnotesize Th.~\ref{t.EgfCycleCK{r-}} Eq.~\eqref{EgfCycleC^K{r-}}}\\\smallskip
$A^{r^-}(q,x,u)$&$\sum_n\sum_{\sigma \in \mathcal{S}_n^{r^-,k}}x^{e(\sigma)}q^{c(\sigma)}\dfrac{u^n}{n!}$ {\footnotesize  $q$-analogue of $A^{r^-}(x,u)$}&{\footnotesize Th.~\ref{t.qegfAr-} Eq.~\eqref{qegfAr-}}\\
 &&{\footnotesize Cor.~\ref{c.qegfAr-} Eq.~\eqref{qegfAr-2}}
  \\
 $\mathcal{S}_n^{r^-,\star,s}$& {\footnotesize the set of $(r^-,\star,s)$-cycle permutations}&\\
  $[A_{n,m}^{r^-}]_s$& $|\{\sigma\in \mathcal{S}_n^{r^-,\star,s}: e(\sigma)=m\}|$ & \\
 $[A_{n}^{r^-}(x)]_s$& $\sum_m [A_{n,m}^{r^-}]_sx^m=\sum_{\sigma\in \mathcal{S}_n^{r^-,\star,s}}x^{e(\sigma)}$ & \\\smallskip\\
 $[A^{r^-}(x,u)]_s$&$\sum_n\sum_{\sigma\in \mathcal{S}_n^{r^-,\star,s}}x^{e(\sigma)}\dfrac{u^n}{n!}=\sum_n[A^{r^-}_{n}(x)]_s\dfrac{u^n}{n!}$ &{\footnotesize Th.~\ref{t.egfAr-s} Eq.~\eqref{egfAr-s}}
  \\
    $[A_{n,m,k}^{r^-}]_s$&  $|\{\sigma\in \mathcal{S}_n^{r^-,k,s}: e(\sigma)=m\}|$ & \\\smallskip\\
 $[A_{n,k}^{r^-}(x)]_s$& $\sum_m [A_{n,m,k}^{r^-}]_sx^m=\sum_{\sigma\in \mathcal{S}_n^{r^-,k,s}}x^{e(\sigma)}$ & \\\smallskip\\
 $[A^{r^-}_k(x,u)]_s$&$\sum_n\sum_{\sigma\in \mathcal{S}_n^{r^-,k,s}}x^{e(\sigma)}\dfrac{u^n}{n!}=\sum_n[A^{r^-}_{n,k}(x)]_s\dfrac{u^n}{n!}$ &{\footnotesize 
 Th.~\ref{t.A(r-)s,k(x,u)} Eq.~\eqref{A(r-)s,k(x,u)}}
  \\\smallskip\\
  
  $\left[A^{r^{-}}(q, x, u)\right]_s$ & $\sum_n\sum_{\sigma\in \mathcal{S}_n^{r^-,\star,s}}x^{e(\sigma)}q^{c(\sigma)}\dfrac{u^n}{n!}$ {\footnotesize  $q$-analogue of $[A^{r^-}(x,u)]_s$ }&{\footnotesize Th.~\ref{t.A(r-)(s)(q,x,u)} Eq.~\eqref{A(r-)(s)(q,x,u)}}\\
 \noalign{\smallskip}\hline\noalign{\smallskip}\caption{Notation.}
\end{longtable}

\begin{theorem}\label{t2.2}
For a fixed integer $r\ge 2$, for all integers $n$ and $m$ such that $0\leq m\leq n$, we have
\begin{equation}\label{recsmallerR}
A^{r^{-}}_{n+1,m}=\sum_{j=1}^{r-1}\binom{n}{j}\sum_{i\geq 0}C_{j+1,i}A^{r^{-}}_{n-j,m-i}+A^{r^{-}}_{n,m}.
\end{equation}
\end{theorem}

\begin{proof}
Consider the integer $n+1$ in an $r^-$-cycle permutation of $\mathcal{S}_{n+1}^{r^-}$.
\begin{enumerate} 
\item If $n+1$ is a fixed point, the remaining elements form a permutation of $[n]$ with $m$ excedances, where the lengths of the cycles are at most $r$. Therefore, there are $A^{r^-}_{n,m}$ possibilities. 

\item Otherwise, we have $\binom{n}{j} C_{j+1,i}$ ways to form a cycle of length $j+1$ with $i$ excedances that contains the integer $n+1$, for $1 \leq j \leq n$. The remaining elements of $[n]$ form a permutation with $m-i$ excedances, whose cycle lengths are at most $r$. Thus, we get $\sum_{j=1}^{r-1} \binom{n}{j} \sum_{i \geq 0} C_{j+1,i} A^{r^-}_{n-j,m-i}$ possibilities. 
\end{enumerate}
\end{proof}

\begin{theorem}\label{t.egfCycleSmallerOrEqualToR}
For a fixed integer $r\ge 2$, the exponential generating function of the $r^-$-associated Stirling Eulerian numbers has the closed form
\begin{equation}\label{egfCycleSmallerOrEqualToR}
A^{r^{-}}(x,u)=\exp{\Biggl(u+x\sum_{j=1}^{r-1}A_{j}(x)\frac{u^{j+1}}{(j+1)!\ }\Biggr)}.
\end{equation}
\end{theorem}

\begin{proof}
Since
\begin{equation*}\label{recAr-n}
A^{r^{-}}_{n+1,m}=\sum_{j=1}^{r-1}\binom{n}{j}\sum_{i\geq 0}C_{j+1,i}A^{r^{-}}_{n-j,m-i}+A^{r^{-}}_{n,m},
\end{equation*}
we have
\begin{equation*}\label{Ar-x}
A^{r^{-}}_{n+1}(x)=\sum_{j=1}^{r-1}\binom{n}{j}C_{j+1}(x)A^{r^{-}}_{n-j}(x)+A^{r^{-}}_{n}(x).
\end{equation*}
We get 
\begin{align*}
\dfrac{\partial}{\partial u}A^{r^{-}}(x,u)&=\sum_{n\geq 0}\sum_{j=1}^{r-1}\binom{n}{j}C_{j+1}(x)A^{r^{-}}_{n-j}(x)\frac{u^{n}}{n!}+A^{r^{-}}(x,u), \\
   &=\sum_{j=1}^{r-1}C_{j+1}(x)\frac{u^{j}}{j!}A^{r^{-}}_{n-j}(x)\frac{u^{n-j}}{(n-j)!\ }+A^{r^{-}}(x,u), \\
  &=\sum_{j=1}^{r-1}C_{j+1}(x)\frac{u^{j}}{j!}A^{r^{-}}(x,u)+A^{r^{-}}(x,u).
\end{align*}
Hence,
\begin{equation*}
\ln{(A^{r^{-}}(x,u))}=\sum_{j=1}^{r-1}C_{j+1}(x)\frac{u^{j+1}}{(j+1)!\ }+u.
\end{equation*}
Finally, we have
\begin{equation*}
A^{r^{-}}(x,u)=\exp{\Bigl(u+\displaystyle{\sum_{j=1}^{r-1}}C_{j+1}(x)\frac{u^{j+1}}{(j+1)!\ }\Bigr)}=\exp{\Bigl(u+x\displaystyle{\sum_{j=1}^{r-1}}A_{j}(x)\frac{u^{j+1}}{(j+1)!\ }\Bigr)}.
\end{equation*}
\end{proof}

\begin{corollary}\label{c.egfCr-2}
For all integers $r\ge 2$, we have \begin{equation}\label{egfCr-2}
C^{r^{-}}(x,u)=\ln{A^{r^{-}}(x,u)}.
\end{equation}
\end{corollary}
\begin{corollary}\label{t.primitive}
The primitive of the exponential generating function of the Eulerian numbers has the closed form
\begin{equation}\label{primitive}
\mathcal{P}(x,u)=\dfrac{\ln(A(x,u))+u(x-1)}{x}.
\end{equation}
\end{corollary}
\begin{proof}
A permutation without restrictions on the cycle length can be considered an ordinary permutation. From Equation~\eqref{egfCycleSmallerOrEqualToR}, we deduce
\begin{align*}
\exp(u+x\sum_{n\geq 1}A_{n}(x)\dfrac{u^{n+1}}{(n+1)!\ })&=A(x,u), \\
u+x\sum_{n\geq 1}A_{n}(x)\dfrac{u^{n+1}}{(n+1)!\ }&=\ln{A(x,u)}, \\
u+x(\mathcal{P}(x,u)-u)&=\ln{A(x,u)}. 
\end{align*}
Thus, we have
\begin{equation*}
\mathcal{P}(x,u)=\dfrac{\ln{A(x,u)+u(x-1)}}{x}.
\end{equation*}
\end{proof}

\begin{theorem}\label{t.EgfCycleCK{r-}}
For fixed integers $r\ge 2$ and $k$, the exponential generating function of the excedance distribution over $(r^-,k)$-cycle permutations has the closed form
\begin{equation}\label{EgfCycleC^K{r-}}
A^{r^{-}}_{k}(x,u)=\dfrac{(\ln{A^{r^-}(x,u)})^k}{k!}=\dfrac{\Biggl(u+x\displaystyle{\sum_{n= 1}^{r-1}}A_{n}(x)\dfrac{u^{n+1}}{(n+1)!\ }\Biggr)^{k}}{k!}.
\end{equation}
\end{theorem}

\begin{proof}
A permutation of $\mathcal{S}_n^{r^-,k}$ is a product of $k$ disjoint circular permutations of length smaller or equal than $r$, hence
\begin{equation*}\label{Ar-k}
A^{r^-}_{n,m,k}=\dfrac{1}{k!}\sum_{\substack{i_1, \ldots , i_k\le r\\
i_1+\cdots+i_k=n}}                   
\sum_{j_{1}+\cdots+j_{k}=m}(i_{1},\ldots , i_{k})!\  
C_{i_{1},j_{1}}\cdots C_{i_{k},j_{k}}. \end{equation*}
We get
\begin{equation*}\label{Ar-kx}
A^{r^-}_{n,k}(x)=\dfrac{1}{k!}\sum_{\substack{i_1, \ldots , i_k\le r\\
i_1+\cdots+i_k=n}}
(i_{1},\ldots ,i_{k})!\  C_{i_{1}}(x)\cdots C_{i_{k}}(x),\end{equation*}
and
\begin{align*}
A^{r^{-}}_{k}(x,u)&=\dfrac{1}{k!}(\sum_{n= 1}^{r}C_{n}(x)\dfrac{u^{n}}{n!})^k=\dfrac{(C^{r^{-}}(x,u))^{k}}{k!},\\
&=\dfrac{(\ln{A}^{r^{-}}(x,u))^{k}}{k!}, \mbox{ from Equation \eqref{egfCr-2},}\\&=\dfrac{1}{k!}\Bigl(u+x\displaystyle{\sum_{n= 1}^{r-1}}A_{n}(x)\dfrac{u^{n+1}}{(n+1)!\ }\Bigr)^{k} ,\mbox{ from Equation \eqref{egfCycleSmallerOrEqualToR}}.
\end{align*}
\end{proof}


\begin{theorem}\label{t.qegfAr-}
For a fixed integer $r\ge 2$, a $q$-analogue exponential generating function of the $r^-$-associated Stirling Eulerian numbers has the closed form
\begin{equation}\label{qegfAr-}
A^{r^{-}}(q,x,u)=\exp{\Biggl( q\Bigl(u+x\sum_{j=1}^{r-1}A_{j}(x)\frac{u^{j+1}}{(j+1)!\ }\Bigr)\Biggr)}.
\end{equation}
\end{theorem}

\begin{proof}
From equation~\eqref{EgfCycleC^K{r-}}, we get
\begin{align*}
\sum_{k\ge 0}A^{r^{-}}_{k}(x,u)q^{k}&=\displaystyle{\sum_{k\geq 0}}\dfrac{\Bigl(u+x\displaystyle{\sum_{j= 1}^{r-1}}A_{j}(x)\dfrac{u^{j+1}}{(j+1)!\ }\Bigr)^{k}}{k!}q^{k},\\&=\exp{\Biggl(q\Bigl(u+x\displaystyle{\sum_{j=1}^{r-1}}A_{j}(x)\frac{u^{j+1}}{(j+1)!\ }\Bigr)\Biggr)}.
\end{align*}
\end{proof}

\begin{corollary}\label{c.qegfAr-}
For all integer $r\ge 2$, we have
\begin{equation}\label{qegfAr-2}
A^{r^{-}}(q,x,u)=\Bigl(A^{r^{-}}(x,u)\Bigr)^{q}.
\end{equation}
\end{corollary}

\begin{theorem}\label{t.egfAr-s}
For a fixed integer $r\ge 2$, the partial differential exponential generating function of the $r^{-}$-associated $s$-Stirling Eulerian numbers has the closed form
\begin{equation}\label{egfAr-s}
\dfrac{\partial ^{s}[A^{r^{-}}(x,u)]_{s}}{\partial u^{s}}=\Biggl(1+x\sum_{i=1}^{r-1}A_{i}(x)\dfrac{u^{i}}{i!}\Biggr)^{s}A^{r^{-}}(x,u).
\end{equation}
\end{theorem}

\begin{proof}
 Considering the first $s-1$ integers and their cycles, we have
\begin{equation*}\label{Ar-s}
[A^{r^{-}}_{n+s,m}]_{s}=\sum_{\substack{i_{1}+\cdots+i_{s-1}=n \\ 0\le i_{1},\ldots , i_{s-1}\le r-1}}(i_{1},\ldots ,i_{s-1})!\ \sum_{j_{1}+\cdots+j_{s}=m}C_{i_{1}+1,j_{1}}\cdots C_{i_{s-1}+1,j_{s-1}} A^{r^{-}}_{n+1-\sum_{\ell=1}^{s-1}i_{\ell},j_{s}}.
\end{equation*}
We get
\begin{equation*}\label{Ar-sx}
[A^{r^{-}}_{n+s}(x)]_{s}=\sum_{\substack{i_{1}+\cdots+i_{s-1}=n \\ 0\le i_{1},\ldots , i_{s-1}\le r-1}}(i_{1},\ldots ,i_{s-1})!\  C_{i_{1}+1}(x)\cdots C_{i_{s-1}+1}(x) A^{r^{-}}_{n+1-\sum_{\ell=1}^{s-1}i_{\ell}}(x),\end{equation*}
and
\begin{align*}
\dfrac{\partial ^{s}[A^{r^-}(x,u)]_{s}}{\partial u^{s}}&=\Bigl(\sum_{i=0}^{r-1}C_{i+1}(x)\dfrac{u^{i}}{i!}\Bigr)^{s-1}\dfrac{\partial A^{r^{-}}(x,u)}{\partial u},\\&=\Bigl(1+\sum_{i=1}^{r-1}C_{i+1}(x)\dfrac{u^{i}}{i!}\Bigr)^{s-1}\Bigl(1+x\sum_{i=1}^{r-1}A_{i}(x)\dfrac{u^{i}}{i!}\Bigr)A^{r^{-}}(x,u),\\
&=\Bigl(1+x\sum_{i=1}^{r-1}A_{i}(x)\dfrac{u^{i}}{i!}\Bigr)^{s}A^{r^{-}}(x,u).
\end{align*}
\end{proof}

\begin{theorem}\label{t.A(r-)s,k(x,u)}
For fixed integers $r,\ k$ and $s$, the partial differential exponential generating function of the  excedance distribution over $(r^-,k,s)-$cycle permutations has the closed form
\begin{equation}\label{A(r-)s,k(x,u)}
\dfrac{\partial^s[A^{r^-}_k(x,u)]_s}{\partial u^s}=\Biggl( 1+x\sum_{i=1}^{r-1}A_{i}(x)\dfrac{u^i}{i!}\Biggr)^s\dfrac{\bigl(\ln{A^{r-}(x,u)}\bigr)^{k-s}}{(k-s)!\ }.
\end{equation}
\end{theorem}

\begin{proof}
Considering the first $s$ integers and their cycles, we have
\begin{equation*}\label{[A^r-_n,m,k]_s}
[A^{r^-}_{n+s,m,k}]_s=\sum_{\substack{i_{1}+\cdots+i_{s}=n \\ 0\le i_{1},\ldots , i_{s}\le r-1}}(i_{1},\ldots , i_{s})! \ \sum_{m_1+\cdots+m_s+m_t=m}C_{i_{1}+1,m_1}\cdots C_{i_{s}+1,m_s}A^{r^-}_{n-\sum_{l=1}^{s}i_{l},m_t,k-s}.
\end{equation*}
We get
\begin{equation*}\label{[A^r-_n,k(x)]_s}
[A^{r^-}_{n+s,k}(x)]_s=\sum_{\substack{i_{1}+\cdots+i_{s} =n\\ 0\le i_{1},\ldots , i_{s}\le r-1}}(i_{1},\ldots ,  i_{s})!\ C_{i_{1}+1}(x)\cdots C_{i_{s}+1}(x)A^{r^-}_{n-\sum_{l=1}^{s}i_{l},k-s}(x),
\end{equation*}
and
\begin{align*}
\dfrac{\partial^s[A^{r^-}_k(x,u)]_s}{\partial u^s}&=\Biggl(\sum_{i=0}^{r-1}C_{i+1}(x)\dfrac{u^{i}}{i!}\Biggr)^sA^{r^-}_{k-s}(x,u) =(1+\sum_{i=1}^{r-1}C_{i+1}(x)\dfrac{u^i}{i!})^s\dfrac{(\ln{A^{r^-}(x,u)})^{k-s}}{(k-s)!\ },\\
        &=\Biggl(1+x\sum_{i=1}^{r-1}A_{i}(x)\dfrac{u^i}{i!}\Biggr)^s\dfrac{(\ln{A^{r^-}(x,u)})^{k-s}}{(k-s)!\ }.
\end{align*}
\end{proof}

\begin{theorem}\label{t.A(r-)(s)(q,x,u)}
For fixed integers $r$ and $s$, a partial differential $q$-analogue exponential generating function of the excedance distribution over $(r^-,\star,s)$-cycle permutations has the closed form
\begin{equation}\label{A(r-)(s)(q,x,u)}
\dfrac{\partial ^{s}[A^{r^-}(q,x,u)]_{s}}{\partial u^s}=q^{s}\Biggl( 1+x\sum_{i=1}^{r-1}A_{i}(x)\dfrac{u^{i}}{i!}\Biggr)^s\Bigl(A^{r^-}(x,u)\Bigr)^q.
\end{equation}
\end{theorem}
 
\begin{proof}
We immediately have the result by multiplying by $q^{k}$ the right member of Equation~\eqref{A(r-)s,k(x,u)} and summing over $k$.
\end{proof}

\section{$r$-cycle permutations}\label{rper}

\begin{longtable}[htb]{cll}
\\\noalign{\smallskip}\hline\noalign{\smallskip} 
$\mathcal{S}_n^{(r)}$& {\footnotesize the set of $r$-cycle permutations}&\\
$A_{n,m}^{(r)}$&  $|\{\sigma\in \mathcal{S}_n^{(r)}: e(\sigma)=m\}|$ & {\footnotesize Th.~\ref{t.recurrenceOflengthr}  Eq.~\eqref{recurrenceOflengthr}}\\
 $A_{n}^{(r)}(x)$& $\sum_m  A_{n,m}^{(r)}x^m=\sum_{\sigma\in \mathcal{S}_n^{(r)}}x^{e(\sigma)}$&\\
 $A^{(r)}(x,u)$& $\sum_{n}A^{(r)}_{n}(x)\dfrac{u^n}{n!}$ the egf &{\footnotesize Th.~\ref{t.egfA(r)} Eq.~\eqref{egfA(r)}}\\
  $\mathcal{S}_n^{(r),k}$& {\footnotesize the set of $((r),k)$-cycle permutations}&\\
$A^{(r)}_{n,m,k}$ &  $|\{\sigma\in \mathcal{S}_n^{(r),k}: e(\sigma)=m\}|$&\\
$A^{(r)}_{n,k}(x)$& $\sum_m A^{(r)}_{n,m,k}x^m=\sum_{\sigma \in \mathcal{S}_n^{(r),k}}x^{e(\sigma)}$ & \\
$A^{(r)}_k(x,u)$&$\sum_n\sum_{\sigma \in \mathcal{S}_n^{(r),k}}x^{e(\sigma)}\dfrac{u^n}{n!}=\sum_nA^{(r)}_{n,k}(x)\dfrac{u^n}{n!}$
&{\footnotesize Th.~\ref{t.A(r)k(x,u)} Eq.~\eqref{A(r)k(x,u)}}\\
$A^{(r)}(q,x,u)$&$\sum_n\sum_{\sigma \in \mathcal{S}_n^{(r),k}}x^{e(\sigma)}q^{c(\sigma)}\dfrac{u^n}{n!}$ {\footnotesize  $q$-analogue of $A^{(r)}(x,u)$}&{\footnotesize Th.~\ref{t.qegfA(r)} Eq.~\eqref{qegfA(r)}}\\
 $\mathcal{S}_n^{(r),\star,s}$& {\footnotesize the set of $((r),\star,s)$-cycle permutations}&\\
  $[A_{n,m}^{(r)}]_s$&  $|\{\sigma\in \mathcal{S}_n^{(r),\star,s}: e(\sigma)=m\}|$& \\
 $[A_{n}^{(r)}(x)]_s$& $\sum_m [A_{n,m}^{(r)}]_sx^m=\sum_{\sigma\in \mathcal{S}_n^{(r),\star,s}}x^{e(\sigma)}$ &
  \\
 $[A^{(r)}(x,u)]_s$&$\sum_n[A^{(r)}_{n}(x)]_s\dfrac{u^n}{n!}$ {\footnotesize  its egf}&{\footnotesize Th.~\ref{t.egfA(r)s} Eq.~\eqref{egfA(r)s}}\\
  
$[A^{(r)}_k(x,u)]_s$&$\sum_n\sum_{\sigma\in\mathcal{S}_n^{(r),\star,s}}x^{e(\sigma)}\dfrac{u^n}{n!}$ &{\footnotesize Th.~\ref{t.A(r,s)k(x,u)} Eq.~\eqref{A(r,s)k(x,u)}}\\
$[A^{(r)}(q,x,u)]_s$&$\sum_n\sum_{\sigma\in\mathcal{S}_n^{(r),k,s}}x^{e(\sigma)}q^{c(\sigma)}\dfrac{u^n}{n!}$ {\footnotesize $q$-analogue of $[A^{(r)}(x,u)]_s$ 
}&{\footnotesize Th.~\ref{t.A(r)(s)(q,x,u)} Eq.~\eqref{A(r)(s)(q,x,u)}}
\\\noalign{\smallskip}\hline\noalign{\smallskip}\caption{Notation.}
\end{longtable}
 
\begin{theorem}\label{t.recurrenceOflengthr}
For a fixed integer $r\geq 2$, for all integers $n\geq m \geq 0$, we have the following formula 
\begin{equation}\label{recurrenceOflengthr}
A^{(r)}_{n+1,m}=\binom{n}{r-1}\sum_{k\geq 1}C_{r,k}A^{(r)}_{n-(r-1),m-k}.
\end{equation}
\end{theorem}


\begin{proof}
First, consider the integer $n+1$ and construct the cycle that contains it in a permutation of $\mathcal{S}_{n+1}^r$. To do this, we choose $r-1$ integers from $[n]$ and form a circular permutation with $n+1$ that has $k$ excedances, where $1\leq k \leq m-1$. There are $\binom{n-1}{k}C_{r,k}$ possibilities for this step. Next, to obtain a permutation of $[n+1]$ with $m$ excedances, where each cycle length is equal to $r$, we consider the remaining $n-(r-1)$ integers and form a permutation of $[n-(r-1)]$ with $m-k$ excedances, ensuring that each cycle length is also equal to $r$. This construction gives us the desired result.
\end{proof}


\begin{theorem}\label{t.egfA(r)}
For a fixed integer $r\geq 2$, the exponential generating function of the numbers $A^{(r)}_{n,m}$ has the closed form
\begin{equation}\label{egfA(r)}
A^{(r)}(x,u)=\exp\Bigl(xA_{r-1}(x)\frac{u^{r}}{r!}\Bigr).
\end{equation}
\end{theorem}

\begin{proof}
From  Equation~\eqref{recurrenceOflengthr}, we get \begin{equation*}\label{A(r)x}
A^{(r)}_{n+1}(x)=\binom{n}{r-1}C_{r}(x)A^{(r)}_{n-r+1}(x).
\end{equation*}
Let us consider the following variable change $r-1=r'$, we have 
\begin{equation*}
A^{(r)}_{n+1}(x)=\binom{n}{r'}C_{r'+1}(x)A^{(r)}_{n-r'}(x).
\end{equation*}
We deduce
\begin{align*}
\dfrac{\partial}{\partial u}A^{(r)}(x,u)&=\sum_{n\geq 0}\binom{n}{r'} C_{r'+1}(x)A^{(r)}_{n-r'}(x)\frac{u^n}{n!}=\sum_{n\geq 0}C_{r'+1}(x)\frac{u^{r'}}{r'!}A^{(r)}_{n-r'}(x)\frac{u^{n-r'}}{(n-r')!\ } ,\\
       &=C_{r'+1}(x)\frac{u^{r'}}{r'!}A^{(r)}(x,u).\end{align*}
Thus,
\begin{equation*}
\dfrac{(A^{r}(x,u))_u'}{A^{(r)}(x,u)}=C_{r'+1}(x)\frac{u^{r'}}{r'!} .
\end{equation*}
Integrating with respect to $u$, we have
\begin{equation*}
A^{(r)}(x,u)=\exp\Bigl(C_{r'+1}(x)\frac{u^{r'+1}}{(r'+1)!\ }\Bigr)=\exp\Bigl(xA_{r'}(x)\frac{u^{r'+1}}{(r'+1)!\ }\Bigr).
\end{equation*}
Thus, $$A^{(r)}(x,u)=\exp\Bigl(xA_{r-1}(x)\frac{u^{r}}{r!}\Bigr).$$
\end{proof}

\begin{theorem}
\label{t.A(r)k(x,u)}
For fixed integers $r\geq 2$ and $k$, the exponential generating function of the excedance distribution over $((r),k)-$cycle permutations has the closed form
\begin{equation}\label{A(r)k(x,u)}
 A^{(r)}_k(x,u)=\dfrac{1}{k!}\left( xA_{r-1}(x)\dfrac{u^r}{r!}\right)^k=\dfrac{\bigl(\ln{A^{(r)}(x,u)}\bigr)^k}{k!}.
\end{equation}
\end{theorem}

\begin{proof}
A permutation of $S^{(r),k}_{n}$ is a product of $k$ disjoints circular permutation of lenght $r$, hence
\begin{equation*}\label{A(r)nmk}
A^{(r)}_{n,m,k}=\dfrac{1}{k!}\binom{n}{r}\cdots \binom{n-(k-1)r}{r}
\sum_{m_{1}+\cdots+m_{k}=m}C_{r,m_{1}}\times\cdots \times C_{r,m_{k}}.
\end{equation*}
We get
\begin{equation*}\label{A(r)nkx}
A^{(r)}_{n,k}(x)=\dfrac{1}{k!}\dfrac{n!}{(r!)^k}(C_{r}(x))^k.
\end{equation*}
We deduce
\begin{equation*}
A^{(r)}_{k}(x,u)=\dfrac{1}{k!}\dfrac{(C_r(x))^{k}u^{kr}}{(r!)^k}=\dfrac{1}{k!}\Bigl(xA_{r-1}(x)\dfrac{u^r}{r!}\Bigr)^k.
\end{equation*}
\end{proof}

\begin{theorem}\label{t.qegfA(r)}
For a fixed integer $r\geq 2$, a $q$-analogue exponential generating function of the number $A^{(r)}_{n,m,k}$ has the closed form
\begin{equation}\label{qegfA(r)}
A^{(r)}(q,x,u)=\exp{\Bigl(qxA_{r-1}(x)\dfrac{u^r}{r!}\Bigr)}.
\end{equation}
\end{theorem}

\begin{proof}
We immediately have the result by multiplying by $q^{k}$ the Equation~\eqref{A(r)k(x,u)} and summing over $k$.
\end{proof}

\begin{corollary}\label{c.Snr(r)}
For fixed integers $r\geq 2$ and $n$, we have
\begin{equation*}\label{Snr(r)}
\sum_{\pi\in \mathcal{S}_{n}^{r,k}} x^{e(\pi)}q^{c(\pi)}=\begin{cases}
      \dfrac{(rk)!\ }{(r!)^k\ k!}x^kq^k A_{r-1}(x)^k, &\mbox{ if\  } n=kr\ ;\\
      0 & \mbox{ otherwise.}
     \end{cases}
\end{equation*}   
\end{corollary}

\begin{corollary}\label{c.Snr(r)1}
For fixed integers $r\geq 2$ and $n$, we have
 \begin{equation*}\label{Snr(r)1}
     | \mathcal{S}_{n}^r|=\begin{cases}
      \dfrac{n!}{r^{n/r}\ (n/r)!\ },&\mbox{ if }  r\mid n\  ;\\
      0& \ otherwise.
     \end{cases}
 \end{equation*}   
\end{corollary}

\begin{corollary}\label{c.egfSnr(r)}
For a fixed integer $r\geq 2$, we have
 \begin{equation*}\label{egfSnr(r)1}
     \sum_{n\ge 0}| \mathcal{S}_{n}^r|\dfrac{u^n}{n!}=\exp{(u^r/r)}.
 \end{equation*}   
\end{corollary}

\begin{theorem}\label{t.egfA(r)s}
For fixed integers $r\geq 2$ and $s$, we have
\begin{equation}\label{egfA(r)s}
\dfrac{\partial ^{s}[A^{(r)}(x,u)]_{s}}{\partial u^{s}}=x^{s}\Biggl( A_{r-1}(x)\dfrac{u^{r-1}}{(r-1)!\ }\Biggr)^{s}A^{(r)}(x,u).
\end{equation}
\end{theorem}

\begin{proof}
Since
\begin{equation*}\label{[A^(r)_n,m]_s}
[A^{(r)}_{n+s,m}]_{s}=\binom{n}{r-1}\cdots \binom{n-(s-1)(r-1)}{r-1}\sum_{m_{1}+\cdots+m_{s}+m_{t}=m}C_{r,{m_{1}}}\cdots C_{r,m_{s}}A^{(r)}_{n-s(r-1),m_{t}},
\end{equation*}
we get 
\begin{equation*}\label{[A(r)nx]s}
[A^{(r)}_{n+s}(x)]_{s}=\binom{n}{r-1}\cdots \binom{n-(s-1)(r-1)}{r-1}(C_{r}(x))^{s}A_{n-s(r-1)}^{(r)}(x).
\end{equation*}
We deduce
\begin{equation*}
\dfrac{\partial^{s}[A^{(r)}(x,u)]_{s}}{\partial u^s}=\Bigl(C_{r}(x)\dfrac{u^{r-1}}{(r-1)!\ }\Bigr)^sA^{(r)}(x,u)
        =x^{s}\Bigl(A_{r-1}(x)\dfrac{u^{r-1}}{(r-1)!\ }\Bigr)^sA^{(r)}(x,u).
\end{equation*}
\end{proof}

\begin{theorem}\label{t.A(r,s)k(x,u)}
For fixed integers $r\geq 2,\ k$ and $s$, the partial differential exponential generating function of the excedance distribution over $((r),k,s)$-cycle permutations has the closed form
\begin{equation}\label{A(r,s)k(x,u)}
\dfrac{\partial ^{s}[A^{(r)}_{k}(x,u)]_{s}}{\partial u^{s}}=x^{s}\Biggl( A_{r-1}(x)\dfrac{u^{r-1}}{(r-1)!\ }\Biggr)^s\dfrac{\bigl(\ln{A^{(r)}(x,u)}\bigr)^{k-s}}{(k-s)!\ }.
\end{equation}
\end{theorem}

\begin{proof}
Let us consider the first $s$ integers and their cycles
\begin{align*}
[A^{(r)}_{n+s,m,k}]_{s}&=\binom{n}{r-1}\cdots \binom{n-(s-1)(r-1)}{r-1}\times\\ &\sum_{m_{1}+\cdots+m_{s}+m_{t}=m}C_{r,m_1}\cdots C_{r,m_s}A^{(r)}_{n-s(r-1),m_{t},k-s}.
\end{align*}
We get
\begin{align*}
\dfrac{\partial^s[A^{(r)}_{k}(x,u)]_s}{\partial u^s}=\Bigl(C_{r}(x)\dfrac{u^{r-1}}{(r-1)!\ }\Bigr)^sA^{(r)}_{k-s}(x,u)=\Biggl(xA_{r-1}(x)\dfrac{u^{r-1}}{(r-1)!\ }\Biggr)^s\dfrac{(\ln{A^{(r)}(x,u))^{k-s}}}{(k-s)!\ }.
\end{align*}
\end{proof}

\begin{theorem}\label{t.A(r)(s)(q,x,u)}
For fixed integers $r\geq 2$ and $s$, a partial differential $q$-analogue exponential generating function of the excedance distribution over $((r),\star,s)$-cycle permutations has the closed form
\begin{equation}\label{A(r)(s)(q,x,u)}
\dfrac{\partial ^{s}[A^{(r)}(q,x,u)]_{s}}{\partial u^s}=q^{s}x^{s}\Biggl( A_{r-1}(x)\dfrac{u^{r-1}}{(r-1)!\ }\Biggr)^s\Bigl(A^{(r)}(x,u)\Bigr)^q.
\end{equation}
\end{theorem}

\begin{proof}
From  Equation~\eqref{A(r,s)k(x,u)}, we get
\begin{align*}
\dfrac{\partial^s[A^{(r)}(q,x,u)]_{s}}{\partial u^s}&=\sum_{k\ge 0}x^{s}\Biggl( A_{r-1}(x)\dfrac{u^{r-1}}{(r-1)!\ }\Biggr)^s\dfrac{(\ln{A^{(r)}(x,u)})^{k-s}}{(k-s)!\ }q^{k-s}q^s, \\
        &=q^sx^s\Biggl( A_{r-1}(x)\dfrac{u^{r-1}}{(r-1)!\ }\Biggr)^s \exp{(q\ln{A^{(r)}(x,u)})}.
\end{align*}
Hence, the result follows.
\end{proof}

\section{$r^+$-cycle permutations}\label{r+per}

\begin{longtable}[htb]{cll}
\\\noalign{\smallskip}\hline\noalign{\smallskip}
$\mathcal{S}_n^{r^+}$& {\footnotesize the set of $r^+$-cycle permutations}&
\\
$A^{r}_{n,m}$& $|\{\sigma\in \mathcal{S}_n^{r^+}: e(\sigma)=m\}|$ &{\footnotesize Th.~\ref{t.Ar+} Eq.~\eqref{Ar+}}
  \\
  &{\footnotesize  \emph{$r$-associated Stirling Eulerian number}}&{\footnotesize Th.~\ref{t.eqRecurrence2>r} Eq.~\eqref{eqRecurrence2>r}}
  \\\\
$A^{r}_n(x)$&$\sum_m A^{r}_{n,m} x^m=\sum_{\sigma\in \mathcal{S}_n^{r^+}}x^{e(\sigma)}$& \\$A^{r}(x,u)$&$\sum_n A^{r}_n(x)\frac{u^n}{n!}$ {\footnotesize the egf}&{\footnotesize Th.~\ref{t.EGFCycle>r} Eq.~\eqref{EGFCycle>r}}\\
 $\mathcal{S}_n^{r^+,k}$& {\footnotesize the set of $(r^+,k)$-cycle permutations}&\\
$A^{r}_{n,m,k}$ &  $|\{\sigma\in \mathcal{S}_n^{r^+,k}: e(\sigma)=m\}|$ &\\$A^{r}_{n,k}(x)$&$\sum_m A^r_{n,m,k}x^m=\sum_{\sigma \in \mathcal{S}_n^{r^+,k}}x^{e(\sigma)}$& 
\\
$A^{r}_k(x,u)$&$\sum_nA^{r}_{n,k}(x)\dfrac{u^n}{n!}$&{\footnotesize Th.~\ref{t.rStirlingNumberOfCycleEqualK} Eq.~\eqref{rStirlingNumberOfCycleEqualK}}\\
$A^r(q,x,u)$&$\sum_k\sum_nA^{r}_{n,k}(x)q^k\dfrac{u^n}{n!}$ {\footnotesize \  $q$-analogue of $A^r(x,u)$}&{\footnotesize Th.~\ref{t.qegfStirling} Eq.~\eqref{qegfStirling}}\\
$\mathcal{S}_n^{r^+,\star,s}$& {\footnotesize the set of $(r^+,\star,s)$-cycle permutations}&\\
 $[A_{n,m}^{r}]_s$& $|\{\sigma\in \mathcal{S}_n^{r^+,\star,s}: e(\sigma)=m\}|$ & \\
 $[A_{n}^{r}(x)]_s$& $\sum_m [A_{n,m}^{r}]_sx^m=\sum_{\sigma\in \mathcal{S}_n^{r^+,\star,s}}x^{e(\sigma)}$ & \\\smallskip\\
$[A^r(x,u)]_s$&$\sum_n\sum_{\sigma\in \mathcal{S}_n^{r^+,\star,s}}x^{e(\sigma)}\dfrac{u^n}{n!}$&{\footnotesize Th.~\ref{t.egfArs} Eq.~\eqref{egfArs}}\\\smallskip\\

$[A^r_k(x,u)]_s$&$\sum_n\sum_{\sigma\in \mathcal{S}_n^{r^+,k,s}}x^{e(\sigma)}\dfrac{u^n}{n!}$&{\footnotesize Th.~\ref{t.RassociatedSstirlingCycleNumberC} Eq.~\eqref{RassociatedSstirlingCycleNumberC}} \\
$[A^r(q,x,u)]_s$&$\sum_k[A^r_k(x,u)]_sq^k$ {\footnotesize  $q$-analogue of $[A^r(x,u)]_s$
}&{\footnotesize Th.~\ref{t.qRassociatedSstirlingCycleNumberC} Eq.~\eqref{qRassociatedSstirlingCycleNumberC}}\\\noalign{\smallskip}\hline\noalign{\smallskip}\caption{Notation.}
\end{longtable}

\begin{theorem}\label{t.Ar+}
For a fixed integer $r$, for all integers $n$ and $m$ such that $0\leq m\leq n$ and $n\ge r+1$, the $r$-associated Stirling Eulerian numbers satisfy the following formula
\begin{equation}\label{Ar+}
A^{r}_{n,m}=mA^{r}_{n-1,m}+(n-m)A^{r}_{n-1,m-1}+\binom{n-1}{r}\sum_{s}C_{r+1,s}A^{r}_{n-r-1,m-s}.
\end{equation}
\end{theorem}

\begin{proof} 
Let $n> r$. Consider the integer $n$ in an $r^+$-cycle permutation of $\mathcal{S}_n^{r^+}$.
\begin{enumerate}
\item If $n$ belongs to a cycle of length $r+1$ with $s$ excedances $(1\leq s < m)$, there are $\binom{n-1}{r}$ ways to choose the remaining $r$ elements and $C_{r+1,s}$ ways to form the specified cycle. We then create a permutation with $m-s$ excedances, where the cycle lengths are greater than $r$ using the remaining $n-r-1$ elements, which gives $A^{r}_{n-r-1,m-s}$ possibilities. Thus, the total number of possibilities is $\binom{n-1}{r}\sum_{s}C_{r+1,s}A^{r}_{n-r-1,m-s}$.
\item Otherwise,  we consider $\tau$ as an $r^+$-cycle permutation in $S_{n-1}^{r^+}$:
\begin{enumerate}
\item If $\tau$ has $m$ excedances, we insert $n$ after an excedance. There are $mA^{r}_{n-1,m}$ possibilities.
\item If $\tau$ has $m-1$ excedances, we insert $n$ after an anti-excedance. There are \mbox{$(n-m)A^{r}_{n-1,m-1}$} possibilities.
\end{enumerate} 
\end{enumerate}
\end{proof}

\begin{theorem}\label{t.eqRecurrence2>r}
For a fixed integer $r$, for all integers $n$ and $m$ such that $0\leq m\leq n$ and $n\ge r+1$, we have
\begin{equation}\label{eqRecurrence2>r}
A^{r}_{n+1,m}=\sum_{j=r}^{n}\binom{n}{j}\sum_{i=1}^{m}C_{j+1,i}A^{r}_{n-j,m-i}.
\end{equation}
\end{theorem}

\begin{proof}
Let $r$ be a fixed integer with $n \geq r$. Consider the integer $n+1$, which belongs to a cycle of length $j+1$ ($r \leq j$) with $i$ ($1 \leq i \leq m$) excedances. There are $\binom{n}{j}$ ways to choose these $j$ elements and $C_{j+1,i}$ ways to form that cycle. We then form a permutation with $m-i$ excedances, where the cycle lengths are greater than $r$, using the remaining $n-j$ elements, resulting in $A^{r}_{n-j,m-i}$ possibilities. Hence, we obtain the result.
\end{proof}

\begin{theorem}\label{t.EGFCycle>r}
For a fixed integer $r\geq 1$, the exponential generating function of the $r$-associated Stirling Eulerian numbers has the closed form
\begin{equation}\label{EGFCycle>r}
A^{r}(x,u)=A(x,u)\exp\Biggl(-u-x\sum_{j=1}^{r-1}A_{j}(x)\dfrac{u^{j+1}}{(j+1)!\ }\Biggr).
\end{equation}
\end{theorem}

\begin{proof}
From Equation~\eqref{eqRecurrence2>r}, we get
\begin{align*}
A^{r}_{n+1}(x)&=\sum_{j=r}^{n}\binom{n}{j}C_{j+1}(x)A^{r}_{n-j}(x)
=\sum_{j=0}^{n}C_{j+1}(x)A^{r}_{n}(x)-\sum_{j=0}^{r-1}C_{j+1}(x)A^{r}_{n}(x), \\
&=\sum_{j=0}^{n}C_{j+1}(x)A^{r}_{n}(x)-A_{n}(x)-\sum_{j=1}^{r-1}C_{j+1}(x)A^{r}_{n}(x).\label{Arnx}
\end{align*}
We deduce
\begin{equation*}
(A^{r}(x,u))_{u}'=\dfrac{\big(A(x,u)\big)_{u}'}{A(x,u)}A^{r}(x,u)-A^{r}(x,u)-\sum_{j=1}^{r-1}C_{j+1}(x)\dfrac{u^{j}}{j!}A^{r}(x,u). \end{equation*}
Thus,
\begin{equation*}
\dfrac{(A^{r}(x,u))_{u}'}{A^{r}(x,u)}=\dfrac{\big(A(x,u)\big)_{u}'}{A(x,u)}-1-\sum_{j=1}^{r-1}C_{j+1}(x)\dfrac{u^{j}}{j!}.
\end{equation*}
Integrating by $u$, we have
\begin{equation*}
\ln{A^{r}(x,u)}=\ln{A(x,u)}-u-\sum_{j=1}^{r-1}C_{j+1}(x)\dfrac{u^{j+1}}{(j+1)!\ }.
\end{equation*}
Hence
\begin{align*}
A^{r}(x,u)&=A(x,u)\exp\Bigl(-u-\sum_{j=1}^{r-1}C_{j+1}(x)\dfrac{u^{j+1}}{(j+1)!\ }\Bigr),\\
&=A(x,u)\exp\Bigl(-u-x\sum_{j=1}^{r-1}A_{j}(x)\dfrac{u^{j+1}}{(j+1)!\ }\Bigr).
\end{align*}
\end{proof}

\begin{corollary}\label{c.egfcr+2}
For a fixed integer $r$, the exponential generating function of the excedance distribution over circular $r^+$-cycle permutations has the closed form  
\begin{equation}\label{egfcr+2}
C^{r^{+}}(x,u)=\ln(A^{r}(x,u)).
\end{equation}
\end{corollary}

\begin{proof}
It follows directly from Equation \eqref{egfcr+}.
\end{proof}

\begin{theorem}\label{t.rStirlingNumberOfCycleEqualK}For  fixed integers $r$ and $k$, the exponential generating function of the excedance distribution over $(r^+,k)$-cycle permutations has the closed form
\begin{equation}\label{rStirlingNumberOfCycleEqualK}
A^{r}_{k}(x,u)=\dfrac{\bigl(\ln(A^{r}(x,u))\bigr)^k}{k!}.
\end{equation}
\end{theorem}

\begin{proof}
A permutation of $\mathcal{S}_n^{r^+,k}$  is a product of $k$ disjoint circular $r^+$-cycle permutations. Hence,  
\begin{equation*} \label{arnmk}
A^{r}_{n,m,k}=\dfrac{1}{k!}\sum_{\substack{i_{1},\ldots,i_{k}> r \\ 
i_{1}+\cdots+i_{k}=n} }\sum_{m_{1}+\cdots+m_{k}=m}(i_{1},\ldots, i_{k})!\  C_{i_{1},m_{1}} \cdots  C_{i_k,m_k}.
 \end{equation*}
The result follows directly.
\end{proof}

\begin{theorem}\label{t.qegfStirling}
For a fixed integer $r$, a $q$-analogue exponential generating function of  the $r$-associated Stirling Eulerian numbers has the closed form
\begin{equation}\label{qegfStirling}
A^{r}(q,x,u)=\Bigl(A^{r}(x,u)\Bigr)^q.
\end{equation}
\end{theorem}

\begin{proof}
From  Equation~\eqref{rStirlingNumberOfCycleEqualK}, we get
\begin{align*}
\sum_{k}A^{r}_{k}(x,u)q^{k}&=\sum_{k}\dfrac{(\ln(A^{r}(x,u)))^k}{k!}q^{k}
=\exp(\ln{A^{r}(x,u)q}),\\&=\Bigl(A^{r}(x,u)\Bigr)^{q}.
\end{align*}
\end{proof}

\begin{theorem}\label{t.egfArs}
For a fixed integer $r\geq 1$, the partial differential  exponential generating function of the excedance distribution over $(r^+,\star,s)$-cycle permutations has the closed form
\begin{equation}\label{egfArs}
\dfrac{\partial ^{s}[A^{r}(x,u)]_{s}}{\partial u^{s}}=x^{s}\Biggl( A(x,u)-\sum_{n=0}^{r-1}A_{n}(x)\dfrac{u^{n}}{n!}\Biggr)^{s}A^{r}(x,u).
\end{equation}
\end{theorem}

\begin{proof}
Let us consider the first $s$ integers and their cycles 
\begin{equation*}\label{[A^r_n,m]_s}
[A^{r}_{n+s,m}]_{s}=\sum_{\substack{i_1, \ldots, i_s\ge r\\i_1+\cdots+i_s=n}}\sum_{m_{1}+\cdots+m_{t}=m}(i_{1},\ldots, i_{s})!\ C_{i_1+1,m_1}
\cdots  C_{i_{s}+1,m_{s}}A^{r}_{n-\sum_{\ell=1}^{s}i_{\ell},m_{t}}. 
\end{equation*}
We get
\begin{equation*}\label{[A^r_n(x)]_s}
[A^{r}_{n+s}(x)]_{s}=\sum_{\substack{i_1, \ldots, i_s\geq r\\i_1+\cdots+i_s=n}}(i_{1},\ldots, i_{s})!\ x^{s}A_{i_{1}}(x)\cdots A_{i_{s}}(x)A^{r}_{n-\sum_{\ell \geq 1}^{s}i_{\ell}}(x).
\end{equation*}
We deduce
\begin{equation*}
\dfrac{\partial^{s}A^{r}(x,u)}{\partial u^{s}}=x^{s}\Bigl(A(x,u)-\sum_{n=0}^{r-1}A_{n}(x)\dfrac{u^{n}}{n!}\Bigr)^{s}A^{r}(x,u).
\end{equation*}
\end{proof}


\begin{theorem}\label{t.RassociatedSstirlingCycleNumberC}
For fixed integers $r\geq 1,\ k$ and $s$, the partial differential exponential generating function of the excedance distribution over $(r^+,k,s)$-cycle permutations has the closed form
\begin{equation}\label{RassociatedSstirlingCycleNumberC}
\dfrac{\partial ^{s}[A^{r}_{k}(x,u)]_{s}}{\partial u^{s}}=x^s\Biggl( A(x,u)-\sum_{n=0}^{r-1}A_{n}(x)\dfrac{u^n}{n!}\Biggr)^s\dfrac{\bigl(\ln{A^{r}(x,u)}\bigr)^{k-s}}{(k-s)!\ }.
\end{equation}
\end{theorem}

\begin{proof}
A permutation of $\mathcal{S}_{n+s}^{r^+,k,s}$  is a product of $k$ disjoint circular $r^+$-cycle permutations. Considering the first $s$ integers and their cycles, we have 
\begin{equation*}\label{ark}
[A^{r}_{n+s,m,k}]_{s}=\sum_{\substack{i_1, \ldots, i_s\ge r\\i_1+\cdots+i_s=n}}\sum_{m_{1}+\cdots+m_{t}=m}(i_{1},\ldots, i_{s})!\ C_{i_1+1,m_1}
\cdots  C_{i_{s}+1,m_{s}}A^{r}_{n-\sum_{\ell=1}^{s}i_{\ell},m_{t},k-s}.
\end{equation*}
The result follows directly.
\end{proof}

\begin{theorem}\label{t.qRassociatedSstirlingCycleNumberC}
For fixed integers $r\geq 1$ and $s$, a partial differential $q$-analogue exponential generating function of the excedance distribution over $(r^+,\star,s)$-cycle permutations has the closed form
\begin{equation}\label{qRassociatedSstirlingCycleNumberC}
\dfrac{\partial ^{s}[A^{r}(q,x,u)]_{s}}{\partial u^s}=q^{s}x^{s}\Biggl( A(x,u)-\sum_{n=0}^{r-1}A_{n}(x)\dfrac{u^n}{n!}\Biggr) ^s\Bigl(A^{r}(x,u)\Bigr)^q.
\end{equation}
\end{theorem}

\begin{proof}
The result follows directly from Equation~\eqref{RassociatedSstirlingCycleNumberC}.
\end{proof}

The following section contains the corollaries of the previous sections, obtained by setting different values for $r$ and $s$ in the $q$-analogues. Extensive results are provided for Brenti's work \cite{Brenti} and for the results of Ksavrelof and Zeng \cite{Ksaverlof}. Various identities are formulated.
 
\section{Excedance distribution by cycle over involutions, permutations and derangements}\label{0+kcycle}

\begin{longtable}[htb]{cll}
\\\noalign{\smallskip}\hline\noalign{\smallskip}
$\mathcal{S}_n^{0^+,k}$& {\footnotesize the set of $(0^+,k)$-cycle permutations}&\\
$A_{n,m,k}$& $|\{\sigma\in \mathcal{S}_n^{0^+,k}: e(\sigma)=m\}|$ &\\$A_{n,k}(x)$&$\sum_mA_{n,m,k}x^m=\sum_{\sigma \in \mathcal{S}_n^{0^+,k}}x^{e(\sigma)}$ &\\$A_k(x,u)$&$\sum_n A_{n,k}(x)\dfrac{u^n}{n!}$&{\footnotesize Cor.~\ref{t.EgfCycleK} Eq.~\eqref{EgfCycleK}}\\
$A(q,x,u)$ & $\sum_n\sum_{\sigma\in \mathcal{S}_n} x^{e(\sigma)}q^{c(\sigma)}\dfrac{u^n}{n!}${\footnotesize \ \ $q$-analogue  of $A(x,u)$}&{\footnotesize Cor.~\ref{t.Brenti0} Eq.~\eqref{Brenti0}}\\
$[A(x,u)]_s$&$\sum_n\sum_{\sigma\in \mathcal{S}_n^{0^+,\star,s}}x^{e(\sigma)}\dfrac{u^n}{n!}$&{\footnotesize Cor.~\ref{t.degfAxus} Eq.~\eqref{degfAxus}}\\
$[A_k(x,u)]_s$&$\sum_n\sum_{\sigma\in \mathcal{S}_n^{0^+,k,s}}x^{e(\sigma)}\dfrac{u^n}{n!}$&{\footnotesize Cor.~\ref{t.sstirlingnumberofcycleC} Eq.~\eqref{sstirlingnumbernumberofcycleC}} \\
$[A(q,x,u)]_s$&$\sum_k[A_k(x,u)]_sq^k$ {\footnotesize $q$-analogue of $[A(x,u)]_s$
}&{\footnotesize Cor.~\ref{t.qAs} Eq.~\eqref{qAs}}\\\noalign{\smallskip}\hline\noalign{\smallskip}\caption{Notation.}
\end{longtable}

\begin{corollary}
A $q$-analogue exponential generating function of the excedance distribution over $2^-$-cycle permutations (involutions) has the closed form
\begin{equation}\label{A(2-)(s)(q,x,u)}
A^{2^-}(q,x,u)=\exp(q(u+xu^2/2)).
\end{equation}
\end{corollary}

\begin{corollary}
For a fixed integer $n$, we have
\begin{align*}
\sum_{\pi\in\mathcal{I}_{2n}}x^{e(\pi)}(q)^{c(\pi)}&=\sum_{j=0}^{n}\dfrac{(2n)!\ }{2^{n-k}(2k)!\ (n-k)!\ }q^{n+k}x^{n-k},\label{even} \\  
\sum_{\pi\in\mathcal{I}_{2n+1}}x^{e(\pi)}(q)^{c(\pi)}&=\sum_{j=0}^{n}\dfrac{(2n+1)!\ }{2^{n-k}(2k+1)!\ (n-k)!\ }q^{n+k+1}x^{n-k}.   
\end{align*}\label{odd}
\end{corollary}

\begin{corollary}
A $q$-analogue exponential generating function of the excedance distribution over $2$-cycle permutations (involution derangements) has the closed form
\begin{equation*}\label{A(2)(s)(q,x,u)}
A^{(2)}(q,x,u)=\exp(qxu^2/2).
\end{equation*}
\end{corollary}

\begin{corollary}
For a fixed integer $n$, we have
\begin{equation*}
\sum_{\pi\in\mathcal{I}_{2n}\cap\mathcal{D}_{2n} }x^{e(\pi)}q^{c(\pi)}=\sum_{j=0}^{n}\dfrac{(2n)!\ }{2^{n}n!}q^nx^n.\label{invder}   
\end{equation*}
\end{corollary}

\begin{corollary}\label{t.EgfCycleK}
For a fixed integer $k$, the exponential generating function of the excedance distribution over $(0^+,k)$-cycle permutations has the closed form
\begin{equation}\label{EgfCycleK}
A_{k}(x,u)=\sum_{n\ge 0}\sum_{{\tiny \sigma\in{\mathcal{S}_n^{0^+,k}}}}x^{e(\sigma)}\dfrac{u^n}{n!}=\dfrac{\bigl(\ln(A(x,u))\bigr)^k}{k!}.
\end{equation}
\end{corollary}

\begin{proof}
The proof follows from Theorem \ref{t.rStirlingNumberOfCycleEqualK}.
\end{proof}

\begin{corollary}[Brenti] \label{t.Brenti0}
A $q$-analogue exponential generating function of the excedance distribution by cycle over permutations has the closed form
\begin{equation}\label{Brenti0}
A(q,x,u)=\Bigl(A(x,u)\Bigr)^{q}.
\end{equation}
\end{corollary} 

\noindent (See \cite[Proposition 7.3]{Brenti})

\begin{proof}
The proof follows from Theorem \ref{t.qegfStirling}.
\end{proof}

\begin{corollary}\label{t.degfAxus}
For a fixed integer $s$, the partial differential exponential generating function of  the excedance distribution over $(0^+,\star,s)$-cycle permutations has the closed form
\begin{equation}\label{degfAxus}
\dfrac{\partial ^{s}[A(x,u)]_{s}}{\partial u^{s}}=\Bigl(A(x,u)\exp(u(x-1))\Bigr)^{s}A(x,u).
\end{equation}
\end{corollary}

\begin{proof}
The proof follows from Theorem \ref{t.egfArs}.
\end{proof}
\begin{corollary}\label{t.sstirlingnumberofcycleC}
For fixed integers $k$ and $s$, the partial differential exponential generating function of  the excedance distribution over $(0^+,k,s)$-cycle permutations has the closed form
\begin{equation}\label{sstirlingnumbernumberofcycleC}
\dfrac{\partial^{s}[A_{k}(x,u)]_{s}}{\partial u^{s}}=\Bigl(A(x,u)\exp(u(x-1))\Bigr)^s\dfrac{\bigl(\ln{A(x,u)}\bigr)^{k-s}}{(k-s)!\ }.
\end{equation}

\end{corollary}
\begin{proof}
It comes from Theorem \ref{t.RassociatedSstirlingCycleNumberC}.
\end{proof}

\begin{corollary}\label{t.qAs}
A partial differential $q$-analogue exponential generating function of the excedance distribution over $(0^+,\star,s)$-cycle permutations has the closed form
\begin{equation}\label{qAs}
\dfrac{\partial ^{s}[A(q,x,u)]_{s}}{\partial u^{s}}=q^{s}\Bigl(A(x,u)\exp(u(x-1))\Bigr)^{s}\Bigl(A(x,u)\Bigr)^{q}.
\end{equation}
\end{corollary}

\begin{proof}
We immediately have the result by multiplying the equation~\eqref{sstirlingnumbernumberofcycleC} by $q^{k}$ and summing over $k$.
\end{proof}

\begin{corollary}\label{Brenti1}
\begin{equation*}
\sum_{\sigma \in S_{n+s}^{0^+,\star,s}}x^{e(\sigma)}(-s)^{c(\sigma)}=(-1)^ss^{n+s}(x-1)^{n}.
\end{equation*}   
\end{corollary}

\begin{proof}
From Corollary \ref{t.qAs}, we have that
\begin{equation*}
\frac{\partial^s[A(q,x,u)]_{s}}{\partial u^s}=q^{s}\Bigl(A(x,u)\exp(u(x-1))\Bigr)^s\Bigl(A(x,u)\Bigr)^q,
\end{equation*}
setting $q=-s$, gives
\begin{align*}
\frac{\partial A(-s,x,u)}{\partial u}&=(-s)^s\Bigl(A(x,u)\exp(u(x-1))\Bigr)^s\Bigl(A(x,u)\Bigr)^{-s}.
\end{align*}
Hence,
\begin{equation}
\frac{\partial^s[A(q,x,u)]_{s}}{\partial u^s}=(-s)^s\exp(su(x-1)). \label{pergen}
\end{equation}
Identifying the coefficients of $\dfrac{u^n}{n!}$ in Equation \eqref{pergen}
 gives the desired proof.
\end{proof}


\begin{corollary}\label{Brenti2}
\begin{equation*}
\sum_{\sigma \in S_{n+s}^{0^+,\star,s}}x^{e(\sigma)}(-s-1)^{c(\sigma)}=(-1)^s(s+1)^{s}\big(xs^{n}-(s+1)^{n}\big)(x-1)^{n-1}.
\end{equation*}
\end{corollary}

\begin{proof}
From Corollary \ref{t.qAs}, we have
\begin{equation*}
\dfrac{\partial^s[A(q,x,u)]_{s}}{\partial u^s}=q^{s}\Bigl(A(x,u)\exp(u(x-1))\Bigr)^s\Bigl(A(x,u)\Bigr)^q.
\end{equation*}
Setting $q=-s-1$, we have
\begin{align*}
\frac{\partial^s [A(-s-1,x,u)]_s}{\partial u^s}&=(-s-1)^s[A(x,u)\exp(u(x-1))]^s(A(x,u))^{-s-1},\\
       &=(-s-1)^s\frac{\exp(su(x-1))}{A(x,u)},\\&=(-s-1)^s\frac{\exp(su(x-1))(x-\exp(u(x-1)))}{x-1},\\
        &=\dfrac{(-s-1)^s}{x-1}\Bigl(x\exp(su(x-1))-\exp((s+1)u(x-1))\Bigr).
\end{align*}
Hence,
\begin{equation}
\frac{\partial^s [A(-s-1,x,u)]_s}{\partial u^s}=\dfrac{(-s-1)^s}{(x-1)}\Bigl(x\exp(su(x-1))-{\exp((s+1)u(x-1))}\Bigr). \label{2ghf} 
\end{equation}
Identifying the coefficients of $\dfrac{u^{n}}{n!}$ in Equation \eqref{2ghf} gives the desired proof.
\end{proof}

\begin{remark}
 If we set $s=1$ in Corollary \ref{Brenti1} and Corollary \ref{Brenti2}, we get Brenti's results \cite[Corollary 7.4]{Brenti}. 
 \begin{align*}
 \sum_{\pi\in\mathcal{S}_n}x^{e(\pi)}(-1)^{c(\pi)}&=-(x-1)^{n-1},\\
 \sum_{\pi\in\mathcal{S}_n}x^{e(\pi)}(-2)^{c(\pi)}&=2(2^{n-1}-x)(x-1)^{n-2}.
 \end{align*}
\end{remark}


\begin{corollary}\label{c.der}
A partial differential $q$-analogue exponential generating function of the excedance distribution over $(1^+,\star,s)$-cycle permutations (derangements) has the closed form
\begin{equation}\label{der}
\dfrac{\partial ^{s}[A^1(q,x,u)]_{s}}{\partial u^{s}}=q^{s}x^s\Bigl(A(x,u)-1\Bigr)^{s}\Bigl(A(x,u)\exp(-u)\Bigr)^{q}.
\end{equation}
\end{corollary}

\begin{proof}
We immediately have the result by setting $r=1$ in equation~\eqref{qRassociatedSstirlingCycleNumberC}.
\end{proof}

\begin{corollary} \label{c.NavFan1}
For fixed integers $n$ and $s$, we have
\begin{equation}\label{NavFan1}
\sum_{\sigma \in \mathcal{S}_{n+s}^{1^+,\star,s}}x^{e(\sigma)}(-s)^{c(\sigma)}=(-1)^s\Bigl(\dfrac{sx}{x-1}\Bigr)^s\sum_{\substack{i_1\ge 1, \ldots , i_s\ge 1\\i_1+\cdots+i_s=n}}(i_1,\ldots , i_s)!\ \prod_{j=1}^{s}(x^{i_j}-1).
\end{equation}
\end{corollary}


\begin{proof}
 From Corollary \ref{c.der} Equation \eqref{der}, setting $q=-s$, we have
\begin{align*}
       \frac{\partial^s [A^1(-s,x,u)]_s}{\partial u^s}&=(-s)^sx^s(A(x,u)-1)^s(A^1(x,u))^{-s},\\
       &=(-s)^sx^s\Bigl(\dfrac{A(x,u)-1}{A(x,u)}\Bigr)^s\exp(su),\\
       &=(-s)^sx^s(1-1/A(x,u))^s\exp(su),\\
       &=(-s)^sx^s\Bigl(1-\dfrac{x-\exp((x-1)u)}{x-1}\Bigr)^s\exp(su),\\
       &=\Bigl(\dfrac{-sx}{x-1}\Bigr)^s(-\exp(u)+\exp(xu))^s,\\
       &=\Bigl(\dfrac{-sx}{x-1}\Bigr)^s\Bigl(\sum_{n\ge 1}(-1+x^n)\dfrac{u^n}{n!}\Bigr)^s. 
\end{align*}
Hence,
\begin{equation}\label{hgf}
\dfrac{\partial^s [A^1(-s,x,u)]_s}{\partial u^s}=(-1)^s\big(\dfrac{sx}{x-1}\big)^s\sum_{\substack{i_1\ge 1, \ldots, i_s\ge 1\\ i_1+\cdots+i_s=n}}(i_1,\ldots, i_s)!\ \prod_{j=1}^{s}(x^{i_j}-1)\dfrac{u^n}{n!}. 
\end{equation}
Identifying the coefficients of $\dfrac{u^n}{n!}$ in Equation \eqref{hgf} gives the desired result.
\end{proof}

\begin{remark}
If we set $s=1$ in Equation~\eqref{NavFan1}, we get Ksavrelof and Zeng's result \cite{Ksaverlof} \begin{equation}
\sum_{\sigma \in \mathcal{D}_{n}}x^{e(\sigma)}(-1)^{c(\sigma)}=-x-x^2-\cdots -x^{n-1}.
\end{equation} 
\end{remark}

\begin{corollary} \label{c.NavFan11}
For fixed integers $n$ and $s$, we have
\begin{equation*}\label{NavFan11}
\sum_{\sigma \in \mathcal{S}_{n+s}^{1^+,\star,s}}x^{e(\sigma)}(-s-1)^{c(\sigma)}=(-1)^s\dfrac{(s+1)^sx^s}{(x-1)^s}\sum_{\ell=0}^{n}\Biggl(\dfrac{x-x^\ell}{x-1} \sum_{\substack{i_1\ge 1, \ldots , i_s\ge 1\\i_1+\cdots+i_s=n-\ell}}(i_1,\ldots , i_s)!  \prod_{j=1}^{s}(x^{i_j}-1)\Biggr).
\end{equation*}
\end{corollary}

\begin{proof}
From Corollary \ref{c.der} Equation \eqref{der}, setting $q=-s-1$, we have
\begin{align*}
       \frac{\partial^s [A^1(-s-1,x,u)]_s}{\partial u^s}&=(-s-1)^sx^s(A(x,u)-1)^s(A^1(x,u))^{-s-1},\\
       &=(-s-1)^sx^s\Bigl(\dfrac{A(x,u)-1}{A(x,u)}\Bigr)^s\exp((s+1)u)A(x,u)^{-1},\\
       &=(-s-1)^sx^s(1-1/A(x,u))^s\exp(su)A(x,u)^{-1}\exp(u),\\
       &=(-s-1)^sx^s\Bigl(1-\dfrac{x-\exp((x-1)u)}{x-1}\Bigr)^s\exp(su) \dfrac{x-\exp((x-1)u)}{x-1}e^u,\\
              &=\dfrac{(-(s+1)x)^s}{(x-1)^{s+1}}(-\exp(u)+\exp(xu))^s(x\exp(u)-\exp(xu)), \\
       &=\dfrac{(-(s+1)x)^s}{(x-1)^{s}}\Bigl(\sum_{n\ge 1}(-1+x^n)\dfrac{u^n}{n!}\Bigr)^s\Bigl(\sum_{n\ge 0}\dfrac{(x-x^n)}{x-1}\dfrac{u^n}{n!}\Bigr). 
\end{align*}
Hence,
\small  
\begin{align} 
      \dfrac{\partial^s [A^1(-s-1,x,u)]_s}{\partial u^s}&=(-1)^s\dfrac{(s+1)^sx^s}{(x-1)^{s}}(\sum_{\substack{i_1\ge 1, \ldots , i_s\ge 1\\\nonumber i_1+\cdots+i_s=n}}(i_1,\ldots , i_s)!\ \prod_{j=1}^{s}(x^{i_j}-1)\dfrac{u^n}{n!}) \\ &(1+\sum_{n\ge 1}\dfrac{(x-x^n)}{x-1}\dfrac{u^n}{n!})\nonumber,\\ 
   &=(-1)^s\dfrac{(s+1)^sx^s}{(x-1)^{s}}\sum_{n\ge 0} \Bigl(\sum_{\ell=0}^{n} \dfrac{(x-x^\ell)}{x-1}\\
   & \Bigl(\sum_{\substack{i_1\ge 1, \ldots , i_s\ge 1\\i_1+\cdots+i_s=n-\ell}}(i_1,\ldots , i_s)!\prod_{j=1}^{s}(x^{i_j}-1)\Bigr)\Bigr)\dfrac{u^n}{n!}. \label{hgf1}
\end{align}

Identifying the coefficients of $\dfrac{u^n}{n!}$ in Equation \eqref{hgf1} gives the desired result.
\end{proof}

\begin{corollary} \label{c.NavFan}
For all integer $n\ge 1$, we have
\begin{equation}
\sum_{\sigma \in \mathcal{D}_{n}}x^{e(\sigma)}(-2)^{c(\sigma)}=\dfrac{-2x(-2^{n-1}x-2^{n-1}x^{n-1}+(x+1)^{n})}{(x-1)^2}.
\label{NavFan}
\end{equation}   
\end{corollary}

\begin{proof}
From Corollary \ref{c.der} Equation \eqref{der}, setting $s=1$ and $q=-2$, we get 
\begin{align*}
       \frac{\partial A^1(-2,x,u)}{\partial u}&=-2x(A(x,u)-1)(A^1(x,u))^{-2},\\
       &=-2x(A(x,u)-1)\left(A(x,u)\exp(-u)\right)^{-2},\\
       &=-2x\left(1/A(x,u)-1/A(x,u)^2\right)\exp(2u),\\
       &=-2x\left(\dfrac{x-\exp((x-1)u)}{x-1}-\dfrac{(x-\exp((x-1)u))^2}{(x-1)^2}\right)\exp(2u).
\end{align*}
 That is,
\begin{equation}\label{4}
      \frac{\partial A^1(-2,x,u)}{\partial u} =\dfrac{-2x}{(x-1)^2}\Bigl(-x\exp(2u)-\exp{(2xu)}+(x+1)\exp((x+1)u)\Bigr).
\end{equation}
Identifying the coefficients of $\dfrac{u^n}{n!}$ in Equation \eqref{4} 
 gives the desired proof.
\end{proof}

\begin{remark}
Note that $\dfrac{-2x(-2^{n-1}x-2^{n-1}x^{n-1}+(x+1)^{n})}{(x-1)^2} \in \mathbb{Z}[x].$ Below we find a few values of Equation \eqref{NavFan}.
    \begin{enumerate}
         \item $\sum_{\sigma \in \mathcal{D}_{2}}x^{e(\sigma)}(-2)^{c(\sigma)}=-2x$.
     \item $\sum_{\sigma \in \mathcal{D}_{3}}x^{e(\sigma)}(-2)^{c(\sigma)}=-2x(x+1)$.
      \item $\sum_{\sigma \in \mathcal{D}_{4}}x^{e(\sigma)}(-2)^{c(\sigma)}=-2x(x-1)^2$. 
      \item $\sum_{\sigma \in \mathcal{D}_{5}}x^{e(\sigma)}(-2)^{c(\sigma)}=-2x(x+1)(x^2-10x+1)$.      
    \end{enumerate}
\end{remark}


\section{$s$-fixed-points-permutations}\label{sfper}

\begin{longtable}[htb]{cll}
\\\noalign{\smallskip}\hline\noalign{\smallskip}
$\mathcal{F}_n^s$& {\footnotesize the set of $s$-fixed-points-permutations}&\\$\mathcal{F}_n^{k,s}$& {\footnotesize the set of $s$-fixed-points-permutations with $k$ cycles}&\\
$[F(x,u)]_s$&$\sum_n\sum_{\sigma\in \mathcal{F}_n^{s}}x^{e(\sigma)}\dfrac{u^n}{n!}$&{\footnotesize Th.~\ref{t.sfpp} Eq.~\eqref{sfpp}}
\\
$[F_k{(x,u)}]_s$&$\sum_n\sum_{\sigma\in \mathcal{F}_n^{k,s}}x^{e(\sigma)}\dfrac{u^n}{n!}$&{\footnotesize Th.~\ref{t.ksfpp} Eq.~\eqref{ksfpp}}\\
$[F(q,x,u)]_s$&$\sum_k[F_k(x,u)]_sq^k$ {\footnotesize $q$-analogue of $[F(x,u)]_s$
}&{\footnotesize Th.~\ref{t.qsfpp} Eq.~\eqref{qsfpp}}\\\noalign{\smallskip}\hline\noalign{\smallskip}\caption{Notation.}
\end{longtable}


\begin{theorem}\label{t.sfpp}
A partial differential exponential generating function of the excedance distribution over $s$-fixed-points-permutations has the closed form
\begin{equation}\label{sfpp}
\dfrac{\partial^{s}[F(x,u)]_{s}}{\partial u^{s}}=\Bigl(A(x,u)\exp(u(x-1))\Bigr)^{s}A(x,u)\exp{(-u)}.
\end{equation}
\end{theorem}

\begin{proof}
Since
\begin{equation*}
[F_{n+s,m}]_{s}=\sum_{\substack{i_1, \ldots , i_s\ge 0\\i_1+\cdots+i_s=n}} (i_{1},\ldots , i_s)!\ 
\sum_{m_{1}+\cdots+m_{s}+m_{t}=m}C_{i_{1}+1,m_{1}}\cdots  C_{i_{s}+1,m_{s}}A^{1}_{n-\sum_{j=1}^{s}i_{j},m_{t}},
\end{equation*} 
we get 
\begin{equation*}
[F_{n+s}(x)]_{s}= \sum_{\substack{i_1, \ldots , i_s\ge 0\\i_1+\cdots+i_s=n}} (i_{1},\ldots , i_s)!\ C_{i_{1}+1}(x)\cdots C_{i_{s}+1}(x)A^{1}_{n-\sum_{j=1}^{s}i_{j}}(x).
\end{equation*}
We deduce  
\begin{align*}
\dfrac{\partial ^{s}[F(x,u)]_{s}}{\partial u^{s}}&=\Bigl(\dfrac{\partial }{\partial u}C(x,u)\Bigr)^{s}A^{1}(x,u),\\&=\Bigl(A(x,u)\exp{(u(x-1))}\Bigr)^{s}A(x,u)\exp(-u).
\end{align*}
\end{proof}


\begin{theorem}\label{t.ksfpp}
Let $k\geq s$ be fixed integers.
The partial differential exponential generating function of the excedance distribution over $s$-fixed-points-permutations having $k$ cycles has the closed form 
\begin{equation}\label{ksfpp}
\dfrac{\partial^s[F_{k}(x,u)]_{s}}{\partial u^s}=\Bigl(A(x,u)\exp(u(x-1))\Bigr)^{s}\dfrac{\bigl(\ln{(A(x,u)\exp{(-u)})}\bigr)^{k-s}}{(k-s)!\ }.
\end{equation}
\end{theorem}

\begin{proof}
Let us consider the  first $s$ integers, we have
\begin{equation*} 
[F_{n+s,m,k}]_{s}=\sum_{\substack{i_1, \ldots, i_s\ge 0\\i_1+\cdots+i_s=n}} (i_{1},\ldots, i_s)!\  \sum_{m_{1}+\cdots+m_{s}+m_{t}=m}C_{i_{1}+1,m_{1}}\cdots  C_{i_s+1,{m_{s}}}A^{1}_{n-\sum_{l=1}^{s}i_{l},m_{t},k-s}.
\end{equation*}
We get \begin{equation*}
[F_{n+s,m}(x)]_{s}=\sum_{\substack{i_1, \ldots , i_s\ge 0\\i_1+\cdots+i_s=n}} (i_{1},\ldots , i_s)!\  C_{i_1+1}(x)\cdots C_{i_s+1}(x)A^{1}_{n-\sum_{l=1}^{s}i_{l},k-s}(x).
\end{equation*}
Hence,
\begin{align*}
\dfrac{\partial^s[F_{k}(x,u)]_s}{\partial u^s}&=\left(\dfrac{\partial C(x,u)}{\partial u}\right)^sA^{1}_{k-s}(x,u), \\
    &=\Bigl(A(x,u)\exp(u(x-1))\Bigr)^{s}\dfrac{\bigl(\ln{(A(x,u)\exp{(-u)})}\bigr)^{k-s}}{(k-s)!\ }.
\end{align*}
\end{proof}

\begin{theorem}\label{t.qsfpp}
A partial differential $q$-analogue exponential generating function of the excedance distribution over $s$-fixed-points-permutations has the closed form
\begin{equation}\label{qsfpp}
\dfrac{\partial ^{s}[F(q,x,u)]_{s}}{\partial u^{s}}=q^{s}\Bigl(A(x,u)\exp(u(x-1))\Bigr)^{s}\Bigl(A(x,u)\Bigr)^{q}\exp(-qu).
\end{equation}
\end{theorem}

\begin{proof}
We immediately have the result by multiplying by $q^{k}$ and summing over $k$ the right member of Equation~\eqref{ksfpp} .
\end{proof}

\begin{corollary}
For  fixed integers $n$ and $s$, we have 
\begin{equation*}
\sum_{\pi\in \mathcal{F}_{n}^s} x^{e(\pi)}(-s)^{c(\pi)}=(-1)^ss^{n}x^{n-s}.
\end{equation*}
\end{corollary}

\begin{corollary}
For  fixed integers $n$ and $s$, we have 
\begin{equation*}
\sum_{\pi\in \mathcal{F}_{n}^s} x^{e(\pi)}(-s-1)^{c(\pi)}=(-1)^s(s+1)^s\dfrac{x(sx+1)^{n-s}-(s+1)^{n-s}x^{n-s}}{x-1}.
\end{equation*}
\end{corollary}

\section{Generalizations}\label{gen}
\begin{longtable}[htb]{cll}
\\\noalign{\smallskip}\hline\noalign{\smallskip}
$\mathcal{S}_n^{(r_{1})\cdots (r_s)}$& {\footnotesize the set of $(r_1)\cdots   (r_s)$-cycle permutations}&\\
$\mathcal{S}_n^{r_{1}\cdots   r_s}$& {\footnotesize the set of $r_1\cdots   r_s$-cycle permutations}&\\
$\mathcal{S}_n^{r_{1}\cdots   r_{s}(r_{s+1})\cdots  (r_{\ell})}$& {\footnotesize the set of $r_{1}\cdots   r_{s}(r_{s+1})\cdots   (r_{\ell})$-cycle permutations }\\
$\mathcal{S}_n^{r_{1}\cdots   r_{s}(r_{s+1})\cdots  (r_{\ell}),k}$& {\footnotesize $\{\sigma\in \mathcal{S}_n^{r_{1}\cdots   r_{s}(r_{s+1})\cdots  (r_{\ell})}: \ c(\sigma)=k\} $}&\\
$\mathcal{S}_n^{r_{1}^*\cdots   r_{\ell}^*}$& {\footnotesize the set of $r_{1}^*\cdots   r_{\ell}^*$-cycle permutations }\\
$[A^{r_1\cdots   r_{s}}_{n+s,m,k}]_s$&$|\{\sigma\in \mathcal{S}_{n+s}^{r_{1}\cdots   r_s,\star,s}:\ e(\sigma)=m\}|$&\\
$[A^{r_1\cdots   r_{s}}_{n+s,k}(x)]_s$&$\sum_{\sigma\in \mathcal{S}_{n+s}^{r_{1}\cdots r_s,k,s}}x^{e(\sigma)}$&\\
$[A^{r_1\cdots   r_{s}}(x,u)]_s$&$\sum_n\sum_{\sigma\in \mathcal{S}_{n}^{r_{1}\cdots   r_s}}x^{e(\sigma)}\dfrac{u^n}{n!}$&{\footnotesize Th.~\ref{t.egfr1s} Eq. \eqref{egfr1s}}\\
$[A^{r_1\cdots   r_{s}}_k(x,u)]_s$&$\sum_n\sum_{\sigma\in \mathcal{S}_{n}^{r_{1}\cdots   r_s},k}x^{e(\sigma)}\dfrac{u^n}{n!}$&{\footnotesize Th.~\ref{t.[A^r1...rl_k(x,u)]_s} Eq.~\eqref{[A^r1...rl_k(x,u)]_s}}\\
$[A^{r_1\cdots   r_{s}}(q,x,u)]_s$&{\footnotesize  $q$-analogue  of $[A^{r_1\cdots  r_{s}}(x,u)]_s$}&{\footnotesize Th.~\ref{t.qsr1} Eq.~\eqref{qsr1}}\\
$A^{(r_{1})\cdots  (r_{\ell})}(x,u)$&$\sum_n\sum_{\sigma\in \mathcal{S}_{n}^{(r_{1})\cdots   (r_\ell)}}x^{e(\sigma)}q^{c(\sigma)}\dfrac{u^n}{n!}$&{\footnotesize Th.~\ref{t.qrell} Eq.~\eqref{qrell}}\\
$[A^{r_{1}\cdots   r_{s}(r_{s+1})\cdots   (r_{\ell})}(q,x,u)]_s$&$\sum_n\sum_{\sigma\in \mathcal{S}_{n}^{r_{1}\cdots  r_{s}(r_{s+1})\cdots  (r_\ell),\star,s}}x^{e(\sigma)}q^{c(\sigma)}\dfrac{u^n}{n!}$&{\footnotesize Th.~\ref{t.q(r1)} Eq.~\eqref{q(r1)}}\\$A^{r_{1}^*\cdots   r_{\ell}^*}(q,x,u)$&$\sum_n\sum_{\sigma\in \mathcal{S}_{n}^{r_{1}^*\cdots   r_\ell^*}}x^{e(\sigma)}q^{c(\sigma)}\dfrac{u^n}{n!}$&{\footnotesize Cor.~\ref{c.r*} Eq.~\eqref{r*} }
\\\noalign{\smallskip}\hline\noalign{\smallskip}\caption{Notation.}
\end{longtable}

\begin{theorem}\label{t.egfr1s}
For a fixed integer $s$, the exponential generating function of the excedance distribution over $(r_1\cdots r_s, \star,s)$-cycle permutations has the differential closed form
\begin{equation}\label{egfr1s}
\dfrac{\partial^{s}[A^{r_{1}\cdots r_s}(x,u)]_{s}}{\partial u^{s}}=x^{s}\Biggl( A_{r_{1}-1}(x)\dfrac{u^{r_{1}-1}}{(r_{1}-1)!\ }\cdots  A_{r_{s}-1}(x)\dfrac{u^{r_{s}-1}}{(r_{s}-1)!\ }\Biggr) A(x,u).
\end{equation} 
\end{theorem}

\begin{proof}
Consider the first $s$ integers in a permutation in $\mathcal{S}_{n+s}^{r_1\cdots r_s,\star,s}$ and their cycles.
We have
\begin{equation*}
    [A^{r_1\cdots  r_{s}}_{n+s,m}]_s=\binom{n}{r_{1}-1,\ldots, r_{s}-1}\sum_{m_1+\cdots  +m_{s}+m_{t}=m}\ C_{r_1,m_{1}}\cdots C_{r_s,m_{s}}A_{n-\sum_{\ell=1}^{s}(r_{\ell}-1),m_{t}}.
\end{equation*}
The result follows directly. 
\end{proof}


\begin{theorem}\label{t.[A^r1...rl_k(x,u)]_s}
The partial differential exponential generating function of the excedance distribution over $(r_1\cdots r_s,k,s)$-cycle permutations has the differential closed form
\begin{equation}\label{[A^r1...rl_k(x,u)]_s}
\dfrac{\partial^{s}[A^{r_{1}\cdots   r_s}_{k}(x,u)]_{s}}{\partial u^{s}}=x^{s}\Biggl( A_{r_{1}-1}(x)\dfrac{u^{r_{1}-1}}{(r_{1}-1)!\ }\cdots   A_{r_{s}-1}(x)\dfrac{u^{r_{s}-1}}{(r_{s}-1)!\ }\Biggr)\dfrac{\bigl(\ln{A(x,u)}\bigr)^{k-s}}{(k-s)!\ }.
\end{equation} 
\end{theorem}

\begin{proof}
Let us consider again the first $s$ integers in permutations of $\mathcal{S}_{n+s}^{r_1\cdots r_{s},k}$, we have
\begin{equation*}
    [A^{r_1\cdots  r_{s}}_{n+s,m,k}]_s=\binom{n}{r_{1}-1,\ldots , r_{s}-1}\sum_{m_1+\cdots+m_{s}+m_{t}=m}C_{r_1,m_{1}}\cdots C_{r_s,m_{s}}A_{n-\sum_{\ell=1}^{s}(r_{\ell}-1),m_{t},k-s}.
\end{equation*}
The result follows directly. 
\end{proof}

\begin{theorem}\label{t.qsr1}
For a fixed integer $s$, a partial differential $q-$analogue exponential generating function of the excedance distribution over 
$r_1\cdots   r_s$-cycle permutations has the closed form
\begin{equation}\label{qsr1}
\dfrac{\partial^s[A^{r_1\cdots   r_s}(q,x,u)]_s}{\partial u^s}=q^sx^{s}\Biggl( A_{r_{1}-1}(x)\dfrac{u^{r_{1}-1}}{(r_{1}-1)!\ }\cdots A_{r_{s}-1}(x)\dfrac{u^{r_{s}-1}}{(r_{s}-1)!\ }\Biggr)\Bigl(A(x,u)\Bigr)^q.
\end{equation}
\end{theorem}

\begin{proof}
We immediately get the result by multiplying by $q^k$ and summing over $k$ the right member of Equation~\eqref{[A^r1...rl_k(x,u)]_s}.
\end{proof}

\begin{theorem}\label{t.qrell}
Let $\ell$ be a fixed integer.
A $q$-analogue exponential generating function of the excedance distribution over $(r_{1})\cdots   (r_{\ell})$-cycle permutations has the closed form
\begin{equation}\label{qrell}
A^{(r_{1})\cdots  (r_{\ell})}(q,x,u)=\exp\Biggl(qx\Bigl(A_{r_{1}-1}(x)\dfrac{u^{r_{1}}}{r_{1}!}+\cdots  +A_{r_{\ell}-1}(x)\dfrac{u^{r_{\ell}}}{r_{\ell}!}\Bigr)\Biggr).
\end{equation}
\end{theorem}

\begin{proof}
A permutation $\sigma$ in $\mathcal{S}_{n}^{(r_1)\cdots  (r_{\ell}),k}$ is the product of $(r_j,k_j)$-cycle permutations $(j=1,\cdots , \ell)$ such that $\sum_jk_j=k$. Thus,
\begin{equation*}
 A^{(r_1)\cdots   (r_\ell)}_{n,m,k}=\sum_{\tiny{i_{1}+\cdots +i_\ell=n}}\sum_{\tiny{
            m_1+\cdots  +m_\ell=m}}
        \sum_{\tiny{k_1+\cdots  + k_{\ell}}=k}(i_1,\ldots , i_\ell)!\  A^{(r_1)}_{i_{1},m_{1},k_{1}}\cdots A^{(r_\ell)}_{i_{\ell},m_{\ell},k_{\ell}}.
\end{equation*}
It follows that,
\begin{align*}
        A^{(r_{1})\cdots   (r_\ell)}(q,x,u)&=A^{(r_1)}(q,x,u)\cdots A^{(r_\ell)}(q,x,u), \\
        &=\exp\Bigl(qxA_{r_1-1}(x)\dfrac{u^{r_1}}{r_1!}\Bigr)\cdots \exp\Bigl(qxA_{r_\ell-1}(x)\dfrac{u^{r_{\ell}}}{r_{\ell}!}\Bigr).
\end{align*}
This gives the result.
\end{proof}

\begin{theorem}\label{t.q(r1)}
Let $\ell$ be a fixed integer. Let $(r_1,\ldots ,r_\ell)$ be the sequence such that $r_i>1,\ i=1,\ldots ,s$ and $1<r_{s+1}<\cdots<r_\ell$. A partial differential $q$-analogue exponential generating function of the excedance distribution over $r_1\cdots r_s(r_{s+1})\cdots  (r_\ell)$-cycle permutations has the closed form
\begin{align}
\nonumber   \dfrac{\partial^s[A^{r_{1}\cdots   r_{s}(r_{s+1})\cdots  (r_{\ell})}(q,x,u)]_{s}}{\partial u^s}&=q^sx^s(A_{r_{1}-1}(x)\dfrac{u^{r_1-1}}{(r_{1}-1)!\ }\cdots \\ & A_{r_{s}-1}(x)\dfrac{u^{r_{s}-1}}{(r_{s}-1)!\ })
    \left(A^{(r_{s+1})\cdots  (r_{\ell})}(x,u)\right)^q.\label{q(r1)}
\end{align}
\end{theorem}

\begin{proof}
Consider the first $s$ integers. 
\begin{align*}
\nonumber [A^{r_1\cdots   r_s(r_{s+1})\cdots   (r_\ell)}_{n+s,m,k}]_{s}=\binom{n}{r_{1}-1,\ldots , r_{s-1}} &\sum_{m_{1}+\cdots  +m_{s}+m_{t}=m}C_{r_{1},m_{1}}\cdots C_{r_{s},m_{s}}\times\\
       &\raggedright A^{(r_{s+1})\cdots  (r_{\ell})}_{n-\sum_{i=1}^{s}r_{i}-1,m_{t},k-s}.
\end{align*}
Therefore, we obtain
   \small 
\begin{align*}
      \nonumber  \dfrac{\partial^s[A^{r_1\cdots   r_s (r_{s+1})\cdots  (r_{\ell})}(q,x,u)]_s}{\partial u^s}&=q^sC_{r_1}(x)\dfrac{u^{r_{1}-1}}{(r_{1}-1)!\ }\cdots C_{r_{s}}(x)\dfrac{u^{r_{s}-1}}{(r_{s}-1)!\ }\times\\ &\Bigl(A^{(r_{s+1})\cdots (r_{\ell})}(x,u)\Bigr)^q,\\ \nonumber 
        &=q^sx^s\Bigl(A_{r_{1}-1}(x)\dfrac{u^{r_1-1}}{(r_{1}-1)!\ }\cdots A_{r_{s}-1}(x)\dfrac{u^{r_{s}-1}}{(r_{s}-1)!\ }\Bigr) \times  \\& \Bigl(A^{(r_{s+1})\cdots (r_{\ell})}(x,u)\Bigr)^q.
\end{align*}
\end{proof}

\begin{corollary}\label{c.r*}
For a fixed integer $\ell$, let $(r_1,\ldots,r_\ell)$ be a sequence such that $1<r_{1}<\cdots<r_{\ell}$. A $q$-analogue exponential generating function of the excedance distribution over $r^{\star}_1\cdots r^{\star}_{\ell}$-cycle permutations has the closed form
\begin{equation}\label{r*}
A^{r^\star_{1}\cdots r^{\star}_{\ell}}(q,x,u)=A(q,x,u)\exp{\Biggl(-qx\sum_{i=1}^{\ell}A_{r_i-1}\dfrac{u^{r_i}}{r_i!}\Biggr)}.
\end{equation}
\end{corollary}


\section{Acknowledgment}
The authors are very grateful to the anonymous referees for their helpful comments and suggestions.

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\noindent 2020 \emph{Mathematics Subject Classification}: Primary 05A15; Secondary 05A05, 05A19, 05A30.

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\noindent \emph{Keywords}:~associated Stirling number, excedance, Eulerian 
number, associated Stirling Eulerian number, generating function, $q$-analogue.

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\vspace*{+.1in}
\noindent
Received  September 26 2023; revised versions received
April 17 2024; September 2 2024; August 11 2025. 
Published in {\it Journal of Integer Sequences},
September 21 2025.

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\noindent
Return to \href{https://cs.uwaterloo.ca/journals/JIS/}{Journal of Integer Sequences home page}.
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