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\begin{document}
\pagenumbering{arabic}
\pagestyle{headings}

\newbox\Adr
\setbox\Adr\vbox{ \centerline{Toufik Mansour} \vskip5pt
\centerline{Department of Mathematics,}
\centerline{University of Haifa, 31905
Haifa, Israel} \centerline{E-mail: {\footnotesize
toufik@math.haifa.ac.il}}

\vskip4mm
\centerline{Yidong Sun$^*$} \vskip2mm
\centerline{Department of Mathematics,}
\centerline{ Dalian Maritime University,
116026 Dalian, P.R. China}
\centerline{E-mail: {\footnotesize
sydmath@yahoo.com.cn}} }

\title{ Excedance numbers for permutations in complex reflection groups}



\author[ Toufik Mansour and Yidong Sun]{\box\Adr}

\subjclass[2000]{Primary 05A05, 05A15.}
\keywords{ Complex reflection group,
excedance, colored permutations}
\thanks{$^*$Corresponding author: Yidong Sun.  }
\thanks{$^*$The author is
supported by The National Science Foundation of China (10726021).}


\begin{abstract}
{Recently, Bagno, Garber and Mansour 
[S\'em. Lotharingien 
Combin.\ {\bf
56} (2007), Art.~B56d] studied a kind of
excedance number on the complex reflection groups and computed its
multidistribution with the number of fixed points on the set of
involutions in these groups. In this note, we consider the similar
problems in more general cases and make a correction of one result
obtained by them.}
\end{abstract}
\maketitle

%===================================================================================

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\font\rms=cmr8
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\markright{\its S\'eminaire Lotharingien de
Combinatoire \bfs 58 \rms (2008), Article~B58b\hfill}
\def\thepage{}
%
%

\section{Introduction}
It is well known that there is a single infinite family of groups
$G_{r,s,n}$ and exactly $34$ other ``exceptional" complex reflection
groups \cite{ST}. The infinite family $G_{r,s,n}$, where $r,s,n$ are
positive integers with $s\mid r$, consists of the groups of $n\times
n$ matrices such that
\begin{itemize}
\item  the entries are either 0 or $r^{\rm th}$ roots of unity;
\item  there is exactly one nonzero entry in each row and each column;
\item  the $(r/s)^{{\rm th}}$ power of the product of the nonzero
entries is 1.
\end{itemize}

The classical Weyl groups appear as special cases: for $r=s=1$ we
have the symmetric group $G_{1,1,n}=S_n$, for $r=2, s=1$ we have the
hyperoctahedral group $G_{2,1,n}=B_n$, and for $r=s=2$ we have the
group of even-signed permutations $G_{2,2,n}=D_n$.

We say that a permutation $\pi \in G_{r,s,n}$ is {\it an involution}
if $\pi^2=1$. More generally, we define
$\mathcal{G}_{r,s,n}^{m}=\{\sigma\in G_{r,s,n}\mid \sigma^m=1\}$.
Recently, Bagno, Garber and Mansour \cite{BGM} studied an excedance
number on the complex reflection groups (see~\cite{SS}) and computed
the number of involutions having specific numbers of fixed points
and excedances. In this note, we consider the similar problems on
the set $\mathcal{G}_{r,s,n}^{m}$.

This paper is organized as follows. In Section~\ref{pre}, we recall
some properties of $G_{r,s,n}$ and define some parameters on
$\grn=G_{r,1,n}$ and hence also on $G_{r,s,n}$. In 
Section~\ref{mainresults}, we present our main results and compute the
corresponding recurrences together with explicit formulas.

%===================================================================================
\section{Preliminaries}\label{pre}

Let $r$ and $n$ be any two positive integers. {\it The group of
colored permutations of $n$ digits with $r$ colors} is the wreath
product $\grn=\mathbb{Z}_r \wr S_n=\mathbb{Z}_r^n \rtimes S_n$
consisting of all the pairs $(z,\tau)$ where $z\in\mathbb{Z}_r^n$
and $\tau \in S_n$. Let $\tau,\tau'\in S_n$,
$z=(z_1,...,z_n)\in\mathbb{Z}_r^n$ and
$z'=(z'_1,...,z'_n)\in\mathbb{Z}_r^n$, the multiplication in $\grn$
is defined by $(z,\tau) \cdot
(z',\tau')=((z_1+z'_{\tau^{-1}(1)},...,z_n+z'_{\tau^{-1}(n)}),\tau
\circ \tau') $, where $+$ is taken modulo $r$.

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\markboth{\SMALL TOUFIK MANSOUR AND YIDONG SUN}{\SMALL EXCEDANCE NUMBERS
FOR PERMUTATIONS IN COMPLEX REFLECTION GROUPS}
%
%

We use some conventions along this paper. For an element
$\sigma=(z,\tau) \in \grn$ with $z=(z_1,...,z_n)$ we write
$z_i(\sigma)=z_i$. For $\sigma=(z,\tau)$, we denote
$|\sigma|=(0,\tau), (0 \in \mathbb{Z}_r^n)$.

A much more natural way to present $\grn$ is the following: consider
the alphabet
$\Sigma=\{1^{[0]},\dots,n^{[0]},1^{[1]},\dots,n^{[1]},\dots,
1^{[r-1]},\dots,n^{[r-1]} \}$ as the set $\Lbrack
n\Rbrack=\{1,\dots,n\}$ colored by the colors $0,\dots,r-1$. Then,
an element of $\grn$ is called a {\it colored permutation}, i.e., a
bijection $\sigma: \Sigma \rightarrow \Sigma$ such that if
$\sigma(i)=k^{[t]}$ then $\sigma(i^{[j]})=k^{[t+j]}$ where $0\leq j
\leq r-1$ and the addition is taken modulo $r$. Occasionally, we
write $j$ bars over a digit $i$ instead of $i^{[j]}$. For example,
an element $(z,\tau)=((1,2,1,2),(3,1,2,4)) \in G_{3,4}$ will be
written as $(\bar{3} \bar{\bar{1}} \bar{2} \bar{\bar{4}})$.

For each $s\mid r$, the {\it complex reflection group} can also be defined as:
$$G_{r,s,n}:=\{\sigma \in G_{r,n} \mid \csum(\sigma)\equiv 0  \; {\rm
mod} \; s\},$$ where ${\rm csum}(\sigma) = \sumlim_{i=1}^n
z_i(\sigma)$.

One can define the following well-known statistics on $S_n$. For any
permutation $\sigma \in S_n$, $i \in \Lbrack n\Rbrack$ is {\it an
excedance of $\sigma$} if and only if $\sigma(i)>i$. We denote the
number of excedances by ${\rm exc}(\sigma)$. Another natural
statistic on $S_n$ is the number of fixed points, denoted by ${\rm
fix}(\sigma)$. We can similarly define some statistics on $\grn$.
The complex reflection group $G_{r,s,n}$ inherits all of them. Given
any ordered alphabet $\Sigma'$, we recall the definition of the {\it
excedance set} of a permutation $\sigma$ on $\Sigma'$:
$${\rm Exc}(\sigma)=\{i \in \Sigma' \mid \sigma(i)>i\},$$ and the {\it excedance
number} is defined to be ${\rm exc}(\sigma)=|{ \rm Exc}(\sigma)|$.

We define the color order on the set
$\Sigma=\{1,\dots,n,\bar{1},\dots,\bar{n},\dots,
1^{[r-1]},\dots,n^{[r-1]} \}$ for $0\leq j<i< r$ by $1^{[i]}<2^{[i]}
< \cdots < n^{[i]} < 1^{[j]} < 2^{[j]} < \cdots < n^{[j]}$. We note
that there are some other possible ways of defining orders on
$\Sigma$, some of them lead to other versions of the excedance
number, see for example \cite{BG}. For example, given the color
order $\bar{\bar{1}} < \bar{\bar{2}} <\bar{\bar{3}} < \bar{1} <
\bar{2} <\bar{3} < 1 < 2 < 3$, we write
$\sigma=(2\bar{1}\bar{\bar{3}}) \in G_{3,3}$ in an extended form
$$\hbox to\hsize{$(\star)\hfill\begin{pmatrix} \bar{\bar{1}} & \bar{\bar{2}} & \bar{\bar{3}} &
\bar{1} & \bar{2}& \bar{3} & 1 & 2 & 3\\
\bar{\bar{2}} & 1 & \bar{3} & \bar{2} & \bar{\bar{1}}  &  3 & 2 &
\bar{1} & \bar{\bar{3}}
\end{pmatrix},\hfill$}$$
which implies that  ${\rm
Exc}(\sigma)=\{\bar{\bar{1}},\bar{\bar{2}},\bar{\bar{3}},\bar{1},\bar{3},1\}$
and ${\rm exc}(\sigma)=6$.

Define ${\rm Exc}_A(\sigma) = \{ i \in \Lbrack n-1\Rbrack \ | \
\sigma(i) > i \}$, where the comparison is with respect to the color
order, and denote ${\rm exc}_A(\sigma) = |{\rm Exc}_A(\sigma)|$. For
instance, if $\sigma=(\bar{1}\bar{\bar{3}}2\bar{\bar{4}}) \in
G_{3,4}$, then ${\rm csum}(\sigma)=5$, ${\rm Exc_A}(\sigma)=\{3\}$
and hence ${\rm exc}_A(\sigma)=1$.

Now we can define the colored excedance number for $\grn$ by $${\rm
exc}^{{\rm Clr}}(\sigma)=r \cdot {\rm exc}_A(\sigma)+{\rm
csum}(\sigma).$$ Let $\Sigma$ be ordered by the color order, and let
${\rm exc}(\sigma)$ denote the number of excedances of $\sigma$ with
respect to the color order. Then it is not difficult to see that
${\rm exc}(\sigma)={\rm exc}^{{\rm Clr}}(\sigma)$
for any $\sigma \in \grn$ (cf.\ \cite{BG}).

For $\sigma=(z,\tau)\in \grn$, $|\sigma|$ is the permutation of
$\Lbrack n\Rbrack$ satisfying $|\sigma|(i)=\tau(i)$. We say that $i
\in \Lbrack n\Rbrack$ is an {\it absolute fixed point} of $\sigma
\in \grn$ if $|\sigma|(i)=i$. We denote the number of absolute fixed
points of $\sigma \in \grn$ by ${\rm fix(\sigma)}$.


\section{Main results and proofs}\label{mainresults}
Recall that $\mathcal{G}_{r,s,n}^{m}=\{\sigma\in
G_{r,s,n}\mid \sigma^m=1\}$. Define
\begin{align*}
H_{r,s,n}^{(m)}(u,v,w)&=\sum_{\sigma\in
\mathcal{G}_{r,s,n}^{m}}u^{{\rm fix}(\sigma)}v^{{\rm
exc}_{A}(\sigma)}w^{{\rm csum}(\sigma)}, \\
\mathcal{H}_{r,s}^{(m)}(x;u,v,w)&=\sum_{n \geq
0}{H_{r,s,n}^{(m)}(u,v,w)\frac{x^n}{n!}}.
%=\sum_{n\geq0}\sum_{\sigma\in
%\mathcal{G}_{r,s,n}^{m}}\left( u^{{\rm fix}(\sigma)}v^{{\rm
%exc}_A(\sigma)}w^{{\rm csum}(\sigma)}\right)\frac{x^n}{n!}.
\end{align*}
It is well known that the Eulerian number, $A_{d-1,k}$, is the
number of permutations on $\Lbrack d-1\Rbrack$ with $k-1$ excedances
for $k\in \Lbrack d-1\Rbrack$, which is also the number of cyclic
permutations in $S_d$ with $k$ excedances. A bijective proof of this
fact is given in \cite[Theorem~1.19]{FS}.

Our main result can be formulated as follows.
\begin{theorem}\label{maintheo}
For any integers $r,m\geq 1$, the generating function
$\mathcal{H}_{r,1}^{(m)}(x;u,v,w)$ is
\begin{eqnarray*}
\exp\left\{\sum_{\{t\mid 0\leq t<r, r\mid tm\}}xuw^t +\sum_{d\mid m,d\geq
2}\frac{x^d}{d!}\sum_{k=1}^{d-1}A_{d-1,k}\sum_{i=0}^k\binom{k}{i}v^{k-i}\sum_{r\mid \frac{tm}{d}}
U_{d-k,t}^{(i)}w^{t}\right\},
\end{eqnarray*}
where $U_{d-k,t}^{(i)}$ is the coefficient of $x^t$ in
$(x+x^2+\cdots+x^{r-1})^i(1+x+\cdots+x^{r-1})^{d-k}$, i.e.,
\begin{eqnarray*}
U_{d-k,t}^{(i)}=\sum_{j=0}^i(-1)^{i-j}\binom{i}{j}\sum_{{\ell}\geq
0}(-1)^{{\ell}}\binom{d+j-k}{{\ell}}\binom{d+j+t-k-{\ell}r-1}{t-{\ell}r}.
\end{eqnarray*}
\end{theorem}
\begin{proof} For any $\pi \in \mathcal{G}_{r,1,n}^{m}$, the length of each
cycle of $\pi$ is a factor of $m$. Therefore there exist
$k_1,k_2,\ldots,k_{d-1}\in \Lbrack n-1\Rbrack$ with $d\mid m$ such that
$k_1,k_2,\ldots,k_{d-1}$ and $n$ form a cycle of $|\pi|$.

If $d=1$, that is, $\pi(n)=n^{[j]}$ for some $j$ with $0\leq j\leq
r-1$, then $\pi^m(n)=n^{[jm]}=n$, which implies that $r\mid jm$. Define
$\pi' \in \mathcal{G}_{r,1,n-1}^m$ by ignoring the last digit of
$\pi$. Then we have
\begin{align*}
{\rm fix}(\pi)&={\rm fix}(\pi')+1,\\
{\rm exc}_A(\pi)&={\rm exc}_A(\pi'),\\
\csum(\pi)&=\csum(\pi')+j.
\end{align*}

If $d\geq 2$, we know that there are $A_{d-1,k}$ cyclic permutations
in $S_d$ with $k$ excedances for $k\in \Lbrack d-1\Rbrack$. For any
cyclic permutation $C$ of length $d$ in $S_d$ with $${\rm
Exc}(C)=\{j\in \Lbrack d-1\Rbrack\mid C(j)>j\}$$ such that ${\rm
exc}(C)=k$, we can color the symbols in $C$ with the color set
$\{[0],[1],\ldots,[r-1]\}$ and obtain the colored cyclic permutation
$C'$. Suppose that ${\rm exc}_A(C')=k-i$. We know that ${\rm
Exc}_A(C')\subseteq{\rm Exc}(C)$, which means that ${\rm
exc}(C)-{\rm exc}_A(C)=i$. In other words, there are $i$ 
symbols in ${\rm Exc}(C)$ with color numbers ranging from $[1]$ to
$[r-1]$. Clearly, there are $\binom{k}{i}$ ways to do this.

Let $t=\csum(C')$ and $t_{\ell}$ be the color number of $\ell\in
\Lbrack d\Rbrack$. Then we have the equation $t=t_1+t_2+\cdots+t_d$
with $0\leq t_1,t_2,\cdots,t_d\leq r-1$ such that
\begin{itemize}
\item $t_j=0$ for $j\in {\rm Exc}(C)$ and $j$ has color number
$[0]$, and 
\item $1\leq t_j\leq r-1$ for all $j\in {\rm Exc}(C)-{\rm
Exc}_A(C')$; there are $i$ such $j$'s.
\end{itemize}
Therefore there are $U_{d-k,t}^{(i)}$ solutions of the
above equation. In total, there are $\binom{k}{i}U_{d-k,t}^{(i)}$
ways to color the symbols in $C$ such that $\csum(C')=t$ and ${\rm
exc}_A(C')=k-i$, where $U_{d-k,t}^{(i)}$ is the coefficient of $x^t$
in $(x+x^2+\cdots+x^{r-1})^i(1+x+\cdots+x^{r-1})^{d-k}$, which can
be expressed as
\begin{align*}
U_{d-k,t}^{(i)}&=[x^t](x+x^2+\cdots+x^{r-1})^i(1+x+\cdots+x^{r-1})^{d-k}\\
&=[x^t]\left(\frac{1-x^r}{1-x}-1\right)^i\left(\frac{1-x^r}{1-x}\right)^{d-k}\\
&=[x^t]\sum_{j=0}^i(-1)^{i-j}\binom{i}{j}\Big(\frac{1-x^r}{1-x}\Big)^{d+j-k}\\
&=\sum_{j=0}^i(-1)^{i-j}\binom{i}{j}\sum_{{\ell}\geq
0}(-1)^{{\ell}}\binom{d+j-k}{{\ell}}\binom{d+j+t-k-{\ell}r-1}{t-{\ell}r}.
\end{align*}
Let $C'=(i_1^{[t_1]},i_2^{[t_2]},\dots,i_d^{[t_d]})$. Then
$C'^d=(i_1^{[t]},i_2^{[t]},\dots,i_d^{[t]})$ with
$t=t_1+t_2+\cdots+t_d$. Hence,
$C'^{m}=(i_1^{[\frac{tm}{d}]},i_2^{[\frac{tm}{d}]},\dots,i_d^{[\frac{tm}{d}]})=1$
implies that $r\mid \frac{tm}{d}$. For any $\pi \in
\mathcal{G}_{r,1,n}^{m}$ such that the symbol $n$ lies in a cycle
$C'$ of length $d\geq 2$ with $d\mid m$ (note that there are
$\binom{n-1}{d-1}$ ways to choose the digits of such a cycle),
define $\pi'' \in \mathcal{G}_{r,1,n-d}^{m}$ in the following way:
write $\pi$ in its complete notation, i.e., as a matrix of two rows,
see $(\star)$. The first row of $\pi''$ is $(1,2,\ldots,n-d)$ while
the second row is obtained from the second row of $\pi$ by ignoring
the digits in $C'$, and ``standardizing" the remaining digits, that is,
by replacing the $i$-th largest of the remaining digits by $i$ (keeping
the colors). For example, if $\pi=(\bar{3} 4
\bar{1} {\bar{2}})\in \mathcal{G}_{2,1,4}^{4}$ and $C'=(1,\bar{3})$,
then $\pi''=(2\bar{1})$. The parameters satisfy
\begin{align*}
{\rm fix}(\pi)&={\rm fix}(\pi''),\\
{\rm exc}_A(\pi)&={\rm exc}_A(\pi'')+{\rm exc}_A(C'),\\
\csum(\pi)&=\csum(\pi'')+\csum(C').
\end{align*}
The above considerations lead to the following recurrence:
%%\begin{small}
\begin{align*}
H_{r,1,n}^{(m)}(u,v,w)&=H_{r,1,n-1}^{(m)}(u,v,w)\sum_{\{t\mid 0\leq t<r,
r\mid tm\}}uw^t  \\
& \kern2cm+ \sum_{d\mid m,d\geq 2}H_{r,1,n-d}^{(m)}(u,v,w)\binom{n-1}{d-1}A_{m,d}(v,w),
\end{align*}
%%\end{small}
where
\begin{equation*}
A_{m,d}(v,w)=\sum_{k=1}^{d-1}A_{d-1,k}\sum_{i=0}^k\binom{k}{i}v^{k-i}\sum_{r\mid \frac{tm}{d}}
U_{d-k,t}^{(i)}w^{t}.
\end{equation*}
Rewriting the recurrence in terms of generating functions, we obtain
that
\begin{align*}
\frac{\partial}{\partial x}\mathcal{H}_{r,1}^{(m)}(x;u,v,w)
&=\sum_{n \geq1}H_{r,1,n}^{(m)}(u,v,w)\frac{x^{n-1}}{(n-1)!}\\
&= \sum_{n \geq
1}\frac{x^{n-1}}{(n-1)!}H_{r,1,n-1}^{(m)}(u,v,w)\sum_{\{t\mid 0\leq t<r,
r\mid tm\}}uw^t\\
&\kern2cm +\sum_{d\mid m,d\geq
2}A_{m,d}(v,w)\frac{x^{d-1}}{(d-1)!}\sum_{n\geq d}\frac{x^{n-d}}{(n-d)!}H_{r,1,n-d}^{(m)}(u,v,w)\\
&=\mathcal{H}_{r,1}^{(m)}(x;u,v,w)\Big(\sum_{\{t\mid 0\leq t<r,
r\mid tm\}}uw^t+\sum_{d\mid m,d\geq
2}A_{m,d}(v,w)\frac{x^{d-1}}{(d-1)!}\Big).
\end{align*}
Thus, the generating function $\mathcal{H}_{r,1}^{(m)}(x;u,v,w)$
satisfies
\begin{equation*}
\frac{\frac{\partial}{\partial
x}\mathcal{H}_{r,1}^{(m)}(x;u,v,w)}{\mathcal{H}_{r,1}^{(m)}(x;u,v,w)}
=\sum_{\{t\mid 0\leq t<r, r\mid tm\}}uw^t+\sum_{d\mid m,d\geq
2}A_{m,d}(v,w)\frac{x^{d-1}}{(d-1)!}.
\end{equation*}
Integrating both sides of the above differential equation 
with respect to $x$, and using the fact that
$\mathcal{H}_{r,1}^{(m)}(0;u,v,w)=1$, we obtain the explicit
expression for $\mathcal{H}_{r,1}^{(m)}(x;u,v,w)$ given in 
Theorem~\ref{maintheo}. This completes the proof of the theorem.
\end{proof}

In particular, if $m=p$ is a prime, then we have the following
corollary.
\begin{corollary}\label{coro3.2}
Let $r\geq 1$, and let $p$ be a prime. The generating function
$\mathcal{H}_{r,1}^{(p)}(x;u,v,w)$ is given by
\begin{equation*}
\exp\left\{ux\lambda_{r,p}(w)
+\frac{x^p}{p!}\sum_{k=1}^{p-1}A_{p-1,k}\sum_{i=0}^k\binom{k}{i}v^{k-i}\sum_{j\geq
0}U_{p-k,jr}^{(i)}w^{jr}\right\},
\end{equation*}
where $\lambda_{r,p}(w)=\sum_{i=0}^{p-1}{w^{\frac{ir}{p}}}$ for
$p\mid r$, and $\lambda_{r,p}(w)=1$ for $p\nmid r$.
\end{corollary}

For the sake of comparison, the cases $p=2$ and $p=3$ in 
Corollary~\ref{coro3.2} generate the following explicit formulas for
$\mathcal{H}_{r,1}^{(2)}(x;u,v,w)$ and
$\mathcal{H}_{r,1}^{(3)}(x;u,v,w)$:
\begin{align*}
\mathcal{H}_{r,1}^{(2)}(x;u,v,w)&= \exp({ux\lambda_{r,2}(w)
+\frac{x^2}{2}(v+(r-1)w^{r})}),\\
\mathcal{H}_{r,1}^{(3)}(x;u,v,w)&= \exp({ux\lambda_{r,3}(w)
+\frac{x^3}{6}B_{3,3}(v,w)}),
\end{align*}
where
$B_{3,3}(v,w)=v^2+v(1+3(r-1)w^r)+(r^2-1)w^{r}+(r-1)(r-2)w^{2r}$.
%%and $A_{4,2}(v,w)=v+(r-1)w^r$,
%%$A_{4,4}(v,w)=v^3+v^2(4+7(r-1)w^r)+v(1+2(r-1)(r+2)w^r+5(r-1)(r-2)w^{2r})+(r^3-1)w^{r}+2(r^2-1)(2r-3)w^{2r}+2\binom{r}{3}w^{3r}$
%%for $2\not|r$

Now let us compute the exponential generating function
$\mathcal{H}_{r,s}^{(m)}(x;u,v,w)$ for the sequence
$\{H_{r,s,n}^{(m)}(u,v,w)\}_{n\geq0}$. For any $\sigma\in
\mathcal{G}_{r,s,n}^{m}$, we have $\csum(\sigma)\equiv 0  \; ({\rm
mod} \; s)$, so we should collect all the terms in which the
exponent of $w$ in $\mathcal{H}_{r,1}^{(m)}(u,v,w)$ is a
multiple of $s$. This observation leads to the following result.

\begin{theorem}\label{theo3.4}
For $r,m,s\geq 1$, let
${\mathcal{H}_{r,1}^{(m)}(x;u,v,yw)}=\sum_{n\geq
0}G_{m,r,n}(x;u,v,w)y^n.$ Then
\begin{equation*}
\mathcal{H}_{r,s}^{(m)}(x;u,v,w)=\sum_{k\geq
0}G_{m,r,sk}(x;u,v,w).
\end{equation*}
\end{theorem}

Now let us focus on the case $m=2$. Recall that
\begin{equation*}
{\mathcal{H}_{r,1}^{(2)}(x;u,v,w)}=
\begin{cases}
e^{ux+\frac{1}{2}x^2(v+(r-1)w^{r})},& {\rm if}\ r\ {\rm odd},\\
e^{ux(1+w^{\frac{r}{2}})+\frac{1}{2}x^2(v+(r-1)w^{r})},& {\rm if}\
r\ {\rm even}.
\end{cases}
\end{equation*}
Then, by Theorem~\ref{theo3.4}, we can compute an explicit formula
for $\mathcal{H}_{r,s}^{(2)}(x;u,v,w)$. Since $s\mid r$, we have two
cases, depending on whether $r$ is odd or even.
\begin{itemize}
\item If $r$ is an odd number, then it is clear that the exponent of $y$ in each term
of the expansion of $\mathcal{H}_{r,1}^{(2)}(x;u,v,yw)$ is always a
multiple of $s$. Hence,
$$\mathcal{H}_{r,s}^{(2)}(x;u,v,w)=\mathcal{H}_{r,1}^{(2)}(x;u,v,w).$$
\item Similarly, if $r$ is an even number and $s\mid\frac{r}{2}$, we have 
$$\mathcal{H}_{r,s}^{(2)}(x;u,v,w)=\mathcal{H}_{r,1}^{(2)}(x;u,v,w).$$
\item Let $r$ be any even number such that $s\nmid\frac{r}{2}$. Since
$e^{ux(1+(yw)^{\frac{r}{2}})}=e^{ux}\sum_{k\geq0}\frac{(ux(yw)^{\frac{r}{2}})^k}{k!}$
and
$e^{\frac{1}{2}x^2(v+(r-1)(yw)^{r})}=e^{\frac{1}{2}x^2v}\sum_{k\geq0}\frac{((r-1)x^2(yw)^r)^k}{2^kk!}$,
then by collecting the coefficients of powers of $y$ in
$\mathcal{H}_{r,1}^{(2)}(x;u,v,w)$, where the exponent of $y$ is a
multiple of $s$, we obtain
$$e^{\frac{1}{2}x^2(v+(r-1)(yw)^{r})}\sum_{k\geq
0}\frac{(ux)^{2k}(yw)^{kr}}{(2k)!}=e^{ux+\frac{1}{2}x^2(v+(r-1)(yw)^{r})}\frac{e^{uxw^{\frac{r}{2}}}+e^{-uxw^{\frac{r}{2}}}}{2}.$$
\end{itemize}
Therefore, the above cases yield the following result.
\begin{proposition}\label{pros1}
We have
\begin{equation*}
{\mathcal{H}_{r,s}^{(2)}(x;u,v,w)}=
\begin{cases}
e^{ux+\frac{1}{2}x^2(v+(r-1)w^{r})},& {\rm if}\ r\ {\rm odd},\\
e^{ux(1+w^{\frac{r}{2}})+\frac{1}{2}x^2(v+(r-1)w^{r})},& {\rm if}\
r\ {\rm even\ and\ }
s\mid\frac{r}{2},\\
e^{ux+\frac{1}{2}x^2(v+(r-1)w^{r})}\frac{e^{uxw^{\frac{r}{2}}}+e^{-uxw^{\frac{r}{2}}}}{2},&
{\rm if}\ r\ {\rm even\ and\ } s\nmid \frac{r}{2}.
\end{cases}
\end{equation*}
\end{proposition}

Note that $\mathcal{H}_{r,s}^{(2)}(x;u,v,w)$ is the generating
function for the number of involutions in
$\mathcal{G}_{r,s,n}^{(2)}$. By expanding generating functions,
Bagno, Garber and Mansour \cite{BGM} obtained explicit formulas
for the number of involutions in $\mathcal{G}_{r,s,n}^{(2)}$. But
the expression in Proposition~5.7 of \cite{BGM} must be corrected 
in the third case, with the correct formula appearing above in the
third case of Proposition~\ref{pros1}.
Hence, Corollaries~$5.8-5.10$ in \cite{BGM} should be, in fact, 
the following three corollaries, respectively.
\begin{corollary}
The polynomial $H_{r,s,n}^{(2)}(u,v,w)$ is given by
\begin{equation*}
\sum_{k_1+2k_2+2k_3=n}\frac{n!}{k_1!(2k_2)!k_3!}\cdot\frac{u^{k_1+2k_2}w^{rk_2}(v+(r-1)w^r)^{k_3}}{2^{k_3}}.
\end{equation*}
\end{corollary}
\begin{corollary} Let $r\geq1$.  The number of colored involutions in
$\mathcal{G}_{r,s,n}^{(2)}$ {\em(}$r$ even, $s \nmid
\frac{r}{2}${\em)} with
exactly $k$ absolute fixed points and ${\rm exc}_A (\pi)=\ell$ is
given by
\begin{equation*}
\sum_{k+2k_3=n,k_1+2k_2=k}\binom{k_3}{\ell}\cdot\frac{n!}{k_1!(2k_2)!k_3!}\cdot
\frac{(r-1)^{k_3-\ell}}{2^{k_3}}.
\end{equation*}
\end{corollary}
\begin{corollary}
The number of involutions  $\pi \in \mathcal{G}_{r,s,n}^{(2)}$ {\em(}$r$
even, $s \nmid \frac{r}{2}${\em)} with ${\rm exc}^{{\rm Clr}}(\pi)=k$ is given by
\begin{equation*}
\sum_{k_1+2k_2+2k_3=n,\
r(k_2+k_3)=k}\frac{n!}{k_1!(2k_2)!k_3!}\cdot\left(\frac{r}{2}\right)^{\frac{k}{r}}.
\end{equation*}
\end{corollary}


\vskip0.5cm
{\bf Acknowledgment.} The authors would like to thank
Eli Bagno and David Garber for reading previous version of the
present paper and for several helpful discussions.

\vskip1cm
%==============================================================================================================
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\vskip0.5cm



\end{document}
