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        {\fancyplain{}{\small Seidel Generation of $q$-derangement numbers}}
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\begin{document}
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\rule{0cm}{25.2mm}\\
{\noindent{\Large \bf A combinatorial interpretation of the Seidel \\
\rule{0mm}{5mm}generation of $q$-derangement numbers}}\\

\vspace{2mm}
\noindent Robert J. {Clarke}${}^1$, Guo-Niu {Han}${}^2$,
 and Jiang {Zeng}${}^3$\\

\noindent{\small
${}^1$Pure Mathematics Department, University of Adelaide,
Adelaide, South Australia 5005\\
 \phantom{${}^3$}rclarke@maths.adelaide.edu.au}\\
\vspace{1mm}
\noindent{\small ${}^2$I.R.M.A., Universit\'e Louis-Pasteur et C.N.R.S.,
67084 Strasbourg Cedex, France\\
\phantom{${}^3$} guoniu@math.u-strasbg.fr}\\
\vspace{1mm}
\noindent{\small
${}^3$Institut Girard Desargues, Universit\'e Claude Bernard (Lyon I)\\
\phantom{${}^3$}F-69622 Villeurbanne Cedex, France\\
 \phantom{${}^3$}zeng@desargues.univ-lyon1.fr}

\vspace{3mm}
\noindent {\small Received September 15, 1997}
\begin{flushright}
{\small {\it AMS Subject Classification}: 05A15, 05A30}
\end{flushright}
\vspace{4mm}
\setlength{\baselineskip}{.9\baselineskip}
\noindent
{\small{\bf Abstract.}
In \cite{DR} Dumont and  Randrianarivony have given several combinatorial
interpretations for the coefficients of the Euler-Seidel matrix associated
to $n!$.
In this paper we consider a $q$-analogue of their results, which
leads to the discovery of a new mahonian statistic ``maf" on the symmetric
group. We then give new proofs and generalizations of some results of
Gessel and Reutenauer \cite{GR} and Wachs \cite{Wa}.
}

\vspace{3mm}
\noindent
{\small {\it Keywords}: Mahonian statistics, permutations, $q$-derangement
numbers, Seidel matrices}

\vspace{5mm}
\normalsize
\section{Introduction}
Euler (see \cite{DR}) considered the {\it difference table}
$(d_n^k)_{0\leq k\leq n}$,
where the generic coefficients $d_n^k$ are defined by
\begin{equation}\label{diff}
d_n^n=n!\quad \textrm{and}\quad d_n^k=d_n^{k+1}-d_{n-1}^k
\qquad (1\leq k\leq n-1).
\end{equation}

Let $a_n^k=d_{n+k}^k$ ($n,\,k\geq 0$). Then the above relations can be
written as
$$
a_0^k=k!\quad \textrm{and}\quad a_n^k=a_{n}^{k-1}+a_{n+1}^{k-1}
\qquad (n,\, k\geq 0).
$$
The matrix $(a_n^k)_{n,\,k\geq 0}$ is
also called  the {\it Seidel matrix} associated to the
sequence $a_n^0$ in the literature (see \cite{Du,DV}).
The first terms of these matrices are as follows:
$$\vcenter{\hbox{$\begin{array}{c|cccccc}
\hbox{$n$}\backslash\hbox{$k$}&0&1&2&3&4&5\\
\hline
0&1& & & & &\\
1&0&1& & &&\\
2&1&1&2& &&\\
3&2& 3& 4 &6&&\\
4&9&11&14&18&24&\\
5&44&53&64&78&96&120\\
\end{array}$}
\smallskip
\hbox{\hskip 3cm $(d_n^k)$}
}
\quad
\longrightarrow
\quad
\vcenter{\hbox{$\begin{array}{c|cccccc}
\hbox{$k$}\backslash\hbox{$n$}&0&1&2&3&4&5\\
\hline
0&1& 0&1 &2 & 9&44\\
1&1&1& 3&11 &53&\\
2&2&4&14&64 &&\\
3&6& 18&78&&&\\
4&24&96&&&&\\
5&120&&&&&\\
\end{array}$}
\smallskip
\hbox{\hskip 3cm $(a_n^k)$}
}
$$

Iterating the difference equation (\ref{diff}) we derive
\begin{equation}\label{derang}
a_n^0=d_n^0=n!\left(1-\frac{1}{1!}+\frac{1}{2!}-\cdots +(-1)^n
\frac{1}{n!}\right),
\end{equation}
which is the {\it classical derangement number} $d_n$, that is,
the number of derangements on $\{1,2,\cdots,n\}$ ({\sl cf}. \cite[p.
67]{Sta}).

In several recent papers \cite{CR,GR,Wa,DW}, the $q$-maj counting of the
derangements on
$\{1,2,\cdots,n\}$ has been studied.
Consider the {\it $q$-derangement numbers} $d_n(q)$ defined by
\begin{equation}
d_n(q)=\mathop{{\sum}}_{\sigma\in{\cal D}_n}q^{\maj\sigma},
\end{equation}
where ${\cal D}_n$ is the set of all derangements on
$\{1,2,\cdots,n\}$.
Then the following $q$-analogue of equation (\ref{derang}) has been obtained:
\begin{equation}
d_n(q) =
[n]_q!\sum_{i=0}^n (-1)^{i}\frac{q^\mychoose{i}{2}}{[i]_q!} \qquad (n\geq
1).
\end{equation}
Here, $[n]_q=1+q+\cdots +q^{n-1}$ is the $q$-analogue of the nonnegative
integer $n$ and
$[n]_q!=[1]_q[2]_q\cdots[n]_q$ is the $q$-analogue of $n!$.

In this paper, we shall put the $q$-derangement numbers in the context of a
Seidel matrix as Dumont and Randrianarivony \cite{DR} did for the ordinary
derangement numbers.
To this end, in section 2 we introduce the notion of $q$-Seidel matrix.
In section 3 we define a new statistic ``maf" on permutations and then
prove bijectively that this is a mahonian statistic.  In section 4 we
consider the $q$-Seidel matrix associated  to the $q$-derangement numbers
and give
combinatorial interpretations for all of the coefficients in this matrix in
terms of the new statistic "maf".
As a consequence we get a new proof of a formula of Gessel and Reutenauer
\cite{GR} and of Wachs \cite{Wa}. Finally we close this paper with some
remarks and open questions.
\bigskip

We will need the following notations and results of $q$-calculus (see
\cite{GaRa}).  The $q$-binomial coefficients are defined by
$$
\mychoose{n}{k}_q = \frac{[n]_q!}{[k]_q![n-k]_q!}\qquad (n\geq k\geq 0).
$$
Define also $(t;q)_n=(1-t)(1-qt)\cdots (1-q^{n-1}t)$ and
$(t;q)_\infty=\lim_{n\to \infty}(t;q)_n$.
Then the two $q$-analogues of the exponential series
 $e^t=\sum_{n\geq 0}t^n/n!$ are defined by
\begin{eqnarray}
e_q(t) &=& \sum_{n\geq 0} \frac{t^n}{
[n]_q!}=\frac{1}{((1-q)t;q)_\infty},\hbox{\label{eq:defe}}\\
E_q(t) &=& \sum_{n\geq 0} \frac{q^\mychoose{n}{2}t^n}{
[n]_q!}=(-(1-q)t;q)_\infty.\hbox{\label{eq:defE}}
\end{eqnarray}
Notice that $e_q(t)\cdot E_q(-t)=1$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{\normalsize $q$-Seidel matrices}
\setcounter{equation}{0}
Let us introduce the following generalization of Seidel matrix.
\begin{defn}
Given a sequence $(a_n(x,q))$ $(n\geq 0)$ of elements in a commutative
ring, we call the $q$-Seidel matrix asssociated to $(a_n(x,q))$  the double
sequence $(a_n^k(x,q))$ $(n\ge 0, k\ge 0)$ given by the recurrence
\begin{equation}\label{seidel}
\begin{cases}
a_n^0(x,q)=a_n(x,q), &\text{$(n\ge 0)$}\cr
a_n^k(x,q)=x q^na_n^{k-1}(x,q)+a_{n+1}^{k-1}(x,q). &\text{$(k\ge 1,n\ge
0)$}\cr
\end{cases}
\end{equation}
Moreover $(a_n^0(x,q))$ is called the initial sequence and $(a_0^n(x,q))$
the final sequence of the $q$-Seidel matrix.
\end{defn}

\begin{lem} We have
\begin{equation}
a_n ^k(x,q) = \sum_{i=0}^k (x q^n)^{k-i}\mychoose{k}{i}_q a_{n+i}^0(x,q).
\label{eq:lem1}
\end{equation}
\end{lem}
\pf Recall that
$$
\mychoose{n}{k}_q =q^{n-1} \mychoose{n-1}{k-1}_q + \mychoose{n-1}{k}_q.
$$
We proceed by induction on $k$.
Clearly (\ref{eq:lem1}) is valid for $k=1$. Suppose (\ref{eq:lem1}) is true
for
$k-1$. We then have
\begin{eqnarray*}
a_n^k(x,q)&=&
\sum_{i=0} ^{k-1} \mychoose{ k-1}{i }_q \Bigl(
(x q^{n})^{k-i} a_{n+i}^0(x,q)+  (x
q^{n+1})^{k-1-i}
 a_{n+1+i}^0(x,q) \Bigr)\cr
&=&(x q^{n})^k a_n^0(x,q)
 + \sum_{i=1} ^{k-1} (x q^{n})^{k-i}\mychoose{ k-1}{i }_q a_{n+i}^0(x,q)\cr
  && \hskip 1cm + \sum_{i=0} ^{k-2} (x q^{n+1})^{k-1-i}\mychoose{ k-1 }{i
}_q
 a_{n+1+i}^0(x,q)+ a_{n+k}^0(x,q)\cr
&=&(x q^{n})^k a_n^0(x,q) + \sum_{i=1} ^{k-1}
(x q^{n})^{k-i}\mychoose{ k }{ i }_q a_{n+i}^0(x,q)+  a_{n+k}^0(x,q).
\end{eqnarray*}
This completes the proof. \qed\medskip

In particular we pass from the initial sequence to the final sequence and
conversely by the {\it Gauss inversion formula} \cite[p. 96]{Ai}:
\begin{eqnarray}
a_0 ^n(x,q)& =&\sum_{i=0} ^n  x^{n-i} \mychoose{ n}{ i}_q a_i
^0(x,q),\hbox{\label{eq:a}}\\
a_n ^0 (x,q)&=& \sum_{i=0} ^n (-x) ^{n-i} q^\mychoose{n-i}{ 2} \mychoose{
n}{i}_q
a_0 ^i(x,q).
\end{eqnarray}

Define the generating functions as follows:
$$
a(t)=\sum_{n\geq 0}a_n^0(x,q)t^n,\qquad \bar a(t)=\sum_{n\geq
0}a_0^n(x,q)t^n,
$$
and
$$
A(t)=\sum_{n\geq 0}a_n^0(x,q)\frac{t^n}{[n]_q!},\quad
\bar A(t)=\sum_{n\geq 0}a_0^n(x,q)\frac{t^n}{ [n]_q!}.
$$

\begin{prop}\label{prop:aA} The generating functions of the
 initial and final sequences are related by the following equations:
\begin{eqnarray}
\bar a(t)&=&\sum_{n\geq 0}a_n^0(x,q)\frac{t^n}{(x t;q)_{n+1}};\\
\overline{A}(t)&=& e_q(x t) A(t).\hbox{\label{eq:AA}}
\end{eqnarray}
\end{prop}
\pf
Note that
$$
\frac{1}{(t;q)_{n+1}}=\sum_{k=0}^\infty\mychoose{n+k}{ k}_q t^k.
$$
Hence
\begin{eqnarray*}
\sum_{n\geq 0}a_n^0(x,q)\frac{t^n}{(x t;q)_{n+1}}&=&\sum_{n,k\geq 0}
\mychoose{n+k}{k}_qa_n^0(x,q)x^k t^{n+k}\cr
&=&\sum_{m\geq 0}t^m\sum_{n=0}^m\mychoose{m}{n}_qx^{m-n}a_n^0(x,q)\cr
&=&\sum_{m\geq 0}a_0^m(x,q)t^m.
\end{eqnarray*}
By (\ref{eq:defe}) we have
\begin{eqnarray*}
e_q(x t) A(t)&=&\sum_{i,j\geq 0} \frac{a_i^0(x,q)t^i}{[i]_q!} \cdot
\frac{x^jt^j}{[j]_q!} \cr
&=&\sum_{i,j\geq 0}\mychoose{i+j}{i}_q a_i^0(x,q) x^{j}\frac{t^{i+j} }{
[i+j]_q!} \cr
&=& \sum_{n\geq 0} \Bigl( \sum_{i=0}^n x^{n-i}\mychoose{n}{i}_q
a_i^0(x,q)\Bigr)
\frac{t^n}{[n]_q!},
\end{eqnarray*}
which completes the proof of (\ref{eq:AA}) in view of (\ref{eq:a}).
\qed\medskip
\remark If $x=q=1$ we get the classical formulas \cite{Du,DV}:
$$
\bar a(t)=\frac{1}{ 1-t}a\left(\frac{t}{ 1-t}\right)\quad \hbox{and}\quad
\bar A(t)=e^t\,A(t).
$$
If $x=0$ we have $\bar A(t)=A(t)$.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{\normalsize A new mahonian statistic ``maf"}
\setcounter{equation}{0}
Let ${S}_n$ be the set of permutations on
$[n]=\{1,\, 2,\, \ldots,\, n\}$.
Recall that $i\in [n]$ is a {\it fixed point}
of $\sigma\in{S}_n$ if $\sigma(i)=i$.
Let $\fix \sigma$ denote the number of fixed points of
$\sigma$. The permutation $\sigma$ has a {\it descent} at
 $i\in\{1,2,\ldots, n-1\}$ if $\sigma(i)>\sigma(i+1)$ and
we call $i$ the {\it descent place} of $\sigma$. The {\it major index}
of $\sigma$, denoted $\maj \sigma$, is the sum of
all the descent places of $\sigma$.
Let $\FIX(\sigma)=\{i \mid \sigma(i)=i\}$ be the set of all fixed points of
$\sigma$
and $\tilde\sigma$ the {\it restriction} of $\sigma$
to $\{1,2,\ldots, n\}\setminus \FIX(\sigma)$.

\begin{defn} If $\sigma\in{S}_n$ with $\FIX(\sigma)=\{i_1,\, i_2,\,
\ldots, i_l\}$,
then the statistic ``maf" is defined by
$$
\maf\sigma = \sum _{j=1}^l (i_j-j) + \maj\tilde\sigma.
$$
\end{defn}

\begin{eg} Let $\sigma=3\, 2\, 1\, 6\, 5\, 9\, 4\, 8\, 7$. Then
$\FIX(\sigma)=\{2,\, 5,\, 8\}$ and $\tilde\sigma=3\, 1\, 6\, 9\, 4\, 7$.
Hence $\fix \sigma=3$, $\maj \sigma=1+2+4+6+8=21$
and $\maf \sigma=(2-1)+(5-2)+(8-3)+(1+4)=14$.
\end{eg}

We now show that the bistatistics $(\fix,\maf)$ and $(\fix,\maj)$ are
equidistributed on the symmetric group ${S}_n$ (Corollary \ref{equi}).
In particular, this shows that $\maf$ is a Mahonian statistic.

Let $\sigma=x_1x_2\ldots x_n\in\sn$. For convenience we put $x_0=-\infty$ and
$x_{n+1}=+\infty$. For $0\leq i\leq n$, a pair $(i,i+1)$ of positions
 is the
\emph{$j$-th slot} of $\sigma$ \emph{provided that} $x_i\ne i$, i.e., $i$
is
not a fixed point of $\sigma$ and that $\sigma$ has $i-j$ fixed points $f$
such that $f<i$.  Clearly we can insert a fixed point into the $j$-th slot
to obtain the permutation
\begin{equation}\label{jslot}
(\sigma,j)=x_1'x_2'\ldots x_i'\ (i+1)\ x_{i+1}'\ldots x_n',
\end{equation}
where $x'=x$ if $x\le i$ and $x'=x+1$ if $x>i$.

More generally, if $\sigma$ is a derangement in $\sn$ and $(i_1,
i_2,\ldots, i_m)$ a sequence of integers such that $0\le i_1\le
i_2\le\cdots\le i_m\le n$,
we can insert $m$ fixed points into the derangement $\sigma$
successively, finally obtaining
$$(\sigma,i_1,\ldots,i_m)=((\sigma,i_1,\ldots,i_{m-1}),i_m).$$
Note that the fixed points of this last permutation are
$i_1+1,i_2+2,\ldots,i_m+m$.
\begin{eg} Let $\sigma=2\,1\,4\,3$ and $(i_1, i_2,\ldots, i_m)=(0, 1, 1, 4)$.
Then we have $(\sigma,0)=1\,3\,2\,5\,4$,
$(\sigma,0,1)=1\,4\,3\,2\,6\,5$, $(\sigma,0,1,1)=1\,5\,3\,4\,2\,7\,6$ and
$(\sigma,0,1,1,4)=1\,5\,3\,4\,2\,7\,6\,8$.
\end{eg}

We can of course undertake the reverse operation. That is, if
a permutation $\sigma$ in $\clS_{m+n}$ has $m$ fixed points we can find a
unique derangement
$\dep(\sigma)\in \clS_n$, called (following Wachs \cite{Wa})
the \emph{derangement part} of $\sigma$, and a unique
sequence of integers $i_1\le\cdots\le i_m$, which we call the \emph{fixed
point sequence} of $\sigma$, such that
$$
\sigma=(\dep(\sigma),i_1,\ldots,i_m).
$$
It is easy to see that
\begin{equation}\label{mafform}
\maf\sigma=\maj\dep(\sigma)+i_1+\cdots+i_m.
\end{equation}

Consider a permutation $\sigma$ with $n$ slots.  The $j$-th slot $(i,\,
i+1)$ of $\sigma$ is said to be \emph{green} if
$\des(\sigma,j)=\des\sigma$, \emph{red} if $\des(\sigma,j)=\des\sigma+1$.
We assign \emph{values} to the green slots of $\sigma$ from right to left,
from 0 to $g$, and to the red slots from left to right, from $g+1$ to $n$.
Denote the value of the $j$-th slot by
$g_j$.  (When we refer to the "largest" slot, we will mean largest in terms
of $j$.)
\medskip

\begin{eg} Let $\sigma=2\,1\,4\,3$.
Then $(\sigma,0)=1\,3\,2\,5\,4$,
$(\sigma,1)=3\,2\,1\,5\,4$,
$(\sigma,2)=2\,1\,3\,5\,4$, $(\sigma,3)=2\,1\,5\,4\,3$, $(\sigma,4)
=2\,1\,4\,3\,5$.  Hence slots 0, 2 and 4
 are green, while 1 and 3 are red. Therefore
\begin{equation}\label{values}
(g_0,\ldots,g_n)=(2,3,1,4,0).
\end{equation}
\end{eg}

It is easy to see that every slot is either green or red. In fact, one can
see that $(i,\, i+1)$ is green if either $x_{i+1}<x_i\le i$,
or $i<x_{i+1}<x_i$, or $x_i\le i<x_{i+1}$. So $(i,\,i+1)$ is red
if either $x_{i+1}\le i<x_i$, or
$i<x_i<x_{i+1}$ or $x_i<x_{i+1}\le i$.  (Expressed in terms of cyclic
intervals ({\sl cf}. \cite{Han}), slot $(i,\,i+1)$ is green if $i+1\in\il
x_i,x_{i+1}\ir$.)

Denote by $d_j$ the number of descents
of $(\sigma,j)$ that lie to the right of $x_i'$ in (\ref{jslot}).
\begin{lem}\label{di}
Let $\sigma$ be a permutation in ${S}_n$.
If the $j$-th slot $(i,\, i+1)$ is green then
$\maj(\sigma, j)-\maj\sigma=d_j$, if $(i,\,i+1)$ is red then
$\maj(\sigma, j)-\maj\sigma=d_j+i$.
\end{lem}
\pf
Let $(i,\,i+1)$ be a green slot.
Since no new descents are formed by inserting a
fixed point into the $j$-th slot of $\sigma$, $\maj(\sigma,j)-\maj\sigma$
equals the number of descents of $\sigma$ that are displaced to the right
when this fixed point is inserted. This number equals $d_j$.
The case in which $(i,\,i+1)$ is red is dealt with
similarly.\qed\medskip

\remark  If $\sigma$ is a derangement in ${S}_n$, the $j$-th
slot of $\sigma$ is just $(j,\,j+1)$ for $0\leq j\leq n$.

\begin{lem}\label{majform} If $\sigma$ is a derangement in ${S}_n$, then
$$\maj(\sigma,j)=\maj\sigma+g_j \quad \hbox{for}\quad 0\leq j\leq n.$$
\end{lem}
\pf
Let $i$ and $j$ be slots.
It follows from Lemma \ref{di}
that if $i$ and $j$ are both green and $i<j$ then
$\maj(\sigma,i)\ge\maj(\sigma,j)$,
 while if $i$ and $j$ are both red and $i<j$ then
$\maj(\sigma,i)\le\maj(\sigma,j)$.
Therefore  $0$ is the green slot of $\sigma$ of highest value, and
if $i$ is red and $j$ is green we have
$\maj(\sigma,i)\geq \maj(\sigma,j)$.
This is because
for any red slot $i$ we have $\maj(\sigma,i)\ge\maj(\sigma,0)$ by
Lemma \ref{di}.
Hence, if $m$ is the largest red slot of $\sigma$,
 i.e., $g_m=n$, for any two slots $i$
and $j$ with $g_i<g_j$ we have
$$\maj\sigma\le\maj(\sigma,i)\le\maj(\sigma,j)\le\maj(\sigma,m).$$
It therefore suffices to show that
$$\maj(\sigma,m)=\maj\sigma+n.$$
Now, consider a green slot $(i,\,i+1)$.
If $i+1$ is a non-excedance place, i.e., $x_{i+1}\le i+1$, then, as
$\sigma$ is a derangment, $x_{i+1}\le i$.  Hence $x_{i+1}<x_i\le i$.  Thus
$i$ is a non-excedance place.
Since $n$ is a non-excedance place and $m+1$, $m+2$, \dots, $n$ are green
slots, we have
$$m+1>x_{m+1}>\cdots>x_n.$$
As the slot $m$ is red, either $m$ is a non-excedance place and $m$ is a
non-descent or $m$ is an excedance place and $m$ is a descent.  In each
case, inserting a fixed point into the $m$-th slot introduces a new descent
for $i=m+1$ and moves $n-(m+1)$ descents one place further to the right.
Hence
$$\maj(\sigma,m)=\maj\sigma+(m+1)+(n-m-1)=\maj\sigma+n,$$
as required. \qed\medskip

\remark Suppose that $\sigma$ is a derangement in $\clS_n$ and $0\le i\le n$.
It follows from Lemmas 3 and 4 that
$d_i=g_i$ if $i$ is green and $d_i=g_i-i$ if $i$ is red.
If $i$ is green then
$$\maj(\sigma,i,i)=\maj(\sigma,i)+g_i.$$
Hence, if $j\le i$, it follows from Lemma \ref{majform} that
$$\maj(\sigma,j,i)=\maj(\sigma,i)+g_j.$$
On the other hand, if $i$ is red, then
$$\maj(\sigma,i,i)=\maj(\sigma,i)+g_i-i.$$
Now one can easily see that, if $k$ is the largest green slot to the left
of slot $i$, $g_k=g_i-i$.  Hence, if $j<i$, it follows again from Lemma
\ref{majform} that
$$\maj(\sigma,j,i)=\maj(\sigma,i)+g_j+1.$$

We are now ready to state the key result of this section.
Let $S(\sigma,m)$ denote
the set of permutations in $\clS_{n+m}$ with derangement part
$\sigma\in {\cal D}_n$. Note that
$$
S(\sigma,m)=\{(\sigma,\,\mathbf{i})\,|\,\mathbf{i}=(i_1,\ldots,i_m)\,\,\,
\hbox{and}\,\,\, 0\leq i_1\leq i_2\leq \cdots \leq i_m\leq n\}.
$$
\begin{thm}\label{bijection}
 There is a bijection $\Psi$ on $S(\sigma, m)$
such that if $\Psi(\sigma,\mathbf{i})=(\sigma,\mathbf{j})$ then
\begin{equation}\label{majmaf}
\maj(\sigma,\mathbf{i})=\maf (\sigma,\mathbf{j}).
\end{equation}
\end{thm}
\pf
We divide the proof into two parts.
\medskip

{\bf The definition of $\Psi$.}
We will define such a bijection $\Psi$ by induction on $m\geq 0$.

First, $\Psi$ is the identity mapping on $S(\sigma,0)$.  Next, we
define $\Psi$ on $S(\sigma, 1)$ by
$$\Psi(\sigma,i)=(\sigma,g_i).$$
Then using equation (\ref{mafform}) and Lemma \ref{majform} we see that
$\Psi$ satisfies equation (\ref{majmaf}).

Let $m>1$ and suppose that
 $\Psi$ has been defined on $S(\sigma, k)$ for $0\leq k\leq m-1$.
Consider $\tau=(\sigma,i_1,\ldots,i_m)$.
Suppose that the $i_m$-th slot of $(\sigma,i_1,\ldots,i_{m-1})$
is green.  Then, if
$$\Psi(\sigma,i_1,\ldots,i_{m-1})=(\sigma,j_2,\ldots,j_m),$$
we define
$$\Psi(\tau)=(\sigma,g_{i_m},j_2,\ldots,j_m).$$
Suppose that the $i_m$-th slot of $(\sigma,i_1,\ldots,i_{m-1})$ is
red. Then the slots $i_1,\ldots,i_m$ cannot be all the same.
Let $k$ be the smallest positive integer such
that $i_{m-k}<i_m$. Thus $i_{m-k}<i_{m-k+1}=\cdots=i_m$. Then, if
$$
\Psi(\sigma,i_1,\ldots,i_{m-k})=(\sigma,j_1,\ldots,j_{m-k}),
$$
we define
$$\Psi(\tau)=(\sigma,
\underbrace{g_{i_m}-i_m,\ldots,g_{i_m}-i_m}_{k-1\mbox{ terms}},
j_1+1,\ldots,j_{m-k}+1,g_{i_m}).$$
The following lemma is easily proved by induction.
\begin{lem} Let $\tau=(\sigma,i_1,\ldots,i_m)$ and
$\Psi(\tau)=(\sigma,j_1,\ldots,j_m)$.

Suppose that at least one of the slots $i_1,\ldots,i_m$ is either green or
is repeated.  Let $i_l$ be the largest such slot.  If $i_l$ is green then
$j_1=g_{i_l}$.  If $i_l$ is red and is repeated then $j_1=g_{i_l}-i_l$.

If on the other hand all of the slots $i_1,\ldots,i_m$ are red and are
distinct, then $j_1=g_{i_1}$.

Suppose that at least one of the slots $i_1,\ldots,i_m$ is red.  If $i_l$
is the largest red slot then $j_m=g_{i_l}$.

If on the other hand all of these slots are green then $j_m=g_{i_1}$.
\end{lem}
It follows from this lemma that $j_1,\ldots,j_m$ as defined above are in
ascending order.

We now show by induction on $m$ that $\Psi$ satisfies equation
(\ref{majmaf}).

If $i_m$ is green, then using Lemma \ref{majform} we have
\begin{eqnarray*}
\maj(\sigma,i_1,\ldots,i_m)&=&\maj(\sigma,i_1,\ldots,i_{m-1})+g_{i_m}\\
&=&\maf(\sigma,j_2,\ldots,j_m)+g_{i_m}\\
&=&\maf(\sigma,g_{i_m},j_2,\ldots,j_m).
\end{eqnarray*}

If $i_m$ is red, let $k$ be the smallest positive integer such that
$i_{m-k}<i_m$, then
\begin{eqnarray*}
\lefteqn{\maj(\sigma,i_1,\ldots,i_m)}\\
&=&\maj(\sigma,i_1,\ldots,i_{m-k})+(m-k)+(k-1)(g_{i_m}-i_m)+g_{i_m}\\
&=&\maf(\sigma,j_1,\ldots,j_{m-k})+(m-k)+(k-1)(g_{i_m}-i_m)+g_{i_m}\\
&=&\maf(\sigma,
\underbrace{g_{i_m}-i_m,\ldots,g_{i_m}-i_m}_{k-1\mbox{ terms}},
j_1+1,\ldots,j_{m-k}+1,g_{i_m}).
\end{eqnarray*}
This is because inserting the first fixed point $i_m$ into
$(\sigma,i_1,\ldots,i_{m-k})$ adds a descent and increases $\maj$ by
$g_{i_m}+(m-k)$.  Inserting each of the remaining fixed points $i_m$ has
the same affect as inserting a fixed point into a green slot of value
$g_{i_m}-i_m$.

\medskip
%==================================

{\bf $\Psi$ is a bijection.}
It remains to show that $\Psi$ is a bijection on $S(\sigma,m)$.  It
suffices to show that $\Psi$ is an injection.

We use induction on $m$. The result is clearly true for $m=0$ and $m=1$.

Let $\tau=(\sigma,i_1,\ldots,i_m)$ and
$\Psi(\tau)=(\sigma,j_1,\ldots,j_m)$.  Suppose that
$\Psi(\tau)=\Psi(\tau')$, where $\tau'=(\sigma,i_1',\ldots,i_m')$.

If both $i_m$ and $i_m'$ are green or red then it is easy to show using the
induction hypothesis that $\tau=\tau'$.  So suppose that $i_m$ is green and
$i_m'$ is red.  Thus $j_1=g_{i_m}$, $j_m=g_{i_m'}$.

Suppose that $i_1,\ldots,i_m$ are all green.  Then $j_m=g_{i_1}$.  Hence
$i_1=i_m'$, contradiction.

Let $i_u$ be the largest red slot amongst $i_1,\ldots,i_m$.  Then
$j_m=g_{i_u}$.  Hence $i_m'=i_u<i_m$.

\emph{Case 1:}  Suppose that one of the slots $i_1',\ldots,i_m'$ is either
green or is repeated.  Let $i_v'$ be the largest such slot.
If $i_v'$ is green, then
$$\Psi(\tau')=(\sigma,g_{i_v'}+(m-v),\ldots,g_{i_m'}).$$
Hence $j_1=g_{i_m}=g_{i_v'}+(m-r)$.  Since $i_m$ and $i_v'$ are both green,
this means that $i_m\le i_v'<i_m'$, contradiction.

If $i_v'$ is red, then
$$\Psi(\tau')=(\sigma,g_{i_v'}-i_v'+(m-v),\ldots,g_{i_m'}).$$
Hence $j_1=g_{i_m}=g_{i_v'}-i_v'+(m-r)$.  But $g_{i_v'}-i_v'$ is the value
of the largest green slot $i_w$ less than $i_v'$.  As $i_m$ is green this
means that $i_m\le i_w<i_v'\le i_m'$, contradiction.

\emph{Case 2:}  Suppose that all of the slots $i_1',\ldots,i_m'$ are red
and distinct.  Then
$$\Psi(\tau')=(\sigma,g_{i_1'}+(m-1),\ldots,i_m').$$
Hence $j_1=g_{i_m}=g_{i_1}'+(m-1)>g_{i-1}'$.  This is a contradiction,
since $i_m$ is green and $i_1'$ is red.
\qed
\medskip

\begin{eg} Let $\sigma=2\,1\,4\,3$ and consider
 $(\sigma, 0,1,1,4)\in S(\sigma, 4)$.
Then the values of the slots of $\sigma$ have been calculated in
(\ref{values}). The bijection $\Psi$ goes as follows:

\noindent since slot 0 is green in $\sigma$ we have
$$
\Psi(\sigma,0)=(\sigma,g_0)=(\sigma,2);
$$
since slot 1 is red we have
$$
\Psi(\sigma, 0,1)=(\sigma, 2+1, g_1)=(\sigma, 3, 3);
$$
again, since slot 1 is red we have
$$
\Psi(\sigma, 0,1,1)=(\sigma, g_1-1, 2+1, g_1)=(\sigma, 2, 3, 3);
$$
Finally, since slot 4 is green we obtain
$$
\Psi(\sigma, 0,1,1,4)=(\sigma, g_4, 2,3,3)
=(\sigma, 0,2,3,3)\in  S(\sigma, 4).
$$
Let $\tau=(\sigma, 0,1,1,4)$ and
$\tau'=(\sigma, 0,2,3,3)$. Then $\tau=1\,5\,3\,4\,2\,7\,6\,8$ and
$\tau'=1\,3\,2\,4\,8\,6\,7\,5$. It is easy to see that
$\maj \tau=12$
and $\maf \tau'=12$. Hence we have checked equation (\ref{majmaf}).
\end{eg}

Using theorem \ref{bijection}, we obtain the following result.
\begin{cor}\label{equi}
$(a)$ There is a bijection $\phi: {S}_n\rightarrow {S}_n$ such
that for any
$\sigma\in {S}_n $ we have
$$
(\fix,\, \maf) \sigma=(\fix,\, \maj)\phi(\sigma).
$$
$(b)$ The bi-statistic $(\fix, \maf)$ is equidistributed with
the bi-statistic $(\fix, \maj)$ on the symmetric group ${S}_n$.
\end{cor}

The following result was first proved by Wachs \cite[corollary 3]{Wa}.
\begin{cor}
Let $\sigma$ be a derangement in $\sn$ and $m\geq 0$. We have
$$
\sum_{\pi\in S(\sigma,\, m)}q^{\maj \pi}=q^{\maj \sigma}\mychoose{m+n}{ n}_q.
$$
\end{cor}
\pf By theorem 5 we have
\begin{eqnarray*}
\sum_{\pi\in S(\sigma,\, m)}q^{\maj \pi}&=&
\sum_{\pi\in S(\sigma,\, m)}q^{\maf \pi}\cr
&=&q^{\maj \sigma}\,\sum_{0\leq i_1\leq \cdots\leq i_m\leq n}
q^{i_1+i_2+\cdots i_m}\cr
&=&q^{\maj \sigma}\mychoose{m+n}{ n}_q.
\end{eqnarray*}
The last line follows from a well-known result \cite[p. 33]{An}. \qed
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{\normalsize $q$-derangement matrices}
\setcounter{equation}{0}
We first prove the following result.
\begin{prop}\label{prop:equi}
Let $(a_n^k(x,q))$  be a $q$-Seidel matrix.
Then the following
three conditions are equivalent:
\begin{eqnarray}
a_n^0(x,q) &=&
[n]_q!\sum_{i=0}^n (-1)^{i}\frac{
q^\mychoose{i}{2}}{[i]_q!},\hbox{\label{prop8:1}}\\
 a_0^n(x,q)&=& [n]_q!\left(1+\sum_{i=1}^n
\frac{(x-1)(x-q)\cdots (x-q^{i-1})}{ [i]_q!}\right),\hbox{\label{prop8:2}}\\
 a_0^n(1,q)&=&[n]_q!\quad \hbox{ and}\quad  a_n^0(x,q) \quad\hbox{is
independent of}\quad  x.\hbox{\label{prop8:3}}
\end{eqnarray}
\end{prop}
\pf
By the $q$-binomial formula \cite[p.7]{GaRa}
$$
\sum_{n=0}^\infty\frac{(a;q)_n}{ (q;q)_n}t^n=\frac{(at;q)_\infty}{
(t;q)_\infty},
$$
we have in view of (\ref{eq:defe}) and (\ref{eq:defE}),
\begin{displaymath}
1+\sum_{n=1}^\infty \frac{(x-1)(x-q)\cdots (x-q^{n-1})}{
[n]_q!}t^n
 = {e_q(x t)E_q(-t)}.
\end{displaymath}
Therefore the generating functions of (\ref{prop8:1}), (\ref{prop8:2})
and (\ref{prop8:3}) are respectively
the following:
\begin{eqnarray}
A(t) &=& \sum_{n\geq 0} a_n^0(x,q)\frac{t^n}{ [n]_q!} = \frac{E_q(-t) }{
1-t},\hbox{\label{prop8:4}}\\
\overline{A}(t) &=&\sum_{n\geq 0} a_0^n(x,q)\frac{t^n}{ [n]_q!} =
\frac{e_q(xt)E_q(-t) }{ 1-t},\hbox{\label{prop8:5}}\\
\overline{A}(t)|_{x=1} &=&\sum_{n\geq 0} a_0^n(1,q)\frac{t^n}{ [n]_q!} =
\frac{1}{ 1-t}.\hbox{\label{prop8:6}}
\end{eqnarray}
So, it suffices to prove that the equivalence of (\ref{prop8:4}),
(\ref{prop8:5}) and (\ref{prop8:6}).  Indeed,

(\ref{prop8:4}) $\Longleftrightarrow$ (\ref{prop8:5}): this follows from
proposition 2;

(\ref{prop8:5}) $\Longrightarrow$ (\ref{prop8:6}): this is obvious;

(\ref{prop8:6}) $\Longrightarrow$ (\ref{prop8:4}): since $A(t)$ is
independent of $x$, equation (\ref{prop8:4})
follows then from (\ref{eq:AA}) by setting $x=1$.
\qed\medskip

\begin{defn} A $q$-derangement matrix is
the $q$-Seidel matrix satisfying any of the three
conditions of proposition 8.
\end{defn}


If $x=1$, then  $a_0^n(x,q) = [n]_q!$ and the $q$-derangement matrix is as
follows~:

{
%\scriptsize
$$
\def\twolines#1#2{\hbox{%
$\matrix{
\noalign{\vskip 3pt}
#1\cr
\noalign{\vskip -1pt}
\quad #2\cr
\noalign{\vskip 3pt}
}$}}
%%%%
\def\twolines#1#2{\binom{#1\quad\hfil}{\hfil\quad #2}}
%%%%
\def\aaa{\twolines{q+2q^2+2q^3}{+2q^4+q^5+q^6}}
\def\aab{\twolines{q+2q^2+2q^3}{+3q^4+2q^5+q^6}}
\def\aac{\twolines{q+2q^2+3q^3}{+4q^4+3q^5+q^6}}
\def\aad{\twolines{q+3q^2+5q^3}{+5q^4+3q^5+q^6}}
%%%%
\begin{array}{c|ccccc}
\hbox{$k$}\backslash\hbox{$n$}&0&1&2&3&4\\
\hline\\
0 & 1      & 0           & q          & q+q^2 & \aaa\cr
1 & 1      & q           & q+q^2+q^3  & \aab  &     \cr
2 & 1+q    & q+2q^2+q^3  & \aac       &       &     \cr
3 & [3]_q! & \aad        &            &       &     \cr
4 & [4]_q! &             &            &       &     \cr
\end{array}
$$
}
$$(a_n^k(1,q))$$

\medskip

Denote by ${S}_n^k$
the set of permutations on $[n+k]$ of which all the fixed points are
included in
$\{n+1, n+2,\ldots, n+k\}$.
In particular ${S}_{n}^0$ is the set of permutations without fixed
points
on $[n]$ and ${S}_{0}^k$ the set of all permutations
on~$[ k]$. The following result generalizes a result of Dumont and 
Randrianarivony \cite{DR}.

\begin{thm}\label{th:main} The coefficients
 $a_n^k(x,q)$ ($n,\,k\geq 0$) in a $q$-derangement matrix
have the following combinatorial interpretation:
\begin{equation}
a_n^k(x,q) = \sum_{\sigma\in{S}_n^k} x^{\fix \sigma}q^{\maf \sigma}.
\end{equation}
\end{thm}
\pf
Notice that ${S}_{n+1}^{k-1} \subset {S}_n^k$. Set
$$
\Delta_n^k={S}_n^k\setminus {S}_{n+1}^{k-1}
=\{ \sigma\in{S}_n^k \mid \sigma(n+1)=n+1\}.
$$
We construct a bijection $\varphi\,:\,\Delta_n^k \rightarrow {
S}_n^{k-1}$
such that for all $\sigma\in \Delta_n^k$,
\begin{eqnarray*}
\maf \sigma &=& n + \maf(\varphi(\sigma)),\cr
\fix\sigma &=& 1 + \fix(\varphi(\sigma)).
\end{eqnarray*}
Indeed, if $\sigma\in\Delta_n^k$ we define
$\varphi(\sigma)$ as the word obtained from $\sigma$ by deleting $n+1$ and
reduce all
the values strictly bigger than $n+1$.
It is readily verified that $\varphi$ is the desired bijection. Therefore
\begin{equation}\label{rmaf}
\sum_{\sigma\in{S}_n^k} x^{\fix\sigma}q^{\maf\sigma}
=x q^n\sum_{\sigma\in{S}_n^{k-1}} x^{\fix\sigma}q^{\maf\sigma}
+\sum_{\sigma\in{S}_{n+1}^{k-1}} \!\!\! x^{\fix\sigma}q^{\maf\sigma},
\end{equation}
which is the recurrence (\ref{seidel}). So it remains to check the initial
condition.
Now ${S}_0^n={S}_n$ and it is well-known \cite{Mac} that
$\sum_{\sigma\in{S}_n} q^{\maj \sigma}=[n]_q!$, so it follows from
corollary \ref{equi}  that
$$
a_0^n(1,q)
=\sum_{\sigma\in{S}_n} q^{\maf \sigma}
=\sum_{\sigma\in{S}_n} q^{\maj \sigma}=[n]_q!.
$$
The theorem follows then from proposition \ref{prop:equi}, since
$a_n^0(x,q)$ is
clearly
independent of $x$.
\qed\medskip

\remark Since $(\fix,\, \maf)$ and $(\fix,\, \maj)$ are not
equidistributed 
on $S_1^2$ we cannot replace $\maf$ by $\maj$ in the above theorem.
\medskip

{}From Corollary \ref{equi}, proposition \ref{prop:equi} and theorem
\ref{th:main} we derive
 the following result.
\begin{cor}
The final sequence of the $q$-derangement matrix
has the following interpretation:
\begin{eqnarray}
a_0^n(x,q)&=&\sum_{\sigma\in{S}_n}  x^{\fix \sigma}q^{\maf \sigma}\\
&=&\sum_{\sigma\in{S}_n}  x^{\fix \sigma}q^{\maj \sigma}\\
&=&[n]_q!\left(1+\sum_{i=1}^n
\frac{(x-1)(x-q)\cdots (x-q^{i-1})}{ [i]_q!}\right).
\end{eqnarray}
\end{cor}
Note that the last equation has been obtained by
Gessel and Reutenauer \cite{GR} and by Wachs \cite{Wa} in the special $x=0$
case using different methods.


\section{\normalsize An open problem about $q$-succession numbers}
\setcounter{equation}{0}
Let $\sigma$ be a permutation in ${S}_n$.  For convenience put
$\sigma(0)=0$.  We say that an element $p$ (with $1\leq p\leq n$) is a
\emph{succession} of $\sigma$ if $\sigma(p)=\sigma(p-1)+1$. The $p$ is
called the {\it succession position}, while $\sigma(p)$ is called the
\emph{succession value}.  Let $\SUC(\sigma)$ be the set of succesion values
of $\sigma$ and let $\suc \sigma$ be the number of successions of $\sigma$.
For example, if
$$\sigma=\left(\begin{array}{ccccccccc}
1&2&3&4&5&6&7&8&9\\
1&4&3&8&9&5&6&7&2
\end{array}
\right),$$
then $\SUC(\sigma)=\{1,9,6,7\}$ and $\suc\sigma=4$.

We use a variant of Foata's first fundamental transformation \cite{Fo}
to show that the statistics $\fix$ and $\suc$ are equidistributed on ${\cal
S}_n$.

Given a permutation $\sigma=\sigma(1)\sigma(2)\cdots \sigma(n)\in S_n$ we set
$\sigma^d=\sigma(2)\cdots\sigma(n)\sigma(1)$.  We call the \emph{standard
form}
of the factorization into cycles of $\sigma$ the unique writing
$\bar\sigma$ such that in each cycle $(a,\sigma(a),\ldots,\sigma^l(a))$ the
maximum $\sigma^l(a)$ is in the last position and the cycles of $\sigma$
are decreasingly ordered according to their maxima. (Note that this is
\emph{not} the usual definition of standard form.)  We define
$\varphi(\sigma)$ as the permutation obtained by erasing the parentheses in
the standard form of
$\bar \sigma^d$.

The following lemma is easy to verify.
\begin{lem}\label{sucfix} The mapping $\varphi$ is a bijection on 
${S}_n$ such that for all $\sigma\in{S}_n$,
$FIX(\sigma)=\SUC(\varphi(\sigma))$ and
$\fix\sigma=\suc\varphi(\sigma)$. 
 Hence the
statistics $\fix$ and $\suc$ are equidistributed on~${ S}_n$.
\end{lem}
For example, if $\sigma=1\,4\,2\,8\,3\,6\,7\,5\,9\in { S}_{9}$, then
$$\sigma^d=4\,2\,8\,3\,6\,7\,5\,9\,1\quad\mbox{ and }\quad
\bar\sigma^d=(1\,4\,3\,8\,9)(5\,6\,7)(2).$$
Erasing the parentheses we obtain the permutation
$\varphi(\sigma)=1\,4\,3\,8\,9\,5\,6\,7\,2$.  We have
$$\FIX(\sigma)=\SUC(\varphi(\sigma))=\{1,6,7,9\}.
$$

Define the statistic
$$\suc' \sigma=
\begin{cases}
\suc \sigma,   &\text{if $\sigma(1)\not=1$,}\cr
\suc \sigma-1, &\text{if $\sigma(1)=1$;}\cr
\end{cases}
$$
and let
\begin{equation}
F_n(x)=\sum_{\sigma\in \sn}x^{\fix \sigma},
\quad S_n(x)=\sum_{\sigma\in \sn}x^{\suc' \sigma}.
\end{equation}
Then, using lemma \ref{sucfix}, we obtain a bijective proof of
the following known results (See \cite{Ro,Ch}).
\begin{prop} We have
\begin{equation}\label{sff}S_{n+1}(x)=F_{n+1}(x)+(1-x)F_n(x),
\end{equation}
and in particular
\begin{equation}\label{sf}
S_{n+1}(0)=d_{n+1}+d_n.
\end{equation}
\end{prop}

Setting $q=1$ in (4.20) we see that
\begin{equation}
\sum_{n\geq 0}F_n(x)\frac{t^n}{n!}=\frac{e^{(x-1)t}}{1-t}.
\end{equation}
Hence, from equation (\ref{sff}), we have
\begin{equation}
\sum_{n\geq 0}S_n(x)\frac{t^n}{n!}=\frac{e^{(x-1)t}}{1-t}
+(1-x)\sum_{n\geq 1}F_{n-1}(x)\frac{t^n}{ n!},
\end{equation}
in which by convention $S_0(x)=F_0(x)=1$. Thus
\begin{equation}
\sum_{n\geq 0}S_n(x)\frac{t^n}{ n!}=
\frac{e^{(x-1)t}}{ 1-t}+(1-x)\int_0^t\frac{e^{(x-1)z}}{ 1-z}dz.
\end{equation}
Let $\cal L$ be the formal Laplace transformation on the ring of formal
power series, that is, ${\cal L}(\sum a_nx^n/n!)=\sum a_nx^n$. Then
\begin{equation}
\sum_{n\geq 0}F_n(x)t^n={\cal L}\left(\frac{e^{(x-1)t}}{ 1-t}\right)
=\sum_{n\geq 0}\frac{n!t^n}{ [1-(x-1)t]^{n+1}}.
\end{equation}
Therefore
\begin{eqnarray}
\sum_{n\geq 0}S_n(x)t^n&=&\sum_{n\geq 0}F_n(x)t^n
+(1-x)\sum_{n\geq 0}F_n(x)t^{n+1}\cr
&=& [1-(x-1)t]\sum_{n\geq 0}F_n(x)t^n\cr
&=& \sum_{n\geq 0}\frac{n!t^n}{ [1-(x-1)t]^n}.
\end{eqnarray}

In the case of $q=1$, using lemma 11, we can restate theorem 9
in terms of successions. Unfortunately, since
the mapping $\varphi$ does not keep track of the maj statistic,
we do not have a full interpretation in the last model.

The distribution of our statistics on ${S}_3$ is as follows:
\begin{center}
\begin{tabular}{r|ccccc}
$\sigma\backslash\hbox{stat}$ &$\maf$ & $\maj$  & $\suc$&$\fix$\\
\hline
1\,2\,3& 0&0  &3&3\\
1\,3\,2& 1&2  &1&1\\
2\,1\,3& 3&1  &0&1\\
2\,3\,1& 2&2  &1&0\\
3\,1\,2& 1&1  &1&0\\
3\,2\,1& 2&3  &0&1
\end{tabular}
\end{center}
\centerline{\bf Statistic distributions on ${S}_3$}
\bigskip

Finally we record two open problems related to our work.

1) Find a mahonian statistic ``mag'' such that $(\suc, \mag)$ is
equidistributed with $(\fix, \maj)$ on the symmetric group ${S}_n$.

2) Generalize the statistic ``maf'' on permutations to {\it words} as in
\cite{CF,Han} for other mahonian statistics.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\end{document}





