\documentclass[12pt]{article}
\def\resp{{\it resp}}
\def\fix{\mathop{fix}\nolimits}
\def\cyc{\mathop{\rm cyc}\nolimits}
\def\arctanh{\mathop{arctanh}\nolimits}
\def\D{\cal D}
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\newtheorem{thm}{Th\'eor\`eme}
\newtheorem{con}{Conjecture}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{prop}[thm]{Proposition}
\newtheorem{cor}[thm]{Corollary}
\newtheorem{defn}{Definition}
\newtheorem{defns}[defn]{Definitions}
\newtheorem{eg}{Exemple}
\newtheorem{egs}[eg]{Examples}
\newtheorem{porism}[thm]{Porism}
\newtheorem{conj}{Conjecture}
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\title{
THE GENERATING FUNCTION FOR THE DIFFERENCE 
IN EVEN AND ODD THREE-LINE LATIN RECTANGLES}
\author{\normalsize Jiang Zeng}
\date{10 octobre, 1995}
\begin{document}
\maketitle

\noindent\textbf{R\'esum\'e.} {\small
On donne une fonction g\'en\'eratrice exponentielle
pour la diff\'erence entre le nombre des rectangles
latin pairs de taille $3\times n$ et le nombre de ceux impairs.
On montre ensuite que cette formmule \'equivaut �\`a une formula,
qui est apparemment diff\'erente, donn\'ee dans \cite{HJ}}.
\medskip

\noindent\textbf{Abstract.} {\small
We give a simple exponential generating function for
the difference between the
number of even reduced Latin rectangles of 
size $3\times n$ and the number of odd ones. 
We show that this formula is equivalent to an apparently
different 
formula given in \cite{HJ}}.
\bigskip

Let ${\cal S}_n$ be the set of permutations of $[n]=\{1, 2, \ldots, n\}$.
For each $\sigma\,\in{\cal S}_n$, we denote by $\cyc\,\sigma$ 
(\resp. $\fix\,\sigma$)
the number of cycles (\resp. fixed points) of $\sigma$. Further
let $L_n$ be the 
generating polynomial for ${\cal S}_n\times {\cal S}_n$ defined by
$$
L_n(\alpha, \beta, p, q, \gamma)=\sum_{\sigma, \tau\in 
{\cal S}_n}\alpha^{\cyc\,\sigma} \beta^{\cyc\,\tau} p^{\fix\,\sigma}
q^{\fix\,\tau} \gamma^{\fix\,\sigma {\tau}^{-1}}.
$$

Let $(x)_n=x(x+1)\cdots(x+n-1)$. 
Using a generalization 
of Cauchy's summation theorem 
for Schur functions and some group representation theories
Andrews et al. \cite{AGJ}
obtained the following expression for the exponential generating
function for the $L_n$ :
\begin{eqnarray*}
(1)\qquad &&1+\sum_{n \geq 1}{{t^n\over n!}L_n(\alpha, \beta, p, q, \gamma)}
=e^{\alpha \beta t(2-p-q-\gamma+pq\gamma)}\\
&&\sum_{m\geq 0}{t^m\over m!}{(\alpha)_m(\beta)_m\over
[1+\beta (1-q)t]^{m+\alpha}[1+\alpha (1-p)t]^{m+\beta}
[1+(1-\gamma)t]^{m+\alpha \beta}}\cdot
\end{eqnarray*}

In \cite{Ze} the author has shown some applications of (1) and 
further derived a combinatorial proof of (1)
by generalizing a standard method due to Gessel~\cite{Ge}. 

Recently Habsieger and Janssen \cite{HJ} raised the problem to
count {\sl signed Latin rectangles}. Recall that 
a {\sl Latin rectangle} of size $k\times n$ is an $k\times n$ array,
such that each row is a permutation of $12\ldots n$, and the product
$�^{-1}�$ of any two distinct row permutations $�$, $�$ is 
{\sl fixed-point}
free, \textit{i.e.} is a {\sl derangement}. A {\sl reduced} Latin rectangle has the additional property that the first
row is the identity permutation. 
A Latin rectangle is called {\sl row-even} if
the product of the signs of its row permutations is $+1$ and {\sl row-odd}
if this product is $-1$. Let $S_{k,n}$  denote the difference between the number
of even reduced Latin rectangles of
size $k\times n$ and the number of odd ones. In \cite{HJ}, 
Habsieger and Janssen proved the following results :
\begin{eqnarray*}
(1.0)\hskip 3cm S_{1,n}&=&1,\hskip 2cm\\
(1.1)\hskip 3cm S_{2,n}&=&(-1)^{n-1}(n-1),\hskip 2cm \\
(1.2)\hskip 3cm S_{3,n}&=&(-1)^{n-1}{(n-2)n!\over 2}
[t^n] e^{2(\arctanh t-t)},\hskip 2cm
\end{eqnarray*}
where $[t^n]F(t)$ stands for the coefficient of $t^n$ in the series
expansion of $F(t)$. Note that in \cite{HJ} the series $\arctanh t$ of (1.2)
was {\sl incorrectly} replaced by $\sinh t$ due to 
an obvious error at the end of its proof \cite[p. 75]{HJ}, where
$(2j+1)$ was {\sl incorrectly} replaced by $(2j+1)!$.

Actually formulas (1.0) and (1.1) are straightforward. 
The purpose of this note is to show that 
(1.2) is a equivalent to a special case of (1).

\begin{prop} We have
$$
1+\sum_{n\geq 1}S_{3,n}{t^n\over n!}=e^{2t}
\left[{(1-t)^2\over 1+t}+{t\over (1+t)^2}\right]. \leqno (1.3)
$$
\end{prop}

\proof Let $\D_n$ be the set of derangements of $[n]$.  
By definition, we have
$$
S_{3,n}=\sum_{\sigma, \tau, \sigma {\tau}^{-1} \in {\cal D}_n}
\varepsilon (\alpha )\varepsilon (\beta ).
$$
As $\varepsilon (\alpha )\varepsilon (\beta )
=(-1)^{n-\cyc \alpha }(-1)^{n-\cyc \beta  }
=(-1)^{\cyc \alpha }(-1)^{\cyc \beta  }$, we conclude that
$$
S_{3,n}=L_n(-1, -1, 0,0,0), \qquad (n\geq 1).
$$
Formula (1.3) follows then from (1). \qed
\medskip

\begin{prop}
 Formulas (1.2) and (1.3) are equivalent.
 \end{prop}
\proof From (1.3) we derive 
$$
\sum_{n\geq 1}(-1)^{n-1}S_{3,n}{t^n\over n!}
=1+{e^{-2t}\over (1-t)^2}(t^2+t^3-1). \leqno (1.4)
$$
Note that
\begin{eqnarray*}
{e^{-2t}\over (1-t)^2}&=&\sum_{n\geq 0}
{(-2)^nt^n\over n!}\cdot \sum_{n\geq 0}(n+1)t^n\\
&=&\sum_{n\geq 0}
\left(\sum_{i=0}^n(n-i+1){(-2)^i\over i!}\right)t^n.
\end{eqnarray*}
Substituting this into (1.4) and equating the coefficients
of $t^n$ yield 
$$
(-1)^{n-1}{S_{3,n}\over n!}=\sum_{i=0}^{n-3}(n-4-i)
{(-2)^{i}\over i!}+{(-2)^{n-1}\over n (n-3)!}.\leqno (1.5)
$$
On the other hand, since 
$$
\arctanh x=\sum_{n\geq 0}{x^{2n+1}\over 2n+1}
={1\over 2}\ln\left({1+x\over 1-x}\right)
$$
therefore
$$
[x^n]e^{2\arctanh x-2x}=[x^n]{1+x\over 1-x}e^{-2x}=
2\sum_{i=0}^{n-1}{(-2)^i\over i!}+{(-2)^i\over n!},
$$
and it follows from (1.2) that
$$
(-1)^{n-1}{S_{3,n}\over n!}=(n-2)\sum_{i=0}^{n-1}{(-2)^i\over i!}
+(-1)^n(n-2){2^{n-1}\over n!}\cdot\leqno (1.6)
$$
It is easy to check that the right hand-sides of formulas (1.5) and 
 (1.6) are equal. Indeed, if we subtract the right-hand side of (1.6) 
from that of (1.5), we obtain
\begin{eqnarray*}
\sum_{i=0}^{n-3}(2+i){(-2)^i\over i!}+{(-2)^{n-2}\over (n-3)!}
&=&2\sum_{i=0}^{n-3}{(-2)^i\over i!}-
2\sum_{i=0}^{n-3}{(-2)^{i-1}\over (i-1)!}+{(-2)^{n-2}\over (n-3)!}\\
&=&2{(-2)^{n-3}\over (n-3)!}+{(-2)^{n-2}\over (n-3)!}\\
&=&0,
\end{eqnarray*}
which was to be proved. \qed

\medskip
Finally, by applying Darboux's method [Co, p. 121] we get from (1.4)
a good asymptotic estimate for $(-1)^{n-1}S_{3, n}$ :
$$
(-1)^{n-1}S_{3, n}\sim \sqrt {2\pi}\ n^{n+3/2}e^{-n-2}
\qquad (n\longrightarrow +\infty).
$$

\section{R\'esum\'e substantiel en fran\c cais}
Un rectangle latin est {\sl pair} selon les lignes si 
le produit des signes des permutation de lignes vaut $+1$
et {\sl impair} selon les lignes si ce produit vaut $-1$.
Notons $S_{k,n}$ la diff\'erence entre le nombre de 
rectangle r\'eduits
pairs de taille $k\times n$ et le nombre de ceux impairs.
Dans \cite{HJ}, Habsieger et Janssen ont \'etabli des 
formules explicites pour
$S_{k,n}$ lorsque $k=1,2,3$ :
\begin{eqnarray*}
(1.0)\hskip 3cm S_{1,n}&=&1,\hskip 2cm\\
(1.1)\hskip 3cm S_{2,n}&=&(-1)^{n-1}(n-1),\hskip 2cm \\
(1.2)\hskip 3cm S_{3,n}&=&(-1)^{n-1}{(n-2)n!\over 2}
[t^n] e^{2(\arctanh t-t)},\hskip 2cm
\end{eqnarray*}

Le but de cette note est de signaler 
une formule simple pour la fonction g\'en\'eratrice exponentielle
de la suite $\{S_{3,n}\}_{n\geq 1}$ :
$$
1+\sum_{n\geq 1}S_{3,n}{t^n\over n!}=e^{2t}
\left[{(1-t)^2\over 1+t}+{t\over (1+t)^2}\right]. \leqno (1.3)
$$
Cette formule est en fait un cas particulier
d'une formule plus g\'en\'erale due \`a
 Gessel~\cite{Ge} ou de
celle de Andrews et al. \cite{AGJ}. On montre ensuite que 
les deux formules (1.2) et (1.3) sont equivalentes.
\vfill
\newpage
\begin{thebibliography}{99}

\bibitem{AGJ} 
G. \textsc{Andrews}, I. P. \textsc{Goulden} and D. M. 
\textsc{Jackson}: 
\emph{Generalization of Cauchy's summation theorem for Schur functions},
 Trans. Amer. Math. Soc., \textbf{310}(1988), 805--820.
 
\bibitem{ Co}
L. \textsc{Comtet}
\emph{Analyse combinatoire}, Vol. 2, Press Univ. de France, Paris, 1970.

\bibitem{Ge}
I. M. \textsc{Gessel}: 
\emph{Counting three-line Latin rectangles}, Combinatoire 
\'Enum\'erative (Montr\'eal, 1985),
 Lecture Notes in Math., vol. 1234, Springer-Verlag, Nex-York, 106--111.


\bibitem{HJ}
L. \textsc{Habsieger} and J. \textsc{Janssen} 
\emph{The difference in even and odd latin rectangles for small cases},
Ann. Sci. Math. Qu\'ebec, \textbf{19}(1995), 69-77.

\bibitem{Ze}
J. \textsc{Zeng}:
\emph{ Counting a pair of permutations and the linearization
coefficients for Jacobi polynomials}, 
Actes de l'Atelier de Combinatoire Franco-Qu\'eb\'ecois (Bordeaux, 6-7 mai 1991),
Publications du LACIM, vol. 10, Universit\'e du  Qu\'ebec \`a Montr\'eal,
Montr\'eal, 1992, 243-257.
\end{thebibliography}
\vskip 40pt
\vbox{\small\sl
\leftline{Jiang ZENG}
\leftline{UFR de Math\'ematiques et Informatique}
\leftline{Universit\'e Louis Pasteur}
\leftline{7, rue Ren\'e Descartes}
\leftline{67084 Strasbourg Cedex, France}
\leftline{e-mail : jzeng@math.u-strasbg.fr}}
\end{document}
\normalsize D\'epartement de Math\'ematique, Universit\'e Louis Pasteur,\\
\normalsize 7, rue Ren\'e Descartes, 67084 Strasbourg C\'edex, France}
\vfill
\eject











