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\title{On the powers  of Motzkin paths} 
\author{Jiang Zeng\\
    \small Institut Girard Desargues, 
    Universite Claude Bernard (Lyon 1),\\
    43 Bd du 11 novembre 1918, 69622 Villeurbanne Cedex, France\\
\small \texttt{zeng@desargues.univ-lyon1.fr}}
\date{}
\begin{document}
\maketitle

%First page headline in AmS-TeX for S\'eminaire Lotharingien de Combinatoire
%--first part
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\font\rms=cmr8 
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\markright{\its S\'eminaire Lotharingien de
Combinatoire \bfs 44 \rms (2000), Article~B44f\hfill}
\def\thepage{}
%

\begin{abstract}
We give a simple combinatorial proof of 
a recent formula of Lascoux for the powers of
 continued fractions. 
\end{abstract}


Given an alphabet $\A$ composed of commuting letters $a$'s, let 
$\l_x(\A)= \prod_{a\in\A} (1+ax)$.
For any  $k\in\R$,
 let $\L^n(k\A)$ and $S^n(k\A)$ be the coefficients 
of $x^n$ in $(\l_x(\A))^k$  
and  $(\l_{-x}(\A))^{-k}$ respectively. 
Recall that 
the {\it weight} of a Motzkin path $\pi$,  noted $w(\pi)$,
 is the product
of the weights of its steps~: 1 for a North-East step,
$\a_i$ for an horizontal step at level $i$, $\y_i$ for a South-East step
between level $i$ and $i-1$.  

Assuming  that one has the following 
continued fraction expansion~:
\begin{eqnarray}
\sum x^n S^n(\A)=
{1\over \dsp (1-\a _0 x)-
{\strut x^2\y_1\over \dsp (1-\a_1x)-
{\strut x^2\y_2\over \dsp (1-\a_2x)-
{\strut x^2\y_3\over \dsp \ddots}}}},
\end{eqnarray}
then Lascoux~\cite{La}, generalizing a result of Lehner, 
proved the following~:
\begin{thm}[Lascoux]
 For any $n\in \N$, any $k\in \R$ , one has
\begin{eqnarray}
 \L^n(k\A) = \sum_{\pi} (-1)^{n-\ell} {k\choose \ell} w(\pi)
\end{eqnarray}
where the sum is over all {\MP} $\pi$ of length $n$, where $\ell$ is the 
 number of ground points (other than the origin) of the path.
\end{thm}

Lascoux's proof of (2) 
is of lots eruditions. Here we show how to derive it directly from 
Flajolet's classical result~\cite{Fl}.


First we note that by definition $S^n(k\A)=(-1)^n\L^n((-k)\A)$, so
(2) can be paraphrased as follows~:
\begin{eqnarray}
S^n(k\A) =\sum_\pi
 {k+\ell -1\choose \ell} w(\pi), 
\end{eqnarray}
where the sum is the same as in (2). 
Since 
the two sides of (3) are
polynomials of $k$, it suffices to prove (3) for $k\in \N$.
In view of Flajolet's result~\cite{Fl},
 which corresponds to the $k=1$ case of (3), the symmetric function 
$S^n(k\A)$ is equal to 
$\sum   w(\pi_1\cdots \pi_k)$, where the sum is over all the $k$-tuples
$(\pi_1, \ldots, \pi_k)$ of {\MP} such that 
the product (or juxtaposition) $\pi_1\cdots \pi_k$ is a Motzkin path 
of length $n$. By convention, an empty {\MP} is of length 0.
Clearly, each  Motzkin path $\pi$ having
$\ell$ ground points (other than the origin) can be 
ontained in such a way from 
${k+\ell -1\choose \ell}$ 
$k$-tuples $(\pi_1, \ldots, \pi_k)$ of {\MP}, which is exactly 
what (3) means.





\begin{thebibliography}{9}
\bibitem{Fl}   P. Flajolet, \emph{Combinatorial aspects 
of continued fractions},
  Discrete Math. {\bf 32} (1980) 125-161. 

\bibitem{La} A. Lascoux, \emph{Motzkin paths and powers 
of continued fractions},  S\'eminaire Lotharingien de Combinatoire
{\bf 44} (2000), Article B44e.
 \end{thebibliography}
\end{document}