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\begin{center}
{\Large{\bf
{A note on a paper of G.~H.~Weiss and M.~Dishon}}}\\
{\sf Helmut Prodinger\\
The John Knopfmacher Centre for Applicable Analysis and Number
Theory\\
 School of Mathematics, University of the Witwatersrand, P.\;O.
Wits\\
 2050 Johannesburg, South Africa\\
\texttt{helmut@gauss.cam.wits.ac.za}
}
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% xxxxxx
Yesterday\footnote{October 16, 2001} I browsed through
old volumes of \textsl{The Fibonacci Quarterly\/} (which I can
recommend to everyone) and found \cite{WeDi76}: In an
attempt to improve on earlier result of Narayana and Kreweras, 
the authors prove that for $r,s\ge1$
\begin{equation*}
[u^rv^s]\frac{1-u-v-\sqrt{1-2(u+v)+(u-v)^2}}{2}=
\frac{1}{r+s-1}\binom{r+s-1}{r}\binom{r+s-1}{s}
\end{equation*}
(the notation $[u^rv^s]f(u,v)$ refers to the coefficient
of $u^rv^s$ in the power series expansion of $f(u,v)$).

However, the really easy method in this context is
the \textsl{Lagrange inversion formula,} as I will
sketch now. Let $S=\big(1-u-v-\sqrt{1-2(u+v)+(u-v)^2}\,\big)/2$,
then $S^2+(u+v-1)S+uv=0$, or
\begin{equation*}
u=\frac{S}{\Phi(S)}\quad\text{with}\quad
\Phi(S)=\frac{v+S}{1-v+S}.
\end{equation*}
Now the Lagrange inversion formula tells us (see, e.\,g.,
\cite{Stanley99}) that
\begin{equation*}
[u^r]S=\frac1r[S^{r-1}]\big(\Phi(S)\big)^r, 
\end{equation*}
or, with $v=St$,
\begin{align*}
[u^rv^s]S&=\frac1r[S^{r-1}v^s]\bigg(\frac{v+S}{1-v+S}\bigg)^r\\
&=\frac1r[S^{r+s-1}t^s]\bigg(\frac{S(1+t)}{1-S(1+t)}\bigg)^r\\
&=\frac1r[t^s](1+t)^r[S^{s-1}]\big(1-S(1+t)\big)^{-r}\\
&=\frac1r[t^s](1+t)^{r+s-1}\binom{r+s-2}{s-1}\\
&=\frac1r\binom{r+s-1}{s}\binom{r+s-2}{s-1},\\
\end{align*}
which is clearly equivalent to the statement to be proved.

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\bibitem{Stanley99}
R.~Stanley, \emph{Enumerative combinatorics. {V}ol. 2}, Cambridge University
  Press, Cambridge, 1999.

\bibitem{WeDi76}
G.~H. Weiss and M.~Dishon, \emph{A method for the evaluation of certain sums
  involving binomial coefficients}, Fibonacci Quarterly \textbf{14} (1976),
  75--77.

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A.~M.~S.~Classification: 05A10

























        
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