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\begin{center}
\vskip 1cm{\LARGE\bf 
An Asymptotic Expansion for the\\
\medskip
Catalan-Larcombe-French Sequence}
\vskip 1cm
\large
Lane Clark\\
Department of Mathematics\\
Southern Illinois University, Carbondale\\
Carbondale, IL 62901--4408\\
USA\\
\href{mailto:lclark@math.siu.edu}{\tt lclark@math.siu.edu} \\
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\vskip .5in

\begin{abstract}
We give an elementary development of a complete asymptotic
expansion for the Catalan-Larcombe-French sequence.
\end{abstract}


\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}{Proposition}[section]
\newtheorem{corollary}{Corollary}[section]
\newtheorem{lemma}{Lemma}[section] 



\def\clktheorem{\bigskip\noindent{\bf Theorem.}\quad}
\def\clktheoremlab#1#2{\bigskip\noindent{\bf Theorem #1}#2.\quad}
\def\endclktheorem{\medskip}
\def\clklemma#1{\bigskip\noindent{\bf Lemma #1.}\quad}
\def\clklemmalab#1#2{\bigskip\noindent{\bf Lemma #1}#2.\quad}
\def\endclklemma{\medskip}
\def\clkremark{\bigskip\noindent{\bf Remark.}\quad}
\def\clkremarks{\bigskip\noindent{\bf Remarks.}\quad}
\def\endclkremark{\medskip}
\def\nowhitespace{\vskip -6truept}
\def\proof{\vskip 6truept\noindent{\bf Proof.}\quad}
\def\endproof{\vskip 6truept}
\def\qed{$\,\,\,\, \blacksquare$}

\def\qed {$\,\,\,\, \blacksquare$}
\def\eee{\hbox{e}\,}
\def\lldots{\ldots\kern -0.8truept}
%
% local definitions to this paper
%
\def\G{{\mathfrak S}}
\def\Q{\mathbb Q}
\def\R{\mathbb R}
\def\C{\mathbb C}
\def\N{\mathbb N}
\def\P{\mathbb P}
\def\Z{\mathbb Z}
\def\Var{\hbox{Var}}
\def\Pr{\hbox{Pr}\,}
\def\des{\hbox{des}}
\def\ddes{\hbox{\smallft des}}
\def\iinv{\hbox{\smallft inv}}
\def\Des{\hbox{Des}}
\def\inv{\hbox{inv}}
\def\Inv{\hbox{Inv}}
\def\dz{\,\text{dz}}

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%  HYPHENATIONS                            -
%-------------------------------------------
\hyphenation{asymptotic companion
   }

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%  DOCUMENT TITLE                          -
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\section{Introduction}

        In their delightful paper, Larcombe and French [3] developed a
number of properties of the sequence (A053175)
$P_0=1$, $P_1=8$, $P_2=80$,
$P_3=896$, $P_4=10816$, $\ldots$ originally discussed by Catalan~[1].
In addition to a generating function, the following formula for $P_n$
was derived
%
\begin{equation}
  P_n = \frac{1}{n!} \, \sum_{p+q=n}
        \binom{2p}{p} \, \binom{2q}{q} \,
        \frac{(2p)! \, (2q)!}{p!\, q!} \qquad (n\in \N)\, .
\end{equation}
%
Recently, Larcombe et al.\  [4] showed that $P_n/2 \binom{2n}{n}^2 \to 1$
as $n\to \infty$ by a rather lengthy analysis.
In this short paper, we give an elementary development of a complete
asymptotic expansion for $P_n/2 \binom{2n}{n}^2$.
We conclude with a table of numerical calculations as a companion of the
theoretical results.

\section{Main result}

    The positive integers are denoted by $\P$; the nonnegative integers
by $\N$; the nonnegative rational numbers by $\Q_0$; 
and the complex numbers by $\C$.
Let $z^0 = (z)_0 \equiv 1$ and $(z)_p = (z)\cdots (z-p+1)$ when
$z\in \C$ and $p\in \P$.
For $z\in \C$ and $p \in \P$, $(z)_p = \sum_{q=0}^p s(p,q) z^q$
where the $s(p,q)$ are the Stirling numbers of the first kind
(see [2; pp.\ 212--214]).
We write $f(z) = O(g(z))$ provided there exist real constants $C$, $D$
with $|f(z)| \le C|g(z)|$ for all $|z| \ge D$.
%The empty sum is $0$ and the empty product is $1$.



        For $0\le p \le n$, let
\[
   a_p = \binom{2p}{p}^2 \, \binom{2n-2p}{n-p}^2 p!\, (n-p)! \, \in \P
\]
hence, $a_p = a_{n-p}$ and, let
\[
   b_p = \binom{2p}{p} \, \binom{2n-2p}{n-p} \in \P
\]
hence, $b_p = b_{n-p}$ and $a_p = n! \, b_p^2 \binom{n}{p}^{-1}$.
For $0\le p \le n-1$,
\[
   \frac{b_{p+1}}{b_p} \, = \,
        \frac{(2p+1)(n-p)}{(p+1)(2n-2p+1)}
\]
then, for $0 \le p \le (n-1)/2$,
\[
   0 &lt; \, \frac{b_{p+1}}{b_p} \, \le 1
\]
hence, for $1 \le p \le (n-1)/2$
%
\begin{equation}
   0 &lt; \, \frac{b_p}{b_0} \,
        = \, \frac{b_p}{b_{p-1}} \, \cdots \,
        \frac{b_1}{b_0} \, \le 1
\end{equation}
%
which is correct for $p=0$ also.



        Fix $s \ge 1$.
For $n = 2m+1 \ge 2s+3$, (1) and symmetry give
\begin{align}
   P_n &amp; = \, \frac{1}{n!} \, \sum_{p=0}^n
          \binom{2p}{p} \, \binom{2n-2p}{n-p} \,
          \frac{(2p)! \, (2n-2p)!}{p!\, (n-p)!}     \notag \\[8pt]
       &amp; = \, \frac{1}{n!} \, \sum_{p=0}^n
           a_p = \, \frac{2}{n!} \, \sum_{p=0}^m a_p
        = 2 \sum_{p=0}^m b_p^2 \binom{n}{p}^{-1}    \notag \\[8pt]
       &amp; = \, 2b_0^2 \left\{
          \sum_{p=0}^s \left( \frac{b_p}{b_0} \right)^2
                \binom{n}{p}^{-1}
        + \sum_{p=s+1}^m \left( \frac{b_p}{b_0} \right)^2
          \binom{n}{p}^{-1} \right\} \, .
\end{align}
%
Now $\binom{n}{p}$ is an increasing sequence of positive integers
for $0 \le p \le (n-1)/2$, so (2) gives
\begin{equation}
  0 &lt; \sum_{p=s+1}^m \left( \frac{b_p}{b_0} \right)^2
          \binom{n}{p}^{-1}
    \le n \binom{n}{s+1}^{-1} = O(n^{-s}) \text{\ \ as $n\to\infty$.}
\end{equation}


        For $0 \le p \le (n-1)/2$,
\[
   \frac{b_p}{b_0} \, = \binom{2p}{p} \,
        \frac{(n)^2_p}{(2n)_{2p}}
\]
hence,
\begin{equation}
  \sum_{p=0}^s \left( \frac{b_p}{b_0} \right)^2 \binom{n}{p}^{-1}
        = \sum_{p=0}^s p! \, \binom{2p}{p}^2
          \, \frac{(n)^3_p}{(2n)^2_{2p}} \, .
\end{equation}


        For $z \in \C$, let
$f_0(z) \equiv 1$ and, for $1\le p \le s$, let
\begin{align}
  f_p(z)
        &amp; = p! \binom{2p}{p}^2 \, \frac{(z)_p^3}{(2z)^2_{2p}}
          = \frac{p!\, \binom{2p}{p}^2 (z)_p}{2^{4p} z^{2p}}
          \, \prod_{j=1}^p
             \left( 1 - \, \frac{2j-1}{2z}\, \right)^{-2} \notag \\[8pt]
        &amp; = \, \frac{p!\, \binom{2p}{p}^2}{2^{4p}}
          \, \sum_{q=0}^p  s(p,q) z^{q-2p}
          \left\{ \prod_{j=1}^p \left( \sum_{r=0}^\infty
             (r+1)(j-0.5)^r z^{-r} \right) \right\}   \notag \\[8pt]
        &amp; = \sum_{r=p}^\infty b(p,r) z^{-r}
            \qquad\qquad (z\in \C; \ |z| \ge s)
\end{align}
%
where $b(p,r) \in \Q_0$ for $r \ge p$ and
$b(p,p) = p! \, \binom{2p}{p}^2 / 2^{4p}$.
For $1 \le p \le s$,
\begin{align*}
&amp; |(z)_p| \le (|z|+s)^p \le 
   \left( \textstyle\frac{3}{2}|z|\right)^p \\[-3pt]
&amp; \kern 4.0in (|z| \ge 2s)                  \\[-3pt]
&amp; |(2z)_{2p}| \ge (2|z|-2s)^{2p}
    \ge |z|^{2p} \, ,
\end{align*}
%
hence,
%
\begin{equation}
   \left| \frac{(z)^3_p}{(2z)_{2p}^2} \right| 
    \le \left( \frac{4}{|z|} \right)^p \qquad\qquad (|z| \ge 2s).
\end{equation}
%
For $r,s \ge p\ge1$, Laurent's Theorem (see [5; V.\ 2, p.\ 6]),
standard estimates for the integral and (7) give
%
\begin{align}
   \big| b(p,r) \big|
    &amp; = \left| \, \frac{1}{2\pi i} \, \oint_{|z|=2s}
        \frac{f_p(z)}{z^{r+1}} \, \dz \, \right|    \notag \\[8pt]
    &amp; \le p! \, \binom{2p}{p}^2 \left( \frac{2}{s} \right)^p
         \, \frac{1}{(2s)^r} \, \le s! \, \binom{2s}{s}^2 \, .
\end{align}
%
Then (8) gives
%
\begin{equation}
\left| \sum_{r=s}^\infty b(p,r) z^{-r} \right|
   \le |z|^{-s} s! \, \binom{2s}{s}^2
   \sum_{t=0}^\infty |z|^{-t} 
   = O \big( |z|^{-s} \big) \, ,
\end{equation}
%
hence, (6,9) give
%
\begin{equation}
   f_p(z) = \sum_{r=p}^{s-1} b(p,r) z^{-r} + O\big( |z|^{-s}\big)
\end{equation}
where $f_s(z) = O\big( |z|^{-s}\big)$.
%\[
%    \left| \frac{(z)_p^3}{(2z)_{2p}^2} \right|
%      \le \left( \frac{2}{|z|} \right)^p \qquad\qquad (|z| \ge 2s)
%\]
%and Laurent's Theorem implies
%\[
%     \left| \sum_{r=s}^\infty b(r,p) z^{-r} \right|
%        \le \, \frac{2^s s! \, \binom{2s}{s}^2 (2s)^s}{|z|^s}
%            \, \sum_{t=0}^\infty
%           \left( \frac{2s}{|z|} \right)^{t}
%           = O(|z|^{-s})
%\]
%hence,
%\[
%   f_p(z) = \sum_{r=p}^{s-1} b(r,p) z^{-r} + O(|z|^{-s}).
%\]
For $s \ge 1$, (10) gives
%
\begin{align}
   g_s(z) &amp; :=  \sum_{p=0}^{s-1} f_p(z)
       \qquad\qquad (z \in \C; \ |z| \ge s)  \notag \\[8pt]
    &amp; \, = \sum_{r=0}^{s-1} c(s,r) z^{-r} + O(|z|^{-s})
\end{align}
%
where $c(s,0)=1$ for $s\ge 1$ and 
$c(s,r) = \sum_{p=1}^r b(p,r) \in \Q_0$
for $1 \le r \le s-1$.
Observe that $c(s+1,r) = c(s,r)$ for $0 \le r \le s-1$.
The analysis for $n = 2m \ge 2s+2$ is identical except that
$P_n$ includes $b_m^2 \binom{n}{m}^{-1}$ not
$2b_m^2 \binom{n}{m}^{-1}$.
Then (3--5,11) give 
the following complete asymptotic expansion for
$P_n / 2 \binom{2n}{n}^2$.


\clktheorem
Fix $s \ge 1$.
There exist effectively calculable nonnegative 
rational numbers $c(s,0)=1$,
$c(s,1)$, $\ldots$ , $c(s,s-1)$ so that
%
\[
  P_n / 2\binom{2n}{n}^2
        = \sum_{r=0}^{s-1} c(s,r) n^{-r}
           + O(n^{-s})  \text{\ \ as $n\to\infty$}. \quad \text{\qed}
\]
%
%Moreover, $c(s+1,r) = c(s,r)$ for $0 \le r \le s-1$.
\endclktheorem


        Let
\[
    h_s(n) = \sum_{r=0}^{s-1} c(s,r) n^{-r}
\]
hence,
%
\begin{align*}
   h_1(n) &amp; = 1, \quad h_2(n) = 1 + \frac{1}{4n} \, , \quad
     h_3(n) = 1 + \, \frac{1}{4n} \, + \, \frac{17}{32n^2}\, , \\[8pt]
   h_4(n) &amp; = 1 + \, \frac{1}{4n} \, + \, \frac{17}{32n^2} \,
                + \, \frac{207}{128n^3}\, \quad\text{\ and} \\[8pt]
   h_5(n) &amp; = 1 + \, \frac{1}{4n} \, + \, \frac{17}{32n^2} \,
                + \, \frac{207}{128n^3}\,
                + \, \frac{14875}{2048n^4}\, .
\end{align*}
%
Let $Q_s(n) = \big[ P_n / 2 \binom{2n}{n}^2 - h_s(n) \big] n^s$.
Our theorem shows $Q_s(n) = O(1)$ as $n\to\infty$.
The table below which gives the first $10$ digits of $Q_2(n)$ and
$Q_3(n)$ for several values of $n$ was found using {\em Mathematica}.
This provides numerical evidence for our theorem.

\begin{center}
\begin{tabular}[h]{rll}
 $n$ \ &amp; \qquad $Q_2(n)$ &amp; \qquad $Q_3(n)$  \\[6pt]
  100 &amp; .5481946735  &amp; 1.6944673581   \\
  200 &amp; .5395231003  &amp; 1.6546200668   \\
  300 &amp; .5367229603  &amp; 1.6418880975   \\
  400 &amp; .5353390483  &amp; 1.6356193435   \\
  500 &amp; .5345137769  &amp; 1.6318884961   \\
  600 &amp; .5339656896  &amp; 1.6294137740   \\
  700 &amp; .5335752174  &amp; 1.6276521934   \\
  800 &amp; .5332829178  &amp; 1.6263343131   \\
  900 &amp; .5330559013  &amp; 1.6253112456   \\
 1000 &amp; .5328744940  &amp; 1.6244940166   \\
 2000 &amp; .5320604149  &amp; 1.6208298850   \\
 3000 &amp; .5317898711  &amp; 1.6196133523
\end{tabular}
\end{center}

\bigskip\bigskip
\noindent{\bf Acknowledgement.}\quad
I wish to thank the referee for comments and suggestions
which have resulted in an improved version of the paper.

\bigskip\bigskip
\centerline{\bf References}

\baselineskip=14truept
\parskip=6truept

\begin{enumerate}

\item
E.\ Catalan,
Sur les Nombres de Segner,
{\it Rend.\ Circ.\ Mat.\ Pal.}
{\bf 1} (1887), 190--201.

\item
L.\ Comtet,
Advanced Combinatorics, D.\ Reidel,
Boston, MA, 1974.

\item
P.J.\ Larcombe and D.R.\ French,
On the `Other' Catalan Numbers: A Historical Formulation Re-examined,
{\it Congr.\ Numer.}
{\bf 143} (2000), 33--64.

\item
P.J.\ Larcombe, D.R.\ French and E.J.\ Fennessey,
The Asymptotic Behaviour of the Catalan-Larcombe-French
Sequence $\{1,8,80,896,10816,\lldots\}$,
{\it Util.\ Math.}
{\bf 60} (2001), 67--77.

\item
A.I. Markushevich,
Theory of Functions of a Complex Variable,
Chelsea Publishing Co., New York, NY, 1985.

\end{enumerate}


\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:
Primary 05A10; Secondary 05A16, 41A60. 

\noindent \emph{Keywords: }  asymptotic expansion,
Catalan-Larcombe-French sequence.


\bigskip
\hrule
\bigskip

\noindent (Concerned with sequence
\seqnum{A053175}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received October 28 2002;
revised version received February 2 2004. 
Published in {\it Journal of Integer Sequences},
March 23 2004.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.math.uwaterloo.ca/JIS/}.
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