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\begin{center}
\vskip 1cm{\LARGE\bf Curious Relations and Identities Involving the \\
\vskip .1in
Catalan Generating Function and Numbers 
}
\vskip 1cm
\large
Asamoah Nkwanta \\
Department of Mathematics \\
Morgan State University \\
Baltimore, MD 21251 \\
USA \\
\href{mailto:Asamoah.Nkwanta@morgan.edu}{\tt Asamoah.Nkwanta@morgan.edu}  \\
\ \\
Akalu Tefera  \\
Department of Mathematics \\
Grand Valley State University \\
Allendale, MI 49401 \\
USA \\
\href{mailto:teferaa@gvsu.edu}{\tt teferaa@gvsu.edu} \\
\end{center}

\vskip .2 in

\begin{abstract}
Riordan matrix methods and manipulation of various generating functions are
used to find curious relations among the Catalan, central binomial, and RNA
generating functions. In addition, the Wilf-Zeilberger method is used to
find\ identities where the gamma function and Catalan numbers are expressed
in terms of the Gauss hypergeometric function. As a consequence of the
identities, new recurrence relations are obtained. In particular, a new
recurrence relation is given for the RNA numbers. Furthermore, other
representations of $\pi $ and the Catalan numbers are obtained.
\end{abstract}

\section{Introduction}

In this paper we use Riordan matrix methods and manipulation of generating
functions to find curious relations among the Catalan, central binomial, and
RNA generating functions. We also give relations between the Fibonacci and
Catalan generating functions. In addition to finding relations involving
generating functions, the Wilf-Zeilberger (WZ) method is used to find\
identities where the gamma function and Catalan numbers \seqnum{A000108} 
\cite{SP1} are expressed in terms of the Gauss hypergeometric function. As a
consequence of the identities, new recurrence relations are obtained.
Furthermore, other representations of $\pi $ and the Catalan numbers 
\seqnum{A000108} \cite{SP1} are presented.

The results of this paper are proved by the WZ method, manipulation of
generating functions, and Riordan matrix multiplication. The notion of a
Riordan matrix is used to prove combinatorial sums and identities (Shapiro
et al.~\cite{Shap2} and Sprugnoli \cite{Sprug}). The WZ method is based on
an algorithm that finds recurrence relations that are satisfied by certain
sums (Wilf and Zeilberger \cite{WIL1} and Petkovsek et al.~\cite{AB}). The
recurrence relations given in this paper by Equations $\left( \text{\ref%
{WZ2eq}}\right) ,\left( \text{\ref{WZ1eq}}\right) ,$ and $\left( \text{\ref%
{WZ-eq}}\right) $ are automatically generated by the Maple\ package EKHAD,
which can be downloaded from Zeilberger's website \cite{ZEIL}. In addition,
the SumTools package in Maple\ that implements Zeilberger's algorithm can
also be used. For an excellent exposition of Zeilberger's algorithm, see
Petkovsek et al.~\cite{AB}.

This paper is arranged as follows. Preliminary material on hypergeometric
functions, the WZ method, the gamma function, generating functions, and
Riordan matrices are given in Section 2. Readers familiar with these topics
may skip this section. In Section 3, Riordan matrix and generating function
methods are utilized to find curious relations among the Catalan, central
binomial, and RNA generating functions. Curious relations involving the
Fibonacci generating function are given in Section 4. The WZ method is used
to find new recurrence relations that are presented in Section 5. The main
result of the paper, Theorem \ref{Thm1}, which is an identity involving the
Catalan numbers \seqnum{A000108} \cite{SP1}, is also given in this
section. In addition, we present in this section other representations of
the irrational number $\pi $ and the ubiquitous Catalan numbers. Some
concluding comments are given in Section 6.

\section{Preliminary material}

Throughout this paper we denote the sets $\{1,2,\dots \}$ \seqnum{A000027}
\cite{SP1} and $\{0,1,\dots \}$ \seqnum{A001477} \cite{SP1} by $\mathbb{N}
$ and $\mathbb{N}_{0}$, respectively. Let $(x)_{k}$ denote the rising
factorial defined by%
\begin{equation*}
(x)_{k}=x(x+1)\cdots (x+k-1)
\end{equation*}%
for $k\in \mathbb{N}$ and $(x)_{0}=1.$ The gamma function denoted by $\Gamma
\left( z\right) $ is defined as%
\begin{equation*}
\Gamma \left( z\right) =\int_{0}^{\infty }t^{z-1}e^{-t}dt,
\end{equation*}%
if ${\Real}\left( z\right) &gt;0$. For $z=n$ where $n\in \mathbb{N}_{0}$%
\begin{equation*}
\Gamma \left( n+1\right) =n!=(1)_{n}.
\end{equation*}%
Some useful properties are $\Gamma \left( 1/2\right) =\sqrt{\pi }$ and 
\begin{equation}
(x)_{k}=\frac{\left( x+k-1\right) !}{\left( x-1\right) !}=\frac{\Gamma
\left( k+x\right) }{\Gamma \left( x\right) }.  \label{gamaeq}
\end{equation}%
The Gauss hypergeometric function ${}_{2}F_{1}$ is defined by 
\begin{equation*}
{}_{2}F_{1}(a,b;c;z)=\sum_{k=0}^{\infty }\frac{(a)_{k}(b)_{k}}{(c)_{k}}\frac{%
z^{k}}{k!}
\end{equation*}%
where $a$, $b$ and $c$ are arbitrary complex constants. If we take $z=1$,
then 
\begin{equation*}
{}_{2}F_{1}(a,b;c;1)=\frac{\Gamma \left( c-a-b\right) \Gamma \left( c\right) 
}{\Gamma \left( c-a\right) \Gamma \left( c-b\right) }.
\end{equation*}%
For more information on the gamma and hypergeometric functions, see
Petkovsek et al.~\cite{AB}.

The Wilf-Zeilberger algorithm was developed by Wilf and Zeilberger \cite%
{WIL1} as a method for certifying the truth of certain combinatorial
identities. The identities involve certain hypergeometric functions. A
discrete function $F\left( n,k\right) $ is called hypergeometric if%
\begin{equation*}
F\left( n+1,k\right) /F\left( n,k\right) \text{ and }F\left( n,k+1\right)
/F\left( n,k\right)
\end{equation*}%
are both rational functions in $n$ and $k$. The WZ proof method is briefly
described as follows. Suppose we want to prove the identity%
\begin{equation*}
\sum_{k=0}^{\infty }U\left( n,k\right) =r\left( n\right) .
\end{equation*}%
If $r\left( n\right) \neq 0,$ then dividing through by $r\left( n\right) $
the given identity is equivalent to%
\begin{equation*}
\sum_{k=0}^{\infty }\left\{ \frac{U\left( n,k\right) }{r\left( n\right) }%
\right\} =1.
\end{equation*}%
Now set $F\left( n,k\right) =U\left( n,k\right) /r\left( n\right) $. We want
to prove $\sum_{k=0}^{\infty }F\left( n,k\right) =1$. To certify the
validity of this, it would suffice to find a function $G\left( n,k\right) $
such that%
\begin{equation*}
F\left( n+1,k\right) -F\left( n,k\right) =G\left( n,k+1\right) -G\left(
n,k\right) .
\end{equation*}%
The production $G$ comes as a result of using the WZ algorithm. The pair of
functions $F$ and $G$ are called WZ pairs. See Wilf and Zeilberger \cite%
{WIL1}, Petkovsek et al.~\cite{AB}, Wilf \cite{WIL}, and Gessel \cite{Gess}
for more information on the WZ algorithm and WZ pairs. See Tefera \cite{Tef}
for a quick review of the WZ method.

A formal power series of the form 
\begin{equation*}
a\left( z\right) =a_{0}+a_{1}z+a_{2}z^{2}+\cdots =\sum_{n\geq 0}a_{n}z^{n}
\end{equation*}%
where $z$ is an indeterminate is called the ordinary \textit{generating
function} of the sequence $\left\{ a_{0},a_{1},\ldots \right\} $. Let%
\begin{equation*}
b\left( z\right) =\sum_{n\geq 0}b_{n}z^{n}=\frac{1}{\sqrt{1-4z}}
\end{equation*}%
denote the central binomial generating function of the sequence $\left\{
b_{0},b_{1},\ldots \right\} $. Then
\begin{equation*}
\left\{ b_{n}\right\}_{n\geq 0}=\left\{ 1,2,6,20,\ldots \right\} =\left\{ 
\binom{2n}{n}\right\}_{n\geq 0}
\end{equation*}%
where $b_{n}$\ denotes the $n$th central binomial coefficient \seqnum{A00984} \cite{SP1}. Let%
\begin{equation}
c\left( z\right) =\sum_{n\geq 0}c_{n}z^{n}=\frac{1-\sqrt{1-4z}}{2z}
\label{cateq}
\end{equation}%
denote the Catalan generating function of the sequence $\left\{
c_{0},c_{1},\ldots \right\} $. Then
\begin{equation}
\left\{ c_{n}\right\} _{n\geq 0}=\left\{ 1,1,2,5,\ldots \right\} =\left\{ 
\frac{1}{1+n}\binom{2n}{n}\right\} _{n\geq 0}  \label{cateq1}
\end{equation}%
where $c_{n}$ denotes the $n$th Catalan number \seqnum{A000108} \cite{SP1}%
. See Stanley [15, 16] for a number of combinatorial and analytical
interpretations of the Catalan numbers \seqnum{A000108} \cite{SP1}. Let%
\begin{equation}
f\left( z\right) =\sum_{n\geq 0}f_{n}z^{n}=\frac{1}{1-z-z^{2}}
\label{fibfg}
\end{equation}%
denote the Fibonacci generating function of the sequence $\left\{
f_{0},f_{1},\ldots \right\} $. Then
\begin{equation*}
\left\{ f_{n}\right\} _{n\geq 0}=\left\{ 1,1,2,3,\ldots \right\} =\left\{ 
\frac{1}{\sqrt{5}}\left( \left( \frac{1+\sqrt{5}}{2}\right) ^{n+1}-\left( 
\frac{1-\sqrt{5}}{2}\right) ^{n+1}\right) \right\} _{n\geq 0}
\end{equation*}%
where $f_{n}$\ denotes the $n$th Fibonacci number \seqnum{A000045} \cite%
{SP1}. We now let%
\begin{equation}
s\left( z\right) =\sum_{n\geq 0}s_{n}z^{n}=\frac{1-z+z^{2}-\sqrt{%
1-2z-z^{2}-2z^{3}+z^{4}}}{2z^{2}}  \label{rnaeq}
\end{equation}%
denote the RNA generating function from discrete mathematical biology
(Waterman [18, 19]). Then $s_{0}=0$, $s_{1}=1$, 
\begin{equation}
s_{n}=\sum_{k\geq 1}\frac{1}{n-k}\binom{n-k}{k}\binom{n-k}{k-1}\text{ }%
\left( k&lt;n\right) ,  \label{rnaeq2}
\end{equation}%
and $\left\{ s_{n}\right\} _{n\geq 0}=\left\{ 0,1,1,1,2,4,8,\ldots \right\} $
where $s_{n}$\ denotes the $n$th RNA number (\seqnum{A004148} \cite{SP1}
and Waterman \cite{WAT}). We now define the Fine generating function, which
we denote by $\Phi \left( z\right) .$ Let%
\begin{equation*}
\Phi \left( z\right) =\sum_{n\geq 0}\Phi _{n}z^{n}=\frac{1-\sqrt{1-4z%
}}{3-\sqrt{1-4z}}\frac{1}{z}.
\end{equation*}%
Then $\left\{ \Phi _{n}\right\} _{n\geq 0}=\left\{ 1,0,1,2,6,18,\ldots
\right\} $ where $\Phi _{n}$\ denotes the $n$th Fine number \seqnum{A000957} \cite{SP1}. See Deutsch and Shapiro \cite{DSHAP1} for more
information on the Fine numbers. For more information on generating
functions, hypergeometric functions, and the WZ method, see 
Petkovsek et al.~\cite{AB}.

The definition of a Riordan matrix and rules for multiplying Riordan
matrices are presented.

\begin{definition}
(Shapiro et al.~\cite{Shap2}, Sprugnoli \cite{Sprug}) An infinite matrix%
\begin{equation*}
L=(l_{n,k})_{n,k\in \mathbb{N}_{0}}
\end{equation*}%
with complex entries in $\mathbb{C}$ is called a \emph{Riordan matrix} if the $%
k^{th}$ column satisfies%
\begin{equation*}
\sum_{n\geq 0}l_{n,k}\ z^{n}=g(z)f(z)^{k}
\end{equation*}%
where $g(z)=g_{0}+g_{1}z+g_{2}z^{2}+\cdots $ and $f(z)=f_{1}z+f_{2}z^{2}+%
\cdots $ belong to the ring of formal power series $\mathbb{C}[[z]],$ and $%
f_{1}\neq 0$ and $g_{0}\neq 0.$
\end{definition}

Riordan matrices are typically written in pair form as $\left(
g(z),f(z)\right) $ where $g(z)$ and $f(z)$ are ordinary generating
functions. Note that Riordan matrices can also be defined for exponential
generating functions. However, the Riordan matrices in this paper are
defined by ordinary generating functions and are the proper Riordan arrays
as given by Sprugnoli \cite{Sprug}. Pascal's triangle \seqnum{A007318} 
\cite{SP1} written in lower-triangular form is an example of a Riordan
matrix and is denoted by%
\begin{equation*}
P=\left( 1/(1-z),z/(1-z)\right) .
\end{equation*}

Let us denote by $L\ast N$, or by simple juxtaposition $LN$, the row-by
column product of two Riordan matrices. If 
\begin{equation*}
L=(l_{n,k})_{n,k\in \mathbb{N}_{0}}=\left( g(z),f(z)\right) \text{ and }%
N=(\nu _{n,k})_{n,k\in \mathbb{N}_{0}}=\left( h(z),l(z)\right)
\end{equation*}%
are Riordan matrices, then%
\begin{align*}
LN=\left( \sum_{j=0}^{n}l_{n,j}\nu _{j,k}\right) _{n,k\in \mathbb{N}_{0}}&amp;
=\left( g(z),f(z)\right) \ast \left( h(z),l(z)\right) \\
&amp; =\left( g(z)h(f(z)),l(f(z))\right) .
\end{align*}%
For more information on Riordan matrices and arrays, see Shapiro et al.~\cite%
{Shap2} and Sprugnoli \cite{Sprug}.

\section{Catalan, central binomial and RNA relations}

Consider the following Riordan matrix relation%
\begin{equation}
B=C_{0}R^{2}E  \label{beqa}
\end{equation}%
where $C_{0}=\left( c(z^{2}),zc(z^{2})\right) $ is Catalan array \seqnum{A053121} \cite{SP1}, $R=\left( s(z),zs(z)\right) $ is the RNA array 
\seqnum{A097724} \cite{SP1} from mathematical biology, and $E=\left(
1/\left( 1-z\right) ,z\right) $ is the lower-triangular array \seqnum{A000012} \cite{SP1} with entries with all ones on and below the main
diagonal. Recall that the generating functions $c\left( z\right) $ and $s(z)$
are, respectively, given by Equations $\left( \text{\ref{cateq}}\right) $
and $\left( \text{\ref{rnaeq}}\right) $. The matrix $B$ \seqnum{A111418} 
\cite{SP1} can also be obtained from%
\begin{equation}
B=P^{2}C_{0}E  \label{beqa3}
\end{equation}%
where $P$ is Pascal matrix \seqnum{A007318} \cite{SP1} (Nkwanta [8, 9]).
By Riordan matrix multiplication, the right side of Equation $\left( \text{%
\ref{beqa3}}\right) $ gives%
\begin{equation}
B=\left( c(z)/\sqrt{1-4z},zc^{2}(z)\right) {\small =}\left( 
\begin{array}{cccccc}
1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\ 
3 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\ 
10 &amp; 5 &amp; 1 &amp; 0 &amp; 0 &amp; \cdots \\ 
35 &amp; 21 &amp; 7 &amp; 1 &amp; 0 &amp; \cdots \\ 
126 &amp; 84 &amp; 36 &amp; 9 &amp; 1 &amp; \cdots \\ 
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots%
\end{array}%
\right) \text{.}  \label{beqa1}
\end{equation}%
The generating function of the left component of $B$ can also be expressed as%
\begin{equation}
c(z)/\sqrt{1-4z}=c^{2}(z)/\left( 1-zc^{2}(z)\right)  \label{beqa5}
\end{equation}%
and is well known as a special case of 
\begin{equation}
B_{k}\left( z\right) =c^{k}(z)/\sqrt{1-4z}=\sum_{n\geq 0}\binom{2n+k%
}{n}z^{n}.  \label{beqa4}
\end{equation}%
For more information on Equation $\left( \text{\ref{beqa4}}\right) $, see
Graham et al.~\cite{GKB}, Riordan \cite{RIO}, and Wilf \cite{WIL}. For more
information and a combinatorial interpretation of $B$, see Nkwanta \cite{NK}%
. The inverse of $B$ is given below%
\begin{equation*}
B^{-1}=\left( \left( 1-z\right) /\left( 1+z\right) ^{2},z/\left( 1+z\right)
^{2}\right) =\left( 
\begin{array}{rrrrrr}
1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; ... \\ 
-3 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; ... \\ 
5 &amp; -5 &amp; 1 &amp; 0 &amp; 0 &amp; ... \\ 
-7 &amp; 14 &amp; -7 &amp; 1 &amp; 0 &amp; ... \\ 
9 &amp; -30 &amp; 27 &amp; -9 &amp; 1 &amp; ... \\ 
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots%
\end{array}%
\right) .
\end{equation*}%
The entries of $B^{-1}$ form monic orthogonal polynomials whose moments are
certain binomial coefficients. The moments are associated with the the
Chebyshev polynomials of the first kind. See Nkwanta and Barnes \cite{NKB}
for more on the connection of $B$ to the Chebyshev polynomials.

By multiplying the matrices on the right side of Equation $\left( \text{\ref%
{beqa}}\right) $ and equating generating function components of the matrix
with Equation $\left( \text{\ref{beqa1}}\right) ,$ the left components give%
\begin{equation}
c(z)/\sqrt{1-4z}=\left( c\left( z^{2}\right) \cdot r\left( z\right) \cdot
t\left( z\right) \right) /\left( 1-zc\left( z^{2}\right) \cdot r\left(
z\right) \cdot t\left( z\right) \right)  \label{beqa2}
\end{equation}%
where $r\left( z\right) =s\left( zc\left( z^{2}\right) \right) $ and%
\begin{equation*}
t\left( z\right) =s\left[ zc\left( z^{2}\right) \cdot s\left( zc\left(
z^{2}\right) \right) \right] .
\end{equation*}%
Recall that $s\left( z\right) $ is the RNA generating function. This leads
to curious relations among the central binomial, Catalan and RNA generating
functions given by Equation $\left( \text{\ref{beqa2}}\right) $. Thus, we
have the following theorem.

\begin{theorem}
\label{RNAgf}
Let $r\left( z\right) =s\left( zc\left( z^{2}\right) \right) $
and $t\left( z\right) =s\left[ zc\left( z^{2}\right) \cdot s\left( zc\left(
z^{2}\right) \right) \right] $ where $c\left( z\right) $\ and $s\left(
z\right) $ are the Catalan and RNA generating functions, respectively. Then
\begin{itemize}
\item[(a)]
$ c\left( z\right)  =  \left( \left( \sqrt{1-4z}\right) \cdot
c\left( z^{2}\right) \cdot r\left( z\right) \cdot t\left( z\right) \right)
/\left( 1-zc\left( z^{2}\right) \cdot r\left( z\right) \cdot t\left(
z\right) \right) $.

\item[(b)]
$b\left( z\right)  =  B_{0}\left( z\right) =\left( c\left(
z^{2}\right) \cdot r\left( z\right) \cdot t\left( z\right) \right) /\left(
c\left( z\right) \cdot \left( 1-zc\left( z^{2}\right) \cdot r\left( z\right)
\cdot t\left( z\right) \right) \right) $.
\end{itemize}
\end{theorem}

\begin{proof}
The theorem follows as a result of Riordan matrix multiplication and
Equation $\left( \text{\ref{beqa}}\right) $.
\end{proof}

As a consequence of Theorem \ref{RNAgf}, the central binomial and Catalan
generating functions are expressed in terms of the RNA generating function.
An alternative expression of the RNA generating function is now given.

\begin{proposition}
Let $w\left( z\right) =\left( s\left( z\right) -1\right) /z$ where $s\left(
z\right) $ is the RNA generating function. Then
\begin{equation}
w\left( z\right) =1/\left( 1-\left( z+z^{2}+z^{3}w\left( z\right) \right)
\right) \text{ or}  \label{rnaeq1}
\end{equation}%
\begin{equation*}
s\left( z\right) =1/\left( 1-z-z^{2}\left( s\left( z\right) -1\right)
\right) .
\end{equation*}
\end{proposition}

\begin{proof}
The proof follows by using Equation $\left( \text{\ref{rnaeq}}\right) $ and
simplifying.
\end{proof}

Note that the constant term of the formal power series of Equation $\left( 
\text{\ref{rnaeq}}\right) $ is removed from Equation $\left( \text{\ref%
{rnaeq1}}\right) $. For more information on the derivation of Equation $%
\left( \text{\ref{rnaeq1}}\right) $, see Nkwanta \cite{NK1}.

\begin{remark}
As a consequence of Equation $\left( \text{\ref{rnaeq1}}\right) $ the RNA
generating function $s\left( z\right) $ can be expressed as the following
continued fraction expansion%
\begin{equation*}
s\left( z\right) =1+\frac{z}{1-z-z^{2}-\frac{z^{3}}{1-z-z^{2}-\frac{z^{3}}{%
1-z-z^{2}-\cdots }}}.
\end{equation*}
\end{remark}

\section{In terms of the Fibonacci generating function}

The following generating function 
\begin{equation}
f\left( z\right) /\left( 1+z^{2}f\left( z\right) \right) =1/\left(
1-z\right) ,  \label{fibgf}
\end{equation}%
where $f\left( z\right) $ is the Fibonacci generating function given by
Equation $\left( \text{\ref{fibfg}}\right) $, leads to curious relations
involving $1/\left( 1-\alpha z\right) $ where $\alpha \neq 0$ and $\alpha
\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$, binomial generating functions, the Catalan generating function $c\left(
z\right) $, and the Fine generating function $\Phi \left( z\right) $.

\begin{proposition}
\label{Fibfg1}%
\begin{itemize}
\item[(a)]
$1/\left( 1-\alpha z\right)  =  f\left( \alpha z\right) /\left(
1+\alpha ^{2}z^{2}f\left( \alpha z\right) \right) $.

\item[(b)] $B_{k}\left( z\right)  =  f\left( zc^{2}\left( z\right) \right)
c^{k}\left( z\right) /\left( 1+z^{2}c^{4}\left( z\right) f\left(
zc^{2}\left( z\right) \right) \right) $.

\item[(c)] $c\left( z\right)  =  f\left( zc\left( z\right) \right) /\left(
1+z^{2}c^{2}\left( z\right) f\left( zc\left( z\right) \right) \right) $

\item[(d)] $\Phi \left( z\right)  =  f\left( z^{2}c^{2}\left( z\right)
\right) /\left( 1+z^{4}c^{4}\left( z\right) f\left( z^{2}c^{2}\left(
z\right) \right) \right) $.
\end{itemize}
\end{proposition}

\begin{proof}
\begin{itemize}

\item[(a)]
Use Equation $\left( \text{\ref{fibgf}}\right) $ and some
simplification.

\item[(b)] Use Equation $\left( \text{\ref{fibgf}}\right) $ and
multiply by $c^{k}\left( z\right) $. Then use Equation $\left( \text{\ref%
{beqa4}}\right) $. 

\item[(c)] For (c) use Equation $\left( \text{\ref{fibgf}}\right) $.
Then, use the fact from Deutsch and Shapiro \cite{DSHAP} that%
\begin{equation*}
c\left( z\right) =1/\left( 1-zc\left( z\right) \right) .
\end{equation*}%

\item[(d)] Use Equation $\left( \text{\ref{fibgf}}\right) $. Then, use the fact
from Deutsch and Shapiro \cite{DSHAP} that%
\begin{equation*}
\Phi \left( z\right) =1/\left( 1-z^{2}c^{2}\left( z\right) \right) .
\end{equation*}
\end{itemize}
\end{proof}

We observe that for $k=0$ and $1$, respectively, the central binomial
generating function $b\left( z\right) $ and Equation $\left( \text{\ref%
{beqa5}}\right) $ are special cases of Proposition \ref{Fibfg1}$\left( \text{%
b}\right) $. As a result of Proposition \ref{Fibfg1}, the binomial, Catalan,
and Fine generating functions are expressed in terms of the Fibonacci
generating function. \ For more information on the derivation of Equation $%
\left( \text{\ref{fibgf}}\right) $, see Jean-Louis and Nkwanta \cite{CANK}.
See Deutsch and Shapiro [2, 3] for more information on the Fine numbers 
\seqnum{A000957} \cite{SP1} and Fine generating function.

\begin{remark}
By combining Theorem \ref{RNAgf} and Proposition \ref{Fibfg1}, we observe
that the generating functions $b\left( z\right) $ and $c\left( z\right) $
can be simultaneously expressed in terms of the Fibonacci and RNA generating
functions.
\end{remark}

\section{Finding recurrence relations}

We now use the Wilf-Zeilberger algorithm to find various recurrence
relations. Applying the WZ algorithm to Equation $\left( \text{\ref{rnaeq2}}%
\right) $ gives the following recurrence relation for the RNA numbers 
\seqnum{A004148} \cite{SP1}.

\begin{proposition}
\label{prop1}For $n\geq 0,$ $s_{n}$ satisfies 
\begin{equation*}
\left( n+5\right) s_{n+4}=\left( 2n+7\right) s_{n+3}+\left( n+2\right)
s_{n+2}+\left( 2n+1\right) s_{n+1}-\left( n-1\right) s_{n}
\end{equation*}%
where $s_{0}=0,s_{1}=1,s_{2}=1,$ and $s_{3}=1$.
\end{proposition}

\begin{proof}
Let $F\left( n,k\right) $ denote the summand of $s_{n}$, that is,

\begin{equation*}
F\left( n,k\right) =\frac{1}{n-k}\binom{n-k}{k}\binom{n-k}{k-1}.
\end{equation*}%
Now, applying the WZ algorithm on $F\left( n,k\right) $ we get%
\begin{equation}
\begin{array}{c}
\left( n+5\right) F\left( n+4,k\right) -\left( 2n+7\right) F\left(
n+3,k\right) -\left( n+2\right) F\left( n+2,k\right) \\ 
\begin{array}{ccccc}
&amp;  &amp;  &amp;  &amp; 
\end{array}%
-\left( 2n+1\right) F\left( n+1,k\right) +\left( n-1\right) F\left(
n,k\right) =G\left( n,k+1\right) -G\left( n,k\right)%
\end{array}
\label{WZ2eq}
\end{equation}%
where $G\left( n,k\right) =F\left( n,k\right) R\left( n,k\right) $ and%
\begin{equation*}
R\left( n,k\right) =-\frac{\left( k-1\right) k\left( n+1-k\right) \left(
n-k\right) }{(n+5-2k)(n+4-2k)^{2}(n+3-2k)^{2}(n+2-2k)^{2}(n+1-2k)}P\left(
n,k\right)
\end{equation*}%
where $P\left( n,k\right) =-\left(
332+304n+196n^{2}+107n^{3}+30n^{4}+3n^{5}\right) $%
\begin{equation*}
+\left( 716+446n+203n^{2}+81n^{3}+12n^{4}\right) k
\end{equation*}%
\begin{equation*}
-\left( 567+183n+47n^{2}+13n^{3}\right) k^{2}
\end{equation*}%
\begin{equation*}
-2\left( -98-2n+n^{2}\right) k^{3}+\left( -25+7n\right) k^{4}
\end{equation*}%
Summing both sides of Equation $($\ref{WZ2eq}$)$ for all values of $k$ we
get the result.
\end{proof}

We now find curious identities involving the Catalan numbers, $\pi $,
hypergeometric functions, and the gamma function. Consider the following
hypergeometric functions%
\begin{equation*}
\text{a) }A\left( n\right) :=\text{ }_{2}F_{1}\left( -1/2,2n+3;2n+\left(
5/2\right) ;-1\right)
\end{equation*}%
\begin{equation*}
\text{b) }M\left( n\right) :=\text{ }_{2}F_{1}\left( 1/2,2n+3;2n+\left(
5/2\right) ;-1\right) .
\end{equation*}%
Then we obtain the following propositions.

\begin{proposition}
Let $D\left( n\right) =A\left( n\right) -2M\left( n\right) .$ Then
\begin{itemize}
\item[(a)]
$D\left( n\right) = \left( 4n+5\right) ^{-1}{}_{2}F_{1}\left(
1/2,2n+3;2n+\left( 7/2\right) ;-1\right) $.

\item[(b)] $D\left( n\right)  =  \sum_{k=1}^{\infty }\frac{\left( -\frac{1}{2%
}\right) _{k}}{\left( 2n+\frac{5}{2}\right) _{k}}\binom{2n+k+1}{k-1}(-1)^{k}$.
\end{itemize}
\end{proposition}

\begin{proof}
Use properties of hypergeometric functions.
\end{proof}

The next proposition follows as a result of the WZ \ algorithm.

\begin{proposition}
\label{prop2}$D\left( n\right) $ satisfies 
\begin{equation}
16\left( n+2\right) ^{2}D\left( n+1\right) -\left( 4n+5\right) \left(
4n+7\right) D\left( n\right) =0.  \label{receq}
\end{equation}
\end{proposition}

\begin{proof}
Let $F\left( n,k\right) $ denote the summand of $D\left( n\right) $, that is,

\begin{equation*}
F\left( n,k\right) =\frac{\left( -\frac{1}{2}\right) _{k}}{\left( 2n+\frac{5%
}{2}\right) _{k}}\binom{2n+k+1}{k-1}(-1)^{k}.
\end{equation*}%
Now, applying the WZ algorithm on $F\left( n,k\right) $ we get%
\begin{equation}
16\left( n+2\right) ^{2}F\left( n+1,k\right) -\left( 4n+5\right) \left(
4n+7\right) F\left( n,k\right) =G\left( n,k+1\right) -G\left( n,k\right)
\label{WZ1eq}
\end{equation}%
where $G\left( n,k\right) =F\left( n,k\right) R\left( n,k\right) $ and%
\begin{equation*}
R\left( n,k\right) =\frac{(2n+3)\left( 4n+5+2k\right) }{\left( k-1\right)
(4n+5)(4n+7)}.
\end{equation*}%
Summing both sides of Equation $\left( \text{\ref{WZ1eq}}\right) $ for all
values of $k$ we get Equation $\left( \text{\ref{receq}}\right) .$
\end{proof}

\begin{corollary}
Solving the recurrence relation given by Equation $\left( \text{\ref{receq}}%
\right) $ gives%
\begin{equation}
D\left( n\right) =\frac{\sqrt{\pi }\Gamma \left( 2n+5/2\right) }{%
2^{2n+4}\left( n+1\right) !^{2}}  \label{Dneq1}
\end{equation}%
where $D\left( 0\right) =\frac{3\pi }{64}.$
\end{corollary}

\begin{proof}
Using a symbolic algebra system (such as Mathematica\ or Maple) to solve
Equation $\left( \text{\ref{receq}}\right) $ gives the result.
\end{proof}

We note here that $D\left( n\right) $ appears below in the numerator of
Equation $\left( \text{\ref{inteq}}\right) $. The motivation for the next
theorem comes from evaluating the following integral%
\begin{equation}
\int_{0}^{1}\frac{2^{2n+5}\left( x^{2}\left( 1-x^{2}\right) ^{2n}\right) }{%
\pi \left( 1+x^{2}\right) ^{2n+3}}dx.  \label{inteq1}
\end{equation}%
It is known by Dana-Picard \cite{DANA} that the value of Expression $\left( 
\text{\ref{inteq1}}\right) $ is the $n$th Catalan number $c_{n}$ \seqnum{A000108} \cite{SP1}. Using Maple\ to integrate Expression $\left( \text{\ref%
{inteq1}}\right) $ gives%
\begin{equation*}
\int_{0}^{1}\frac{2^{2n+5}\left( x^{2}\left( 1-x^{2}\right) ^{2n}\right) }{%
\pi \left( 1+x^{2}\right) ^{2n+3}}dx=\frac{\Gamma \left( n+1/2\right) 4^{n}}{%
\sqrt{\pi }\Gamma \left( n+2\right) }
\end{equation*}%
where by using properties of the gamma function the right side simplifies to 
$c_{n}$. Consequently,%
\begin{equation}
\frac{c_{n}}{D\left( n\right) }=\left( \frac{\left( 2n\right) !}{\left(
n+1\right) n!^{2}}\right) \left( \frac{2^{2n+4}\left( n+1\right) !^{2}}{%
\sqrt{\pi }\Gamma \left( 2n+5/2\right) }\right) =\frac{2^{2n+4}\left(
n+1\right) \Gamma \left( 2n+1\right) }{\sqrt{\pi }\Gamma \left(
2n+5/2\right) }.  \label{inteq0}
\end{equation}%
Thus, we find that%
\begin{equation}
\int_{0}^{1}\frac{2^{2n+5}\left( x^{2}\left( 1-x^{2}\right) ^{2n}\right) }{%
\pi \left( 1+x^{2}\right) ^{2n+3}}dx=\frac{2^{2n+4}\left( n+1\right) D\left(
n\right) \Gamma \left( 2n+1\right) }{\sqrt{\pi }\Gamma \left( 2n+5/2\right) }%
.  \label{inteq}
\end{equation}%
The right side of Equation $\left( \text{\ref{inteq}}\right) $ is indeed $%
c_{n}$. This can also be directly obtained by using Equation $\left( \text{%
\ref{Dneq1}}\right) $ and \ some simplification. Given Equation $\left( 
\text{\ref{Dneq1}}\right) $ and rearranging and simplifying the right hand
side of Equation $\left( \text{\ref{inteq}}\right) $ leads to the following
simple identity connecting the Catalan numbers \seqnum{A000108} \cite{SP1}
and the gamma function. For $n\in \mathbb{N}_{0}$%
\begin{equation*}
\text{ }c_{n}=\frac{\Gamma \left( 2n+1\right) }{\Gamma \left( n+1\right)
\Gamma \left( n+2\right) }\text{.}
\end{equation*}

We now note here that by using the Wolfram Alpha Widget: Definite Integral
Calculator \cite{WIG} to integrate Expression $\left( \text{\ref{inteq1}}%
\right) $ gives%
\begin{equation}
\frac{2^{2n+5}n\left( n+1\right) D\left( n\right) \Gamma \left( 2n\right) }{%
\sqrt{\pi }}\text{, if }\Real\left( n\right) &gt;-1/2,  \label{inteq2}
\end{equation}%
which does not simplify to $c_{n}.$ As a consequence of this, we observe
from Equation $\left( \text{\ref{inteq}}\right) $ that the Wolfram
calculator does not include $\Gamma \left( 2n+5/2\right) $ in the
denominator of Expression $\left( \text{\ref{inteq2}}\right) $. This has
been reported to Wolfram by the authors. We have been informed by Wolfram
that the appropriate computer technicians are investigating this issue with
their calculator.

We now rearrange the right side of Equation $\left( \text{\ref{inteq}}
\right) $ as Equation $\left( \text{\ref{id1}}\right) $ in the theorem
below. This theorem leads to curious relations where the Catalan numbers
\seqnum{A000108} \cite{SP1} are expressed in terms of $\pi $,
hypergeometric functions, and the gamma function. The proof of the theorem
follows as a result of the WZ algorithm.

\begin{theorem}
\label{Thm1}For $n\in \mathbb{N}_{0},$ 
\begin{equation}
\sum_{k=1}^{\infty }\frac{2^{6n+6}\left( -\frac{1}{2}\right) _{k}}{\left(
4n+3\right) \left( 2n+\frac{5}{2}\right) _{k}}\binom{4n+2}{2n+2}^{-1}\binom{%
2n+k+1}{k-1}(-1)^{k}=\pi c_{n}.  \label{id1}
\end{equation}
\end{theorem}

\begin{proof}
Divide both sides of Equation $\left( \text{\ref{id1}}\right) $ by $\pi
c_{n} $ and let $H\left( n\right) $ be the resulting left hand side and $%
F\left( n,k\right) $ be its summand, that is,

\begin{equation*}
F\left( n,k\right) =\frac{2^{6n+6}\left( -\frac{1}{2}\right) _{k}}{\pi
c_{n}\left( 4n+3\right) \left( 2n+\frac{5}{2}\right) _{k}}\binom{4n+2}{2n+2}%
^{-1}\binom{2n+k+1}{k-1}(-1)^{k}.
\end{equation*}%
Then proving Equation $\left( \text{\ref{id1}}\right) $ is equivalent to
showing $H\left( n\right) =1$ for $n\in \mathbb{N}_{0}$. Now, applying the
WZ algorithm on $F\left( n,k\right) $ we get 
\begin{equation}
F\left( n+1,k\right) -F\left( n,k\right) =G\left( n,k+1\right) -G\left(
n,k\right)  \label{WZ-eq}
\end{equation}%
where $G\left( n,k\right) =F\left( n,k\right) R\left( n,k\right) $ and%
\begin{equation*}
R\left( n,k\right) =-\frac{k-1}{(2n+3)(5+2k+4n)}.
\end{equation*}%
Summing both sides of Equation $\left( \text{\ref{WZ-eq}}\right) $ for all
values of $k$ we get%
\begin{equation*}
H\left( n+1\right) -H\left( n\right) =0.
\end{equation*}%
Since $H\left( 0\right) =1$,we have $H\left( n\right) =1$ for $n\in \mathbb{N%
}_{0}$.
\end{proof}

\begin{corollary}
For $n\in \mathbb{N}_{0}$%
\begin{equation*}
\frac{1}{c_{n}}\sum_{k=1}^{\infty }\frac{2^{6n+6}\left( -\frac{1}{2}\right)
_{k}}{\left( 4n+3\right) \left( 2n+\frac{5}{2}\right) _{k}}\binom{4n+2}{2n+2}%
^{-1}\binom{2n+k+1}{k-1}(-1)^{k}=\pi .
\end{equation*}
\end{corollary}

\begin{corollary}
For $n\in \mathbb{N}_{0}$%
\begin{equation*}
\frac{1}{\pi }\sum_{k=1}^{\infty }\frac{2^{6n+6}\left( -\frac{1}{2}\right)
_{k}}{\left( 4n+3\right) \left( 2n+\frac{5}{2}\right) _{k}}\binom{4n+2}{2n+2}%
^{-1}\binom{2n+k+1}{k-1}(-1)^{k}=c_{n}.
\end{equation*}
\end{corollary}

Theorem \ref{Thm1} can be rearranged and written in terms of the $\pi $, $%
c_{n}$, and $D\left( n\right) $ as the following identity.

\begin{proposition}
\label{prop3}For $n\in \mathbb{N}_{0}$%
\begin{equation*}
\frac{2^{6n+6}D\left( n\right) }{\left( 4n+3\right) \pi c_{n}}=\binom{4n+2}{%
2n+2}=\frac{4n+1}{n+1}\binom{4n}{2n}.
\end{equation*}
\end{proposition}

\begin{proof}
Using Equations $\left( \text{\ref{cateq1}}\right) $ and $\left( \text{\ref%
{Dneq1}}\right) $, $\Gamma \left( 1/2\right) =\sqrt{\pi }$ and some
simplification gives%
\begin{eqnarray}
\frac{2^{6n+6}D\left( n\right) }{\left( 4n+3\right) \pi c_{n}} &amp;=&amp;\frac{%
2^{6n+6}}{\left( 4n+3\right) \pi }\cdot \frac{\left( n+1\right) n!n!}{\left(
2n\right) !}\cdot \frac{\sqrt{\pi }\Gamma \left( 2n+5/2\right) }{%
2^{2n+4}\left( n+1\right) !^{2}}  \label{gamaeq2} \\
&amp;=&amp;\frac{\Gamma \left( 2n+5/2\right) }{\sqrt{\pi }}\cdot \frac{2^{4n+2}}{%
\left( 4n+3\right) \left( n+1\right) \left( 2n\right) !}  \notag \\
&amp;=&amp;\frac{\Gamma \left( 2n+5/2\right) }{\Gamma \left( 1/2\right) }\cdot \frac{%
2^{4n+2}}{\left( 4n+3\right) \left( n+1\right) \left( 2n\right) !}.  \notag
\end{eqnarray}%
Rewriting%
$\Gamma \left( 2n+5/2\right)$  as 
$\Gamma \left( 2\left( n+1\right) +1/2\right)$,
applying Equation $\left( \text{\ref{gamaeq}}\right)$, and using 
\begin{equation*}
\frac{\Gamma \left( n+1/2\right) }{\Gamma \left( 1/2\right) }=(1/2)_{n}=%
\frac{\left( 2n\right) !}{n!4^{n}}
\end{equation*}%
gives 
\begin{equation}
\frac{\Gamma \left( 2n+5/2\right) }{\Gamma \left( 1/2\right) }=\frac{\Gamma
\left( 2\left( n+1\right) +1/2\right) }{\Gamma \left( 1/2\right) }=\left(
1/2\right) _{2n+2}=\frac{\left( 4n+4\right) !}{\left( 2n+2\right) !4^{2n+2}}.
\label{gamaeq1}
\end{equation}%
Substituting Equation $\left( \text{\ref{gamaeq1}}\right) $ into Equation $%
\left( \text{\ref{gamaeq2}}\right) $ yields%
\begin{equation*}
\frac{2^{6n+6}D\left( n\right) }{\left( 4n+3\right) \pi c_{n}}=\frac{\left(
4n+4\right) !}{\left( 2n+2\right) !4^{2n+2}}\cdot \frac{2^{4n+2}}{\left(
4n+3\right) \left( n+1\right) \left( 2n\right) !}.
\end{equation*}%
Now apply properties of factorials and binomial coefficients and some
simplification we obtain%
\begin{equation*}
\frac{2^{6n+6}D\left( n\right) }{\left( 4n+3\right) \pi c_{n}}=\frac{\left(
4n+2\right) !}{\left( 2n+2\right) !\left( 2n\right) !}=\binom{4n+2}{2n+2},
\end{equation*}%
which proves the result.
\end{proof}

We note that Equation $\left( \text{\ref{gamaeq1}}\right) $ gives an exact
formula for $\Gamma \left( 2n+5/2\right) .$ This can also be obtained by
direct computation by using Proposition \ref{prop3} and Equation $\left( 
\text{\ref{Dneq1}}\right) .$

\section{Conclusion}

We now give some concluding comments. The binomial, Catalan, and Fine
generating functions are expressed in terms of the Fibonacci generating
function by Proposition \ref{Fibfg1}. This result may lead to new properties
of these generating functions, which in turn may lead to new properties of
the counting numbers associated the generating functions. The recurrence
relations given by Propositions \ref{prop1} and \ref{prop2} are open for
combinatorial interpretations. In particular, finding an RNA interpretation
of Proposition \ref{prop1} would be of interest to those who study discrete
and combinatorial mathematical biology. The Wolfram calculator anomaly given
by Expression $\left( \text{\ref{inteq2}}\right) $ is what sparked the
authors' interest in looking at relations among hypergeometric functions, the
gamma function, and the Catalan numbers \seqnum{A000108} \cite{SP1}. This
generated further investigation and analysis of $D\left( n\right) $, which
subsequently led to the derivation of Equation $\left( \text{\ref{inteq}}%
\right) .$ As a result of the right side of Equation $\left( \text{\ref%
{inteq}}\right) $ and using the Wilf-Zeilberger algorithm we were able to
find other representations of the irrational number $\pi $ and the
ubiquitous Catalan numbers $c_{n}$ \seqnum{A000108} \cite{SP1}. Thus,
Theorem \ref{Thm1}, which is the main result of the paper, demonstrates the
roles in which symbolic computation, numerical computation, and experimental
mathematics play in the discovery of new mathematical results.

\section{Acknowledgement}

The authors would like to thank the anonymous referee for providing useful
comments and suggestions. The authors would also like to thank Guoping Zhang
for useful comments and discussions on early drafts of the manuscript.

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\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 05A15; Secondary 05A19.

\noindent \emph{Keywords: } 
WZ algorithm, WZ method, Riordan array, gamma function, generating function, Catalan number.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000012},
\seqnum{A000027},
\seqnum{A000045},
\seqnum{A000108},
\seqnum{A000957},
\seqnum{A001477},
\seqnum{A004148},
\seqnum{A007318},
\seqnum{A039598},
\seqnum{A053121},
\seqnum{A097724}, and
\seqnum{A111418}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  July 14 2013;
revised version received   October 6 2013.
Published in {\it Journal of Integer Sequences}, October 13 2013.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


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