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\begin{center}
\vskip 1cm{\LARGE\bf 
Two Catalan-type Riordan Arrays and their \\
\vskip .03in
Connections to the Chebyshev Polynomials of \\
\vskip .10in
the First Kind
}
\vskip 1cm
\large
Asamoah Nkwanta and Earl R. Barnes\\
Department of Mathematics\\
Morgan State University\\
Baltimore, MD 21251\\
USA\\
\href{mailto:Asamoah.Nkwanta@morgan.edu}{\tt Asamoah.Nkwanta@morgan.edu} \\
\href{mailto:Earl.Barnes@morgan.edu}{\tt Earl.Barnes@morgan.edu} \\
\end{center}

\vskip .2 in

\begin{abstract}
Riordan matrix methods and properties of generating functions are used to
prove that the entries of two Catalan-type Riordan arrays are linked to the
Chebyshev polynomials of the first kind. The connections are that the rows
of the arrays are used to expand the monomials $\left( 1/2\right) \left(
2x\right) ^{n}$ and $\left( 1/2\right) \left( 4x\right) ^{n}$ in terms of
certain Chebyshev polynomials of degree $n$. In addition, we find new
integral representations of the central binomial coefficients and Catalan
numbers.
\end{abstract}

\section{Introduction}

The Chebyshev polynomials have applications in numerical analysis and
approximation theory, Fourier series, combinatorics, and other areas of
mathematics. Riordan arrays are important for proving combinatorial
identities and sums, and have applications in combinatorics and graph
theory, combinatorial number theory, algebra, and special functions. Riordan
array connections to the Chebyshev polynomials of the first kind have been
given by Barry \cite{PBAR1}, Barry and Hennessy \cite{PBAR2}, Luzon and
Moron \cite{LUZ}, and others. In this paper, we establish that the entries
of two special Riordan matrices, we call Catalan-type Riordan arrays, are
linked to the Chebyshev polynomials of the first kind.

The following two infinite lower-triangular arrays, respectively denoted by $%
\mathbf{A}$ and $\mathbf{B}$, are proved by Riordan matrix multiplication to
be linked to the Chebyshev polynomials of the first kind: 
\begin{equation}
\mathbf{A}=\left( 
\begin{array}{cccccc}
1 & 0 & 0 & \cdots & 0 & \cdots \\ 
2 & 1 & 0 & \cdots & 0 & \cdots \\ 
6 & 4 & 1 & \cdots & 0 & \cdots \\ 
20 & 15 & 6 & \ddots & 0 & \cdots \\ 
70 & 56 & 28 & \cdots & 1 & \cdots \\ 
\vdots & \vdots & \vdots & \cdots & \vdots & \ddots \\ 
\binom{2n}{n} & \binom{2n}{n-1} & \binom{2n}{n-2} & \cdots & \binom{2n}{n-k}
& \cdots \\ 
\vdots & \vdots & \vdots & \cdots & \vdots & \cdots%
\end{array}%
\right)  \label{array0}
\end{equation}

and%
\begin{equation}
\mathbf{B}=\left( 
\begin{array}{cccccc}
1 & 0 & 0 & \cdots & 0 & \cdots \\ 
3 & 1 & 0 & \cdots & 0 & \cdots \\ 
10 & 5 & 1 & \cdots & 0 & \cdots \\ 
35 & 21 & 7 & \ddots & 0 & \cdots \\ 
126 & 84 & 36 & \cdots & 1 & \cdots \\ 
\vdots & \vdots & \vdots & \cdots & \vdots & \ddots \\ 
\binom{2n+1}{n} & \binom{2n+1}{n-1} & \binom{2n+1}{n-2} & \cdots & \binom{%
2n+1}{n-k} & \cdots \\ 
\vdots & \vdots & \vdots & \cdots & \vdots & \vdots%
\end{array}%
\right) .  \label{array11}
\end{equation}%
The connections are that the rows of the arrays are used to expand the
monomials $\left( 1/2\right) \left( 2x\right) ^{n}$ and $\left( 1/2\right)
\left( 4x\right) ^{n}$ in terms of certain Chebyshev polynomials of degree $%
n $. Finding rows of other infinite lower-triangular arrays that have
similar connections to the Chebyshev polynomials or other special functions
would be of interest. Before proving the connections for $\mathbf{A}$ and $%
\mathbf{B}$, we mention some properties of the Chebyshev polynomials of the
first kind.

The Chebyshev polynomials of the first kind, denoted by $T_{n}\left(
x\right) $, are defined by the relation 
\begin{equation*}
T_{n}\left( x\right) =\cos n\theta \text{ where }x=\cos \theta ,\text{ }%
-1\leq x\leq 1
\end{equation*}%
and given by the recursion%
\begin{equation*}
T_{n+1}\left( x\right) =2xT_{n}\left( x\right) -T_{n-1}\left( x\right) \text{
}\left( n\geq 1\right)
\end{equation*}%
with initial conditions $T_{0}\left( x\right) =1$ and $T_{1}\left( x\right)
=x$. The first few higher order polynomials are 
\begin{equation*}
\begin{array}{ccc}
T_{2}\left( x\right) & = & 2x^{2}-1 \\ 
T_{3}\left( x\right) & = & 4x^{3}-3x \\ 
T_{4}\left( x\right) & = & 8x^{4}-8x^{2}+1.%
\end{array}%
\end{equation*}%
The bivariate generating function of $T_{n}\left( x\right) $, denoted by $%
T\left( x,z\right) ,$ is defined by 
\begin{equation*}
T\left( x,z\right) :=\sum_{n\geq 0}T_{n}\left( x\right) z^{n}=\frac{1-xz}{%
1+z^{2}-2xz}.
\end{equation*}%
The shifted Chebyshev polynomials of the first kind, denoted by $T_{n}^{\ast
}\left( x\right) $, are defined by 
\begin{equation*}
T_{n}^{\ast }\left( x\right) :=T_{n}\left( 2x-1\right) ,\text{ }0\leq x\leq 1
\end{equation*}%
and given by the recursion%
\begin{equation*}
T_{n+1}^{\ast }\left( x\right) =2\left( 2x-1\right) T_{n}^{\ast }\left(
x\right) -T_{n-1}^{\ast }\left( x\right) \text{ }\left( n\geq 1\right)
\end{equation*}%
with initial conditions $T_{0}^{\ast }\left( x\right) =1$ and $T_{1}^{\ast
}\left( x\right) =2x-1$. The first few higher order shifted polynomials are 
\begin{equation*}
\begin{array}{ccc}
T_{2}^{\ast }\left( x\right) & = & 8x^{2}-8x+1 \\ 
T_{3}^{\ast }\left( x\right) & = & 32x^{3}-48x^{2}+18x-1 \\ 
T_{4}^{\ast }\left( x\right) & = & 128x^{4}-256x^{3}+160x^{2}-32x+1.%
\end{array}%
\end{equation*}%
The bivariate generating function of $T_{n}^{\ast }\left( x\right) $,
denoted by $T^{\ast }\left( x,z\right) ,$ is defined by 
\begin{equation*}
T^{\ast }\left( x,z\right) :=\sum_{n\geq 0}T_{n}^{\ast }\left( x\right)
z^{n}=\frac{1-2xz+z}{1+2z+z^{2}-4xz}.
\end{equation*}

\begin{remark}
By the property $T_{n}^{\ast }\left( x^{2}\right) =T_{2n}\left( x\right) ,$
one can show that%
\begin{equation*}
T^{\ast }\left( x^{2},z\right) =E\left( x,z\right)
\end{equation*}%
where $E\left( x,z\right) $ denotes the generating function of the even
coefficients of $T\left( x,z\right) $, see Equation $\left( \text{\ref{even0}%
}\right) $.
\end{remark}

More information on properties of the Chebyshev polynomials can be found in
Mason and Handscomb \cite{MAS}, Lanczos \cite{LAN}, Rivlin \cite{RIV}, and
Gradshteyn and Ryzhik \cite{GRAD}.

We now show the connections for a few rows of each array. The arrays $%
\mathbf{A}$ and $\mathbf{B}$ are linked to the Chebyshev polynomials of the
first kind as a result of the even and odd coefficients of $T\left(
x,z\right) $ occurring in the expansion of $\left( 1/2\right) \left(
2x\right) ^{n}$ (see Lanczos \cite{LAN}, Table VI). The connections are
illustrated by, respectively, multiplying $\mathbf{A}$ and $\mathbf{B}$ by
the column vectors containing the entries of the sequences of even
coefficients $\left\{ T_{0}\left( x\right) ,T_{2}\left( x\right) ,\ldots
\right\} $ and odd coefficients $\left\{ T_{1}\left( x\right) ,T_{3}\left(
x\right) ,\ldots \right\} $ of $T\left( x,z\right) $. We denote $%
T_{2n}\left( x\right) $ by $T_{2n}$ and make the adjustment $T_{0}=1/2$, for
the even coefficients. Then, for the first four rows of the product we get
the following column vector%
\begin{equation*}
\left( 
\begin{array}{cccc}
1 & 0 & 0 & 0 \\ 
2 & 1 & 0 & 0 \\ 
6 & 4 & 1 & 0 \\ 
20 & 15 & 6 & 1%
\end{array}%
\right) \left( 
\begin{array}{c}
T_{0} \\ 
T_{2} \\ 
T_{4} \\ 
T_{6}%
\end{array}%
\right) =\left( 
\begin{array}{c}
T_{0} \\ 
2T_{0}+T_{2} \\ 
6T_{0}+4T_{2}+T_{4} \\ 
20T_{0}+15T_{2}+6T_{4}+T_{6}%
\end{array}%
\right) =\left( 
\begin{array}{c}
1/2 \\ 
2x^{2} \\ 
8x^{4} \\ 
32x^{6}%
\end{array}%
\right) .
\end{equation*}%
We write this vector as%
\begin{equation*}
\left( 1/2\right) \left( \left( 2x\right) ^{0}\text{ }\left( 2x\right) ^{2}%
\text{ }\left( 2x\right) ^{4}\text{ }\left( 2x\right) ^{6}\right) ^{T}.
\end{equation*}%
For the odd coefficients we denote $T_{2n+1}\left( x\right) $ by $T_{2n+1}$.
Then, for the first four rows of the product in this case we get the
following column vector%
\begin{equation*}
\left( 
\begin{array}{cccc}
1 & 0 & 0 & 0 \\ 
3 & 1 & 0 & 0 \\ 
10 & 5 & 1 & 0 \\ 
35 & 21 & 7 & 1%
\end{array}%
\right) \left( 
\begin{array}{c}
T_{1} \\ 
T_{3} \\ 
T_{5} \\ 
T_{7}%
\end{array}%
\right) =\left( 
\begin{array}{c}
T_{1} \\ 
3T_{1}+T_{3} \\ 
10T_{1}+5T_{3}+T_{5} \\ 
35T_{1}+21T_{3}+7T_{5}+T_{7}%
\end{array}%
\right) =\left( 
\begin{array}{c}
x \\ 
4x^{3} \\ 
16x^{5} \\ 
64x^{7}%
\end{array}%
\right) .
\end{equation*}%
We write this vector as%
\begin{equation*}
\left( 1/2\right) \left( \left( 2x\right) ^{1}\text{ }\left( 2x\right) ^{3}%
\text{ }\left( 2x\right) ^{5}\text{ }\left( 2x\right) ^{7}\right) ^{T}.
\end{equation*}%
Thus, it appears that the column vector whose entries are $\left( 1/2\right)
\left( 2x\right) ^{n}$ $\left( n\geq 0\right) $ can be expressed as linear
combinations of the columns of $\mathbf{A}$ and $\mathbf{B}$ where $%
T_{n}\left( x\right) $ are the coefficients of the linear combinations.

The shifted Chebyshev polynomials of the first kind are linked to $\mathbf{A}
$ as a result of the sequence of coefficients $\left\{ T_{0}^{\ast }\left(
x\right) ,T_{1}^{\ast }\left( x\right) ,\ldots \right\} $ of $T^{\ast
}\left( x,z\right) $ occurring in the expansion of $\left( 1/2\right) \left(
4x\right) ^{n}$ (see Lanczos \cite{LAN}, Table VIII). This connection is
illustrated by multiplying the array by the column vector containing the
coefficients of $T^{\ast }\left( x,z\right) $. Here we denote $T_{n}^{\ast
}\left( x\right) $ by $T_{n}^{\ast }$ and make the adjustment $T_{0}^{\ast
}=1/2$. Then, for the first four rows of the product we get%
\begin{equation*}
\left( 
\begin{array}{cccc}
1 & 0 & 0 & 0 \\ 
2 & 1 & 0 & 0 \\ 
6 & 4 & 1 & 0 \\ 
20 & 15 & 6 & 1%
\end{array}%
\right) \left( 
\begin{array}{c}
T_{0}^{\ast } \\ 
T_{1}^{\ast } \\ 
T_{2}^{\ast } \\ 
T_{3}^{\ast }%
\end{array}%
\right) =\left( 
\begin{array}{c}
T_{0}^{\ast } \\ 
2T_{0}^{\ast }+T_{1}^{\ast } \\ 
6T_{0}^{\ast }+4T_{1}^{\ast }+T_{2}^{\ast } \\ 
20T_{0}^{\ast }+15T_{1}^{\ast }+6T_{2}^{\ast }+T_{3}^{\ast }%
\end{array}%
\right) =\left( 
\begin{array}{c}
1/2 \\ 
2x \\ 
8x^{2} \\ 
32x^{3}%
\end{array}%
\right) .
\end{equation*}%
We write this vector as%
\begin{equation*}
\left( 1/2\right) \left( \left( 4x\right) ^{0}\text{ }\left( 4x\right) ^{1}%
\text{ }\left( 4x\right) ^{2}\text{ }\left( 4x\right) ^{3}\right) ^{T}.
\end{equation*}%
Thus, it appears that the column vector whose entries are $\left( 1/2\right)
\left( 4x\right) ^{n}$ $\left( n\geq 0\right) $ can be expressed as a linear
combination of the columns of $\mathbf{A}$ where $T_{n}^{\ast }\left(
x\right) $ are the coefficients of the linear combination. Note, when using
the shifted Chebyshev polynomials we do not distinguish between the even and
odd coefficients of $T^{\ast }\left( x,z\right) $.

The above illustrations suggest that the Chebyshev connections can be
established by matrix multiplication. The connections are mentioned by
Lanczos \cite{LAN} and a trigonometric proof is given by expanding $x^{n}$
in terms of the shifted Chebyshev polynomials $T_{k}^{\ast }\left( x\right) $
for $0\leq k\leq n$. We derive these expansions as well as the expansions of 
$x^{2n}$ and $x^{2n+1}.$ The monomial $x^{2n}$ is expanded in terms of $%
T_{k}\left( x\right) $ for $0\leq k\leq 2n$ and $x^{2n+1}$ in terms of $%
T_{k}\left( x\right) $ for $0\leq k\leq 2n+1.$ The approach we follow to
prove the connections makes use of Riordan matrix methods and properties of
generating functions.

As a by-product of the connections, we derive integral representations of
the central binomial coefficients \seqnum{A00984} \cite{SP1} and Catalan
numbers \seqnum{A000108} \cite{SP1} (see Equations $\left( \text{\ref%
{iega1}}\right) $, $\left( \text{\ref{iega2}}\right) $, $\left( \text{\ref%
{iega23}}\right) $, $\left( \text{\ref{iega3}}\right) $, $\left( \text{\ref%
{iega4}}\right) $, $\left( \text{\ref{iega5}}\right) $, $\left( \text{\ref%
{iega6}}\right) $, and $\left( \text{\ref{iega7}}\right) $). The integrals
do not seem to explicitly appear in Gradshteyn and Ryzhik \cite{GRAD}. They
do not appear in Stanley's textbook \cite{STAN,STAN1} that contains
a large number of combinatorial and analytical interpretations of the
Catalan numbers, nor in Koshy's book \cite{KOSHY} that contains a wide
variety of applications of the Catalan numbers. In addition, the
representations we give differ from the representations given by Penson and
Sixdeniers \cite{PEN}, Sofo \cite{SOFO}, Dana-Picard \cite{DP1}, \cite{DP2},
Aigner \cite{AIGNER}, and Yuan \cite{YUAN1}, \cite{YUAN}. Thus, it appears
that the\ integral representations are new.

This paper is arranged as follows. The definition of a Riordan matrix is
given in Section 2. We show that $\mathbf{A}$ and $\mathbf{B}$ are
Catalan-type Riordan arrays in Section 3. The inverses of $\mathbf{A}$ and $%
\mathbf{B}$ are also mentioned in this section and are linked to certain
families of monic orthogonal polynomials. In addition, a combinatorial
interpretation of $\mathbf{B}$ is briefly mentioned. We prove that $\mathbf{A%
}$ and $\mathbf{B}$ are linked to the Chebyshev polynomials of the first
kind in Section 4. Theorems \ref{rthm3} and \ref{rthm4} of this section are
the main results of this paper. The integral representations of the central
binomial coefficients and Catalan numbers are given in Section 5. We then
conclude this paper by suggesting possible problems for future research in
Section 6. The material in Sections 1, 2 and 3 is known in the areas of
numerical approximation theory, enumerative combinatorics and Riordan array
theory, and special functions. We make the paper self contained and the
results in Sections 4 and 5 accessible to a wider audience by including this
material. An interesting feature of this paper is that it brings together
ideas from several areas of mathematics.

\section{Riordan Matrix}

Let $%
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
$ denote the natural numbers (including $0$) and $\mathbb{C}$ the complex
numbers. Then, an infinite matrix$\ L=$ $\left( \ell _{n,k}\right) _{n,k\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
}$ with entries in $\mathbb{C}$ is called a \textit{Riordan matrix} if the $%
kth$ column satisfies 
\begin{equation*}
\sum_{n\geq 0}\ell _{n,k}z^{n}:=g\left( z\right) \left( f\left( z\right)
\right) ^{k}
\end{equation*}%
where%
\begin{equation*}
g\left( z\right) =1+g_{1}z+g_{2}z^{2}+\cdots ,\text{ and }f\left( z\right)
=f_{1}z+f_{2}z^{2}+f_{3}z^{3}+\cdots 
\end{equation*}%
belong to the ring of formal power series $\mathbb{C}\left[ \left[ z\right] %
\right] $ and $f_{1}\neq 0$. Note that $g_{0}=1$ is for convenience and not
a necessary condition for the definition. A formal power series of the form 
\begin{equation*}
b\left( z\right) =b_{0}+b_{1}z+b_{2}z^{2}+\cdots =\sum_{n\geq 0}b_{n}z^{n}
\end{equation*}%
where $z$ is an indeterminate is called the \textit{ordinary generating
function} of the sequence $\left\{ b_{n}\right\} $. Riordan matrices are
typically denoted by pairs of generating functions as $L=\left( g\left(
z\right) ,f\left( z\right) \right) $. The matrices $L$ can be defined by
either ordinary or exponential generating functions. However, the matrices
presented in this paper are defined by ordinary generating functions. We
note here that the proper Riordan arrays given by Sprugnoli \cite{SPRU1} are
what we call Riordan arrays or matrices.

Two important results for multiplying Riordan matrices are now given.

\begin{theorem}
\label{rtm1}$\left( \text{\cite{SGWW,NKLWS}}\right) $ If $L=\left(
\ell _{n,k}\right) _{n,k\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
}=\left( g\left( z\right) ,f\left( z\right) \right) $ is a Riordan matrix
and $h\left( z\right) $ is the generating function of the sequence
associated with the entries of the column vector $h=\left( h_{k}\right)
_{k\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
}$, then the product of $L$ and $h\left( z\right) ,$ defined by $L\otimes
h\left( z\right) =g\left( z\right) h\left( f\left( z\right) \right) ,$ is
the generating function of the sequence associated with the entries of the
column vector $\left( \sum_{k=0}^{n}\ell _{n,k}h_{k}\right) _{n\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
}.$
\end{theorem}

Let $L\ast N$, or by simple juxtaposition $LN$, denote the row-by-column
product of two Riordan matrices $L$ and $N$.

\begin{theorem}
\label{rtm2}$\left( \text{\cite{SGWW,SPRU1}}\right) $ If%
\begin{equation*}
L=\left( \ell _{n,k}\right) _{n,k\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
}=\left( g\left( z\right) ,f\left( z\right) \right) \text{ and }N=\left(
v_{n,k}\right) _{n,k\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
}=\left( h\left( z\right) ,l\left( z\right) \right) \text{ }
\end{equation*}%
are Riordan matrices, then $L\ast N$ is%
\begin{equation*}
L\ast N=\left( \sum_{j=0}^{n}\ell _{n,j}v_{j,k}\right) _{n,k\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
}=\left( g\left( z\right) h\left( f\left( z\right) \right) ,l\left( f\left(
z\right) \right) \right) ,
\end{equation*}%
and the set $\mathbf{R}$ of all Riordan matrices is a group under the
operation of matrix multiplication denoted by $\left( \mathbf{R},\ast
\right) $.
\end{theorem}

The notation $\left( \mathbf{R},\ast \right) $ denotes the \textit{Riordan
group}. The identity element of the group is $e=\left( 1,z\right) $. This is
the usual unit diagonal matrix. The inverse element of the group is%
\begin{equation}
L^{-1}=\left( g\left( z\right) ,f\left( z\right) \right) ^{-1}=\left(
1/g\left( \overline{f}\left( z\right) \right) ,\overline{f}\left( z\right)
\right)  \label{inveq}
\end{equation}%
where $\overline{f}\left( z\right) $ is the compositional inverse of $%
f\left( z\right) .$

See Shapiro et al. \cite{SGWW} and Sprugnoli \cite{SPRU1} for more
information on the group and Riordan matrices.

\section{Catalan Type Riordan Matrices and Chebyshev Polynomials}

Riordan matrices were first introduced by Shapiro et al. in 1991 \cite{SGWW}%
.. It turns out that the arrays $\mathbf{A}$ and $\mathbf{B}$ given in $1956$
by Lanczos \cite{LAN} are the Riordan matrices (\seqnum{A094527} \cite%
{SP1}) and (\seqnum{A111418} \cite{SP1}), respectively. Array $\mathbf{B}$
can be obtained by multiplying the matrices%
\begin{equation}
\mathbf{CE}=\left( 
\begin{array}{ccccc}
1 & 0 & 0 & 0 & \cdots \\ 
2 & 1 & 0 & 0 & \cdots \\ 
5 & 4 & 1 & 0 & \cdots \\ 
14 & 14 & 6 & 1 & \cdots \\ 
\vdots & \vdots & \vdots & \vdots & \ddots%
\end{array}%
\right) \left( 
\begin{array}{ccccc}
1 & 0 & 0 & 0 & \cdots \\ 
1 & 1 & 0 & 0 & \cdots \\ 
1 & 1 & 1 & 0 & \cdots \\ 
1 & 1 & 1 & 1 & \cdots \\ 
\vdots & \vdots & \vdots & \vdots & \ddots%
\end{array}%
\right)  \label{array2}
\end{equation}%
where $\mathbf{C}$ denotes the Shapiro-Catalan matrix (\seqnum{A039598} 
\cite{SP1,SHA}, and $\mathbf{E}$ the matrix with all 1's on and
below the main diagonal and 0's everywhere else (\seqnum{A000012} \cite%
{SP1}). Note, the Chebyshev connection to $\mathbf{B}$ where $\mathbf{B}$ is
a Riordan matrix is mentioned in Nkwanta \cite{NKW2} but not proved. Array $%
\mathbf{A}$ can be obtained by multiplying the matrices%
\begin{equation}
\mathbf{PT}=\left( 
\begin{array}{ccccc}
1 & 0 & 0 & 0 & \cdots \\ 
1 & 1 & 0 & 0 & \cdots \\ 
1 & 2 & 1 & 0 & \cdots \\ 
1 & 3 & 3 & 1 & \cdots \\ 
\vdots & \vdots & \vdots & \vdots & \ddots%
\end{array}%
\right) \left( 
\begin{array}{ccccc}
1 & 0 & 0 & 0 & \cdots \\ 
1 & 1 & 0 & 0 & \cdots \\ 
3 & 2 & 1 & 0 & \cdots \\ 
7 & 6 & 3 & 1 & \cdots \\ 
\vdots & \vdots & \vdots & \vdots & \ddots%
\end{array}%
\right)  \label{array3}
\end{equation}%
where $\mathbf{P}$ denotes the Pascal Triangle (\seqnum{A007318} \cite%
{SP1}) written in lower triangular matrix form and $\mathbf{T}$ the matrix (%
\seqnum{A094531} \cite{SP1}) whose leftmost column entries record the
count of the number of king walks down a chessboard \cite{GTSH,SGWW}. The two matrix products are confirmed below by Proposition \ref{maxprod}.

\begin{definition}
The \textit{Catalan generating function, denoted by }$c\left( z\right) $, is
defined as%
\begin{equation}
c\left( z\right) :=\left( 1-\sqrt{1-4z}\right) /2z=\sum_{n\geq 0}c_{n}z^{n}%
\text{ }  \label{catgf}
\end{equation}%
where%
\begin{equation*}
c_{n}=\left( 1/\left( n+1\right) \right) \binom{2n}{n}
\end{equation*}%
denotes the $nth$ \textit{Catalan number }(\seqnum{A000108} \cite{SP1}).
\end{definition}

The following lemma involving $c\left( z\right) $ is useful for proving
Proposition \ref{maxprod}, Theorems \ref{rthm3} and \ref{rthm4}, and
Propositions \ref{Invprop} and \ref{relprop}.

\begin{lemma}
\label{catlemma}$\left( \text{\cite{DEU}}\right) $%
\begin{equation*}
\begin{array}{cccc}
1.\text{ }1/\sqrt{1-4z} & = & c\left( z\right) /\left( 1-zc^{2}\left(
z\right) \right) & =c\left( z\right) /\left( 2-c\left( z\right) \right) \\ 
&  &  &  \\ 
2.\text{ }zc^{2}\left( z\right) 
\begin{array}{cc}
& 
\end{array}
& = & c\left( z\right) -1. & 
\end{array}%
\end{equation*}
\end{lemma}

\begin{proposition}
\label{maxprod}$\left( \text{\cite{NKW}}\right) $, $\left( \text{\cite{NKW1}}%
\right) $, $\left( \text{\cite{SP1}}\right) $%
\begin{equation*}
\mathbf{P\ast T}=\mathbf{A}\text{ and }\mathbf{C\ast E}=\mathbf{B}.
\end{equation*}
\end{proposition}

\begin{proof}
(Sketch) The Riordan pair forms of $\mathbf{C}$ and $\mathbf{E}$ are $%
\mathbf{C}=\left( c^{2}\left( z\right) ,zc^{2}\left( z\right) \right) $ and $%
\mathbf{E}=\left( 1/\left( 1-z\right) ,z\right) $. Applying Theorem \ref%
{rtm2} and Lemma \ref{catlemma}(1), then the Riordan pair form of $\mathbf{B}
$ is%
\begin{equation}
\mathbf{B}=\left( c\left( z\right) /\sqrt{1-4z},zc^{2}\left( z\right)
\right) =\mathbf{C\ast E}.  \label{Beq}
\end{equation}%
The Riordan pair forms of $\mathbf{P}$ and $\mathbf{T}$ are $\mathbf{P}%
=\left( 1/\left( 1-z\right) ,z/\left( 1-z\right) \right) $ and%
\begin{equation*}
\mathbf{T}=\left( 1/\sqrt{1-2z-3z^{2}},\left( 1-z-\sqrt{1-2z-3z^{2}}\right)
/2z\right) \text{.}
\end{equation*}%
Applying Theorem \ref{rtm2}, then the Riordan pair form of $\mathbf{A}$ is%
\begin{equation}
\mathbf{A}=\left( 1/\sqrt{1-4z},c\left( z\right) -1\right) =\mathbf{P\ast T}.
\label{Aeq}
\end{equation}
\end{proof}

Thus, $\mathbf{A}$ and $\mathbf{B}$ are Catalan-type Riordan arrays since
they involve $c\left( z\right) $. Details of the computations involved in
the proofs are omitted and left for the reader as exercises and motivation
for Theorem \ref{rtm2}.

\begin{remark}
Arrays $\mathbf{A}$ and $\mathbf{B}$ can also be obtained from their
formation rules (or dot diagrams). A formation rule, denoted by $\left[ Z%
\mathbf{;}A\right] $ where $Z$ corresponds to the rule for obtaining the
leftmost or zeroth column entries and $A$ for obtaining all other column
entries, is a recurrence relation that defines the way entries of a Riordan
matrix are computed. The formation rules of $\ \mathbf{A}$ and $\mathbf{B}$
are, respectively, $\left[ 2,2;1,2,1\right] $ $\left( \text{\cite{SP1}}%
\right) $ and $\left[ 3,1;1,2,1\right] $ \cite{NKW,NKW1}.
See the cited references for more information on the
formation rules and dot diagrams of Riordan matrices.
\end{remark}

Arrays $\mathbf{A}$ and $\mathbf{B}$ are special cases of the generalized
Riordan array of the form%
\begin{equation*}
\mathbf{L=}\left( \frac{1-\lambda z-\mu z^{2}}{1+az+bz^{2}},\frac{z}{%
1+az+bz^{2}}\right)
\end{equation*}%
given by Barry and Hennessy \cite{PBAR2}. The array $\mathbf{L}$ is
associated with a certain tridiagonal matrix called a Stieltjes matrix and
has the special property that the entries of $\mathbf{L}^{-1}$ are monic
orthogonal polynomials whose moments are the entries of the leftmost column
of $\mathbf{L}$.

We now give the inverses of $\mathbf{A}$ and $\mathbf{B}$ and mention that
the entries of $\mathbf{A}^{-1}$ and $\mathbf{B}^{-1}$ are monic orthogonal
polynomials whose moments are certain binomial coefficients.

\begin{proposition}
\label{Invprop}The inverses of $\mathbf{A}$ and $\mathbf{B}$ are,
respectively,%
\begin{equation*}
\mathbf{A}^{-1}=\left( \left( 1-z\right) /\left( 1+z\right) ,z/\left(
1+z\right) ^{2}\right) =\left( 
\begin{array}{rrrrrr}
1 & 0 & 0 & 0 & 0 & ... \\ 
-2 & 1 & 0 & 0 & 0 & ... \\ 
2 & -4 & 1 & 0 & 0 & ... \\ 
-2 & 9 & -6 & 1 & 0 & ... \\ 
2 & -16 & 20 & -8 & 1 & ... \\ 
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots%
\end{array}%
\right)
\end{equation*}%
and%
\begin{equation*}
\mathbf{B}^{-1}=\left( \left( 1-z\right) /\left( 1+z\right) ^{2},z/\left(
1+z\right) ^{2}\right) =\left( 
\begin{array}{rrrrrr}
1 & 0 & 0 & 0 & 0 & ... \\ 
-3 & 1 & 0 & 0 & 0 & ... \\ 
5 & -5 & 1 & 0 & 0 & ... \\ 
-7 & 14 & -7 & 1 & 0 & ... \\ 
9 & -30 & 27 & -9 & 1 & ... \\ 
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots%
\end{array}%
\right) .
\end{equation*}
\end{proposition}

Details of the computations of the inverses are omitted and left for the
reader as exercises and motivation for Equation $\left( \text{\ref{inveq}}%
\right) $.

\begin{remark}
\label{reminv}Arrays $\mathbf{A}^{-1}$ and $\mathbf{B}^{-1}$ are,
respectively, the coefficient arrays of certain families of monic orthogonal
polynomials for which the central binomial coefficients%
\begin{equation*}
\left\{ \binom{2n}{n}\right\} _{n\geq 0}=\left\{ 1,2,6,20,70,252,\ldots
\right\} \text{ }(\text{\seqnum{A00984} \cite{SP1}})
\end{equation*}%
and binomial coefficients%
\begin{equation*}
\left\{ \binom{2n+1}{n+1}\right\} _{n\geq 0}=\left\{
1,3,10,35,126,462,\ldots \right\} \text{ }(\text{\seqnum{A001700} \cite%
{SP1}})
\end{equation*}%
are the moments associated with the polynomials.
\end{remark}

The results mentioned by Remark \ref{reminv} can be confirmed by
Propositions 9-10, 16 and 20 of Barry and Hennessy \cite{PBAR2}.

Arrays $\mathbf{A}$ and $\mathbf{B}$ are of combinatorial interest. $\mathbf{%
A}$ is of interest given that: 1) the columns involve the binomial
coefficients (see \seqnum{A000984}, \seqnum{A001791}, \seqnum{%
A002694} \cite{SP1} for the first few columns) and 2) the first column
contains the central binomial coefficients. $\mathbf{B}$ is of interest
given that: 1) the columns here also involve the binomial coefficients (see 
\seqnum{A001700}, \seqnum{A002054}, \seqnum{A003516}, and 
\seqnum{A030053} \cite{SP1} for the first few columns), 2) the first
column contains the sequence $\left\{ 1,5,21,84,330,\ldots \right\} $ (%
\seqnum{A002054} \cite{SP1}) that goes back to Cayley \cite{CAY}, and 3) $%
\mathbf{B}$ has a lattice path interpretation.

The entries of $\mathbf{B}$ count certain unit-step lattice paths of length $%
n$ and height $k$ in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
^{3}$ \cite{NKW}. The path step directions are illustrated below.%
\begin{eqnarray*}
&&\frame{$%
\begin{array}{lll}
\left( 0,0,1\right) =\text{N (North)} &  & \left( 0,0,-1\right) =\text{S
(South)} \\ 
\left( 0,1,0\right) =\text{E (East)} &  & \left( 0,-1,0\right) =\text{W
(West)} \\ 
\left( 1,0,0\right) =\text{F (Forward)} &  & 
\end{array}%
$} \\
&&%
\begin{array}{cccc}
&  &  & 
\end{array}%
\text{Fig. 1\textbf{\ }Unit Step Directions}
\end{eqnarray*}%
All of the paths begin at the origin $\left( 0,0,0\right) $ and are
considered to be in the three-dimensional Euclidean space, never passing
below the $\left( x,y\right) $ plane. The length of each path is the number
of unitary steps, and the height corresponds to the $z$ value of the path's
end point $\left( x,y,z\right) $. The paths counted by the entries of $%
\mathbf{B}$ are denoted as $NSEW\widetilde{F}$ paths where $\widetilde{F}$
denotes that all $F$ steps are restricted to height zero$.$ For example, $%
NESW\widetilde{F}\widetilde{F}NN$ denotes one of the $12,376=\binom{17}{6}$
unit-step $NSEW\widetilde{F}$ paths of length $n=8$ ending at height $k=2$.

See Nkwanta \cite{NKW} for more information on the lattice path
interpretation of $\mathbf{B}$ and more general classes of other related
higher dimensional lattice paths. See, respectively, Sloane \cite{SP1} and
Nkwanta \cite{NKW1} for information on the constructions of the arrays $%
\mathbf{A}$ and $\mathbf{B}$. Also, see \ Sloane \cite{SP1} for information
on connections between $\mathbf{A}$ and topics from combinatorics. See Barry
and Hennessy \cite{PBAR2} for more information on moments, monic orthogonal
polynomials, and Riordan arrays.

\section{Riordan Matrix Proof of the Chebyshev Connection}

Let $O\left( x,z\right) $ and $E\left( x,z\right) $ denote, respectively,
the bivariate generating functions of the even and odd coefficients of $%
T\left( x,z\right) $. Then,%
\begin{equation}
E\left( x,z\right) =\left( \frac{T\left( x,z\right) +T\left( x,-z\right) }{2}%
\right) =\frac{1+z^{2}-2x^{2}z^{2}}{1+2z^{2}+z^{4}-4x^{2}z^{2}}
\label{even0}
\end{equation}%
and%
\begin{equation}
O\left( x,z\right) =\left( \frac{T\left( x,z\right) -T\left( x,-z\right) }{2}%
\right) =\frac{xz\left( 1-z^{2}\right) }{1+2z^{2}+z^{4}-4x^{2}z^{2}}.
\label{odd}
\end{equation}%
If $T_{0}=1/2,$ then%
\begin{equation}
E\left( x,z\right) -1/2=\frac{1-z^{4}}{2\left(
1+2z^{2}+z^{4}-4x^{2}z^{2}\right) }.  \label{even}
\end{equation}%
Let%
\begin{equation*}
L\left( x,\sqrt{z}\right) =E\left( x,\sqrt{z}\right) -1/2
\end{equation*}%
denote the generating function of the even coefficients where all the odd
coefficients with zero are removed. That is, applying $\sqrt{z}$ removes the
aerated form of the generating function. This means that the sequence $%
\left\{ T_{0},0,T_{2},0,T_{4},\ldots \right\} $ associated with the
generating function given by Equation $\left( \text{\ref{even}}\right) $ is
converted to the sequence $\left\{ T_{0},T_{2},T_{4},\ldots \right\} $. Then,%
\begin{equation}
L\left( x,\sqrt{z}\right) =\left( 1/2\right) \sum_{n\geq 0}T_{2n}\left(
x\right) z^{n}=\frac{1-z^{2}}{2\left( 1+2z+z^{2}-4x^{2}z\right) }
\label{even1}
\end{equation}%
is the generating function of the sequence $\left\{ T_{0},T_{2},T_{4},\ldots
\right\} $. Let%
\begin{equation*}
M^{\ast }\left( x,z\right) =T^{\ast }\left( x,z\right) -1/2
\end{equation*}%
denote the generating function of the shifted polynomials with $T_{0}^{\ast
}=1/2$. Then,%
\begin{equation}
M^{\ast }\left( x,z\right) =\left( 1/2\right) \sum_{n\geq 0}T_{n}^{\ast
}\left( x\right) z^{n}=\frac{1-z^{2}}{2\left( 1+2z+z^{2}-4xz\right) }
\label{shif}
\end{equation}%
is the generating function of the sequence $\left\{ T_{0}^{\ast
},T_{1}^{\ast },T_{2}^{\ast },\ldots \right\} $. Here we observe that
Equations $\left( \text{\ref{even1}}\right) $ and $\left( \text{\ref{shif}}%
\right) $ differ only by the form of $x$. Let%
\begin{equation*}
N\left( x,\sqrt{z}\right) =O\left( x,\sqrt{z}\right) /\sqrt{z}
\end{equation*}%
denote the generating function of the odd coefficients where all the even
coefficients with zero are removed. Applying and dividing by $\sqrt{z}$
removes the aerated form of the generating function. In this case the
sequence $\left\{ 0,T_{1},0,T_{3},\ldots \right\} $ associated with the
generating function given by Equation $\left( \text{\ref{odd}}\right) $ is
converted to the sequence $\left\{ T_{1},T_{3},T_{5},\ldots \right\} $. Then,%
\begin{equation}
N\left( x,\sqrt{z}\right) =\sum_{n\geq 0}T_{2n+1}\left( x\right) z^{n}=\frac{%
x\left( 1-z\right) }{1+2z+z^{2}-4x^{2}z}  \label{shif1}
\end{equation}%
is the generating function of the sequence $\left\{ T_{1},T_{3},T_{5},\ldots
\right\} $. See Sprugnoli \cite{SPRU2} for more information on the technique
involving $\sqrt{z}$.

The generating functions given above lead to the following theorems.

\begin{theorem}
\label{rthm3}Consider the Catalan-type Riordan arrays 
\begin{equation*}
\mathbf{A}=\left( 1/\sqrt{1-4z},c\left( z\right) -1\right) \text{ and }%
\mathbf{B}=\left( c\left( z\right) /\sqrt{1-4z},zc^{2}\left( z\right)
\right) \text{ }
\end{equation*}%
where $c\left( z\right) $ is the Catalan generating function. Let $L\left( x,%
\sqrt{z}\right) $ be the generating function of the sequence of even
coefficients $\left\{ T_{0},T_{2},T_{4},\ldots \right\} $ of\ \ $T\left(
x,z\right) $. Let $N\left( x,\sqrt{z}\right) $ be the generating function of
the sequence of odd coefficients $\left\{ T_{1},T_{3},T_{5},\ldots \right\} $
of\ \ $T\left( x,z\right) $. Then,%
\begin{equation*}
\begin{array}{ccc}
1.\text{ }\mathbf{A}\otimes L\left( x,\sqrt{z}\right) & = & \frac{1}{2\left(
1-4x^{2}z\right) } \\ 
&  &  \\ 
2.\text{ }\mathbf{B}\otimes N\left( x,\sqrt{z}\right) & = & \frac{x}{%
1-4x^{2}z}.%
\end{array}%
\end{equation*}%
Moreover, for $n\geq 0,$ the column vector whose entries are $\left(
1/2\right) \left( 2x\right) ^{2n}$ can be expressed as a linear combination
of the columns of $\mathbf{A}$ where the even coefficients of $\ T\left(
x,z\right) $ are the coefficients of the linear combination. The column
vectors whose entries are $\left( 1/2\right) \left( 2x\right) ^{2n+1}$ can
be expressed as a linear combination of the columns of $\mathbf{B}$ where
the odd coefficients of $\ T\left( x,z\right) $ are the coefficients of the
linear combination.
\end{theorem}

\begin{proof}
The proof of part (1) of Theorem \ref{rthm3} is as follows. By Equations $%
\left( \text{\ref{Aeq}}\right) $ and $\left( \text{\ref{even1}}\right) $,
and Theorem \ref{rtm1} we obtain 
\begin{eqnarray*}
\mathbf{A}\otimes L\left( x,\sqrt{z}\right) &=&\left( 1/\sqrt{1-4z},c\left(
z\right) -1\right) \otimes \frac{1-z^{2}}{2\left( \left( 1+z\right)
^{2}-4x^{2}z\right) } \\
&=&\left( 1/\sqrt{1-4z}\right) \left( \frac{1-\left( c\left( z\right)
-1\right) ^{2}}{2\left( c^{2}\left( z\right) -4x^{2}\left( c\left( z\right)
-1\right) \right) }\right) \text{.}
\end{eqnarray*}%
Applying Lemma \ref{catlemma}(2) to the denominator and simplifying gives%
\begin{eqnarray*}
\mathbf{A}\otimes L\left( x,\sqrt{z}\right) &=&\left( 1/\sqrt{1-4z}\right)
\left( \frac{2c\left( z\right) -c^{2}\left( z\right) }{2c^{2}\left( z\right)
-c^{2}\left( z\right) 4x^{2}z}\right) \\
&=&\left( 1/\sqrt{1-4z}\right) \left( \frac{2-c\left( z\right) }{2c\left(
z\right) \left( 1-4x^{2}z\right) }\right) .
\end{eqnarray*}%
From Lemma \ref{catlemma}(1) we have 
\begin{equation*}
\left( 1/\sqrt{1-4z}\right) \left( 2-c\left( z\right) \right) =c\left(
z\right) .
\end{equation*}%
Making this substitution and simplifying we get%
\begin{equation*}
\mathbf{A}\otimes L\left( x,\sqrt{z}\right) =\frac{1}{2\left(
1-4x^{2}z\right) }.
\end{equation*}%
Now, expanding the resulting generating function gives 
\begin{equation*}
\frac{1}{2\left( 1-4x^{2}z\right) }=\left( 1/2\right) \left(
1+4x^{2}z+16x^{4}z^{2}+64x^{6}z^{3}+\cdots \right) .
\end{equation*}%
Thus, the coefficients form the column vector with entries $\left(
1/2\right) \left( 2x\right) ^{2n},n\geq 0.$

The linear combination follows as a consequence of the matrix multiplication
and rearrangement of the resulting column vector. Thus, for $n,k\geq 0$ we
get%
\begin{equation*}
T_{0}\left( 
\begin{array}{c}
1 \\ 
2 \\ 
6 \\ 
20 \\ 
\vdots \\ 
\binom{2n}{n} \\ 
\vdots%
\end{array}%
\right) +T_{2}\left( 
\begin{array}{c}
0 \\ 
1 \\ 
4 \\ 
15 \\ 
\vdots \\ 
\binom{2n}{n-1} \\ 
\vdots%
\end{array}%
\right) +\cdots +T_{2k}\left( 
\begin{array}{c}
\binom{0}{-k} \\ 
\binom{2}{1-k} \\ 
\binom{4}{2-k} \\ 
\binom{6}{3-k} \\ 
\vdots \\ 
\binom{2n}{n-k} \\ 
\vdots%
\end{array}%
\right) +\cdots =\left( 
\begin{array}{c}
1/2 \\ 
2x^{2} \\ 
8x^{4} \\ 
32x^{6} \\ 
\vdots \\ 
2^{2k-1}x^{2k} \\ 
\vdots%
\end{array}%
\right) .
\end{equation*}%
This proves the first part of the theorem.

The proof of part (2) of Theorem \ref{rthm3} is as follows. By Lemma \ref%
{catlemma}(1), Equations $\left( \text{\ref{Beq}}\right) $ and $\left( \text{%
\ref{shif1}}\right) $, and Theorem \ref{rtm1} we obtain%
\begin{eqnarray*}
\mathbf{B}\otimes N\left( x,\sqrt{z}\right) &=&\left( c^{2}\left( z\right)
/\left( 1-zc^{2}\left( z\right) \right) ,zc^{2}\left( z\right) \right)
\otimes \frac{x\left( 1-z\right) }{\left( 1+z\right) ^{2}-4x^{2}z} \\
&=&\left( c^{2}\left( z\right) /\left( 1-zc^{2}\left( z\right) \right)
\right) \left( \frac{x\left( 1-zc^{2}\left( z\right) \right) }{\left(
1+zc^{2}\left( z\right) \right) ^{2}-4x^{2}zc^{2}\left( z\right) }\right) \\
&=&\frac{xc^{2}\left( z\right) }{\left( 1+zc^{2}\left( z\right) \right)
^{2}-4x^{2}zc^{2}\left( z\right) }.
\end{eqnarray*}%
Applying Lemma \ref{catlemma}(2) to the denominator here and simplifying we
get%
\begin{eqnarray*}
\mathbf{B}\otimes N\left( x,\sqrt{z}\right) &=&\left( \frac{xc^{2}\left(
z\right) }{c^{2}\left( z\right) -c^{2}\left( z\right) 4x^{2}z}\right) \\
&=&\frac{x}{1-4x^{2}z}.
\end{eqnarray*}%
Now, expanding this generating function gives 
\begin{equation*}
\frac{x}{1-4x^{2}z}=x\left( 1+4x^{2}z+16x^{4}z^{2}+64x^{6}z^{3}+\cdots
\right) .
\end{equation*}%
The coefficients here form the column vector with entries $\left( 1/2\right)
\left( 2x\right) ^{2n+1},n\geq 0.$

Similarly, the linear combination follows here as a consequence of the
matrix multiplication and rearrangement of the resulting column vector.
\end{proof}

\begin{theorem}
\label{rthm4}Consider the Catalan-type Riordan array 
\begin{equation*}
\mathbf{A}=\left( 1/\sqrt{1-4z},c\left( z\right) -1\right) \text{ }
\end{equation*}%
where $c\left( z\right) $ is the Catalan generating function. Let $M^{\ast
}\left( x,z\right) $ be generating function of the sequence of coefficients $%
\left\{ T_{0}^{\ast },T_{1}^{\ast },T_{2}^{\ast },\ldots \right\} $ of $\
T^{\ast }\left( x,z\right) $. Then,%
\begin{equation*}
\begin{array}{ccc}
\mathbf{A}\otimes M^{\ast }\left( x,z\right) & = & \frac{1}{2\left(
1-4xz\right) }.%
\end{array}%
\end{equation*}%
Moreover, for $n\geq 0,$ the column vector whose entries are $\left(
1/2\right) \left( 4x\right) ^{n}$ can be expressed as a linear combination
of the columns of $\mathbf{A}$ where the coefficients of $\ T^{\ast }\left(
x,z\right) $ are the coefficients of the linear combination.
\end{theorem}

\begin{proof}
By Equations $\left( \text{\ref{Aeq}}\right) $ and $\left( \text{\ref{shif}}%
\right) $, Theorem \ref{rtm1}, and simplifying we obtain%
\begin{eqnarray*}
\mathbf{A}\otimes M^{\ast }\left( x,z\right) &=&\left( 1/\sqrt{1-4z},c\left(
z\right) -1\right) \otimes \frac{1-z^{2}}{2\left( \left( 1+z\right)
^{2}-4xz\right) } \\
&=&\left( 1/\sqrt{1-4z}\right) \left( \frac{2c\left( z\right) -c^{2}\left(
z\right) }{2c^{2}\left( z\right) -c^{2}\left( z\right) 4xz}\right) \\
&=&\left( 1/\sqrt{1-4z}\right) \left( \frac{2-c\left( z\right) }{2c\left(
z\right) \left( 1-4xz\right) }\right) .
\end{eqnarray*}%
Now, similar to the proof of Theorem \ref{rthm3}(1) we use Lemma \ref%
{catlemma}(1) and simplify to get%
\begin{equation*}
\mathbf{A}\otimes M^{\ast }\left( x,z\right) =\frac{1}{2\left( 1-4xz\right) }%
..
\end{equation*}%
Expanding this generating function gives 
\begin{equation*}
\frac{1}{2\left( 1-4xz\right) }=\left( 1/2\right) \left(
1+4xz+16x^{2}z^{2}+64x^{3}z^{3}+\cdots \right) .
\end{equation*}%
The coefficients form the column vector with entries $\left( 1/2\right)
\left( 4x\right) ^{n},n\geq 0.$

The linear combination follows here similar to that given in the proof of
Theorem \ref{rthm3}.
\end{proof}

\begin{corollary}
\label{cor1}The powers of $x$ can be expressed in terms of $\ T_{n}\left(
x\right) $ and $T_{n}^{\ast }\left( x\right) $ as follows:%
\begin{equation*}
\begin{array}{cc}
\text{1. } & x^{2n}=\left( 2/4^{n}\right) \left( T_{0}\binom{2n}{n}+T_{2}%
\binom{2n}{n-1}+\cdots +T_{2n}\right) \\ 
\text{2. } & x^{n}=\left( 2/4^{n}\right) \left( T_{0}^{\ast }\binom{2n}{n}%
+T_{1}^{\ast }\binom{2n}{n-1}+\cdots +T_{n}^{\ast }\right) \\ 
\text{3. } & x^{2n+1}=\left( 1/4^{n}\right) \left( T_{1}\binom{2n+1}{n}+T_{3}%
\binom{2n+1}{n-1}+\cdots +T_{2n+1}\right) .%
\end{array}%
\end{equation*}
\end{corollary}

Another related connection involving $\mathbf{A}$ is now given.

\begin{proposition}
\label{relprop}Let%
\begin{equation*}
\mathbf{D=}\left( \left( 1-z\right) /\left( 1+z\right) ,2z/\left( 1+z\right)
^{2}\right) =\left( 
\begin{array}{rrrrrr}
1 & 0 & 0 & 0 & 0 & ... \\ 
-2 & 2 & 0 & 0 & 0 & ... \\ 
2 & -8 & 4 & 0 & 0 & ... \\ 
-2 & 18 & -24 & 8 & 0 & ... \\ 
2 & -32 & 80 & -64 & 16 & ... \\ 
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots%
\end{array}%
\right) .
\end{equation*}%
Then,%
\begin{equation*}
\mathbf{A}\otimes \left( \mathbf{D}\otimes \left( 1/\left( 1-xz\right)
\right) \right) =1/\left( 1-2xz\right) .
\end{equation*}
\end{proposition}

\begin{proof}
(Sketch) By Theorem \ref{rtm1}, Equation $\left( \text{\ref{Aeq}}\right) $,
and simplifying we get%
\begin{eqnarray*}
\mathbf{A}\otimes \left( \mathbf{D}\otimes \left( 1/\left( 1-xz\right)
\right) \right) &=&\mathbf{A}\otimes \left( \left( \left( 1-z\right) /\left(
1+z\right) \right) \cdot \frac{1}{1-\left( 2xz/\left( 1+z\right) ^{2}\right) 
}\right) \\
&=&\mathbf{A}\otimes \frac{1-z^{2}}{1+\left( 1-x\right) 2z+z^{2}} \\
&=&\left( 1/\sqrt{1-4z},c\left( z\right) -1\right) \otimes \frac{1-z^{2}}{%
1+\left( 1-x\right) 2z+z^{2}}.
\end{eqnarray*}%
Now, again by Theorem \ref{rtm1}, applying Lemma \ref{catlemma} multiple
times to the numerator and denominator of the product, and simplifying gives
the result%
\begin{eqnarray*}
\left( 1/\sqrt{1-4z},c\left( z\right) -1\right) \otimes \frac{1-z^{2}}{%
1+\left( 1-x\right) 2z+z^{2}} &=&\frac{c\left( z\right) }{c\left( z\right)
-x2zc\left( z\right) } \\
&=&\frac{1}{1-2xz}.
\end{eqnarray*}%
Expanding the generating function gives 
\begin{equation*}
\frac{1}{1-2xz}=1+2xz+4x^{2}z^{2}+8x^{3}z^{3}+\cdots .
\end{equation*}%
The coefficients form the column vector with entries $\left( 2x\right)
^{n},n\geq 0.$
\end{proof}

As a consequence of this result, the rows of $\mathbf{A}$ can be used to
expand the monomial $\left( 2x\right) ^{n}$ in terms of certain polynomials $%
q_{n}\left( x\right) $ of degree $n$ where the entries of the Riordan matrix 
$\mathbf{D}$ are the coefficients of $q_{n}\left( x\right) $. For instance,
for the first four rows%
\begin{equation*}
\left( 
\begin{array}{cccc}
1 & 0 & 0 & 0 \\ 
2 & 1 & 0 & 0 \\ 
6 & 4 & 1 & 0 \\ 
20 & 15 & 6 & 1%
\end{array}%
\right) \left( 
\begin{array}{c}
q_{0} \\ 
q_{1} \\ 
q_{2} \\ 
q_{3}%
\end{array}%
\right) =\left( 
\begin{array}{c}
q_{0} \\ 
2q_{0}+q_{1} \\ 
6q_{0}+4q_{1}+q_{2} \\ 
20q_{0}+15q_{1}+6q_{2}+q_{3}%
\end{array}%
\right) =\left( 
\begin{array}{c}
1 \\ 
2x \\ 
4x^{2} \\ 
8x^{3}%
\end{array}%
\right)
\end{equation*}%
where $q_{0}=q_{0}\left( x\right) =1$, $q_{1}=q_{1}\left( x\right) =2x-2,$ 
\begin{equation*}
q_{2}=q_{2}\left( x\right) =4x^{2}-8x+2\text{, and }q_{3}=q_{3}\left(
x\right) =8x^{3}-24x^{2}+18x-2.
\end{equation*}%
We write the resulting vector as%
\begin{equation*}
\left( \left( 2x\right) ^{0}\text{ }\left( 2x\right) ^{1}\text{ }\left(
2x\right) ^{2}\text{ }\left( 2x\right) ^{3}\right) ^{T}.
\end{equation*}%
Thus, the column vector whose entries are $\left( 2x\right) ^{n}$ $\left(
n\geq 0\right) $ can be expressed as a linear combination of the columns of $%
\mathbf{A}$ where the polynomials $q_{n}\left( x\right) $ are the
coefficients of the linear combination. Moreover, the powers of $x$ can be
expressed in terms of the polynomials.

\section{Integral Representations}

A fundamental problem in numerical analysis is to approximate a continuous
function $f\left( x\right) $ by an $nth$ degree polynomial%
\begin{equation*}
p\left( x\right) =a_{0}+a_{1}x+a_{2}x^{2}+\cdots +a_{n}x^{n}
\end{equation*}%
over the interval $\left[ a,b\right] $. A continuous function $f\left(
x\right) $ can be expanded over $\left[ -1,1\right] $ in terms of the
Chebyshev polynomials to obtain a Fourier-Chebyshev expansion of the form%
\begin{equation*}
f\left( x\right) =\sum_{n=0}^{\infty }a_{n}T_{n}\left( x\right)
=a_{0}T_{0}\left( x\right) +a_{1}T_{1}\left( x\right) +a_{2}T_{2}\left(
x\right) +\cdots \text{ }
\end{equation*}%
where the coefficients $a_{n}$ are given by%
\begin{equation*}
a_{0}=\frac{1}{\pi }\int_{-1}^{1}\frac{f\left( x\right) }{\sqrt{1-x^{2}}}dx,%
\text{ and }a_{n}=\frac{2}{\pi }\int_{-1}^{1}\frac{f\left( x\right)
T_{n}\left( x\right) }{\sqrt{1-x^{2}}}dx,\text{ }n\geq 1\text{ \cite{RIV}.}
\end{equation*}%
From a practical point of view, there still remains the challenge of
accurately evaluating these integrals by using numerical procedures. On the
other hand, the integrals can be easily evaluated for the functions given by
Corollary \ref{cor1}. Moreover, the coefficients $a_{n}$ are the entries of
the Catalan-type arrays $\mathbf{A}$ and $\mathbf{B}$ and they can be read
from the columns of the arrays.

We now use properties of Chebyshev polynomials and derive certain definite
integrals. Consider the orthogonality condition 
\begin{equation*}
\int_{-1}^{1}\frac{T_{m}\left( x\right) T_{n}\left( x\right) }{\sqrt{1-x^{2}}%
}dx=\left\{ 
\begin{array}{c}
0,\text{ if }m\neq n \\ 
\frac{\pi }{2},\text{ if }m=n\neq 0 \\ 
\pi ,\text{ if }m=n=0%
\end{array}%
\right.
\end{equation*}%
for the Chebyshev polynomials of the first kind \cite{RIV,GRAD}.
Recall that%
\begin{equation*}
T_{0}\left( x\right) =T_{0}=1/2\text{.}
\end{equation*}%
Then, from part (1) of Corollary \ref{cor1}, multiplying both sides of 
\begin{equation*}
x^{2n}=\left( 2/4^{n}\right) \left( T_{0}\binom{2n}{n}+T_{2}\binom{2n}{n-1}%
+\cdots +T_{2n}\right)
\end{equation*}%
by $T_{0}\left( x\right) $ and the weight function $1/\sqrt{1-x^{2}}$,
integrating the expression over the interval $\left[ -1,1\right] $ with
respect to $x$, applying the orthogonality condition, and simplifying gives
integral representations of the binomial coefficients and Catalan numbers.
The representations, respectively, are%
\begin{equation}
\frac{4^{n}}{\pi }\int_{-1}^{1}\frac{x^{2n}}{\sqrt{1-x^{2}}}dx=\left(
n+1\right) c_{n}  \label{iega1}
\end{equation}%
and 
\begin{equation}
\frac{4^{n}}{\pi \left( n+1\right) }\int_{-1}^{1}\frac{x^{2n}}{\sqrt{1-x^{2}}%
}dx=c_{n}.  \label{iega2}
\end{equation}%
From part (3) of Corollary \ref{cor1}, multiplying both sides of%
\begin{equation*}
x^{2n+1}=\left( 1/4^{n}\right) \left( T_{1}\binom{2n+1}{n}+T_{3}\binom{2n+1}{%
n-1}+\cdots +T_{2n+1}\right)
\end{equation*}%
by $T_{1}\left( x\right) $ and the weight function $1/\sqrt{1-x^{2}}$,
integrating the expression over the interval $\left[ -1,1\right] $ with
respect to $x$, applying the orthogonality condition, and simplifying we get
two more representations. The representations are%
\begin{equation}
\frac{\left( n+1\right) 2^{2n+1}}{\pi \left( 2n+1\right) }\int_{-1}^{1}\frac{%
x^{2n+2}}{\sqrt{1-x^{2}}}dx=\frac{\left( n+1\right) }{2n+1}\binom{2n+1}{n}%
=\left( n+1\right) c_{n}  \label{iega23}
\end{equation}%
and%
\begin{equation}
\frac{2^{2n+1}}{\pi \left( 2n+1\right) }\int_{-1}^{1}\frac{x^{2n+2}}{\sqrt{%
1-x^{2}}}dx=\frac{1}{2n+1}\binom{2n+1}{n}=c_{n}.  \label{iega3}
\end{equation}%
Now, consider the orthogonality condition 
\begin{equation*}
\int_{0}^{1}\frac{T_{j}^{\ast }\left( x\right) T_{k}^{\ast }\left( x\right) 
}{\sqrt{x-x^{2}}}dx=\left\{ 
\begin{array}{c}
0,\text{ if }j\neq k \\ 
\frac{\pi }{2},\text{ if }j=k\neq 0 \\ 
\pi ,\text{ if }j=k=0%
\end{array}%
\right.
\end{equation*}%
for the shifted Chebyshev polynomials of the first kind \cite{FOX}. Recall
that%
\begin{equation*}
T_{0}^{\ast }\left( x\right) =T_{0}^{\ast }=1/2\text{.}
\end{equation*}%
Then, from part (2) of Corollary \ref{cor1}, multiplying both sides of%
\begin{equation}
x^{n}=\left( 2/4^{n}\right) \left( T_{0}^{\ast }\binom{2n}{n}+T_{1}^{\ast }%
\binom{2n}{n-1}+\cdots +T_{n}^{\ast }\right)  \label{monialeq}
\end{equation}%
by $T_{0}^{\ast }\left( x\right) $ and the weight function $1/\sqrt{x-x^{2}}$%
, integrating the expression over the interval $\left[ 0,1\right] $ with
respect to $x$, applying the orthogonality condition, and then simplifying
also gives integral representations of the central binomial coefficients and
Catalan numbers. The representations, respectively, are%
\begin{equation}
\frac{4^{n}}{\pi }\int_{0}^{1}\frac{x^{n}}{\sqrt{x-x^{2}}}dx=\left(
n+1\right) c_{n},  \label{iega4}
\end{equation}%
and%
\begin{equation}
\frac{4^{n}}{\pi \left( n+1\right) }\int_{0}^{1}\frac{x^{n}}{\sqrt{x-x^{2}}}%
dx=c_{n}.  \label{iega5}
\end{equation}%
Similarly, multiplying both sides of Equation $\left( \text{\ref{monialeq}}%
\right) $ by $T_{1}^{\ast }\left( x\right) $ gives the following
representations for $n\geq 1$,%
\begin{equation}
\frac{4^{n}\left( n+1\right) }{\pi n}\int_{0}^{1}\frac{2x^{n+1}-x^{n}}{\sqrt{%
x-x^{2}}}dx=\frac{\left( n+1\right) }{n}\binom{2n}{n-1}=\left( n+1\right)
c_{n},  \label{iega6}
\end{equation}%
and%
\begin{equation}
\frac{4^{n}}{\pi n}\int_{0}^{1}\frac{2x^{n+1}-x^{n}}{\sqrt{x-x^{2}}}dx=\frac{%
1}{n}\binom{2n}{n-1}=c_{n}.  \label{iega7}
\end{equation}

The Wolfram Alpha Widget \cite{WOLF} was used to confirm all of the
integrals that were derived above analytically. The Wolfram definite
integral calculator is a computational tool useful for experimental
mathematics. It is interesting that the calculator computes the integrals
and gives the results \ in terms of the gamma function $\Gamma \left(
n\right) $, $n>0$ (integer). The exact formulae for the central binomial
coefficients and Catalan numbers can be easily derived from certain
properties of $\Gamma \left( n\right) $ for the results given by the
calculator. See Koshy \cite{KOSHY} and Dana-Picard \cite{DP1} for
interesting relationships between the Catalan numbers and $\Gamma \left(
n\right) $. For more information on experimental mathematics, see Bailey and
Borwein \cite{BAIL,BAIL1}.

\section{\textbf{Concluding Comments}}

A few problems for possible future research projects are suggested. Finding
other Riordan arrays that are linked to the Chebyshev polynomials of the
first kind as a result of the coefficients of $T\left( x,z\right) $ and $%
T^{\ast }\left( x,z\right) $ occurring in the expansion of a continuous
function $f\left( x\right) $ is of interest. This may lead to finding new
techniques for improving the numerical approximation of $f\left( x\right) $
by Chebyshev polynomials. Finding more algebraic properties as a result of
the connection may also help with obtaining better approximations. See
Rivlin \cite{RIV} for more information on Chebyshev polynomials and methods
for approximating continuous functions. Finding the inverse of other Riordan
arrays that are linked to monic orthogonal polynomials and moments are also
of interest. By Proposition \ref{relprop} the array $\mathbf{A}$ is linked
to a certain class of polynomials $q_{n}\left( x\right) $. So, finding other
connections among other classes of polynomials, special functions, and $%
\mathbf{A}$ and $\mathbf{B}$ are of interest. Finding combinatorial
interpretations of $\mathbf{A}$ and $\mathbf{B}$ in terms of Chebyshev
polynomials is of interest as well as finding connections between the
polynomials and the lattice paths given by Nkwanta \cite{NKW}. Finding
Chebyshev polynomial interpretations of the matrix multiplication of the
arrays by column vectors made up of the coefficients of the generating
functions associated with the Chebyshev and shifted Chebyshev polynomials is
of interest. See Shapiro \cite{SHA1} and Benjamin et al. \cite{BEN1}, \cite%
{BEN2} for combinatorial interpretations of the Chebyshev polynomials.
Finding connections of $\mathbf{A}$ and $\mathbf{B}$ to the Chebyshev
polynomials of the second kind is of interest. See Barry \cite{PBAR1}, Barry
and Hennessy \cite{PBAR2}, and Luzon and Moron \cite{LUZ} for Riordan array
connections to the Chebyshev polynomials of the second kind. Finding other
Chebyshev, Riordan array connections may lead to integral representations of
other important and interesting counting sequences.

\section{\textbf{Acknowledgment}}

The authors would like to thank the referee for providing useful comments
and suggesting Propositions \ref{Invprop} and \ref{relprop}, and Remark \ref%
{reminv}. We also would like to thank Leon Woodson for useful discussions on
early drafts of the manuscript.

\begin{thebibliography}{99}
\bibitem{AIGNER} M. Aigner, \textit{A Course in Enumeration},
Springer-Verlag, 2007.

\bibitem{BAIL} D. H. Bailey and J. M. Borwein, Exploratory experimentation
and computation, \textit{Notices Amer. Math. Soc.}\textbf{\ 58} (2011),
1410--1419.

\bibitem{BAIL1} D. H. Bailey and J. M. Borwein, Experimental mathematics:
examples, methods, and implications, \textit{Notices Amer. Math. Soc.}
\textbf{\ 52} (2005), 502--514.

\bibitem{PBAR1} P. Barry, Symmetric third-order recurring sequences,
Chebyshev polynomials, and Riordan arrays, \textit{J. of Integer Sequences} 
\textbf{12} (2009), \href{http://www.cs.uwaterloo.ca/journals/JIS/VOL12/Barry4/barry64.html}{Article 09.8.6}.

\bibitem{PBAR2} P. Barry and A. Hennessy, Meixner-type results for Riordan
arrays and associated integer sequences, \textit{J. of Integer Sequences} 
\textbf{13} (2010), \href{http://www.cs.uwaterloo.ca/journals/JIS/VOL13/Barry5/barry96s.html}{Article 10.9.4}.

\bibitem{BEN1} A. T. Benjamin, L. Ericksen, P. Jayawant, and M. Shattuck,
Combinatorial trigonometry with Chebyshev polynomials, \textit{J. 
Statistical Planning and Inference} \textbf{140} (2010) 2157--2160.

\bibitem{BEN2} A. T. Benjamin and D. Walton, Combinatorially composing
Chebyshev polynomials, \textit{J. Statistical Planning and Inference}
\textbf{140} (2010) 2161--2167.

\bibitem{CAY} A. Cayley, On partitions of a polygon,\textit{\ Proc. London
Math. Soc.} \textbf{22} (1891) 237--262.

\bibitem{DP1} T. Dana-Picard, Integral presentations of Catalan numbers, 
\textit{Int. J. Math. Educ. in Science and
Technology} \textbf{41} (2010) 63--69.

\bibitem{DP2} T. Dana-Picard, Integral presentations of Catalan numbers and
Wallis formula, \textit{Int. J. Math. Educ. in
Science and Technology} \textbf{42} (2011) 122--129.

\bibitem{DEU} E. Deutsch and L. W. Shapiro, Seventeen Catalan identities, 
\textit{Bull. Inst. Combin. Appl.} 
\textbf{31} (2001) 31--38.

\bibitem{FOX} L. Fox and D. F. Mayers, \textit{Computing Methods for
Scientists and Engineers}, Oxford University Press, 1968.

\bibitem{GTSH} S. Getu and L. W. Shapiro, Lattice paths and Bessel
functions, \textit{Congressus Numerantium} \textbf{108} (1995) 161--169.

\bibitem{GRAD} I. S. Gradshteyn and I. M. Ryzhik, \textit{Tables of
Integrals, Series and Products}, 7th Ed., ed. A. Jeffrey and D.
Zwillinger, Academic Press, 2007.

\bibitem{KOSHY} T. Koshy, \textit{Catalan Numbers with Applications,} Oxford
University Press, 2009.

\bibitem{LAN} C. Lanczos, \textit{Applied Analysis}, Prentice-Hall,
1956.

\bibitem{LUZ} A. Luzon and M. A. Moron, Recurrence relations for polynomial
sequences via Riordan matrices, \textit{Linear Algebra Appl.} \textbf{433}
(2010) 1422--1446.

\bibitem{MAS} J. C. Mason and D. C. Handscomb, \textit{Chebyshev Polynomials}%
, CRC Press, 2003.

\bibitem{NKW} A. Nkwanta, Riordan matrices and higher-dimensional lattice
walks, \textit{J. Statistical Planning and Inference} \textbf{140}
(2010) 2321--2334.

\bibitem{NKW1} A. Nkwanta, A Riordan matrix approach to unifying a selected
class of combinatorial arrays, \textit{Congressus Numerantium} \textbf{160}
(2003) 33--45.

\bibitem{NKW2} A. Nkwanta, Lattice paths, generating functions, and the
Riordan group,\textit{\ }Ph.D. Dissertation, Howard University, Washington
D.C., 1997.

\bibitem{NKLWS} A. Nkwanta and L. W. Shapiro, Pell walks and Riordan
matrices, \textit{Fibonacci Quart.} \textbf{43} (2005) 170--180.

\bibitem{PEN} K. A. Penson and J. M. Sixdeniers, Integral representations of
Catalan and related numbers, \textit{J. of Integer Sequences} \textbf{4}
(2001) \href{http://www.cs.uwaterloo.ca/journals/JIS/VOL4/SIXDENIERS/Catalan.html}{Article 01.2.5}.

\bibitem{RIV} T. J. Rivlin, \textit{The Chebyshev Polynomials: From
Approximation Theory to Algebra and Number Theory, 2nd Edition,} John Wiley,
1990.

\bibitem{SHA} L. W. Shapiro, A Catalan triangle, \textit{Discrete Math.} 
\textbf{14} (1976) 83--90.

\bibitem{SHA1} L. W. Shapiro, A combinatorial proof of a Chebyshev
polynomial identity, \textit{Discrete Math.} \textbf{34} (1981) 203--206.

\bibitem{SGWW} L. W. Shapiro, S. Getu, W. J. Woan, and L. Woodson, The
Riordan group, \textit{Discrete Appl. Math.} \textbf{34} (1991) 229--239.

\bibitem{SP1} N. J. A. Sloane, \textit{The On-line Encyclopedia of Integer
Sequences}, published online at
\url{http://oeis.org}.

\bibitem{SOFO} A. Sofo, Derivatives of Catalan related sums, \textit{JIPAM.
J. Inequal. Pure Appl. Math.} \textbf{10} (2009).

\bibitem{SPRU1} R. Sprugnoli, Riordan arrays and the Abel-Gould identity, 
\textit{Discrete Math.} \textbf{142} (1995) 213--233.

\bibitem{SPRU2} R. Sprugnoli, An Introduction to Mathematical Methods in
Combinatorics, Available online at 
\url{http://www.dsi.unifi.it/~resp/Handbook.pdf}, 2006.

\bibitem{STAN} R. P. Stanley, \textit{Enumerative Combinatorics}, Vol. 2,
Cambridge Studies in Advanced Mathematics \textbf{62}, Cambridge University
Press, 1999.

\bibitem{STAN1} R. P. Stanley, \textit{Enumerative Combinatorics}, Vol. 2,
Catalan Addendum, published online at 
\url{http://www-math.mit.edu/~rstan/ec/catadd.pdf}, October 22 2011.

\bibitem{WOLF} Wolfram Alpha Widget,
published online at \url{http://www.wolfram.com/widgets}, 2011.

\bibitem{YUAN1} Q. Yuan, Walks on graphs and tensor products, Paper posted
to Annoying Precision, 
\url{http://qchu.wordpress.com/2010/03/07/walks-on-graphs-and-tensor-products/},
March 7 2010.

\bibitem{YUAN} Q. Yuan, The Catalan numbers, regular languages, and
orthogonal polynomials, Paper posted to Annoying Precision, archived at \\
\href{http://qchu.wordpress.com/2009/06/07/the-catalan-numbers-regular-languages-and-orthogonal-polynomials/}{\tt http://qchu.wordpress.com/2009/06/07/}{the-catalan-numbers-regular-languages -and-orthogonal-polynomials/}, June 7 2009.
\end{thebibliography}

\bigskip
\hrule
\bigskip

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

\noindent \emph{Keywords: } 
Riordan array, Chebyshev polynomial, generating function, Catalan
number, central binomial coefficient, definite integral.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000012},
\seqnum{A000108},
\seqnum{A000984},
\seqnum{A001700},
\seqnum{A001791},
\seqnum{A002054},
\seqnum{A002694},
\seqnum{A003516},
\seqnum{A007318},
\seqnum{A030053},
\seqnum{A039598},
\seqnum{A094527},
\seqnum{A094531}, and
\seqnum{A111418}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  December 8 2011;
revised version received  February 6 2012.
Published in {\it Journal of Integer Sequences}, March 11 2012.

\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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