\documentclass[12pt,reqno]{article}

\usepackage[usenames]{color}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{amscd}

\usepackage[colorlinks=true,
linkcolor=webgreen,
filecolor=webbrown,
citecolor=webgreen]{hyperref}

\definecolor{webgreen}{rgb}{0,.5,0}
\definecolor{webbrown}{rgb}{.6,0,0}

\usepackage{color}
\usepackage{fullpage}
\usepackage{float}

\usepackage{psfig}
\usepackage{graphics,amsmath,amssymb}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{latexsym}
\usepackage{epsf}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{.1in}
\setlength{\evensidemargin}{.1in}
\setlength{\topmargin}{-.5in}
\setlength{\textheight}{8.9in}

\newcommand{\seqnum}[1]{\href{http://www.research.att.com/cgi-bin/access.cgi/as/~njas/sequences/eisA.cgi?Anum=#1}{\underline{#1}}}

\begin{document}

\begin{center}
\epsfxsize=4in
\leavevmode\epsffile{logo129.eps}
\end{center}

\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{rema}[theorem]{Remark}
\newenvironment{remark}{\begin{rema}\normalfont\quad}{\end{rema}}

\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{xca}[theorem]{Exercise}

\begin{center}
\vskip 1cm{\LARGE\bf Notes on a Family of Riordan Arrays and  \\
\vskip .1in
Associated Integer Hankel Transforms} \vskip 1cm \large
Paul Barry\\
School of Science\\
Waterford Institute of Technology\\
Ireland\\
\href{mailto:pbarry@wit.ie}{\tt pbarry@wit.ie} \\
\ \\
Aoife Hennessy \\
Department of Computing, Mathematics and Physics\\
Waterford Institute of Technology\\
Ireland\\
\href{mailto:aoife.hennessy@gmail.com}{\tt aoife.hennessy@gmail.com} \\
\end{center}
\vskip .2 in

\begin{abstract} We examine a set of special Riordan arrays, their inverses and associated Hankel transforms.
\end{abstract}

\section{Introduction}
In this note we explore the properties of a simply defined family of Riordan arrays \cite{SGWW}.
The inverses of these arrays are closely related to well-known Catalan-defined matrices. This motivates us to study the Hankel transforms \cite{Layman} of the images
of some well-known families of sequences under the inverse matrices. This follows a general principle which states that the Hankel transform of
 the images of ``simple'' sequences under certain Catalan-defined matrices can themselves be ``simple'' in structure. We give several examples of this
phenomenon in this note.

Special sequences will be referred to by their A-number in the On-Line
Encyclopedia of Integer Sequences, \cite{Sloane}.
\section{The matrix $\mathbf{M}$}
Throughout this note, we let the matrix $\mathbf{M}$ be the Riordan array \seqnum{A158454}
$$\mathbf{M}=\left(\frac{1}{1-x^2},\frac{x}{(1+x)^2}\right).$$ We shall also consider the related matrices
$$\mathbf{\tilde{M}}=\left(\frac{1}{1-x^2},\frac{-x}{(1-x)^2}\right) \quad \text{and} \quad \mathbf{M}^+=\left(\frac{1}{1-x^2},\frac{x}{(1-x)^2}\right).$$
The matrix $\mathbf{M}$ begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 & 0 & 0
&
0 & 0 & \ldots \\0 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 1 & -2 &
1 & 0 & 0 & 0 & \ldots \\ 0 & 4 & -4 & 1 & 0 & 0 & \ldots \\ 1
&
-6 & 11 & -6 & 1 & 0 & \ldots \\0 & 9 & -24 & 22 & -8 & 1
&\ldots\\ \vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right)\end{displaymath} while $\mathbf{\tilde{M}}$ begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 & 0 & 0
&
0 & 0 & \ldots \\0 & -1 & 0 & 0 & 0 & 0 & \ldots \\ 1 & -2 &
1 & 0 & 0 & 0 & \ldots \\ 0 & -4 & 4 & -1 & 0 & 0 & \ldots \\ 1
&
-6 & 11 & -6 & 1 & 0 & \ldots \\0 & -9 & 24 & -22 & 8 & -1
&\ldots\\ \vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath}
The entries of $\mathbf{M}^+$ are the absolute value of entries of both these matrices.
\noindent We can calculate the general term $M_{n,k}$ of $\mathbf{M}$ as follows.
\begin{eqnarray*}
M_{n,k}&=&[x^n]\frac{1}{1-x^2}\frac{x^k}{(1+x^2)^k}\\
&=&[x^{n-k}](1-x)^{-1} (1+x)^{-(2k+1)}\\
&=&[x^{n-k}] \sum_{j=0}^{\infty}x^j \sum_{i=0}^{\infty} \binom{-(2k+1)}{i}x^i\\
&=&[x^{n-k}] \sum_{j=0}^{\infty} \sum_{i=0}^{\infty} \binom{2k+i}{i}(-1)^i x^{i+j}\\
&=& \sum_{j=0}^{n-k} \binom{n+k-j}{n-k-j}(-1)^{n-k-j}\\
&=& \sum_{j=0}^{n-k} \binom{n+k-j}{2k} (-1)^{n-k-j}.\end{eqnarray*}

\noindent An alternative expression for $M_{n,k}$ can be obtained by noticing that
$$\mathbf{M}=\left(\frac{1}{1-x^2},\frac{x}{(1+x)^2}\right)=\left(\frac{1}{1-x},x\right)\cdot \left(\frac{1}{1+x},\frac{x}{(1+x)^2}\right).$$
The general term of $\left(\frac{1}{1-x},x\right)$ is $[k \le n]\cdot 1$, while that of $\left(\frac{1}{1+x},\frac{x}{(1+x)^2}\right)$ is
$(-1)^{n-k}\binom{n+k}{2k}$. Thus we obtain the alternative expression
$$M_{n,k}=\sum_{j=0}^n (-1)^{j-k}\binom{k+j}{2k}.$$
This translates the combinatorial identity
$$\sum_{j=0}^n (-1)^{j-k}\binom{k+j}{2k}=\sum_{j=0}^{n-k} \binom{n+k-j}{2k} (-1)^{n-k-j}.$$
\noindent Yet another expression for $M_{n,k}$ can be obtained by observing that we have the factorization
$$\mathbf{M}=\left(\frac{1}{1-x^2},x\right) \cdot \left(1, \frac{x}{(1+x)^2}\right).$$ This leads to the expression
$$M_{n,k}=\sum_{j=0}^n \binom{j+k-1}{j-k}(-1)^{j-k} \frac{1+(-1)^{n-j}}{2}.$$
\noindent The row sums of $\mathbf{M}$ are the periodic sequence
$$1,1,0,1,1,0,1,1,0,\ldots$$ with generating function $\frac{1+x}{1-x^3}$, while the diagonal sums are the alternating sign version
of \seqnum{A078008} given by
$$1,0,2,-2,6,-10,22,\ldots$$ with generating function $\frac{1+x}{(1-x)(1+2x)}$.

\noindent Of particular note are the images of the Catalan numbers $C_n$ \seqnum{A000108} and the central binomial numbers
$\binom{2n}{n}$ \seqnum{A000984} under this matrix. Letting
$$c(x)=\frac{1-\sqrt{1-4x}}{2x}$$ be the generating function of the Catalan numbers $C_n$,
we have
$$\left(\frac{1}{1-x^2},\frac{x}{(1+x)^2}\right)\cdot c(x)=\frac{1}{1-x^2}\frac{1-\sqrt{1-4\frac{x}{(1+x)^2}}}{2\frac{x}{(1+x)^2}}=\frac{1}{1-x}.$$
Thus the image of $C_n$ under the matrix $\mathbf{M}$ is the all $1$'s sequence
$$1,1,1,\ldots$$ The image of $C_{n+1}$ is also interesting. We get the sequence $1,2,2,2,\ldots$ with generating
function $\frac{1+x}{1-x}$. This can be generalized to the following result: The image of $C_{n+k}$ under the matrix
$\mathbf{M}$ has generating function $\frac{\sum_{j=0}^k a_{k,j} x^j}{1-x}$, where $a_{n,k}$ is the general term of the
matrix $(c(x),xc(x)^2)$. We have $a_{n,k}=\binom{2n}{n-k}\frac{2k+1}{n+k+1}$. A consequence of this is the fact that the image by
$\mathbf{M}$ of the Hankel matrix with general term $C_{n+k}$ is a matrix whose rows tend to $\binom{2n}{n}$, where the first row is $C_n$\,:
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 1 & 2 & 5
&
14 & 42 & \ldots \\1 & 2 & 5 & 14 & 42 & 132 & \ldots \\ 1 & 2 &
6 & 19 & 62 & 207 & \ldots \\ 1 & 2 & 6 & 20 & 69 & 242 & \ldots \\ 1
&
2 & 6 & 20 & 70 & 251 & \ldots \\1 & 2 & 6 & 20 & 70 & 252
&\ldots\\ \vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath} This matrix is in fact equal to
$$\mathbf{M}\cdot (c(x),xc(x)^2)\cdot (c(x),xc(x)^2)^T=\left(\frac{1}{1-x},x\right)\cdot(c(x),xc(x)^2)^T$$ since
$(C_{n+k})=\mathbf{LDL}^T$ \cite{PPWW} where $\mathbf{L}=(c(x),xc(x)^2)$ and $\mathbf{D}=\mathbf{I}$ in the case of the
Catalan numbers.

\noindent Now taking $\binom{2n}{n}$ with generating function $\frac{1}{\sqrt{1-4x}}$, we obtain
$$\left(\frac{1}{1-x^2},\frac{x}{(1+x)^2}\right)\cdot \frac{1}{\sqrt{1-4x}}=\frac{1}{1-x^2}\frac{1}{\sqrt{1-4\frac{x}{(1+x)^2}}}=\frac{1}{(1-x)^2}.$$
Thus the image of $\binom{2n}{n}$ under $\mathbf{M}$ is $n+1$ or the counting numbers \seqnum{A000027}
$$1,2,3,4,5,\ldots$$
\section{The inverse matrix $\mathbf{M}^{-1}$}
Since
$$\mathbf{M}=\left(\frac{1}{1-x},x\right)\cdot \left(\frac{1}{1+x},\frac{x}{(1+x)^2}\right)$$ we see that
$$\mathbf{M}^{-1}=\left(\frac{1}{1+x},\frac{x}{(1+x)^2}\right)^{-1} (1-x,x)=(c(x),xc(x)^2)(1-x,x).$$
Thus
$$\mathbf{M}^{-1}=(c(x)(1-xc(x)^2),xc(x)^2)=(1-x^2 c(x)^4, xc(x)^2).$$
The general term of $\mathbf{M}^{-1}$ is given by
$$M_{n,k}^{-1}=\sum_{j=0}^n (-1)^{j+k}\binom{1}{j-k}\binom{2n}{n-j}\frac{2j+1}{n+j+1}.$$ We observe that
$$\mathbf{M}^{-1}=(c(x)(1-xc(x)^2),xc(x)^2)=(c(x),xc(x)^2)-(xc(x)^3,xc(x)^2)$$ and hence
\begin{eqnarray*}M_{n,k}^{-1}&=&\frac{2k+1}{n+k+1}\binom{2n}{n-k}-\frac{2k+3}{n+k+2}\binom{2n}{n-k-1}\\
&=&\frac{2k+1}{n+k+1}\binom{2n}{n-k}-\frac{2k+3}{n+k+2}\frac{n-k}{n+k+1}\binom{2n}{n-k}\\
&=&\frac{2(2k^2+4k-n+1)}{(k+n+1)(k+n+2)}\binom{2n}{n-k}.\end{eqnarray*}
Thus if we apply the matrix $\mathbf{M}^{-1}$ to the sequence $a_n$ we obtain the image sequence
$b_n$ given by
\begin{eqnarray*}b_n &=&\sum_{k=0}^n \sum_{j=0}^n (-1)^{j+k}\binom{1}{j-k}\binom{2n}{n-j}\frac{2j+1}{n+j+1} a_k\\
&=&\sum_{k=0}^n \frac{2k+1}{n+k+1}\binom{2n}{n-k}a_k-\sum_{k=0}^n \frac{2k+3}{n+k+2}\binom{2n}{n-k-1}a_k\\
&=&\sum_{k=0}^n\frac{2(2k^2+4k-n+1)}{(k+n+1)(k+n+2)}\binom{2n}{n-k}a_k.\end{eqnarray*}

\noindent We can obtain a further expression for this image by observing that
$$\mathbf{M}^{-1}=\left(1,\frac{x}{(1+x)^2}\right)^{-1}\cdot (1-x^2,x)=(1,xc(x)^2)\cdot (1-x^2,x).$$
We let $T_{n,k}$ be the general term of the matrix $(1,xc(x)^2)$. We have $T_{0,0}=1$, and
$$T_{n,k}=\binom{2n-1}{n-k}\frac{2k}{n+k}$$ otherwise.
Then
$$b_n = \sum_{k=0}^n T_{n,k}(a(k)-a^{(2)}(k)),$$
where if $k < 2$, $a^{(2)}(k)=0$, otherwise $a^{(2)}(k)=a(k-2).$

The production matrix \cite{ProdMat_0} of $\mathbf{M}^{-1}$ begins
\begin{displaymath}\left(\begin{array}{ccccccc} 0 & 1 & 0 & 0
&
0 & 0 & \ldots \\-1 & 2 & 1 & 0 & 0 & 0 & \ldots \\ -2 & 1 &
2 & 1 & 0 & 0 & \ldots \\ -2 & 0 & 1 & 2 & 1 & 0 & \ldots \\ -2
&
0 & 0 & 1 & 2 & 1 & \ldots \\-2 & 0 & 0 & 0 & 1 & 2
&\ldots\\ \vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath} We note that the production matrix of
the array $$\left(\frac{1}{1+x},\frac{x}{(1+x)^2}\right)^{-1}=(c(x),xc(x)^2)$$ is given by the
related tri-diagonal matrix
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 1 & 0 & 0
&
0 & 0 & \ldots \\1 & 2 & 1 & 0 & 0 & 0 & \ldots \\ 0 & 1 &
2 & 1 & 0 & 0 & \ldots \\ 0 & 0 & 1 & 2 & 1 & 0 & \ldots \\ 0
&
0 & 0 & 1 & 2 & 1 & \ldots \\0 & 0 & 0 & 0 & 1 & 2
&\ldots\\ \vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right)\end{displaymath} which is the Jacobi-Stieltjes matrix for the
orthogonal signed Morgan-Voyce polynomials $b_n(x)=\sum_{k=0}^n (-1)^{n-k}\binom{n+k}{2k}x^k$ (see \seqnum{A085478}).
\section{$\mathbf{M}^{-1}$ and Hankel transforms}
We can characterize the image of the family of sequences $rn+1$ under the transformation matrix
$\mathbf{M}^{-1}$ as follows. Here, $r \in \mathbf{Z}$.
\begin{proposition}\label{Prop} The image of $rn+1$ under the matrix $\mathbf{M}^{-1}$ is
$$(rn+1)C_n$$ with Hankel transform given by
$$[x^n]\frac{1-(r-1)^2 x}{1-2x+(r-1)^2x^2}.$$ \end{proposition}
\begin{proof}
The image of $rn+1$ has generating function given by
\begin{eqnarray*} g(r,x)&=&\mathbf{M}^{-1} \frac{(r-1)x+1}{(1-x)^2}\\
&=&(c(x)(1-xc(x)^2),xc(x)^2)\cdot \frac{(r-1)x+1}{(1-x)^2}\\
&=& c(x)(1-xc(x)^2)\frac{(r-1)xc(x)^2+1}{(1-xc(x)^2)^2}\\
&=& c(x)\frac{(r-1)xc(x)^2+1}{1-xc(x)^2}\\
&=& \frac{(1-r)(1-4x)-(1-r+2x(r-2))\sqrt{1-4x}}{2x(1-4x)}\end{eqnarray*}
which is the generating function of $(rn+1)C_n$.
Then the orthogonal measure associated to $g(x,r)$ (i.e., the measure for which the numbers
$(rn+1)C_n$ are the moments) is given by
$$w(x,r)=\frac{1}{2\pi} \frac{x(r-1)-2(r-2)}{\sqrt{x(4-x)}}.$$ This leads to a modified
Jacobi polynomial. Using the techniques of \cite{Gautschi} (see also \cite{CRI, RPB}) we see that
the $\beta$ coefficients corresponding to the relevant three term recurrence are given by
$$-r^2+2r+1, \frac{-3r^+6r+1}{(r^2-2r-1)^2}, -\frac{- (r^2 - 27r - 1)7(r^4 - 47r^3 - 27r^2 + 127r + 1)}{(37r^2 - 67r - 1)^2},\ldots$$
Thus the Hankel transform is given by \cite{Krat1, Krat2}
$$1, - r^2 + 27r + 1, - 37r^2 + 67r + 1, r^4 - 47r^3 - 27r^2 + 127r + 1,\ldots$$ which is the expansion of
$$\frac{1-(r-1)^2 x}{1-2x+(r-1)^2x^2}$$ as required.
\end{proof}
Similarly, we can look at the image of the power sequences $n\to r^n$ under $\mathbf{M}^{-1}$. We find
\begin{proposition} The Hankel transform of the image of $r^n$ under the matrix $\mathbf{M}^{-1}$ is given by
$$[x^n]\frac{1-(r-1)^2x}{1-(r^2-r+2)x+(r-1)^2x^2}.$$
\end{proposition}
\section{The matrix $\mathbf{\tilde{M}}$}
The general term $\tilde{M}_{n,k}$ of the matrix $\mathbf{\tilde{M}}$ is given by
$$\tilde{M}_{n,k}=\sum_{j=0}^n (-1)^{n-j+k}\binom{k+j}{2k}=\sum_{j=0}^{n-k}(-1)^{j-k}\binom{n+k-j}{2k}.$$
The row sums of this matrix are the periodic sequence $1,-1,0,1,-1,0,\ldots$ with generating function
$\frac{1-x}{1-x^3}$. The diagonal sums are the sequence \seqnum{A078014} with generating function
$\frac{1-x}{1-x+2x^3}$.
The image of $C_n$ under $\mathbf{\tilde{M}}$ is
$$1,-1,1,-1,\ldots$$ or $(-1)^n$. Similarly the image of $\binom{2n}{n}$ is $(-1)^n (n+1).$
The image of the Catalan Hankel matrix $(C_{n+k})$ under $\mathbf{\tilde{M}}$ begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 1 & 2 & 5
&
14 & 42 & \ldots \\-1 & -2 & -5 & -14 & -42 & -132 & \ldots \\ 1 & 2 &
6 & 19 & 62 & 207 & \ldots \\ -1 & -2 & -6 & -20 & -69 & -242 & \ldots \\ 1
&
2 & 6 & 20 & 70 & 251 & \ldots \\-1 & -2 & -6 & -20 & -70 & -252
&\ldots\\ \vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath}
\noindent The inverse matrix $\mathbf{\tilde{M}}^{-1}$ has general term
$$\sum_{j=0}^n (-1)^j\binom{1}{j-k}\binom{2n}{n-j}\frac{2j+1}{n+j+1}.$$
\noindent We have the following proposition concerning the image of $r^n$ under the inverse of the matrix $\mathbf{\tilde{M}}$.
\begin{proposition} The Hankel transform of the image of the sequence $n \to r^n$ under the matrix
$\mathbf{\tilde{M}}^{-1}$ is given by
$$[x^n] \frac{1-x}{1+2rx+x^2}.$$
\end{proposition}


\section{The positive matrix $\mathbf{M}^+=\left(\frac{1}{1-x^2},\frac{x}{(1-x)^2}\right)$}
The matrix $\mathbf{M}^+$ begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 & 0 & 0
&
0 & 0 & \ldots \\0 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 2 &
1 & 0 & 0 & 0 & \ldots \\ 0 & 4 & 4 & 1 & 0 & 0 & \ldots \\ 1
&
6 & 11 & 6 & 1 & 0 & \ldots \\0 & 9 & 24 & 22 & 8 & 1
&\ldots\\ \vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath}
\noindent It has general term
$$M_{n,k}^+=\sum_{j=0}^n (-1)^{n-j}\binom{k+j}{2k}=\sum_{j=0}^{n-k} \binom{n+k-j}{2k}(-1)^j = \sum_{j=0}^n \binom{j+k-1}{j-k} \frac{1+(-1)^{n-j}}{2}.$$
The row sums of this matrix are easily seen to have generating function $\frac{1-x}{(1+x)(1-3x+x^2)}$. These are the squared Fibonacci numbers
$F(n+1)^2$, \seqnum{A007598}. Thus we have
$$F(n+1)^2=\sum_{k=0}^n \sum_{j=0}^n (-1)^{n-j}\binom{k+j}{2k},$$ for instance. The diagonal sums are
the Jacobsthal numbers variant \seqnum{A078008}, with general term $\frac{2^n}{3}+2\frac{(-1)^n}{3}.$
The image of the Catalan numbers $C_n$ by this matrix are an alternating sum of the large Schr\"oder numbers $S_n$ (\seqnum{A006318}) given by
$$\sum_{k=0}^n (-1)^{n-k} S_k.$$ This has generating function
$$\frac{1-x-\sqrt{1-6x+x^2}}{2x(1+x)}.$$
Likewise, the image of the central binomial coefficients $\binom{2n}{n}$ is the sequence starting $1, 2, 11, 52, 269, \ldots$ \seqnum{A026933} with
generating function
$$\frac{1}{(1+x)\sqrt{1-6x+x^2}}.$$ This is therefore
$$\sum_{k=0}^n (-1)^{n-k} D_k$$ where $D_n$ is the sequence of central Delannoy numbers \seqnum{A001850}.

Again, for appropriate families of sequences, the images under the inverse of this matrix can have interesting Hankel transforms. For instance, we have the following result.
\begin{proposition} The Hankel transform of the image under ${\mathbf{M}^+}^{-1}$ of the generalized Fibonacci numbers with general term
$$\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n-k}{k}r^k$$ is given by
$$[x^n]\frac{1+(r-1)x}{1+2x+(r-1)^2x^2}.$$
\end{proposition}
\noindent Finally, we note that the production matrix of the inverse ${\mathbf{M}^+}^{-1}$ takes the form
\begin{displaymath}\left(\begin{array}{ccccccc} 0 & 1 & 0 & 0
&
0 & 0 & \ldots \\-1 & -2 & 1 & 0 & 0 & 0 & \ldots \\ 2 & 1 &
-2 & 1 & 0 & 0 & \ldots \\ -2 & 0 & 1 & -2 & 1 & 0 & \ldots \\ 2
&
0 & 0 & 1 & -2 & 1 & \ldots \\-2 & 0 & 0 & 0 & 1 & -2
&\ldots\\ \vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath}
\section{A one-parameter family of Riordan arrays}
We note that
$$\mathbf{M}=\left(\frac{1}{1-x^2},\frac{x}{(1+x)^2}\right)=\left(\frac{1}{(1-x)(1+x)},\frac{x}{(1+x)^2}\right)$$ is the element
$\mathbf{M}_1$ of the family of Riordan arrays $\mathbf{M}_r$ defined by
$$\mathbf{M}_r=\left(\frac{1}{(1-x)(1+rx)},\frac{x}{(1+rx)^2}\right).$$ The general term of $\mathbf{M}_r$ is given by
$$\sum_{j=0}^{n-k} \binom{n+k-j}{2k} (-r)^{n-k-j}.$$

\noindent We can generalize the results of the foregoing to this family.
For instance, the row sums of $\mathbf{M}_r^{-1}$ are precisely $r^n C_n$, while the image of the sequence
$r^n \binom{2n}{n}$ under the matrix $\mathbf{M}_r$ has general term $\frac{1-r^{n+1}}{1-r}=\sum_{k=0}^n r^k$.


\noindent The production matrix  of $\mathbf{M}_r^{-1}$ begins
\begin{displaymath}\left(\begin{array}{ccccccc} r-1 & 1 & 0 & 0
&
0 & 0 & \ldots \\r^2-r-1 & 2r & 1 & 0 & 0 & 0 & \ldots \\ -r-1 & r^2 &
2r & 1 & 0 & 0 & \ldots \\ -r-1 & 0 & r^2 & 2r & 1 & 0 & \ldots \\ -r-1
&
0 & 0 & r^2 & 2r & 1 & \ldots \\-r-1 & 0 & 0 & 0 & r^2 & 2r
&\ldots\\ \vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath}
\section{Acknowledgements}
We are grateful to an anonymous reviewer whose careful reading of the original manuscript has resulted in some important clarifications.
Some are incorporated in the above text. We single out two further results here. Firstly, it can be noted that
\begin{eqnarray*}M_{n,k}^{-1}&=&[x^n](1-x^2c(x)^2)(xc(x)^2)^k\\
&=&\frac{k}{n}\binom{2n}{n-k}-\frac{k+2}{n}\binom{2n}{n-k-2}.\end{eqnarray*}
Thus for $n,k\ge 1$ we obtain the identity
$$\frac{2k+1}{n+k+1}\binom{2n}{n-k}-\frac{2k+3}{n+k+2}\binom{2n}{n-k-1}=\frac{k}{n}\binom{2n}{n-k}-\frac{k+2}{n}\binom{2n}{n-k-2}.$$
Secondly, Proposition \textbf{\ref{Prop}} can be extended as follows\,:
\begin{proposition} The image of the sequence $(rn+s)_{n=0}^{\infty}$ under the matrix $M^{-1}$ is
$$(rn+s)C_n,$$ with Hankel transform given by
$$[x^n]\frac{s-(r-s)^2x}{1-2sx+(r-s)^2x^2}.$$ \end{proposition}
\begin{thebibliography}{9}


\bibitem{CRI} A. Cvetkovi\'c, P. Rajkovi\'c and M. Ivkovi\'c,
    \\ \href{http://www.cs.uwaterloo.ca/journals/JIS/VOL5/Ivkovic/ivkovic3.html} {Catalan numbers, the Hankel transform and Fibonacci
    numbers}, \emph{J.
    Integer Seq.} {\bf 5} (2002), Article 02.1.3.

\bibitem{ProdMat_0} E. Deutsch, L. Ferrari, and S. Rinaldi,
    Production matrices, \emph{Adv. in Appl.
    Math.}
    \textbf{34} (2005),
    101--122.

\bibitem{Gautschi} W. Gautschi, \emph{Orthogonal Polynomials:
    Computation and Approximation}, Clarendon Press, Oxford,
    2003.

\bibitem{Krat1} C. Krattenthaler, Advanced determinant
    calculus, math.CO/9902004. Available electronically at
    \href{http://arxiv.org/abs/math/9902004}{http://arxiv.org/abs/math/9902004} .

\bibitem{Krat2} C. Krattenthaler, Advanced determinant
    calculus: A complement, {\it Linear Algebra
    Appl.} \textbf{411} (2005), 68--166.

\bibitem{Layman} J.~W. Layman, \href{http://www.cs.uwaterloo.ca/journals/JIS/VOL4/LAYMAN/hankel.html} {The Hankel transform and some of its
properties}, \emph{J. Integer Seq.} \textbf{4} (2001), Article
01.1.5.

\bibitem{PPWW} P. Peart and W-J. Woan, \href{http://www.cs.uwaterloo.ca/journals/JIS/VOL3/PEART/peart1.html} {Generating functions via
    Hankel and Stieltjes matrices}, \emph{J. Integer
    Seq.} \textbf{3} (2000),
    Article 00.2.1.

\bibitem{RPB} P.~M. Rajkovi\'c, M.~D. Petkovi\'c and P. Barry, The
    Hankel transform of the sum of consecutive generalized
    Catalan numbers, {\it
    Integral Transforms Spec. Funct.} \textbf{18} (2007),
    285--296.

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

\bibitem {Sloane} N.~J.~A.~Sloane, {\em \htmladdnormallink {The On-Line Encyclopedia of Integer Sequences}{http://www.research.att.com/~njas/sequences}},
published electronically at \texttt{http://www.research.att.com/$\sim$njas/sequences/}.

\end{thebibliography}

\bigskip
\hrule
\bigskip
\noindent 2000 {\it Mathematics Subject Classification}: Primary
11B83; Secondary  11B65, 11Y55, 42C05.

\noindent \emph{Keywords:} Riordan array, integer sequence, production matrix, Hankel transform, orthogonal polynomial.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000027},
\seqnum{A000108},
\seqnum{A000984},
\seqnum{A001850},
\seqnum{A007598},
\seqnum{A026933},
\seqnum{A078008},
\seqnum{A078014},
\seqnum{A085478}, and
\seqnum{A158454}.)


\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received March 19 2009;
revised version received  July 6 2009.
Published in {\it Journal of Integer Sequences}, July 6 2009.

\bigskip
\hrule
\bigskip

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


\end{document}

                                                                                


.
