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\begin{center}
\vskip 1cm{\LARGE\bf A Note on a One-Parameter Family of \\
\vskip .1in
Catalan-Like Numbers} \vskip 1cm \large
Paul Barry\\
School of Science\\
Waterford Institute of Technology\\
Ireland\\
\href{mailto:pbarry@wit.ie}{\tt pbarry@wit.ie} \\
\end{center}
\vskip .2 in

\begin{abstract} We study a family of sequences of Catalan-like numbers based on the series reversion process. Properties of these sequences are derived,
 including continued fraction expansions, associated orthogonal polynomials and associated Aigner matrices, which turn out to be
 Riordan arrays.
\end{abstract}

\section{Introduction}
The purpose of this note is to explore some properties of the family of sequences obtained by reverting the expression
$$\frac{x(1+rx)}{1+2rx+r(r+1)x^2}$$ where $r$ is an integer parameter. The analysis involves elements of the theory of
Riordan arrays \cite{SGWW}, orthogonal polynomials \cite{Chihara, Gautschi}, continued fractions \cite{Wall_CA}, the Deleham construction (see \seqnum{A084938}) and Hankel transforms \cite{Layman}. The overall context of this note is that of
``Catalan-like'' numbers, a notion defined and developed by Martin Aigner \cite {Aigner_Cat, Aigner_recurrent}. See also
\cite{Ferrari_1, Ferrari_2, Zhizheng_Aigner}.
\noindent In the sequel, $[x^n]$ denotes the operator that extracts the coefficient of $x^n$ in a power series, $\text{Rev}$ denotes the operation of
reverting a sequence (thus $\bar{f}(x)=\text{Rev}f(x)$ satisfies $f(\bar{f}(x))=x$), $[P]$ is the Iverson operator \cite{Concrete} equal to $0$ if $P$ is false, and $1$ if $P$ is true, and
$$c(x)=\frac{1-\sqrt{1-4x}}{2x}=
\cfrac{1}{1-
\cfrac{x}{1-
\cfrac{x}{1-
\cfrac{x}{1-\cdots}}}}$$ is the g.f. of the Catalan numbers \seqnum{A000108}. In addition, $(g,f)$ will denote a Riordan array whose $k$-th column
has generating function $g(x)f(x)^k$. In the next section, we shall provide an introduction to the Riordan group.
\noindent We recall that a number sequence $a_n$ is ``Catalan-like'' if none of the Hankel determinants
$|a_{i+j}|_{i,j=0}^n$ is zero, while a lower-triangular matrix $(a_{n,k})$ is called an \emph{Aigner matrix} if
there exist two sequences $s_n$ and $t_n$ such that
$$a_{0,0}=1,\quad a_{0,k}=0 \quad(k>0)$$ and
$$a_{n,k}=a_{n-1,k-1}+s_k a_{n-1,k}+t_{k+1} a_{n-1,k+1}\quad (n,k \ge 1).$$
The (integer) \emph{Hankel transform} of the sequence $a_n$ is the sequence $h_n$ with general term
$$h_n=|a_{i+j}|_{i,j=0}^n.$$
\section{The Riordan group}
The \emph{Riordan group} \cite{SGWW}, \cite{Spru}, is a set of
infinite lower-triangular integer matrices, where each matrix is
defined by a pair of generating functions
$g(x)=1+g_1x+g_2x^2+\ldots$ and $f(x)=f_1x+f_2x^2+\ldots$ where
$f_1\ne 0$ \cite{Spru}. The associated matrix is the matrix whose
$i$-th column is generated by $g(x)f(x)^i$ (the first column being
indexed by 0). The matrix corresponding to the pair $g, f$ is
denoted by $(g, f)$ or $\cal{R}$$(g,f)$. The group law is then given
by
\begin{displaymath} (g, f)*(h, l)=(g(h\circ f), l\circ
f).\end{displaymath} The identity for this law is $I=(1,x)$ and the
inverse of $(g, f)$ is $(g, f)^{-1}=(1/(g\circ \bar{f}), \bar{f})$
where $\bar{f}$ is the compositional inverse of $f$.

A Riordan array of the form $(g(x),x)$, where $g(x)$ is the
generating function of the sequence $a_n$, is called the
\emph{sequence array} of the sequence $a_n$. Its general term is
$a_{n-k}$. Such arrays are also called \emph{Appell} arrays as they form the elements of the
Appell subgroup.
\newline\newline If $\mathbf{M}$ is the matrix $(g,f)$, and
$\mathbf{a}=(a_0,a_1,\ldots)'$ is an integer sequence with ordinary
generating function $\cal{A}$ $(x)$, then the sequence
$\mathbf{M}\mathbf{a}$ has ordinary generating function
$g(x)$$\cal{A}$$(f(x))$. The (infinite) matrix $(g,f)$ can thus be considered to act on the ring of
integer sequences $\mathbf{Z}^\mathbf{N}$ by multiplication, where a sequence is regarded as a
(infinite) column vector. We can extend this action to the ring of power series
$\mathbf{Z}[[x]]$ by
$$(g,f):\cal{A}(\mathnormal{x}) \longrightarrow \mathnormal{(g,f)}\cdot \cal{A}\mathnormal{(x)=g(x)}\cal{A}\mathnormal{(f(x))}.$$
\begin{example} The binomial matrix $\mathbf{B}$ is the element
$(\frac{1}{1-x},\frac{x}{1-x})$ of the Riordan group. It has general
element $\binom{n}{k}$. More generally, $\mathbf{B}^m$ is the
element $(\frac{1}{1-m x},\frac{x}{1-mx})$ of the Riordan group,
with general term $\binom{n}{k}m^{n-k}$. It is easy to show that the
inverse $\mathbf{B}^{-m}$ of $\mathbf{B}^m$ is given by
$(\frac{1}{1+mx},\frac{x}{1+mx})$.
\end{example}

The row sums of the matrix $(g, f)$ have generating function
$$(g,f)\cdot \frac{1}{1-x}=\frac{g(x)}{1-f(x)}$$
 while the diagonal sums of $(g, f)$ (sums of left-to-right diagonals in the North East direction) have generating
function $g(x)/(1-xf(x))$. These coincide with the row sums of the ``generalized'' Riordan array $(g,xf)$.i\newline\newline
The bi-variate generating function of the Riordan array $(g,f)$ is given by $\frac{g(x)}{1-yf(x)}$.
\section{The sequences $a_n(r)$}
The reversion of the function $$\frac{x(1+rx)}{1+2rx+r(r+1)x}$$ is obtained by solving the equation
$$\frac{u(1+ru)}{1+2ru+r(r+1)u}=x$$ for the unknown $u$. We obtain
$$u=\frac{\sqrt{1-4rx^2}+2rx-1}{2r(1-(r+1)x)}.$$
We now define
\begin{eqnarray*} a_n(r)&=& [x^{n+1}]\text{Rev}\frac{x(1+rx)}{1+2rx+r(r+1)x^2}\\
&=&[x^{n+1}]\frac{\sqrt{1-4rx^2}+2rx-1}{2r(1-(r+1)x)}\\
&=&[x^n]\frac{\sqrt{1-4rx^2}+2rx-1}{2xr(1-(r+1)x)}\\
&=&[x^n]\frac{c(rx^2)}{1-rxc(rx^2)}\\
&=&[x^n](c(rx^2),rxc(rx^2))\cdot \frac{1}{1-x}\\
&=&[x^n](c(rx^2),xc(rx^2))\cdot \frac{1}{1-rx}\\
&=& \sum_{k=0}^n \frac{k+1}{n+k+2}\binom{n}{\frac{n-k}{2}}(1+(-1)^{n-k})r^{\frac{n-k}{2}}r^k\\
&=& \sum_{k=0}^n \frac{k+1}{n+k+2}\binom{n}{\frac{n-k}{2}}(1+(-1)^{n-k})r^{\frac{n+k}{2}}\\
&=& \sum_{k=0}^n [n\le 2k]\binom{n}{k}\frac{2k-n+1}{k+1} r^k.\end{eqnarray*}
The final equalities above follow from the fact that the general term of the Riordan array
$$(c(rx^2),xc(rx^2))$$ is given by
$$\frac{k+1}{n+k+2}\binom{n}{\frac{n-k}{2}}(1+(-1)^{n-k})r^{\frac{n-k}{2}}.$$
\noindent A short table of these sequences is given below.
\begin{center}\begin{tabular}{|c|c|c|} \hline $r$ & A-number & Reversion of\\ \hline
$1$ & \seqnum{A001405} & $\frac{x(1+x)}{1+2x+2x^2}$ \\ \hline
$2$ & \seqnum{A151281} & $\frac{x(1+2x)}{1+4x+6x^2}$\\ \hline
$3$ & \seqnum{A151162} & $\frac{x(1+3x)}{1+6x+12x^2}$ \\ \hline
$4$ & \seqnum{A151254} & $\frac{x(1+4x)}{1+8x+20x^2}$ \\ \hline
$5$ & \seqnum{A156195} & $\frac{x(1+5x)}{1+10x+30x^2}$ \\ \hline
$6$ & \seqnum{A156361} & $\frac{x(1+6x)}{1+12x+42x^2}$ \\ \hline
\end{tabular}\end{center}
For example, $a_n(1)$ is the sequence of central binomial numbers $\binom{n}{\lfloor \frac{n}{2} \rfloor}$,
while $a_n(3)$  counts the number of walks within $\mathbf{N}^3$ (the first octant of $\mathbf{Z}^3$) starting at $(0,0,0)$ and consisting
of $n$ steps taken from $\{(-1, 0,0), (1, 0, 0), (1, 0, 1), (1, 1, 0)\}$ \cite{Bostan}.
\section{Continued fractions}
We have
\begin{eqnarray*} \frac{c(rx^2)}{1-r x c(rx^2)}&=&\frac{1}{\frac{1}{c(rx^2)}-rx}\\
&=&\frac{1}{1-rx+(\frac{1}{c(rx^2)}-1)}\\
&=&\frac{1}{1-rx+\frac{1-c(rx^2)}{c(rx^2)}}\\
&=&\frac{1}{1-rx-r x^2c(rx^2)}\\
&=&\cfrac{1}{1-rx-
\cfrac{rx^2}{1-
\cfrac{rx^2}{1-
\cfrac{rx^2}{1-\cdots}}}}.\end{eqnarray*}
\noindent Thus the generating function of $a_n(r)$ is given by the continued fraction above. An immediate consequence of this is that $a_n(r)$ has Hankel transform $r^{\binom{n+1}{2}}$ \cite{Krat1, Wall_CA}. This proves that these numbers are
``Catalan-like'' ($r \ne 0$).
\noindent We note that the above implies that the bi-variate generating function of the Riordan array
$$(c(rx^2),xc(rx^2))$$ is given by
$$\cfrac{1}{1-xy-
\cfrac{rx^2}{1-
\cfrac{rx^2}{1-
\cfrac{rx^2}{1-\cdots}}}}$$ while that of the generalized Riordan array
$$(c(rx^2),rxc(rx^2))$$ is given by
$$\cfrac{1}{1-rxy-
\cfrac{rx^2}{1-
\cfrac{rx^2}{1-
\cfrac{rx^2}{1-\cdots}}}}.$$
\noindent There is another link to continued fractions, via the \emph{Deleham construction}. For the purposes of this note, we define this as follows.
Given two sequences $r_n$ and $s_n$,we use the notation
$$ r \quad \Delta \quad s = [r_0,r_1,r_2,\ldots] \quad \Delta \quad [s_0,s_1,s_2,\ldots]$$ to denote the
number triangle whose bi-variate generating function is given by
$$\cfrac{1}{1-
\cfrac{(r_0x+s_0 xy)}{1-
\cfrac{(r_1x+s_1 xy)}{1-
\cfrac{(r_2x+s_2 xy)}{1-\cdots}}}}.$$
In this instance, we follow Deleham in \seqnum{A120730} by taking
$r_n$ to be the sequence that begins $$0,1,-1,0,0,1,-1,0,0,1,-1,0,\ldots$$ and $s_n$ to be the sequence
that begins $$1,0,0,-1,1,0,0,-1,1,0,0,-1,1,0,\ldots$$ (extending periodically).
We thus arrive at the number triangle with bi-variate generating function
$$\cfrac{1}{1-
\cfrac{xy}{1-
\cfrac{x}{1+
\cfrac{x}{1+
\cfrac{xy}{1-
\cfrac{xy}{1-
\cfrac{x}{1+
\cfrac{x}{1+
\cfrac{xy}{1-\cdots}}}}}}}}}.$$ This is the triangle with general term
$$[n\le 2k]\binom{n}{k}\frac{2k-n+1}{k+1}.$$ A consequence of the fact that
$$a_n(r)=\sum_{k=0}^n [n\le 2k]\binom{n}{k}\frac{2k-n+1}{k+1} r^k$$ is that the generating function
of $a_n(r)$ may also be expressed as
$$\cfrac{1}{1-
\cfrac{rx}{1-
\cfrac{x}{1+
\cfrac{x}{1+
\cfrac{rx}{1-
\cfrac{rx}{1-
\cfrac{x}{1+
\cfrac{x}{1+
\cfrac{rx}{1-\cdots}}}}}}}}}.$$ For example, the generating function of the central binomial numbers $\binom{n}{\lfloor \frac{n}{2} \rfloor}$
can be expressed as
$$\cfrac{1}{1-
\cfrac{x}{1-
\cfrac{x}{1+
\cfrac{x}{1+
\cfrac{x}{1-
\cfrac{x}{1-
\cfrac{x}{1+
\cfrac{x}{1+
\cfrac{x}{1-\cdots}}}}}}}}}=\cfrac{1}{1-x-
\cfrac{x^2}{1-
\cfrac{x^2}{1-
\cfrac{x^2}{1-\cdots}}}}.$$
Similarly the generating function of the sequence with g.f. given by $\frac{c(2x^2}{1-2xc(2x^2)}$, or \seqnum{A151281} (the number of walks within $\mathbf{N}^2$ (the first quadrant of $\mathbf{Z}^2$) starting at $(0,0)$ and consisting
of $n$ steps taken from $\{(-1, 0), (1, 0), (1,1)\}$), can be expressed as a continued fraction as follows\,:

$$\cfrac{1}{1-
\cfrac{2x}{1-
\cfrac{x}{1+
\cfrac{x}{1+
\cfrac{2x}{1-
\cfrac{2x}{1-
\cfrac{x}{1+
\cfrac{x}{1+
\cfrac{2x}{1-\cdots}}}}}}}}}=\cfrac{1}{1-2x-
\cfrac{2x^2}{1-
\cfrac{2x^2}{1-
\cfrac{2x^2}{1-\cdots}}}}.$$
\section{$LDL^t$ decomposition and orthogonal polynomials}
We let $\mathbf{H}=\mathbf{H}(r)$ denote the Hankel matrix of the sequence $a_n(r)$, with general element
$a_{i+j}(r)$. The theory of ``Catalan-like'' numbers ensures us that
$$\mathbf{H}=\mathbf{L}\mathbf{D}\mathbf{L}^t$$ where $\mathbf{D}$ is a diagonal matrix, and $\mathbf{L}=\mathbf{L}(r)$ is a lower-triangular matrix
with $1$'s on the diagonal. Moreover, $\mathbf{L}^{-1}$ is the coefficient array of a family of orthogonal polynomials \cite{PPWW, WW}. Given that
the Hankel transform of $a_n(r)$ is $r^{\binom{n+1}{2}}$, it is clear from the theory that in fact
$$\mathbf{D}=\text{Diag}\{1,r,r^2,r^3,\ldots\}.$$
We obtain
$$\mathbf{L}=\left(\frac{c(rx^2)}{1-rxc(rx^2)}, xc(rx^2)\right),$$ and
$$\mathbf{L}^{-1}=\left(\frac{1-rx}{1+rx^2},\frac{x}{1+rx^2}\right).$$ This latter matrix has general term
$$(-r)^{\lfloor \frac{n-k+1}{2} \rfloor}\binom{n-\lfloor \frac{n-k+1}{2} \rfloor}{\lfloor \frac{n-k}{2} \rfloor}.$$ The matrix
$\mathbf{L}^{-1}$ is the coefficient array of a set of generalized Chebyshev polynomials of the third kind.
To see this, we first let
$$U_n(x;r)=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n-k}{k} (-r)^k (2x)^{n-2k}.$$ $U_n(x;1)$ corresponds to the usual Chebyshev polynomials of the second kind. The family of polynomials
$$U_n(x/2;r)$$ has coefficient array
$$\left(\frac{1}{1+rx^2}, \frac{x}{1+rx^2}\right).$$ The \emph{generalized Chebyshev polynomials of the third kind} can then be defined to be
$$V_n(x;r)=U_n(x;r)-rU_{n-1}(x;r).$$ (Again, $r=1$ corresponds to the usual Chebyshev polynomials of the third kind \cite{Boya}).
\noindent Then $\mathbf{L}^{-1}$ is the coefficient array of the orthogonal polynomials $V_n(x/2;r)$. We can easily verify that these
polynomials satisfy the three-term recurrence
$$V_{n+2}(x/2;r)=xV_{n+1}(x/2;r)-rV_{n}(x/2;r).$$
\begin{example} We take the case $r$=3. Then we get
\begin{displaymath}\mathbf{L}(3)=\left(\begin{array}{ccccccc} 1 & 0 & 0 & 0
&
0 & 0 &
\ldots \\3 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 12 & 3 & 1 & 0 & 0 &
0 & \ldots \\ 45 & 15 & 3 & 1 & 0 & 0 & \ldots \\ 180 & 54 & 18 & 3
&
1 & 0 &
\ldots \\702 & 234 & 63 & 21 & 3 & 1 &\ldots\\ \vdots & \vdots &
\vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right)\end{displaymath} and
\begin{displaymath}\mathbf{L}(3)^{-1}=\left(\begin{array}{ccccccc} 1 & 0 & 0 & 0
&
0 & 0 &
\ldots \\-3 & 1 & 0 & 0 & 0 & 0 & \ldots \\ -3 & -3 & 1 & 0 & 0 &
0 & \ldots \\ 9 & -6 & -3 & 1 & 0 & 0 & \ldots \\ 9 & 18 & -9 & -3
&
1 & 0 &
\ldots \\-27 & 27 & 27 & -12 & -3 & 1 &\ldots\\ \vdots & \vdots &
\vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath}
\end{example}
\noindent $\mathbf{L}(3)$ is an Aigner matrix for the sequences $3,0,0,0,\ldots$ and
$3,3,3,\ldots$. Thus for instance
$$ 702=3\cdot 180+3\cdot 54,\qquad 234=180+3\cdot 18.$$
The generalized Chebyshev polynomials of the third kind associated to $a_n(3)$ begin
$$1, x-3, x^2-3x-3, x^3-3x^2-6x-9, \ldots$$
\section{Moment representation}
By means of the Stieltjes transform \cite{Gautschi}, we can establish that
$$a_n(r)=\frac{1}{2\pi} \int_{-2\sqrt{r}}^{2\sqrt{r}} x^n \frac{\sqrt{4r-x^2}}{r(r+1-x)}dx + \frac{r-1}{r}(r+1)^n.$$
\section{Production matrices} It is instructive to examine the production matrices \cite{ProdMat_0, ProdMat} of some of the Riordan arrays involved in this note.
The Riordan array $$(c(rx^2),xc(rx^2))$$ has production matrix
\begin{displaymath}\left(\begin{array}{ccccccc} 0 & 1 & 0 & 0
&
0 & 0 &
\ldots \\r & 0 & 1 & 0 & 0 & 0 & \ldots \\ 0 & r & 0 & 1 & 0 &
0 & \ldots \\ 0 & 0 & r & 0 & 1 & 0 & \ldots \\ 0 & 0 & 0 & r
&
0 & 1 &
\ldots \\0 & 0 & 0 & 0 & r & 0 &\ldots\\ \vdots & \vdots &
\vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right)\end{displaymath} while the generalized Riordan array
$$(c(rx^2),rxc(rx^2))$$ has production matrix
\begin{displaymath}\left(\begin{array}{ccccccc} 0 & r & 0 & 0
&
0 & 0 &
\ldots \\1 & 0 & r & 0 & 0 & 0 & \ldots \\ 0 & 1 & 0 & r & 0 &
0 & \ldots \\ 0 & 0 & 1 & 0 & r & 0 & \ldots \\ 0 & 0 & 0 & 1
&
0 & r &
\ldots \\0 & 0 & 0 & 0 & 1 & 0 &\ldots\\ \vdots & \vdots &
\vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath} \noindent
The Riordan array
$$\mathbf{L}=\left(\frac{c(rx^2)}{1-rxc(rx^2)}, xc(rx^2)\right)$$ has production matrix
\begin{displaymath}\left(\begin{array}{ccccccc} r & 1 & 0 & 0
&
0 & 0 &
\ldots \\r & 0 & 1 & 0 & 0 & 0 & \ldots \\ 0 & r & 0 & 1 & 0 &
0 & \ldots \\ 0 & 0 & r & 0 & 1 & 0 & \ldots \\ 0 & 0 & 0 & r
&
0 & 1 &
\ldots \\0 & 0 & 0 & 0 & r & 0 &\ldots\\ \vdots & \vdots &
\vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right)\end{displaymath} which confirms that its first column, which is $a_n(r)$, has
generating function given by the continued fraction
$$\cfrac{1}{1-rx-
\cfrac{rx^2}{1-
\cfrac{rx^2}{1-\cdots}}}.$$
\noindent
$\mathbf{L}$ is thus the Aigner matrix for the sequences $s_n$ given by $r,0,0,0,\ldots$ and
$t_n$ given by $r,r,r,\ldots$.
\section{$a_n(r)$ as row sums}
Following \cite{Ferrari_2}, it is possible to exhibit the sequences $a_n(r)$ as the row sums of a given number triangle.
In effect, the matrix $\mathbf{L}\mathbf{B}^{-1}$, where $\mathbf{B}$ is the Binomial matrix with
general term $\binom{n}{k}$ (Pascal's triangle, \seqnum{A007318}), has row sums equal to $a_n(r)$.
This matrix is the inverse of
$$\left(\frac{1}{1-x},\frac{x}{1-x}\right)\cdot \left(\frac{1-rx}{1+rx^2},\frac{x}{1+rx^2}\right)=\left(\frac{1-(r+1)x}{1-2x+(r+1)x^2}, \frac{x(1-x)}{1-2x+(r+1)x^2}\right).$$
We may verify this algebraically: the row sums of
$$\mathbf{L}\mathbf{B}^{-1}=\left(\frac{c(rx^2)}{1-rxc(rx^2)}\frac{1}{1+xc(rx^2)},\frac{xc(rx^2)}{1+xc(rx^2)}\right)$$ do have
generating function $\frac{c(rx^2)}{1-rxc(rx^2)}$.
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\end{thebibliography}

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

\noindent \emph{Keywords:} Integer sequence, Catalan-like, Riordan arrays, Hankel transform, orthogonal polynomials,
Chebyshev polynomials.

\bigskip
\hrule
\bigskip
\noindent (Concerned with sequences
\seqnum{A000108}, \seqnum{A001405},
\seqnum{A007318},
\seqnum{A084938},
\seqnum{A120730},
\seqnum{A151281},
\seqnum{A151162},
\seqnum{A151254},
\seqnum{A156195}, and
\seqnum{A156361}.)

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\vspace*{+.1in}
\noindent
Received March 11 2009;
revised version received  June 6 2009.
Published in {\it Journal of Integer Sequences}, June 6 2009.

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