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\begin{center}
\vskip 1cm{\LARGE\bf On the Central Coefficients of \\
\vskip .1in
Riordan Matrices
}
\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 use the Lagrange-B\"urmann inversion theorem to characterize the generating function of the central coefficients of the elements of the Riordan group of matrices. We apply this result to calculate the generating function of the central elements of a number of explicit Riordan arrays, defined by rational expressions, and in two cases we use the generating functions thus found to calculate the Hankel transforms of the central elements, which are themselves expressible as combinatorial polynomials. We finally look at two cases of Riordan arrays defined by non-rational expressions. The last example uses our methods to calculate the generating function of $\binom{3n}{n}$.
\end{abstract}

\section{Introduction}
The central coefficients of many common (ordinary) Riordan arrays play an important role in combinatorics. For instance, the central terms of the Pascal triangle, defined by the Riordan array $\left(\frac{1}{1-t},\frac{t}{1-t}\right)$, are the central binomial coefficients $\binom{2n}{n}$ \seqnum{A000984}, with a multitude of combinatorial interpretations. Similarly, the central elements of the Delannoy triangle,
$\left(\frac{1}{1-t},\frac{t(1+t)}{1-t}\right)$ are the central Delannoy numbers $\sum_{k=0}^n \binom{n}{k}^2 2^k$ \seqnum{A001850}, again with many combinatorial interpretations. It is interesting therefore to be able to give the generating function of such central terms in a systematic way.
In two recent papers \cite{Barry_Bell, Kruchinin}, it has been shown how to find the generating function of the central elements of elements of the Bell subgroup of the Riordan group. For instance, the Pascal triangle $\left(\frac{1}{1-t},\frac{t}{1-t}\right)$ is an element of this subgroup. In this note, we shall extend this result to arbitrary elements of the Riordan group.
We first recall the result concerning the Bell subgroup that we wish to extend. We use the notation $\bar{f}(t)$ to denote the compositional inverse of the power series $f$ (where $f(0)=0$). This is the unique function such that $f(\bar{f}(t))=\bar{f}(f(t))=t$.
\begin{theorem}\label{laterthm} Let $(d(t),td(t))$ be an element of the Bell subgroup of the Riordan group of matrices $\mathcal{R}$. If $d_{n,k}$ denotes the $(n,k)$-th element of this matrix, then we have
$$d_{2n,n}=(n+1)[t^n]\frac{1}{t}\overline{\left(\frac{t}{d(t)}\right)}(t).$$
\end{theorem}

\begin{corollary} If $d_{n,k}$ denotes the general term of the Bell matrix $(d(t),td(t))$, then the generating function of the central terms $d_{2n,n}$ is given by
$$\frac{d}{dt} \overline{\left(\frac{t}{d(t)}\right)}.$$
\end{corollary}

\begin{example} The most well-known example of a Bell matrix (see later) is the Binomial matrix $\mathbf{B}$ with general element
$d_{n,k}=\binom{n}{k}$. As a member of the Riordan group of matrices, this is
$$\mathbf{B}=\left(\frac{1}{1-t}, \frac{t}{1-t}\right).$$
Then the generating function of $d_{2n,n}=\binom{2n}{n}$ is given by
$$\frac{d}{dt} \overline{t(1-t)}=\frac{d}{dt} \frac{1-\sqrt{1-4t}}{2} = \frac{1}{\sqrt{1-4t}}.$$
\end{example}
The next section provides a brief review of integer sequences, the Lagrange-B\"urmann inversion theorem, and Riordan arrays, and may be skipped by those familiar with these notions. We then announce and prove our general result, and finally we look at some examples.

Many interesting
examples of sequences and Riordan arrays can be found in Neil Sloane's On-Line
Encyclopedia of Integer Sequences (OEIS), \cite{SL1, SL2}. 
Sequences are frequently referred to by their A-number in the OEIS.

\section{Integer sequences, Lagrange Inversion and Riordan arrays}
For an integer sequence $a_n$, that is, an element of $\mathbb{Z}^\mathbb{N}$, the power series
$f(t)=\sum_{k=0}^{\infty}a_k t^k$ is called the \emph{ordinary generating function} or g.f. of the sequence.
$a_n$ is thus the coefficient of $t^n$ in this series. We denote this by
$a_n=[t^n]f(t)$. For instance, $F_n=[t^n]\frac{t}{1-t-t^2}$ is the $n$-th Fibonacci number \seqnum{A000045}, while
$C_n=[t^n]\frac{1-\sqrt{1-4t}}{2t}$ is the $n$-th Catalan number \seqnum{A000108}. The properties and examples of use of the notation
$[t^n]$ can be found in \cite{Merlini_MC}.

For a power series
$f(t)=\sum_{n=0}^{\infty}a_n t^n$ with $f(0)=0$ and $f'(0)\neq 0$ we define the \emph{reversion} or \emph{compositional inverse} of $f$
to be the
power series $\bar{f}(t)$ (also written as $f^{[-1]}(t)$) such that $f(\bar{f}(t))=t$. We sometimes write
$\bar{f}= \text{Rev}f$.

The following is the version of Lagrange inversion that we shall need \cite{Merlini_When, Stanley_2}.
\begin{theorem} (Lagrange-B\"urmann Inversion). Suppose that a formal power series $w = w(t)$
is implicitly defined by the relation $w = t \phi(w)$, where $\phi(t)$ is a formal power series such that
$\phi(0) \ne 0$. Then, for any formal power series $F(t)$,
$$[t^n]F(w(t))=\frac{1}{n}[t^{n-1}] F'(t)(\phi(t))^n.$$
\end{theorem}

In the sequel, we shall be interested in the Hankel transform of the central terms that we shall meet. The Hankel transform \cite{Kratt, Layman} of a sequence $a_n$ is the sequence $h_n$ where $h_n=|a_{i+j}|_{0 \le i,j \le n}$. If the g.f. of $a_n$ is expressible as a continued fraction of the form
$$\cfrac{\mu_0}{1-\alpha_0 t-
\cfrac{\beta_1 t^2}{1-\alpha_1 t -
\cfrac{\beta_2 t^2}{1-\alpha_2 t -
\cfrac{\beta_3 t^2}{1-\alpha_3 t -\cdots}}}},$$
then the Hankel transform of $a_n$ is given by \cite{Kratt}
$$h_n=\mu_0^{n+1}\beta_1^n\beta_2^{n-1}\cdots\beta_{n-1}^2\beta_n.$$

\noindent
The \emph{Riordan group} \cite{CMS, SGWW, Spru}, is a set of
infinite lower-triangular integer matrices, where each matrix is
defined by a pair of generating functions
$d(t)=1+d_1t+d_2t^2+\ldots$ and $h(t)=h_1 t+h_2 t^2+\ldots$ where
$h_1\ne 0$ \cite{Spru}. The corresponding matrix is the matrix whose
$i$-th column is generated by $d(t)h(t)^i$ (the first column being
indexed by $0$). The matrix corresponding to the pair $d, h$ is
denoted by $(d, h)$. The group law, which corresponds to matrix multiplication, is then given
by
\begin{displaymath} (d, h)\cdot(f, g)=(d, h)(f, g)=(d(f\circ h), g\circ
h).\end{displaymath} The identity for this law is $I=(1,t)$ and the
inverse of $(d, h)$ is
$$(d, h)^{-1}=(1/(d\circ \bar{h}), \bar{h}),$$
where $\bar{h}$ is the compositional inverse of $h$.
We denote by $\mathcal{R}$ the group of Riordan matrices.
If $\mathbf{M}$ is the matrix $(d,h)$, and
$\mathbf{a}=(a_0,a_1,\ldots)^T$ is an integer sequence with ordinary
generating function $\cal{A}$$(t)$, then the sequence
$\mathbf{M}\mathbf{a}$ has ordinary generating function
$d(t)$$\cal{A}$$(h(t))$. The (infinite) matrix $(d,h)$ can thus be considered to act on the ring of
integer sequences $\mathbb{Z}^\mathbb{N}$ by multiplication, where a sequence is regarded as a
(infinite) column vector. We can extend this action to the ring of power series
$\mathbb{Z}[[t]]$ by
$$(d,h):\cal{A}(\mathnormal{t}) \mapsto \mathnormal{(d,h)}\cdot
\cal{A}\mathnormal{(t)=d(t)}\cal{A}\mathnormal{(h(t))}.$$
\begin{example} The so-called \emph{binomial matrix} $\mathbf{B}$ \seqnum{A007318} is the element
$(\frac{1}{1-t},\frac{t}{1-t})$ of the Riordan group. It has general
element $\binom{n}{k}$, and hence as an array coincides with Pascal's triangle. More generally, $\mathbf{B}^m$ is the
element $(\frac{1}{1-m t},\frac{t}{1-mt})$ of the Riordan group,
with general term $\binom{n}{k}m^{n-k}$. We find that the
inverse $\mathbf{B}^{-m}$ of $\mathbf{B}^m$ is given by
$(\frac{1}{1+mt},\frac{t}{1+mt})$.
\end{example}

The \emph{Bell} subgroup of $\mathcal{R}$ is the set of matrices of the form
$$(d(t),td(t)).$$ Note that
$$(d(t),td(t))^{-1}=\left(\frac{\overline{td}}{t},\overline{td}\right).$$ An interesting sequence characterization of
Bell matrices may be found in \cite{Jin}.
\section{The main result}
The main result of this note is the following.
\begin{theorem}\label{laterthm1} Let $(d(t),h(t))=(d(t),t f(t))$ be an element of the  Riordan group of matrices $\mathcal{R}$ (with $f(0) \ne 0$).  Let $d_{n,k}$ denote the $(n,k)$-th element of this matrix, and let $v(t)$ denote the power series (with $v(0)=0$)
$$v(t)=\overline{\left(\frac{t}{f(t)}\right)}.$$ Then the generating function of the central term sequence $d_{2n,n}$ is given by

$$\frac{d(v(t))}{f(v(t))} \frac{d}{dt} v(t).$$
\end{theorem}
\begin{proof}
We let $d_{n,k}$ denote the general element of the Riordan array $(d(t),tf(t))$.
Then we have
\begin{eqnarray*}
d_{2n,n}&amp;=&amp;[t^{2n}] d(t)(t f(t))^n\\
&amp;=&amp;[t^{2n}] t^n d(t)f(t)^{n}\\
&amp;=&amp;[t^n] \frac{d(t)}{f(t)} f(t)^{n+1}\\
&amp;=&amp;(n+1) \frac{1}{n+1}[t^n] F'(t)f(t)^{n+1},\end{eqnarray*}
where $F'(t)=\frac{d(t)}{f(t)}$.
Applying the Lagrange-B\"urmann inversion theorem, we obtain
\begin{eqnarray*}d_{2n,n}&amp;=&amp;(n+1) [t^{n+1}] F(v(t))\\
&amp;=&amp; [t^n] F'(v(t)) v'(t) \\
&amp;=&amp; [t^n] \frac{d(v(t))}{f(v(t))} \frac{d}{dt} v(t).\end{eqnarray*}
\end{proof}


\section{Examples}
\begin{example} We look at the matrix
$$\left(\frac{1}{1-2t},\frac{t(1-2t)}{1-3t}\right).$$
The central elements of this matrix begin $1, 3, 15, 90, 579, 3858,\ldots$.
In this case, we have
$$v(t)=\overline{\left(\frac{t}{f(t)}\right)}=\overline{\left(\frac{t(1-3t)}{1-2t}\right)}.$$
Thus $v(t)$ is the appropriate solution (with $v(0)=0$) of the equation
$$\frac{v(1-3v)}{1-2v}=t.$$
\noindent We find that
$$v(t)=\frac{1+2t-\sqrt{1-8t+4t^2}}{6}.$$
We also have that
$$\frac{d(t)}{f(t)}=\frac{1-3t}{(1-2t)^2}.$$
The generating function of $d_{2n,n}$ is thus given by
$$\frac{d(v(t))}{f(v(t))} \frac{d}{dt} v(t)=\frac{1+4t+\sqrt{1-8t+4t^2}}{2 \sqrt{1-8t+4t^2}}.$$
We can generalize this result to the case of the Riordan array
$$\left(\frac{1}{1-rt}, \frac{t(1-rt)}{1-(r+1)t}\right).$$ We find in this case that the central term sequence
$d_{2n,n}$ is generated by
$$\frac{1+rt+\sqrt{1-2(r+2)t+r^2t^2}}{2\sqrt{1-2(r+2)t+r^2t^2}}.$$
The general element of $\left(\frac{1}{1-rt}, \frac{t(1-rt)}{1-(r+1)t}\right)$ is given by
$$d_{n,k}=[t^n] \frac{1}{1-rt}\left(\frac{t(1-rt)}{1-(r+1)t}\right)^k,$$ which gives us the expression
$$d_{2n,n}=\sum_{k=0}^n \binom{n-1}{k}\binom{2n-k-1}{n-k}(-r)^k (r+1)^{n-k}.$$
The bivariate generating function
$$\frac{1+rt+\sqrt{1-2(y+2)t+y^2t^2}}{2\sqrt{1-2(y+2)t+y^2t^2}}$$ is the generating function of the lower-triangular matrix with
general element
$$\binom{n}{k}\binom{2n-k-1}{n-k},$$ and so we also have
$$d_{2n,n}=\sum_{k=0}^n \binom{n}{k}\binom{2n-k-1}{n-k} r^k.$$
\noindent We note that this matrix begins
\begin{displaymath}\left(\begin{array}{ccccc} 1 &amp; 0 &amp; 0 &amp; 0
&amp; \cdots \\1 &amp; 1 &amp; 0 &amp; 0  &amp; \cdots \\ 3 &amp; 4 &amp;
1 &amp; 0  &amp; \cdots \\ 10 &amp; 18 &amp; 9 &amp; 1 &amp; \cdots \\  \vdots &amp;
\vdots &amp; \vdots &amp; \vdots  &amp;
\ddots\end{array}\right).\end{displaymath}
\noindent We finish this example by offering the following conjecture for the form of the Hankel transform $h_n$ of $d_{2n,n}$ \cite{Layman}.
$$h_n=(r+1)^{\lceil \frac{n^2}{2} \rceil} \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} 4^k r^k \sum_{j=0}^{n-2k} \binom{2k+j+1}{j}\binom{n-j-1}{n-2k-j}.$$
\end{example}
\begin{example} In this example, we start with the Riordan array
$$\left(\frac{1}{1-2t}, \frac{t(1-3t)}{1-2t}\right).$$
Then $$ \frac{v(1-2v)}{1-3v}=t$$ gives us
$$v(t)=\frac{1-3t-\sqrt{1-14t+9t^2}}{4},$$ while in this case we have
$$\frac{d(t)}{f(t)}=\frac{1}{1+3t}.$$
We obtain that the sequence of central terms $d_{2n,n}$, which begins
$$1, 7, 69, 763, 8881, 106407, 1298949, 16065483, \ldots,$$  has generating function
$$\frac{1}{\sqrt{1-14t+9t^2}}.$$
This sequence is \seqnum{A084774}. It corresponds to the central coefficients of $(1+7t+10t^2)^n$.
We now look at the general case of the matrix
$$\left(\frac{1}{1-rt}, \frac{t(1-(r+1)t}{1-rt}\right).$$ We obtain
$$v(t)=\frac{1-(1+r)t-\sqrt{1-2(1+3r)t+(r+1)^2 t^2}}{2r},$$ and
$$\frac{d(t)}{f(t)}=\frac{1}{1+(1+r)t}.$$ \noindent We find that the central term sequence $d_{2n,n}$ is generated by
$$\frac{1}{\sqrt{1-2(3r+1)t+(r+1)^2t^2}}.$$
We can then show \cite{Noe} that
$$d_{2n,n}=[t^n](1+(3r+1)t+r(r+1)t^2)^n,$$ and
$$d_{2n,n}=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k}\binom{2k}{k}(3r+1)^{n-2k}(r(2r+1))^k.$$
\noindent We have
$$\frac{1}{\sqrt{1-2(3r+1)t+(r+1)^2t^2}}=
\cfrac{1}{1-(3r+1)t-
\cfrac{2r(r+1)t^2}{1-(3r+1)t-
\cfrac{r(r+1)t^2}{1-(3r+1)t-\cdots}}}.$$
\noindent
This shows that the Hankel transform of $d_{2n,n}$ in this instance is given by
$$h_n = 2^n (r(r+1))^{\binom{n+1}{2}}.$$
\end{example}
\begin{example}
We finish this section by looking at the Pascal-like triangles \cite{Barry_Pascal} given by
$$\left(\frac{1}{1-t},\frac{t(1+rt)}{1-t}\right).$$
We have
$$\frac{v(1-v)}{1+rv}=t,$$ and hence
$$v(t)=\frac{1-rt+\sqrt{1-2(r+2)t+r^2t^2}}{2},$$ and
$$\frac{d(t)}{f(t)}=\frac{1}{1+rt}.$$
We find that the generating function of $d_{2n,n}$ is given by
$$\frac{1}{\sqrt{1-2(r+2)t+r^2t^2}},$$ which \cite{Noe} shows that
$$d_{2n,n}=[t^n](1+(r+2)t+(r+1)t^n)^n,$$ with
$$d_{2n,n}=\sum_{k=0}^n \binom{n}{k}\binom{2n-k}{n}r^k.$$
\noindent We have
$$\frac{1}{\sqrt{1-2(r+2)t+r^2t^2}}=
\cfrac{1}{1-(r+2)t-
\cfrac{2(r+1)t^2}{1-(r+2)t-
\cfrac{(r+1)t^2}{1-(r+2)t-\cdots}}}.$$
\noindent
This shows that the Hankel transform of $d_{2n,n}$ in this instance is given by
$$h_n = 2^n (r+1)^{\binom{n+1}{2}}.$$

\end{example}
\section{Non-rational terms}
In this section, we look at two simple examples of Riordan arrays defined by non-rational terms. In the first, the generating function of the central terms is elementary, while this is not the case for the central elements of the second array.
\begin{example}
We consider the Riordan array
$$\left(\frac{1}{\sqrt{1-4t}}, \frac{t}{c(t)}\right),$$
where $$c(t)=\frac{1-\sqrt{1-4t}}{2t}$$ is the generating function of the Catalan numbers \seqnum{A000108}.
This matrix begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 &amp; 0 &amp;
0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\2 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 6 &amp; 1
&amp; 1 &amp; 0 &amp; 0 &amp;
0 &amp; \ldots \\ 20 &amp; 3 &amp; 0 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 70 &amp; 10 &amp; 1
&amp; -1 &amp; 1 &amp; 0 &amp; \ldots \\252 &amp; 35 &amp; 4 &amp; 0 &amp; -2 &amp; 1
&amp;\ldots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath}


\noindent We have
$$v c(v)=t,$$ and hence
$$v(t)=t(1-t).$$
We have
$$\frac{d(t)}{f(t)}=\frac{1-\sqrt{1-4t}}{2t\sqrt{1-4t}}.$$
\noindent
We find that
$$\frac{d(v(t))}{f(v(t))}=\frac{1}{(1-t)(1-2t)}.$$
\noindent But $$(t(1-t))'=1-2t,$$ and so we find that the generating function of $d_{2n,n}$ in this case is $\frac{1}{1-t}$.
Thus the central terms of the Riordan array $\left(\frac{1}{\sqrt{1-4t}}, \frac{t}{c(t)}\right)$ are all equal to $1$.
\end{example}
\begin{example} For this example, we look at the array
$$\left(\frac{1}{\sqrt{1-4t}}, tc(t)\right).$$
\noindent This is the matrix \seqnum{A092392} with general element $$d_{n,k}=\binom{2n-k}{n}.$$
\noindent Thus we have
$$d_{2n,n}=\binom{3n}{2n}=\binom{3n}{n},$$ which begins
$$1, 3, 15, 84, 495, 3003, 18564, 116280, \ldots.$$ \noindent This is \seqnum{A005809}.
Solving the equation
$$\frac{v}{c(v)}=t,$$ we find that
$$v(t)=\frac{2}{\sqrt{3}} \sqrt{t} \sin\left(\frac{1}{3}\arcsin\left(\frac{1}{2} \sqrt{27t}\right)\right).$$
\noindent
This is the g.f. of the sequence that begins
$$0,1, 1, 3, 12, 55, 273, 1428, \ldots,$$ or the sequence $\frac{1}{2n+1} \binom{3n}{n}$ with a $0$ at the beginning.
Thus $v'(t)$ is the g.f. of $\frac{n+1}{2n+1}\binom{3n}{n}$, \seqnum{A174687}. We have
$$\frac{d(t)}{f(t)}=\frac{1}{\sqrt{1-4t}}\frac{1}{c(t)}=\frac{1-\sqrt{1-4t}}{2t\sqrt{1-4t}},$$
from which it follows that we have
$$\frac{d(v(t))}{f(v(t))}=\frac{1}{2}\left(1+\frac{3^{1/4}} {\sqrt{\sqrt{3}-8\sqrt{t}\sin\left(\frac{1}{3}\arcsin\left(\frac{\sqrt{27t}}{2}\right)\right)}}\right).$$
\noindent Thus the generating function of the central elements $\binom{3n}{n}$ of the Riordan array $$\left(\frac{1}{\sqrt{1-4t}}, tc(t)\right)$$ is given by $$\frac{1}{\sqrt{3}}\left(1+\frac{3^{1/4}} {\sqrt{\sqrt{3}-8\sqrt{t}\sin\left(\frac{1}{3}\arcsin\left(\frac{\sqrt{27t}}{2}\right)\right)}}\right)\frac{d}{dt}\sqrt{t} \sin\left(\frac{1}{3}\arcsin\left(\frac{1}{2} \sqrt{27t}\right)\right).$$
\noindent We see that even in the case of this relatively simply defined Riordan array, the generating function of the central terms is far from elementary.
\end{example}
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\noindent 2010 {\it Mathematics Subject Classification}: Primary
11B83; Secondary 05A15, 11C20,  15B36.
\noindent \emph{Keywords:} Integer sequence, central coefficients, Riordan group, Lagrange inversion, Hankel transform.

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\noindent (Concerned with sequences
\seqnum{A000045},
\seqnum{A000108},
\seqnum{A000984},
\seqnum{A001850},
\seqnum{A005809},
\seqnum{A007318},
\seqnum{A084774},
\seqnum{A092392}, and
\seqnum{A174687}.)

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\noindent
Received January 2 2013;
revised version received May 2 2013.
Published in {\it Journal of Integer Sequences}, May 8 2013.

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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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