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\begin{center}
\vskip 1cm{\LARGE\bf On the Central Coefficients of Bell Matrices} \vskip 1cm
\large
Paul Barry\\
School of Science\\
Waterford Institute of Technology\\
Ireland\\
\href{mailto:pbarry@wit.ie}{\tt pbarry@wit.ie} \\
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\begin{abstract} 
We use the Lagrange inversion theorem to characterize the central
coefficients of matrices in the Bell subgroup of the Riordan group of
matrices. We give examples of how by using different means of
calculating these coefficients we can deduce the generating functions
of interesting sequences.
\end{abstract}

\section{Introduction}

\begin{example} The most well-known example of a Bell matrix (see later) is the Binomial matrix $\mathbf{B}$ with general element
$T_{n,k}=\binom{n}{k}$. As a member of the Riordan group of matrices, this is
$$\mathbf{B}=\left(\frac{1}{1-x}, \frac{x}{1-x}\right).$$  By the \emph{central coefficients} of this matrix we understand the terms
$T_{2n,n}$. In this case, we have
$$T_{2n,n}=\binom{2n}{n}=[x^n] \frac{1}{\sqrt{1-4x}}.$$
\noindent A natural question to ask is how does the defining element
 $$\frac{1}{1-x}$$ of $\mathbf{B}$ relate to $\frac{1}{\sqrt{1-4x}}$ ? The answer is in fact quite simple. In this case, we form the associated Bell matrix
 $$ (1-x, x(1-x))$$ and take its inverse. This is $$(c(x), xc(x)),$$ where $c(x)=\frac{1-\sqrt{1-4x}}{2x}$ is the generating function of the Catalan numbers. Then we have
 $$\binom{2n}{n}=(n+1)[x^n] c(x),$$ or equivalently,
 $$ [x^n]c(x)=\frac{1}{n+1}\binom{2n}{n}=\frac{1}{n+1}T_{2n,n}.$$

 We now note that $$xc(x)=\overline{x(1-x)}(x)=\overline {\left(\frac{x}{\frac{1}{1-x}}\right)}(x),$$ where
 $\,\bar{}\,$ denotes the reversion or compositional inverse.
 Finally, we have
 $$T_{2n,n}=\binom{2n}{n}=(n+1)[x^n]\frac{1}{x}\overline {\left(\frac{x}{\frac{1}{1-x}}\right)}(x).$$
\end{example}
The above result can be generalized to all elements of the Bell subgroup of the Riordan group of matrices. We have the following
theorem, which we shall prove in section $\mathbf{3}$.
\begin{xrefthm}{laterthm} Let $(g(x),xg(x))$ be an element of the Bell subgroup of the Riordan group of matrices $\mathcal{R}$. If $T_{n,k}$ denotes the $n,k$-th element of this matrix, then we have
$$T_{2n,n}=(n+1)[x^n]\frac{1}{x}\overline{\left(\frac{x}{g(x)}\right)}(x).$$
\end{xrefthm}

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

For a power series
$f(x)=\sum_{n=0}^{\infty}a_n x^n$ 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 $\bar{f}(x)$ (also written as $f^{[-1]}(x)$) such that $f(\bar{f}(x))=x$. We sometimes write
$\bar{f}= \text{Rev}f$.

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
$g(x)=1+g_1x+g_2x^2+\ldots$ and $f(x)=f_1x+f_2x^2+\ldots$ where
$f_1\ne 0$ \cite{Spru}. The corresponding 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 $(g, f)$. The group law, which corresponds to matrix multiplication, is then given
by
\begin{displaymath} (g, f)\cdot(h, l)=(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 $f$.
We denote by $\mathcal{R}$ the group of Riordan matrices.
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}$$(f(x))$. The (infinite) matrix $(g,f)$ can thus be considered to act on the ring of
integer sequences $\mathbb{Z}^\mathbb{N}$ by multiplication, where a sequence is regarded as a
(infinite) column vector. We can extend this action to the ring of power series
$\mathbb{Z}[[x]]$ by
$$(g,f):\cal{A}(\mathnormal{x}) \mapsto \mathnormal{(g,f)}\cdot
\cal{A}\mathnormal{(x)=g(x)}\cal{A}\mathnormal{(f(x))}.$$
\begin{example} The so-called \emph{binomial matrix} $\mathbf{B}$ \seqnum{A007318} is the element
$(\frac{1}{1-x},\frac{x}{1-x})$ 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 $(\frac{1}{1-m x},\frac{x}{1-mx})$ 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 \emph{Bell} subgroup of $\mathcal{R}$ is the set of matrices of the form
$$(g(x),xg(x)).$$ Note that
$$(g(x),xg(x))^{-1}=\left(\frac{\overline{xg}}{x},\overline{xg}\right).$$
We shall call a member of the Bell subgroup of $\mathcal{R}$ a Bell matrix. An interesting sequence characterization of
Bell matrices may be found in \cite{Jin}.
\section{Proof of the theorem}
We restate the theorem and prove it in this section.
\begin{theorem}\label{laterthm} Let $(g(x),xg(x))$ be an element of the Bell subgroup of the Riordan group of matrices $\mathcal{R}$. If $T_{n,k}$ denotes the $n,k$-th element of this matrix, then we have
$$T_{2n,n}=(n+1)[x^n]\frac{1}{x}\overline{\left(\frac{x}{g(x)}\right)}(x).$$
\end{theorem}
\begin{proof}
We let $T_{n,k}$ denote the general element of the Bell matrix $(g(x),xg(x))$.
Then we have
\begin{eqnarray*}
T_{2n,n}&amp;=&amp;[x^{2n}] g(x)(xg(x))^n\\
&amp;=&amp;[x^{2n}] x^n g(x)^{n+1}\\
&amp;=&amp;[x^n] g(x)^{n+1}\\
&amp;=&amp;(n+1) \frac{1}{n+1}[x^n]g(x)^{n+1}.\end{eqnarray*}

We now recall the Lagrange inversion theorem \cite{Merlini_When}. This says that if we have
$$h(w)=\frac{w}{g(w)}, \quad g(0)\neq 0,$$ then
$$[x^n]\bar{h}(x)=\frac{1}{n}[w^{n-1}]g(w)^n.$$ Letting $n \rightarrow n+1$ gives
$$[x^{n+1}]\overline{\left(\frac{x}{g(x)}\right)}(x)=\frac{1}{n+1}[w^n]g(w)^{n+1},$$ or
$$[x^{n}]\frac{1}{x}\overline{\left(\frac{x}{g(x)}\right)}(x)=\frac{1}{n+1}[w^n]g(w)^{n+1}.$$

Thus we deduce that
$$T_{2n,n}=(n+1) \frac{1}{n+1}[x^n]g(x)^{n+1}=(n+1)[x^{n}]\frac{1}{x}\overline{\left(\frac{x}{g(x)}\right)}(x).$$
\end{proof}

\begin{corollary}
The central coefficients $T_{2n,n}$ of the Bell matrix $(g(x), xg(x))$ are given by $(n+1)$ times the elements of the first column of the Bell matrix
$$\left(\frac{1}{g(x)},\frac{x}{g(x)}\right)^{-1}.$$
\end{corollary}
\begin{proof} By the theory of the Bell subgroup, we have

$$\left(\frac{1}{g(x)},\frac{x}{g(x)}\right)^{-1}=\left(\frac{1}{x}\overline{\left(\frac{x}{g(x)}\right)},\,\overline{\left(\frac{x}{g(x)}\right)}\right).$$
Since $T_{2n,n}=(n+1)[x^n]\frac{1}{x}\overline{\left(\frac{x}{g(x)}\right)}$ by the Theorem, the result follows.
\end{proof}

\section{Examples}
In this section, we show how, by calculating $T_{2n,n}$ in different ways, we can show that certain sequences of interest have a generating function of a given form. Normally these sequences will have the form $\frac{1}{n+1}T_{2n,n}$. We start with the best-known example, the Catalan numbers.
\begin{example} \textbf{The Catalan numbers} $C_n=\frac{1}{n+1}\binom{2n}{n}$.

For this example, we consider, as in the introduction, the Riordan array
$$\left(\frac{1}{1-x},\frac{x}{1-x}\right).$$ We can calculate $T_{2n,n}$ directly as follows.
\begin{eqnarray*}
T_{2n,n}&amp;=&amp;[x^{2n}]\frac{1}{1-x}\frac{x^n}{(1-x)^n}\\
&amp;=&amp;[x^n](1-x)^{-n-1}\\
&amp;=&amp;[x^n]\sum_{k=0}^{\infty} \binom{-n-1}{k}(-1)^k x^k\\
&amp;=&amp;[x^n]\sum_{k=0} \binom{n+k}{k}x^k\\
&amp;=&amp;\binom{2n}{n},\end{eqnarray*} as expected.
Now by the theorem, we look at the first column of the matrix $(1-x,x(1-x))^{-1}$. We obtain
$$T_{2n,n}=(n+1)[x^n]\frac{1}{x}\overline{\left(\frac{x}{\frac{1}{1-x}}\right)}=(n+1)[x^n]\frac{1}{x}\,\overline{x(1-x)}(x).$$
To carry out the reversion we solve $u(1-u)=x$ to obtain the result (with $u(0)=0$) given by
$$u=\frac{1-\sqrt{1-4x}}{2}.$$
Thus $$T_{2n,n}=(n+1)[x^n]\frac{1-\sqrt{1-4x}}{2x}.$$
Since $T_{2n,n}=\binom{2n}{n}$, we obtain
$$C_n = \frac{1}{n+1}\binom{2n}{n}=[x^n]\frac{1-\sqrt{1-4x}}{2x}.$$
\end{example}
\begin{example} \textbf{The generalized Catalan numbers} $\frac{1}{2n+1}\binom{3n}{n}$. We wish to prove the following result about the generating function of $\frac{1}{2n+1}\binom{3n}{n}$:
$$\frac{1}{2n+1}\binom{3n}{n}=[x^n]\frac{2}{\sqrt{3x}}\sin\left(\frac{\arcsin\left(\sqrt{\frac{27x}{4}}\right)}{3}\right).$$
In order to do this, we calculate the central coefficients $T_{2n,n}$ of the Catalan triangle $(c(x),xc(x))$ \seqnum{A033184} in two ways. First of all, by the theorem, they are given by $(n+1)$ times the first column of
$$(1/c(x),x/c(x))^{-1}.$$ Thus
\begin{eqnarray*}T_{2n,n}&amp;=&amp;(n+1)[x^n]\frac{1}{x}\overline{\left(\frac{x}{c(x)}\right)}(x)\\
&amp;=&amp;(n+1)[x^n]\frac{1}{x}\overline{\left(\frac{x(1+\sqrt{1-4x})}{2}\right)}(x).\end{eqnarray*}
Taking the solution $$u=\frac{2\sqrt{x}}{\sqrt{3}}\sin\left(\frac{\arcsin\left(\sqrt{\frac{27x}{4}}\right)}{3}\right)$$ of the equation $$\frac{u(1+\sqrt{1-4u})}{2}=x,$$ we obtain
$$T_{2n,n}=(n+1)[x^n]\frac{2}{\sqrt{3x}}\sin\left(\frac{\arcsin\left(\sqrt{\frac{27x}{4}}\right)}{3}\right).$$
Secondly, we now note that directly, we have
$$T_{2n,n}=[x^{2n}]c(x)(xc(x))^n=[x^n]c(x)^{n+1}.$$
Then by Formula $(2.4)$ of \cite{Merlini_When}, used backwards, we obtain
$$T_{2n,n}=\frac{n+1}{n+1+2n}\binom{n+1+2n}{n}=\frac{n+1}{3n+1}\frac{(3n+1)(3n)!}{(2n+1)(2n)!n!}=\frac{n+1}{2n+1}\binom{3n}{n}.$$
The result follows immediately by comparing the two expressions for $T_{2n,n}$.
\end{example}
\begin{example}
We wish to show that the sequence \seqnum{A001002} with general term
$$\frac{1}{n+1}\sum_{k=0}^n \binom{n+k}{k}\binom{k}{n-k}$$ has g.f. given by
$$\frac{1}{3x}\left(4 \sin\left(\frac{\arcsin\left(\frac{27x+11}{6}\right)}{3}\right)-1\right).$$
To this end, we consider the central coefficients of the Riordan array
$$\left(\frac{1}{1-x-x^2},\frac{x}{1-x-x^2}\right).$$ By the theorem, $T_{2n,n}$ for this array is given by
$(n+1)$ times the first column of
$$((1-x-x^2),x(1-x-x^2))^{-1}.$$ Thus
\begin{eqnarray*}T_{2n,n}&amp;=&amp;(n+1)[x^n]\frac{1}{x}\overline{(x(1-x-x^2))}(x)\\
&amp;=&amp;(n+1)[x^n]\frac{1}{x}\frac{1}{3}\left(4 \sin\left(\frac{\arcsin\left(\frac{27x+11}{6}\right)}{3}\right)-1\right).\end{eqnarray*}
On the other hand, we have
\begin{eqnarray*}T_{2n,n}&amp;=&amp;[x^{2n}]\frac{1}{1-x-x^2}\frac{x^n}{(1-x-x^2)^n}\\&amp;=&amp;[x^n]\frac{1}{(1-x-x^2)^{n+1}}\\
&amp;=&amp;[x^n]\sum_{k=0}\binom{-n-1}{k}(-x)^k(1+x)^k\\
&amp;=&amp;[x^n]\sum_{k=0}\binom{n+k}{k}x^k \sum_{j=0}^k \binom{k}{j}x^j\\
&amp;=&amp;\sum_{k=0}\binom{n+k}{k}\binom{k}{n-k}.\end{eqnarray*}
This is
\seqnum{A038112}. A comparison of both expressions for $T_{2n,n}$ now yields the result.
\end{example}
\begin{example}
For our next example, we will use the following result.
\begin{lemma} Let $f(x)=\frac{x}{1+ax^2+bx^2}$. Then
\begin{enumerate}
\item $$\bar{f}(x)=\frac{1-ax-\sqrt{1-2ax+(a^2-4b)x^2}}{2bx},$$
\item $$[x^n]\bar{f}(x)=\sum_{k=0}\binom{n-1}{2k} C_k a^{n-2k-1}b^k.$$
\end{enumerate}
\end{lemma}
\begin{proof}
The first result follows by solving the equation
$$\frac{u}{1+au+bu^2}=x$$ and taking the determination for which
$u(0)=0$.
The second result is a direct application of Lagrange inversion. We have
$$[x^n]\bar{f}=\frac{1}{n}[x^{n-1}](1+ax+bx^2)^n.$$
Expansion of the binomial  now gives the result \cite{Tri}.
\end{proof}
We note that this implies that
$$[x^n]\frac{1}{x}\,\bar{f}(x)=\sum_{k=0}\binom{n}{2k} C_k a^{n-2k}b^k.$$
We now wish to study the sequence \seqnum{A007440} with general term
$$\frac{1}{n+1}\sum_{k=0}\binom{n+1}{k}\binom{k}{n-k}(-1)^k=\sum_{k=0}\binom{n}{k}\binom{k}{n-k}\frac{(-1)^k}{n-k+1}.$$
For this, we consider the matrix
$$({1-x-x^2},x(1-x-x^2)).$$ For this matrix, $T_{2n,n}$ is given by $(n+1)$
times the first column of
$$\left(\frac{1}{1-x-x^2},\frac{x}{1-x-x^2}\right)^{-1}.$$ Equivalently
$$T_{2n,n}=(n+1)[x^{n}]\frac{1}{x}\overline{\left(\frac{x}{1-x-x^2}\right)}(x).$$
We obtain
$$T_{2n,n}=(n+1)[x^n]\frac{\sqrt{1+2x+5x^2}-x-1}{2x^2}=[x^n]\frac{\sqrt{1+2x+5x^2}-x-1}{2x^2\sqrt{1+2x+5x^2}}.$$
By the lemma, we also have
$$T_{2n,n}=(n+1)\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k}C_k (-1)^{n-k}. $$ This is essentially \seqnum{A104506}.
We now calculate $T_{2n,n}$ directly.
\begin{eqnarray*}T_{2n,n}&amp;=&amp;[x^{2n}](1-x-x^2)x^n(1-x-x^2)^n\\
&amp;=&amp;[x^n](1-x(1+x))^{n+1}\\
&amp;=&amp;[x^n] \sum_{k=0}^{n+1}\binom{n+1}{k}(-1)^k x^k \sum_{j=0}^k \binom{k}{j}x^j\\
&amp;=&amp;\sum_{k=0}\binom{n+1}{k}\binom{k}{n-k}(-1)^k.\end{eqnarray*}
Thus \begin{eqnarray*}
\frac{1}{n+1}\sum_{k=0}\binom{n+1}{k}\binom{k}{n-k}(-1)^k&amp;=&amp;\sum_{k=0}\binom{n}{k}\binom{k}{n-k}\frac{(-1)^k}{n-k+1}\\
&amp;=&amp;\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k}C_k (-1)^{n-k}\\
&amp;=&amp;[x^n]\frac{\sqrt{1+2x+5x^2}-x-1}{2x^2}.\end{eqnarray*}
This is the shifted reversion of the Fibonacci numbers.
We remark in passing that the Hankel transform \cite{Layman} of the Fibonacci reversion is $-(-1)^{\binom{n+1}{2}}\, F(n)$. It is also noteworthy that the exponential generating function of
$T_{2n,n}=\sum_{k=0}\binom{n+1}{k}\binom{k}{n-k}(-1)^k$ is given by
$$\frac{d}{dx}e^{-x}I_1(2ix)/i,\quad i=\sqrt{-1}.$$ The Hankel transform of $T_{2n,n}$ begins
$$1, -4, -4, 11, 11, -29, -29, 76, 76, -199, -199,...$$ where the sequence \seqnum{A002878} that begins
$1,4,11,29,76,....$ has general term $L_{2n+1}$.
\end{example}

\begin{example} In our last example we wish to show that the sequence \seqnum{A078531} with general term
$$\frac{4^n}{n+1} \binom{\frac{3n-1}{2}}{n}$$ has g.f. given by $$\frac{1}{x}\left(\frac{1}{12}+\frac{1}{6}\sin\left(\frac{\arcsin(216x^2-1)}{3}\right)\right).$$
To this end we consider the central coefficients of the Riordan array
$$\left(\frac{1}{\sqrt{1-4x}},\,\frac{x}{\sqrt{1-4x}}\right).$$ The central coefficients $T_{2n,n}$ of this matrix will then be
given by the first column of the Riordan array
$$\left(\sqrt{1-4x},x\sqrt{1-4x}\right)^{-1}.$$ We find that
$$T_{2n,n}=(n+1)[x^n]\frac{1}{x}\left(\frac{1}{12}+\frac{1}{6}\sin\left(\frac{\arcsin(216x^2-1)}{3}\right)\right).$$
Now we can also calculate $T_{2n,n}$ directly as:
\begin{eqnarray*}T_{2n,n}&amp;=&amp;[x^{2n}]\frac{1}{\sqrt{1-4x}}\frac{x^n}{\sqrt{1-4x}^n}\\
&amp;=&amp;[x^n](1-4x)^{-\frac{n+1}{2}}\\
&amp;=&amp;[x^n]\sum_{k=0}\binom{-\frac{n+1}{2}}{k}(-1)^k 4^k x^k\\
&amp;=&amp;[x^n]\sum_{k=0}\binom{\frac{n+1}{2}+k-1}{k} 4^k x^k\\
&amp;=&amp;\binom{\frac{3n-1}{2}}{n}4^n.\end{eqnarray*}
Comparison of the expressions for $T_{2n,n}$ now gives the result.

Note that $T_{2n,n}=(n+1)$\seqnum{A078531}$(n)=$\seqnum{A091527}$(n)$.
\end{example}
\section{Acknowledgements} I am grateful to an anonymous referee whose careful reading of this manuscript, along with cogent remarks, has hopefully improved the readability and usefulness of this paper.
\begin{thebibliography}{9}

\bibitem{Tri} P. Barry,
On integer-sequence-based constructions of generalized Pascal triangles,
\emph{J. Integer Seq.},
{\bf 9} (2006), 
\href{http://www.cs.uwaterloo.ca/journals/JIS/VOL9/Barry/barry91.html}{Article 06.2.4}.

\bibitem{CMS} C. Corsani, D. Merlini, and R. Sprugnoli,
Left-inversion of combinatorial sums, \emph{Discrete Math.},
\textbf{180} (1998), 107--122.

\bibitem{Jin} Sung-Tae Jin, A characterization of the Riordan Bell
subgroup by C-sequences, \emph{Korean J. Math.},
\textbf{17} (2009),  147--154.

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

\bibitem{Merlini_When} D. Merlini, R. Sprugnoli, and M.~C.~Verri,
Lagrange inversion: when and how, \emph{Acta Appl. Math.}, \textbf{94}
(2006),  233--249.

\bibitem{Merlini_MC} D. Merlini, R. Sprugnoli, and M.~C.
Verri, The method of coefficients, \emph{Amer. Math. Monthly},
\textbf{114} (2007),  40--57.


\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{SL1} N. J. A.~Sloane, \emph{The
On-Line Encyclopedia of Integer Sequences}. Published electronically
at \href{http://oeis.org}{http://oeis.org}, 2011.

\bibitem{SL2} N. J. A.~Sloane, The On-Line Encyclopedia of Integer
Sequences, \emph{Notices Amer. Math. Soc.}, \textbf{50} (2003),  912--915.

\bibitem{Spru} R. Sprugnoli, Riordan arrays and combinatorial sums,
\emph{Discrete Math.}, \textbf{132} (1994), 267--290.



\end{thebibliography}

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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, Bell subgroup, Bell matrix.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000045},
\seqnum{A000108},
\seqnum{A001002},
\seqnum{A002878},
\seqnum{A007318},
\seqnum{A007440},
\seqnum{A033184},
\seqnum{A038112},
\seqnum{A078531},
\seqnum{A091527},
\seqnum{A104506}, and
\seqnum{A174687}.)

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\vspace*{+.1in}
\noindent
Received June 29 2010;
revised version received  November 11 2010; March 22 2011.
Published in {\it Journal of Integer Sequences}, March 26 2011.

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