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\vskip 1cm{\LARGE\bf On Integer-Sequence-Based Constructions of \\
\vskip .2cm
Generalized Pascal Triangles} \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 introduce an integer sequence based construction
of invertible centrally symmetric number triangles, which generalize
Pascal's triangle. We characterize the row sums and central
coefficients of these triangles, and examine other properties. Links
to the Narayana numbers are explored. Use is made of the Riordan
group to elucidate properties of a special one-parameter subfamily.
An alternative exponential approach to constructing generalized
Pascal triangles is briefly explored.
\end{abstract}
\section{Introduction}
In this article, we look at two methods of using given integer
sequences to construct generalized Pascal matrices. In the first
method, we look at the number triangle associated with the square
matrix $\mathbf{BDB}'$, where $\mathbf{B}$ is the binomial matrix
$\binom{n}{k}$ and $\mathbf{D}$ is the diagonal matrix defined by
the given integer sequence. We study this construction in some
depth, and characterize the sequences related to the central
coefficients of the resulting triangles in a special case. We
study the cases of the Fibonacci and Jacobsthal numbers in
particular. The second construction is defined in terms of a
generalization of $\exp(\mathbf{M})$, where $\mathbf{M}$ is a
sub-diagonal matrix defined by the integer sequence in question.
Our look at this construction is less detailed. It is a measure of
the ubiquity of the Narayana numbers that they arise in both
contexts. 

The plan of the article is as follows. We begin with an
introductory section, where we define what this article will
understand as a generalized Pascal matrix, as well as looking at
the binomial transform, the Riordan group, and the Narayana
numbers, all of which will be used in subsequent sections. The
next preparatory section looks at the reversion of the expressions
$\frac{x}{1+{\alpha}x+{\beta}x^2}$ and $\frac{x(1-ax)}{1-bx}$,
which are closely related to subsequent work. We then introduce
the first family of generalized Pascal triangles, and follow this
by looking at those elements of this family that correspond to the
``power'' sequences $n\to r^n$, while the section after that takes
the specific cases of the Fibonacci and Jacobsthal numbers. We
close the study of this family by looking at the generating
functions of the columns of these triangles in the general case.

The final section briefly studies an alternative construction
based on a generalized matrix exponential construction.
\section{Preliminaries}
Pascal's triangle, with general term $\binom{n}{k}$, $n,k\geq 0$,
has fascinated mathematicians by its wealth of properties since its
discovery \cite{Pas}. Viewed as an infinite lower-triangular matrix, it is
invertible, with an inverse whose general term is given by
$(-1)^{n-k}\binom{n}{k}$. Invertibility follows from the fact that
$\binom{n}{n}=1$. It is \emph{centrally symmetric}, since by definition,
$\binom{n}{k}=\binom{n}{n-k}$. All the terms of this matrix are
integers.

By a generalized Pascal triangle we shall understand a lower-triangular
infinite integer matrix $T=T(n,k)$ with $T(n,0)=T(n,n)=1$ and
$T(n,k)=T(n,n-k)$. We shall index all matrices in this paper beginning at
the $(0,0)$-th element.

We shall use transformations that operate on integer sequences
during the course of this note. An example of such a
transformation that is widely used in the study of integer
sequences is the so-called \emph{binomial transform} \cite{Bin},
which associates to the sequence with general term $a_n$ the
sequence with general term $b_n$ where
\begin{equation}b_n=\sum_{k=0}^n{{n}\choose{k}}a_k.\end{equation}
If we consider the sequence with general term $a_n$ to be the vector
$\mathbf{a}=(a_0,a_1,\ldots)$
 then we obtain the binomial transform of the sequence by multiplying this (infinite)
 vector by the lower-triangle matrix $\mathbf{B}$ whose $(n,k)$-th element is equal to
$\binom{n}{k}$:
\begin{displaymath}\mathbf{B}=\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 2 &amp; 1
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 3 &amp; 3 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 4 &amp;
6 &amp; 4 &amp; 1 &amp; 0 &amp; \ldots
\\1 &amp; 5 &amp; 10 &amp; 10 &amp; 5 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath} This
transformation is invertible, with
\begin{equation}a_n=\sum_{k=0}^n
{{n}\choose{k}}(-1)^{n-k}b_k.\end{equation} We note that
$\mathbf{B}$ corresponds to Pascal's triangle. Its row sums are
$2^n$, while its diagonal sums are the Fibonacci numbers $F(n+1)$.
If $\mathbf{B}^m$ denotes the $m-$th power of $\mathbf{B}$, then the
$n-$th term of $\mathbf{B}^m\mathbf{a}$ where $\mathbf{a}=\{a_n\}$
is given by $\sum_{k=0}^n m^{n-k}{{n}\choose{k}}a_k$.

If $\mathcal{A}(x)$ is the ordinary generating function of the
sequence $a_n$, then the generating function of the transformed
sequence $b_n$ is $\frac{1}{1-x}\mathcal{A}(\frac{x}{1-x})$.
 The binomial transform is an element of the Riordan
group, which can be defined as follows.

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 $f, g$ 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 $f$.

If $\mathbf{M}$ is the matrix $(g,f)$, and $\mathbf{a}=\{a_n\}$ 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))$.

\begin{example} As an
example, the Binomial matrix $\mathbf{B}$ is the element
$(\frac{1}{1-x},\frac{x}{1-x})$ of the Riordan group. More
generally, $\mathbf{B}^k$ is the element
$(\frac{1}{1-kx},\frac{x}{1-kx})$ of the Riordan group. It is easy
to show that the inverse $\mathbf{B}^{-k}$ of $\mathbf{B}^k$ is
given by $(\frac{1}{1+kx},\frac{x}{1+kx})$. \end{example}

The row sums of the matrix $(g, f)$ have generating function
$g(x)/(1-f(x))$ while the diagonal sums of $(g, f)$ have generating
function $g(x)/(1-xf(x))$.

We shall frequently refer to sequences by their sequence number in
the On-Line Encylopedia of Integer Sequences \cite{SL1}, \cite{SL2}.
For instance, Pascal's triangle is \seqnum{A007318} while the Fibonacci numbers are \seqnum{A000045}.

\begin{example} An example of a well-known centrally symmetric
invertible triangle that is not an element of the Riordan group is
the Narayana triangle $\mathbf{\tilde{N}}$, defined by
\begin{displaymath}
\tilde{N}(n,k)=\frac{1}{k+1}\binom{n}{k}\binom{n+1}{k}=\frac{1}{n+1}\binom{n+1}{k+1}\binom{n+1}{k}\end{displaymath}
for $n,k\ge 0$.
Other expressions for $\tilde{N}(n,k)$ are given by
\begin{displaymath}
\tilde{N}(n,k)={\binom{n}{k}}^2-\binom{n}{k+1}\binom{n}{k-1}=\binom{n+1}{k+1}\binom{n}{k}
-\binom{n+1}{k}\binom{n}{k+1}.\end{displaymath}

This triangle begins
\begin{displaymath}\mathbf{\tilde{N}}=\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 3 &amp; 1 &amp;
0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 6 &amp; 6 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 10 &amp; 20
&amp; 10 &amp; 1 &amp; 0 &amp; \ldots
\\1 &amp; 15 &amp; 50 &amp; 50 &amp; 15 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath} We shall
characterize this matrix in terms of a generalized matrix
exponential construction later in this article. Note that in the
literature, it is often the triangle
$\tilde{N}(n-1,k-1)=\frac{1}{n}\binom{n}{k}\binom{n}{k-1}$ that is
referred to as the Narayana triangle. Alternatively, the triangle
$\tilde{N}(n-1,k)=\frac{1}{k+1}\binom{n-1}{k}\binom{n}{k}$ is
referred to as the Narayana triangle. We shall denote this latter
triangle by $N(n,k)$. We then have
\begin{displaymath}\mathbf{\mathbf{N}}=\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 1 &amp; 0 &amp;
0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 3 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 6 &amp; 6 &amp;
1 &amp; 0 &amp; 0 &amp; \ldots
\\1 &amp; 10 &amp; 20 &amp; 10 &amp; 1 &amp; 0 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath}
Note that for $n,k\ge 1$, $N(n,k)=\frac{1}{n}\binom{n}{k}\binom{n}{k+1}$. We have,
for instance,
\begin{displaymath}\tilde{N}(n-1,k-1)=\frac{1}{n}\binom{n}{k}\binom{n}{k-1}={\binom{n}{k}}^2-\binom{n-1}{k}\binom{n+1}{k}=
\binom{n}{k}\binom{n-1}{k-1}-\binom{n}{k-1}\binom{n-1}{k}.\end{displaymath}
The last expression represents a $2\times2$ determinant of adjacent
elements in Pascal's triangle. The Narayana triangle is
\seqnum{A001263}.

A related identity is the following, \cite{Coker}, \cite{Chen}:
\begin{equation}\label{CokerChen}
\sum_{k=0}^{n-1}\frac{1}{n}\binom{n}{k}\binom{n}{k+1}x^k=\sum_{k=0}^{\lfloor
\frac{n-1}{2}\rfloor}\binom{n-1}{2k}c(k)x^k(1+x)^{n-2k-1}\end{equation}
where $c(n)$ is the $n$-th Catalan number
$c(n)=\binom{2n}{n}/(n+1)$, \seqnum{A000108}. This identity can be
interpreted in terms of Motzkin paths, where by a \emph{Motzkin
path} of length $n$ we mean a lattice path in $\mathbf{Z}^2$ between
$(0,0)$ and $(n,0)$ consisting of up-steps $(1,1)$, down-steps
$(1,-1)$ and horizontal steps $(1,0)$ which never goes below the
$x$-axis. Similarly, a \emph{Dyck} path of length $2n$ is a lattice
path in $\mathbf{Z}^2$ between $(0,0)$ and $(2n,0)$ consisting of
up-paths $(1,1)$ and down-steps $(1,-1)$ which never go below the
$x$-axis. Finally, a (large) \emph{Schr\"oder} path of length $n$ is
a lattice path from $(0,0)$ to $(n,n)$ containing no points above
the line $y=x$, and composed only of steps $(0,1)$, $(1,0)$ and
$(1,1)$.

For instance, the number of Schr\"oder paths from $(0,0)$ to $(n,n)$
is given by the large Schr\"oder numbers $1,2,6,22,90,\ldots$ which
correspond to $z=2$ for the \emph{Narayana polynomials}
\cite{Sulanke1}, \cite{Sulanke2}
\begin{displaymath} N_n(z)=\sum_{k=1}^n
\frac{1}{n}\binom{n}{k-1}\binom{n}{k}z^k.\end{displaymath}


\end{example}
\section{On the series reversion of $\frac{x}{1+{\alpha}x+{\beta}x^2}$ and $\frac{x(1-ax)}{1-bx}$}
A number of the properties of the triangles that we will study are related to the
special cases of the series reversions of $\frac{x}{1+{\alpha}x+{\beta}x^2}$ and $\frac{x(1-ax)}{1-bx}$
where $b=a-1$, $\alpha=a+1$ and $\beta=b+1$. We shall develop results relating to these reversions in full
generality in this section and specialize later at the appropriate places.

Solving the equation $$\frac{y}{1+\alpha{y}+\beta{y^2}}=x$$
yields $$y_1=\frac{1-\alpha{x}-\sqrt{1-2\alpha{x}+(\alpha^2-4\beta)x^2}}{2\beta{x}}$$
while solving the equation $$\frac{y(1-ay)}{1-by}=x$$ leads to
$$y_2=\frac{1+bx-\sqrt{(1+bx)^2-4ax}}{2a}.$$
We shall occasionally use the notation $y_1(\alpha, \beta)$ and $y_2(a,b)$ where relevant for
these functions. Note for instance that $\frac{y_2(1,0)}{x}=\frac{1-\sqrt{1-4x}}{2x}$ is the generating function of the Catalan numbers.
\begin{proposition} Let $\alpha=a+1$, $\beta=b+1$, and assume that $b=a-1$ (and hence,
$\beta=\alpha-1$). Then $$\frac{y_2}{x}-y_1=1.$$ \end{proposition}
\begin{proof} Straight-forward calculation. \end{proof}
Note that $1$ is the generating function of $0^n=1,0,0,0,\ldots$.
\begin{example} Consider the case $a=2$, $b=1$. Let $\alpha=3$ and $\beta=2$,  so we are considering
$\frac{x}{1+3x+2x^2}$ and $\frac{x(1-2x)}{1-x}$. We obtain
\begin{eqnarray*}y_1(3,2)&amp;=&amp;\frac{1-3x-\sqrt{1-6x+x^2}}{4x}\\
\frac{y_2(2,1)}{x}&amp;=&amp;\frac{1+x-\sqrt{1-6x+x^2}}{4x}\\
\frac{y_2(2,1)}{x}-y_1(3,2)&amp;=&amp;1.\end{eqnarray*} Thus $y_1(3,2)$ is the
generating function for $0,1,3,11,45,197,903,4279,\ldots$ while
$\frac{y_2(2,1)}{x}$ is the generating function for
$1,1,3,11,45,197,903,4279,\ldots$. These are the little Schr\"oder
numbers \seqnum{A001003}.
\end{example}
\begin{example} We consider the case $a=1$, $b=1-r$, that is, the case of $\frac{x(1-x)}{1-(1-r)x}$.
We obtain
\begin{eqnarray*}\frac{y_2(1,1-r)}{x}&amp;=&amp;\frac{1-(r-1)x-\sqrt{(1+(1-r)x)^2-4x}}{2x}\\
&amp;=&amp;\frac{1-(r-1)x-\sqrt{1-2(r+1)x+(r-1)^2x^2}}{2x}\end{eqnarray*}
\end{example}
\begin{example}\label{Pairing} We calculate the expression $\frac{y_2(1,1-r)}{rx}-\frac{1-r}{r}$. We get
\begin{eqnarray*}\frac{y_2(1,1-r)}{rx}-\frac{1-r}{r}&amp;=&amp;\frac{1-(r-1)x-\sqrt{1-2(r+1)x+(r-1)^2x^2}}{2rx}+\frac{2(r-1)x}{2rx}\\
&amp;=&amp;\frac{1+(r-1)x-\sqrt{1-2(r+1)x+(r-1)^2x^2}}{2rx}\\
&amp;=&amp;\frac{y_2(r,r-1)}{x}.\end{eqnarray*}
In
other words, $$\frac{y_2(r,r-1)}{x}=\frac{y_2(1,1-r)}{rx}-\frac{1-r}{r}.$$
A well-known example of this is the case of the large Schr\"oder numbers with generating function
$\frac{1-x-\sqrt{1-6x+x^2}}{2x}$ and the little Schr\"oder numbers with generating function
$\frac{1+x-\sqrt{1-6x+x^2}}{4x}$. In this case, $r=2$. Generalizations of this ``pairing'' for $r&gt;2$ will be studied in a later section.
For $r=1$ both sequences coincide with the Catalan numbers $c(n)$.
\end{example}
\begin{proposition}\label{BinInc} The binomial transform of
$$y_1=\frac{1-\alpha{x}-\sqrt{1-2\alpha{x}+(\alpha^2-4\beta)x^2}}{2\beta{x^2}}$$
is
$$\frac{1-(\alpha+1){x}-\sqrt{1-2(\alpha+1)x+((\alpha+1)^2-4\beta)x^2}}{2\beta{x^2}}.$$
\end{proposition}
\begin{proof} The binomial transform of $y_1$ is
\begin{eqnarray*}&amp; &amp;\frac{1}{1-x}\left\{1-\frac{{\alpha}x}{1-x}-\sqrt{1-\frac{2{\alpha}x}{1-x}+({\alpha}^2-4{\beta})\frac{x^2}{(1-x)^2}}\right\}/(2{\beta}\frac{x^2}{(1-x)^2})\\
&amp;=&amp;(1-x-{\alpha}x-\sqrt{(1-x)^2-2{\alpha}x(1-x)+({\alpha}^2-4{\beta})x^2})/(2{\beta}x^2)\\
&amp;=&amp;(1-({\alpha}+1)x-\sqrt{1-2({\alpha}+1)x+({\alpha}^2+2{\alpha}+1-4{\beta}^2)x^2})/(2{\beta}x^2)\\
&amp;=&amp;\frac{1-(\alpha+1){x}-\sqrt{1-2(\alpha+1)x+((\alpha+1)^2-4\beta)x^2}}{2\beta{x^2}}.
\end{eqnarray*}
\end{proof}
\begin{example} The binomial transform of $1,3,11,45,197,903,\ldots$ with generating function
\\$\frac{1-3x-\sqrt{1-6x+x^2}}{4x^2}$ is $1,4,18,88,456,2464,13736,\ldots$, \seqnum{A068764}, with
generating function \\$\frac{1-4x-\sqrt{1-8x+8x^2}}{4x^2}$. Thus the binomial
transform links the series reversion of $x/(1+3x+2x^2)$ to that of $x/(1+4x+2x^2)$.
 We note that this can be interpreted in the context of Motzkin paths as an incrementing of
 the colours available for the H$(1,0)$ steps.
\end{example}

We now look at the general terms of the sequences generated by $y_1$ and $y_2$.
We use the technique of Lagrangian inversion for this. We begin with $y_1$. In order
to avoid notational overload, we use $a$ and $b$ rather than $\alpha$ and $\beta$,
hoping that confusion won't arise.

Since for $y_1$ we have $y=x(1+ay+by^2)$ we can apply Lagrangian inversion to get the
following expression for the general term of the sequence generated by $y_1$:
$$[t^n]y_1=\frac{1}{n}[t^{n-1}](1+at+bt^2)^n.$$
At this point we remark that there are many ways to develop the trinomial expression,
and the subsequent binomial expressions. Setting these different expressions equal
for different combinations of $a$ and $b$ and different relations between $a$ and
$b$ can lead to many interesting combinatorial identities, many of which can be
interpreted in terms of Motzkin paths. We shall confine ourselves to the derivation
of two particular expressions. First of all,
\begin{eqnarray*}[t^n]y_1&amp;=&amp;\frac{1}{n}[t^{n-1}](1+at+bt^2)^n\\
&amp;=&amp;\frac{1}{n}[t^{n-1}]\sum_{k=0}^n \binom{n}{k}(at+bt^2)^k\\
&amp;=&amp;\frac{1}{n}[t^{n-1}]\sum_{k=0}^n \binom{n}{k}t^k\sum_{j=0}^{k}\binom{k}{j}
a^jb^{k-j}t^{k-j}\\
&amp;=&amp;\frac{1}{n}[t^{n-1}]\sum_{k=0}^n\sum_{j=0}^k\binom{n}{k}\binom{k}{j}a^jb^{k-j}t^{2k-j}\\
&amp;=&amp;\frac{1}{n}\sum_{k=0}^n\binom{n}{k}\binom{k}{n-k-1}a^{2k-n+1}b^{n-k-1}\\
&amp;=&amp;\frac{1}{n}\sum_{k=0}^n\binom{n}{k}\binom{k}{2k-n+1}a^{2k-n+1}b^{n-k-1}.\end{eqnarray*}
Of the many other possible expressions for $[t^n]y_1$, we cite the
following examples:
\begin{eqnarray*}[t^n]y_1&amp;=&amp;\frac{1}{n}\sum_{k=0}^n\binom{n}{k}\binom{k+1}{2k-n-1}b^{2k-n+1}b^{n-k-1}\\
&amp;=&amp;\frac{1}{n}\sum_{k=0}^n\binom{n}{k}\binom{k}{2k-n+1}b^{n-k-1}a^{2k-n+1}\\
&amp;=&amp;\frac{1}{n}\sum_{k=0}^n\binom{n}{k}\binom{n-k}{k-1}b^{k-1}a^{n-2k+1}\\
&amp;=&amp;\frac{1}{n}\sum_{k=0}^n\binom{n}{k+1}\binom{n-k-1}{k+1}b^ka^{n-2k}.
\end{eqnarray*}
We shall be interested at a later stage in generalized Catalan sequences. The following
interpretation of $[t^n]y_1$ is therefore of interest.
\begin{proposition} $$[t^n]y_1=\sum_{k=0}^{\lfloor \frac{n-1}{2} \rfloor}
\binom{n-1}{2k}c(k)a^{n-2k-1}b^k.$$\end{proposition}
\begin{proof}
\begin{eqnarray*}[t^n]y_1&amp;=&amp;\frac{1}{n}[t^{n-1}](1+at+bt^2)^n\\
&amp;=&amp;\frac{1}{n}[t^{n-1}](at+(1+bt^2))^n\\
&amp;=&amp;\frac{1}{n}[t^{n-1}]\sum_{j=0}^na^jt^j(1+bt^2)^{n-j}\\
&amp;=&amp;\frac{1}{n}[t^{n-1}]\sum_{j=0}^n\sum_{k=0}^{n-j}\binom{n}{j}\binom{n-j}{k}a^jb^kt^{2k+j}\\
&amp;=&amp;\frac{1}{n}\sum_{k=0}\binom{n}{n-2k-1}\binom{2k+1}{k}a^{n-2k-1}b^k\\
&amp;=&amp;\frac{1}{n}\sum_{k=0}^n\binom{n}{2k+1}\binom{2k+1}{k}a^{n-2k-1}b^k\\
&amp;=&amp;\frac{1}{n}\sum_{k=0}\frac{n}{2k+1}\binom{n-1}{n-2k-1}\frac{2k+1}{k+1}\binom{2k}{k}a^{n-2k-1}b^k\\
&amp;=&amp;\sum_{k=0}\binom{n-1}{2k}c(k)a^{n-2k-1}b^k.\end{eqnarray*}
\end{proof}
\begin{corollary}
\begin{eqnarray*}c(n)&amp;=&amp;0^n+\sum_{k=0}^{\lfloor \frac{n-1}{2} \rfloor}
\binom{n-1}{2k}c(k)2^{n-2k-1}\\
c(n+1)&amp;=&amp;\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}
\binom{n-1}{2k}c(k)2^{n-2k}.\end{eqnarray*}
\end{corollary}
\begin{proof} The sequence $c(n)-0^n$, or $0,1,2,5,14,\ldots$, has generating function
$$\frac{1-\sqrt{1-4x}}{2x}-1=\frac{1-2x-\sqrt{1-4x}}{2x}$$ which corresponds to
$y_1(2,1)$.\end{proof}
This is the formula of Touchard \cite{Touchard}, with adjustment for the first term.
\begin{corollary}$$[t^n]y_1(r+1,r)=\sum_{k=0}^{n-1}\frac{1}{n}\binom{n}{k}\binom{n}{k+1}r^k.$$
\end{corollary}
\begin{proof} By the proposition, we have
$$[t^n]y_1(r+1,r)=\sum_{k=0}^{\lfloor \frac{n-1}{2} \rfloor}
\binom{n-1}{2k}c(k)(r+1)^{n-2k-1}r^k.$$
The result then follows from identity (\ref{CokerChen}).
\end{proof}
This therefore establishes a link to the Narayana numbers.
\begin{corollary} Let $s_n(a,b)$ be the sequence with general term
$$s_n(a,b)=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}
\binom{n}{2k}c(k)a^{n-2k}b^k.$$ Then the binomial transform of this sequence
is the sequence $s_n(a+1,b)$ with general term
$$s_n(a+1,b)=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}
\binom{n}{2k}c(k)(a+1)^{n-2k}b^k.$$
\end{corollary}
\begin{proof} This is a re-interpretation of the results of Proposition \ref{BinInc}.
\end{proof}
 We now take a quick look at $[t^n]y_2$. In this case, we have
$$y=x\frac{1-by}{1-ay}$$ so we can apply Lagrangian inversion.
Again, various expressions arise depending
on the order of expansion of the binomial expressions involved. For instance,
\begin{eqnarray*}[t^n]y_2&amp;=&amp;\frac{1}{n}[t^{n-1}]\left(\frac{1-bt}{1-at}\right)^n\\
&amp;=&amp;\frac{1}{n}[t^{n-1}](1-bt)^n(1-at)^{-n}\\
&amp;=&amp;\frac{1}{n}[t^{n-1}]\sum_{k=0}^n\sum_{j=0}\binom{n}{k}\binom{n+j-1}{j}a^j(-b)^kt^{k+j}\\
&amp;=&amp;\frac{1}{n}\sum_{k=0}^n\binom{n}{k}\binom{2n-k-2}{n-1}a^{n-k-1}(-b)^k.\end{eqnarray*}
A more interesting development is given by the following.
\begin{eqnarray*}
[t^n]\frac{y_2}{x}&amp;=&amp;[t^{n+1}]y_2\\
&amp;=&amp;\frac{1}{n+1}[t^n](1-bt)^{n+1}(1-at)^{-(n+1)}\\
&amp;=&amp;\frac{1}{n+1}[t^n]\sum_{k=0}^{n+1}\binom{n+1}{k}(-bt)^{n+1-k}\sum_{j=0}\binom{-n-1}{j}(-at)^j\\
&amp;=&amp;\frac{1}{n+1}[t^n]\sum_{k=0}^{n+1}\sum_{j=0}\binom{n+1}{k}\binom{n+j}{j}(-b)^{n-k+1}a^jt^{n+1-k+j}\\
&amp;=&amp;\frac{1}{n+1}\sum_{j=0}\binom{n+1}{j+1}\binom{n+j}{j}(-b)^{n-j}a^j\\
&amp;=&amp;\sum_{j=0}^n\frac{1}{j+1}\binom{n}{j}\binom{n+j}{j}a^j(-b)^{n-j}.
\end{eqnarray*}
An alternative expression obtained by developing for $k$ above is given by
$$[t^n]\frac{y_2}{x}=\sum_{k=0}^{n+1}\frac{1}{n-k+1}\binom{n}{k}\binom{n+k-1}{k-1}a^{k-1}(-b)^{n-k+1}.$$
Note that the underlying matrix with general element $\frac{1}{k+1}\binom{n}{k}\binom{n+k}{k}$ is
\seqnum{A088617}, whose general element gives the number of  Schr\"oder paths  from $(0,0)$ to
$(2n,0)$, having $k$ $U(1,1)$ steps.
Recognizing that $\sum_{j=0}^n\frac{1}{j+1}\binom{n}{j}\binom{n+j}{j}a^j(-b)^{n-j}$ is a convolution, we can also write
\begin{eqnarray*}
[t^n]\frac{y_2}{x}&amp;=&amp;\sum_{k=0}^n\frac{1}{k+1}\binom{n}{k}\binom{n+k}{k}a^k(-b)^{n-k}\\
&amp;=&amp;\sum_{k=0}^n\frac{1}{n-k+1}\binom{n}{n-k}\binom{2n-k}{n-k}a^{n-k}(-b)^k\\
&amp;=&amp;\sum_{k=0}^n\frac{1}{n-k+1}\binom{n}{k}\binom{2n-k}{n}a^{n-k}(-b)^k\\
&amp;=&amp;\sum_{k=0}^n\frac{1}{n-k+1}\binom{2n-k}{k}\binom{2n-k-k}{n-k}a^{n-k}(-b)^k\\
&amp;=&amp;\sum_{k=0}^n\binom{2n-k}{k}\frac{1}{n-k+1}\binom{2n-2k}{n-k}a^{n-k}(-b)^k\\
&amp;=&amp;\sum_{k=0}^n\binom{2n-k}{k}c(n-k)a^{n-k}(-b)^k\\
&amp;=&amp;\sum_{k=0}^n\binom{n+k}{2k}c(k)a^k(-b)^{n-k}.\end{eqnarray*}
Again we note that the matrix with general term
$\binom{n}{k}\binom{2n-k}{k}\frac{1}{n-k+1}$ is \seqnum{A060693},
whose general term  counts the number of Schr\"oder paths  from
$(0,0)$ to $(2n,0)$, having $k$ peaks. $\binom{n+k}{2k}c(k)$ is
another expression for \seqnum{A088617}. Gathering these results
leads to the next proposition.
\begin{proposition} $[t^n]\frac{y_2(a,b)}{x}$ is given by the equivalent expressions
\begin{eqnarray*}&amp;&amp;\sum_{k=0}^n\frac{1}{k+1}\binom{n}{k}\binom{n+k}{k}a^k(-b)^{n-k}\\
&amp;=&amp;\sum_{k=0}^n\binom{n+k}{2k}c(k)a^k(-b)^{n-k}\\
&amp;=&amp;\sum_{k=0}^n\binom{2n-k}{k}c(n-k)a^{n-k}(-b)^k.\end{eqnarray*}
\end{proposition}
We summarize some of these results in \textbf{Table 1}, where $c_n=c(n)=\frac{1}{n+1}\binom{2n}{n}$, and
$P(x)=1-2(r+1)x+(r-1)^2x^2$, and $N(n,k)=\frac{1}{n}\binom{n}{k}\binom{n}{k+1}$. We use the terms ``Little sequence'' and
``large sequence'' in analogy with the Schr\"oder numbers. In \cite{Shapiro} we note that the terms
``Little Schr\"oder'', ``Big Schr\"oder'' and ``Bigger Schr\"oder'' are used. For instance, the numbers $1,3,11,45,\ldots$ appear there
as the ``Bigger Schr\"oder'' numbers.
\begin{center}
\textbf{Table 1. \ Summary of section results  }
\begin{tabular}{|c|c|c|}\hline
Large sequence, $S_n$ &amp; Little sequence, $s_n$ &amp; Larger sequence $s_n-0^n$  \\
e.g. $1,2,6,22,90,\ldots$ &amp; e.g. $1,1,3,11,45,\ldots$ &amp; e.g. $0,1,3,11,45,\ldots$\\\hline
$\frac{x(1-x)}{1-(1-r)x}$&amp;$\frac{x(1-rx)}{1-(r-1)x}$&amp;$\frac{x}{1+(r+1)x+rx^2}$\\\hline
$\frac{1-(r-1)x-\sqrt{P(x)}}{2x}$&amp;$\frac{1+(r-1)x-\sqrt{P(x)}}{2rx}$&amp;$\frac{1-(r+1)x-\sqrt{P(x)}}{2rx}$\\\hline
$a_0=1,a_n=\frac{1}{n}\sum_{k=0}^n\binom{n}{k}\binom{n}{k-1}r^k$&amp;$a_0=1,a_n=\sum_{k=0}^nN(n,k)r^k$&amp;$\sum_{k=0}^{n-1}N(n,k)r^k$\\\hline
$\sum_{k=0}^n\binom{n+k}{2k}c_k(r-1)^{n-k}$&amp;$\sum_{k=0}^n\binom{n+k}{2k}c_kr^k(1-r)^{n-k}$&amp;$\sum_{k=0}\binom{n-1}{2k}c_k(r+1)^{n-2k-1}r^k$\\\hline
$\sum_{k=0}^n\binom{2n-k}{k}c_{n-k}(r-1)^k$&amp;$\sum_{k=0}^n\binom{2n-k}{k}c_{n-k}r^{n-k}(1-r)^k$&amp;
-\\
\hline
\end{tabular}\end{center}

\begin{center}
\textbf{Table 2. \ Little and Large sequences in OEIS  }
\end{center}
\begin{center}
\begin{tabular}{|c|c|c|c|}\hline
$r$&amp;$s_n$&amp;$S_n$&amp;Triangle\\\hline
$1$&amp;A000984&amp;A000984&amp;A007318\\\hline
$2$&amp;A001003&amp;A006318&amp;A008288\\\hline
$3$&amp;A007564&amp;A047891&amp;A081577\\\hline
$4$&amp;A059231&amp;A082298&amp;A081578\\\hline
$5$&amp;A078009&amp;A082301&amp;A081579\\\hline
$6$&amp;A078018&amp;A082302&amp;A081580\\\hline
$7$&amp;A081178&amp;A082305&amp;\\\hline
$8$&amp;A082147&amp;A082366&amp;\\\hline
$9$&amp;A082181&amp;A082367&amp;\\\hline
$10$&amp;A082148&amp;&amp;\\\hline
\end{tabular}\end{center}

Note that by Example \ref{Pairing} we can write
$$s_n=\frac{1}{r}S_n+\frac{(r-1)0^n}{r}.$$
\section{Introducing the family of centrally symmetric invertible triangles}
The motivation for the construction that follows comes from the
following easily established proposition.
\begin{proposition}
\begin{displaymath}\binom{n}{k}=\sum_{j=0}^{n-k}\binom{k}{j}\binom{n-k}{j}=
\sum_{j=0}^{k}\binom{k}{j}\binom{n-k}{j}.\end{displaymath}
\end{proposition}
\begin{proof} We consider identity $5.23$ of \cite{Conc}:
\begin{displaymath}\binom{r+s}{r-p+q}=\sum_{j}\binom{r}{p+j}\binom{s}{q+j}\end{displaymath}
itself a consequence of Vandermonde's convolution identity. Setting
$r=k$, $s=n-k$, $p=q=0$, we obtain
\begin{displaymath}\binom{n}{k}=\sum_{j}\binom{k}{j}\binom{n-k}{j}.\end{displaymath}
\end{proof}

Now let $a_n$ represent a sequence of integers with $a_0=1$. We
define an infinite array of numbers for $n,k\geq 0$ by
\begin{displaymath}T(n,k)=\sum_{j=0}^{n-k}\binom{k}{j}\binom{n-k}{j}a_j.\end{displaymath}
and call it \emph{the triangle associated with the sequence $a_n$ by this
construction}. That it is a number triangle follows from the next
proposition.
\begin{proposition} The matrix with general term $T(n,k)$ is an
integer-valued centrally symmetric invertible lower-triangular matrix.
\end{proposition}
\begin{proof} All elements in the sum are integers, hence $T(n,k)$ is an integer for all $n,k\geq 0$. $T(n,k)=0$ for $k&gt;n$ since then $n-k&lt;0$ and hence the
sum is $0$. We have
\begin{displaymath}T(n,n)=\sum_{j=0}^{n-n}\binom{n}{j}\binom{n-n}{j}a_j=\sum_{j=0}^0\binom{n}{j}\binom{0}{j}a_j=\binom{n}{0}\binom{0}{0}a_0=1\end{displaymath}
which proves that the matrix is invertible. Finally, we have
\begin{eqnarray*} T(n,n-k)&amp;=&amp;\sum_{j=0}^{n-(n-k)}\binom{n-k}{j}\binom{n-(n-k)}{j}a_j\\
&amp;=&amp;\sum_{j=0}^{k}\binom{n-k}{j}\binom{k}{j}a_j\\
&amp;=&amp;T(n,k).\end{eqnarray*}\end{proof} It is clear that Pascal's
triangle corresponds to the case where $a_n$ is the
sequence $1,1,1,\ldots$.

Occasionally we shall use the above construction on sequences $a_n$
for which $a_0=0$. In this case we still have a centrally symmetric
triangle, but it is no longer invertible, since for example
$T(0,0)=0$ in this case.

By an abuse of notation, we shall often use $T(n,k;a_n)$ to denote
the triangle associated to the sequence $a_n$ by the above
construction, when explicit mention of $a_n$ is required.

The associated square symmetric matrix with general term
$$T_{sq}(n,k)=\sum_{j=0}^n\binom{k}{j}\binom{n}{j}a_j$$
is easy to describe. We let
$\mathbf{D}=\mathbf{D}(a_n)=diag(a_0,a_1,a_2,\ldots)$. Then
$$\mathbf{T}_{sq}=\mathbf{BDB}'$$
is the square symmetric (infinite) matrix associated to our
construction. Note that when $a_n=1$ for all $n$, we get the square
Binomial or Pascal matrix $\binom{n+k}{k}$.

 Among the attributes of the triangles that
we shall construct that interest us, the family of central sequences
(sequences associated to $T(2n,n)$ and its close relatives) will be
paramount. The central binomial coefficients $\binom{2n}{n}$,
\seqnum{A000984}, play an important role in combinatorics. We begin
our examination of the generalized triangles by characterizing their
`central coefficients' $T(2n,n)$. We obtain
\begin{eqnarray*}T(2n,n)&amp;=&amp;\sum_{j=0}^{2n-n}\binom{2n-n}{j}\binom{n}{j}a_j\\
&amp;=&amp;\sum_{j=0}^n {\binom{n}{j}}^2a_j.\end{eqnarray*}

For the case of Pascal's triangle with $a_n$ given by $1,1,1,\ldots$
we recognize the identity $\binom{2n}{n}=\sum_{j=0}^n
{\binom{n}{j}}^2$. In like fashion, we can characterize $T(2n+1,n)$,
for instance.
\begin{eqnarray*}T(2n+1,n)&amp;=&amp;\sum_{j=0}^{2n+1-n}\binom{2n+1-n}{j}\binom{n}{j}a_j\\
&amp;=&amp;\sum_{j=0}^{n+1} \binom{n+1}{j}\binom{n}{j}a_j\end{eqnarray*}
which generalizes the identity $\binom{2n+1}{n}=\sum_{j=0}^{n+1}
\binom{n+1}{j}\binom{n}{j}$. This is \seqnum{A001700}.
We also have
\begin{eqnarray*}T(2n-1,n-1)&amp;=&amp;\sum_{j=0}^{2n-1-n+1}\binom{2n-1-n+1}{j}\binom{n-1}{j}a_j\\
&amp;=&amp;\sum_{j=0}^{n} \binom{n-1}{j}\binom{n}{j}a_j.\end{eqnarray*} This
generalizes the equation $\binom{2n-1}{n-1}+0^n=\sum_{j=0}^n
\binom{n-1}{j}\binom{n}{j}$. See \seqnum{A088218}.

In order to generalize the Catalan numbers $c(n)$, \seqnum{A000108},
in our context, we note that $c(n)=\binom{2n}{n}/(n+1)$ has the
alternative representation
\begin{displaymath}c(n)=\binom{2n}{n}-\binom{2n}{n-1}=\binom{2n}{n}-\binom{2n}{n+1}.\end{displaymath}
This motivates us to look at $T(2n,n)-T(2n,n-1)=T(2n,n)-T(2n,n+1)$. We obtain
\begin{eqnarray*}
T(2n,n)-T(2n,n-1)&amp;=&amp;\sum_{j=0}^n
\binom{n}{j}^2a_j-\sum_{j=0}^{2n-n+1}\binom{n-1}{j}\binom{2n-n+1}{j}a_j\\
&amp;=&amp;\sum_{j=0}^n
\binom{n}{j}^2a_j-\sum_{j=0}^{n+1}\binom{n-1}{j}\binom{n+1}{j}a_j\\
&amp;=&amp;\delta_{n,0}a_n+\sum_{j=0}^n(
\binom{n}{j}^2-\binom{n-1}{j}\binom{n+1}{j})a_j\\
 &amp;=&amp;\delta_{n,0}a_0+\sum_{j=0}^n \tilde{N}(n-1,j-1)a_j
\end{eqnarray*}
where we use the formalism $\binom{n-1}{n+1}=-1$, for $n=0$, and $\binom{n-1}{n+1}=0$
for $n&gt;0$. We assume that $\tilde{N}(n,-1)=0$ and $\tilde{N}(-1,k)=\binom{1}{k}-\binom{0}{k}$
in the above.
For instance, in the case of Pascal's triangle, where $a_n=1$ for
all $n$, we retrieve the Catalan numbers. We have also established a
link between these generalized Catalan numbers and the Narayana
numbers. We shall use the notation
$$c(n;a(n))=T(2n,n)-T(2n,n-1)=T(2n,n)-T(2n,n+1)$$ for this sequence, which we regard as
a sequence of \emph{generalized Catalan numbers}.
\begin{example} We first look at the case $a_n=2^n$. Thus
\begin{displaymath} T(n,k)=\sum_{j=0}^{n-k}
\binom{k}{j}\binom{n-k}{j}2^j\end{displaymath} with matrix
representation
\begin{displaymath}\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 3 &amp; 1 &amp;
0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 5 &amp; 5 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 7 &amp; 13 &amp;
7 &amp; 1 &amp; 0 &amp; \ldots
\\1 &amp; 9 &amp; 25 &amp; 25 &amp; 9 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath} which is the
well-known Delannoy number triangle \seqnum{A008288}. We have
\begin{displaymath} T(n,k)=\sum_{j=0}^k \binom{k}{j}\binom{n-j}{k}.\end{displaymath} We shall generalize this
identity later in this note.

As a Riordan array, this is given by
\begin{displaymath}(\frac{1}{1-x},\frac{x(1+x)}{1-x}).\end{displaymath}
Anticipating the general case, we examine the row sums of this
triangle, given by
\begin{displaymath}\sum_{k=0}^n\sum_{j=0}^{n-k}\binom{k}{j}\binom{n-k}{j}2^j.\end{displaymath}
Using the formalism of the Riordan group, we see that this sum has
generating function given by
\begin{displaymath}\frac{\frac{1}{1-x}}{1-\frac{x(1+x)}{1-x}}=\frac{1}{1-2x-x^2}.\end{displaymath}
In other words, the row sums in this case are the numbers
$Pell(n+1)$, \seqnum{A000129}, \cite{Pell}. We look at the inverse
binomial transform of these numbers, which has generating function
\begin{displaymath}
\frac{1}{1+x}\frac{1}{1-2\frac{x}{1+x}-\frac{x^2}{(1+x)^2}}=\frac{1+x}{1-2x^2}.\end{displaymath}
This is the generating function of the sequence
$1,1,2,2,4,4,\ldots$, \seqnum{A016116}, which is the doubled
sequence of $a_n=2^n$.

Another way to see this result is to observe that we have the
factorization
\begin{displaymath}(\frac{1}{1-x},\frac{x(1+x)}{1-x})=(\frac{1}{1-x},\frac{x}{1-x})(1,\frac{x(1+2x)}{1+x})\end{displaymath}
where $(\frac{1}{1-x},\frac{x}{1-x})$ represents the binomial
transform. The row sums of the Riordan array
$(1,\frac{x(1+2x)}{1+x})$ are $1,1,2,2,4,4,\ldots$.

For this triangle, the central numbers $T(2n,n)$ are the well-known
central Delannoy numbers $1,3,13,63,\ldots$ or \seqnum{A001850},
with ordinary generating function $\frac{1}{\sqrt{1-6x+x^2}}$ and
exponential generating function $e^{3x}I_0(2\sqrt{2}x)$ where $I_n$
is the $n$-th modified Bessel function of the first kind
\cite{ModBessel}. They represent the coefficients of $x^n$ in the
expansion of $(1+3x+2x^2)^n$. We have
$$ T(2n,n;2^n)=\sum_{k=0}^n\binom{n}{k}^2 2^k=\sum_{k=0}^n\binom{n}{k}\binom{n+k}{k}.$$

The numbers $T(2n+1,n)$ in this case are \seqnum{A002002}, with
generating function $(\frac{1-x}{\sqrt{1-6x+x^2}}-1)/(2x)$ and
exponential generating function
$e^{3x}(I_0(2\sqrt{2}x)+\sqrt{2}I_1(2\sqrt{2}x))$. We note that
$T(2n-1,n-1)$ represents the coefficient of $x^n$ in
$((1-x)/(1-2x))^n$. It counts the number of peaks in all Schr\"oder paths
from $(0,0)$ to $(2n,0)$.

The numbers $T(2n,n)-T(2n,n-1)$ are $1,2,6,22,90,394,1806,\ldots$ or
the large Schr\"oder numbers. These are the series reversion of $\frac{x(1-x)}{1+x}$.
Thus the generating function of the sequence $\frac{1}{2}(T(2n,n;2^n)-T(2n,n-1;2^n))$ is
$$y_2(1,-1)=\frac{1-x-\sqrt{1-6x+x^2}}{2x}.$$

We remark that in \cite{Del}, the author states that ``The Schr\"oder numbers bear the
same relation to the Delannoy numbers as the Catalan numbers do to
the binomial coefficients.'' This note amplifies on this statement, defining
generalized Catalan numbers for a family of number triangles.

\end{example}

\begin{example} We take the case $a_n=(-1)^n$. Thus
\begin{displaymath} T(n,k)=\sum_{j=0}^{n-k}
\binom{k}{j}\binom{n-k}{j}(-1)^j\end{displaymath} with matrix
representation
\begin{displaymath}\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 0 &amp; 1 &amp;
0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; -1 &amp; -1 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; -2 &amp; -2 &amp;
-2 &amp; 1 &amp; 0 &amp; \ldots
\\1 &amp; -3 &amp; -2 &amp; -2 &amp; -3 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath} As a Riordan
array, this is given by
\begin{displaymath}(\frac{1}{1-x},\frac{x(1-2x)}{1-x}).\end{displaymath}
Again, we look at the row sums of this
triangle, given by
\begin{displaymath}\sum_{k=0}^n\sum_{j=0}^{n-k}\binom{k}{j}\binom{n-k}{j}(-1)^j.\end{displaymath}
Looking at generating functions, we see that this sum has
generating function given by
\begin{displaymath}\frac{\frac{1}{1-x}}{1-\frac{x(1-2x)}{1-x}}=\frac{1}{1-2x+2x^2}.\end{displaymath}
In other words, the row sums in this case are the numbers
$1,2,2,0,-4,-8,-8,\ldots$ with exponential generating function $\exp(x)(\sin(x)+\cos(x))$, \seqnum{A009545}.
Taking the inverse binomial transform of these
numbers, we get the generating function \begin{displaymath}
\frac{1}{1+x}\frac{1}{1-2\frac{x}{1+x}+2\frac{x^2}{(1+x)^2}}=\frac{1+x}{1+x^2}.\end{displaymath}
This is the generating function of the sequence $1,1,-1,-1,1,1,\ldots$
which is the doubled sequence of $a_n=(-1)^n$.

Another way to see this result is to observe that we have the
factorization
\begin{displaymath}(\frac{1}{1-x},\frac{x(1-2x)}{1-x})=(\frac{1}{1-x},\frac{x}{1-x})(1,\frac{x(1-x)}{1+x})\end{displaymath}
where $(\frac{1}{1-x},\frac{x}{1-x})$ represents the binomial
transform. The row sums of the Riordan array
$(1,\frac{x(1-x)}{1+x})$ are $1,1,-1,-1,1,1,\ldots$.

The central terms $T(2n,n)$ turn out to be an `aerated' signed
version of $\binom{2n}{n}$ given by $1,0,-2,0,6,0,-20,\ldots$ with
ordinary generating function $\frac{1}{\sqrt{1+4x^2}}$ and
exponential generating function $I_0(2\sqrt{-1}x)$. They represent the
coefficients of $x^n$ in $(1-x^2)^n$.
We have
$$ T(2n,n;(-1)^n)=\sum_{k=0}^n\binom{n}{k}^2 (-1)^k=\sum_{k=0}^n\binom{n}{k}\binom{n+k}{k}(-1)^k2^{n-k}.$$
The terms
$T(2n+1,n)$ turn out to be a signed version of $\binom{n}{\lfloor
n/2 \rfloor}$, namely $$1,-1,-2,3,6,-10,-20,35,70,\ldots$$ with
ordinary generating function $(\frac{1+2x}{\sqrt{1+4x^2}}-1)/(2x)$
and exponential generating function
$I_0(2\sqrt{-1}x)+\sqrt{-1}I_1(2\sqrt{-1}x)$.

The generalized Catalan numbers $T(2n,n)-T(2n,n-1)$ are the numbers
$$1,-1,0,1,0,-2,0,5,0,-14,0,\ldots$$ with generating function
$y_2(1,2)=\frac{1+2x-\sqrt{1+4x^2}}{2x}$. This is the series reversion of $\frac{x(1-x)}{1-2x}$.

We note that the sequence $T(2(n+1),n)-T(2(n+1),n+1)$ is
$(-1)^{n/2}c(n/2)(1+(-1)^n)/2$ with exponential generating function
$I_1(2\sqrt{-1}x)/(\sqrt{-1}x)$.
\end{example}

\section{A one-parameter sub-family of triangles} The above examples
motivate us to look at the one-parameter subfamily given by the set
of triangles defined by the power sequences $n\to r^n$, for $r \in
\mathbf{Z}$. The case $r=1$ corresponds to Pascal's triangle, while
the case $r=0$ corresponds to the `partial summing' triangle with
$1$s on and below the diagonal.
\begin{proposition} The matrix associated to the sequences $n\to r^n$, $r \in \mathbf{Z}$, is given by
the Riordan array \begin{displaymath}
\left(\frac{1}{1-x},\frac{x(1+(r-1)x)}{1-x}\right).\end{displaymath}\end{proposition}
\begin{proof} The general term $T(n,k)$ of the above matrix is given by
\begin{eqnarray*} T(n,k)&amp;=&amp;[x^n](1+(r-1)x)^kx^k(1-x)^{-(k+1)}\\
&amp;=&amp;[x^{n-k}](1+(r-1)x)^k(1-x)^{-(k+1)}\\
&amp;=&amp;[x^{n-k}]\sum_{j=0}^k \binom{k}{j}(r-1)^jx^j\sum_{i=0}\binom{k+i}{i}x^i\\
&amp;=&amp;[x^{n-k}]\sum_{j=0}^k\sum_{i=0}\binom{k}{j}\binom{k+i}{i}(r-1)^jx^{i+j}\\
&amp;=&amp;\sum_{j=0}^k\binom{k}{j}\binom{k+n-k-j}{n-k-j}(r-1)^j\\
&amp;=&amp;\sum_{j=0}^k\binom{k}{j}\binom{n-j}{k}(r-1)^j\\
&amp;=&amp;\sum_{j=0}^k\binom{k}{j}\binom{n-k}{j}r^j.\end{eqnarray*}
where the last equality is a consequence of identity (3.17) in \cite{SpruId}.
\end{proof}
\begin{corollary}The row sums of the triangle defined by $n\to r^n$ are the binomial transforms of the
doubled sequence $n \to 1,1,r,r,r^2,r^2,\ldots$, i.e., $n\to r^{\lfloor \frac{n}{2} \rfloor}$. \end{corollary}
\begin{proof} The row sums of $(\frac{1}{1-x},\frac{x(1+(r-1)x)}{1-x})$ are the binomial
transform of the row sums of its product with the inverse of the binomial matrix. This product is
\begin{displaymath}(\frac{1}{1+x},\frac{x}{1+x})(\frac{1}{1-x},\frac{x(1+(r-1)x)}{1-x})=(1,\frac{x(1+rx)}{1+x}).\end{displaymath}
The row sums of this product have generating function given by
\begin{displaymath}\frac{1}{1-\frac{x(1+rx)}{1+x}}=\frac{1+x}{1-rx^2}.\end{displaymath}
This is the generating function of $1,1,r,r,r^2,r^2\ldots$ as required.
\end{proof}
We note that the generating function for the row sums of the triangle corresponding to $r^n$ is $\frac{1}{1-2x-(r-1)x^2}$.

We now look at the term $T(2n,n)$ for this subfamily.
\begin{proposition} $T(2n,n;r^n)$ is the coefficient of $x^n$ in $(1+(r+1)x+rx^2)^n$.
\end{proposition}
\begin{proof} We have $(1+(r+1)x+rx^2)=(1+x)(1+rx)$. Hence
\begin{eqnarray*}[x^n](1+(r+1)x+rx^2)^n&amp;=&amp;[x^n](1+x)^n(1+rx)^n\\
&amp;=&amp;[x^n]\sum_{k=0}^n\sum_{j=0}^n \binom{n}{k}\binom{n}{j}r^jx^{k+j}\\
&amp;=&amp;\sum_{j=0}^n \binom{n}{n-j}\binom{n}{j}r^j\\
&amp;=&amp;\sum_{j=0}^n \binom{n}{j}^2r^j.\end{eqnarray*}
\end{proof}
\begin{corollary} The generating function of $T(2n,n;r^n)$ is
$$\frac{1}{\sqrt{1-2(r+1)x+(r-1)^2x^2}}.$$
\end{corollary}
\begin{proof}
Using Lagrangian inversion, we can show that
$$[x^n](1+ax+bx^2)^n=[t^n]\frac{1}{\sqrt{1-2at+(a^2-4b)t^2}}$$ (see
exercises $5.3$ and $5.4$ in \cite{Wilf}). Then
\begin{eqnarray*}
[x^n](1+(r+1)x+rx^2)^n&amp;=&amp;[t^n]\frac{1}{\sqrt{1-2(r+1)t+((r+1)^2-4r)t^2}}\\
&amp;=&amp;[t^n]\frac{1}{\sqrt{1-2(r+1)t+(r-1)^2t^2}}
\end{eqnarray*}
\end{proof}
\begin{corollary} $$\sum_{k=0}^n \binom{n}{k}^2r^k=\sum_{k=0}^n\binom{n}{2k}\binom{2k}{k}(r+1)^{n-2k}r^k=
\sum_{k=0}^n\binom{n}{k}\binom{n-k}{k}(r+1)^{n-2k}r^k.$$ \end{corollary}
\begin{proof} This follows since the coefficient of $x^n$ in $(1+ax+bx^2)^n$ is given
by \cite{Noe} $$\sum_{k=0}^n\binom{n}{2k}\binom{2k}{k}a^{n-2k}b^k=\sum_{k=0}^n\binom{n}{k}\binom{n-k}{k}a^{n-2k}b^k.$$
Hence each term is equal to $T(2n,n;r^n)$.
\end{proof}
We now look at the sequence $T(2n-1,n-1)$.
\begin{proposition} $T(2n-1,n-1;r^n)$ is the coefficient of $x^n$ in $(\frac{1-(r-1)x}{1-rx})^n$
\end{proposition}
\begin{proof}
We have $\frac{1-(r-1)x}{1-rx}=\frac{1-rx+x}{1-rx}=1+\frac{x}{1-rx}$. Hence
\begin{eqnarray*} [x^n](\frac{1-(r-1)x}{1-rx})^n&amp;=&amp;[x^n](1+\frac{x}{1-rx})^n\\
&amp;=&amp;[x^n]\sum_{k=0}^n\sum_{k=0}^n\binom{n}{k}x^k\sum_{j=0}\binom{k+j-1}{j}r^jx^j\\
&amp;=&amp;\sum_{j=0}\binom{n}{n-j}\binom{n-1}{j}r^j\\
&amp;=&amp;\sum_{j=0}\binom{n}{j}\binom{n-1}{j}r^j.\end{eqnarray*}
\end{proof}
\begin{corollary}
$$\sum_{k=0}^n\binom{n}{k}\binom{n-1}{k}r^k=\sum_{k=0}^n\binom{n}{k}\binom{n+k-1}{k}(1-r)^{n-k}r^k
=\sum_{k=0}^n\binom{n}{k}\binom{2n-k-1}{n-k}(1-r)^kr^{n-k}.$$
\end{corollary}
\begin{proof} The coefficient of $x^n$ in $(\frac{1-ax}{1-bx})^n$ is seen to be
$$\sum_{k=0}^n\binom{n}{k}\binom{n+k-1}{k}(-a)^{n-k}r^k
=\sum_{k=0}^n\binom{n}{k}\binom{2n-k-1}{n-k}(-a)^kr^{n-k}.$$
Hence all three terms in the statement are equal to $T(2n-1,n-1;r^n)$.
\end{proof}
We can generalize the results seen above for $T(2n,n)$, $T(2n+1,n)$, $T(2n-1,n-1)$
and $T(2n,n)-T(2n,n-1)$ as follows.
\begin{proposition} Let $T(n,k)=\sum_{k=0}^{n-k}\binom{k}{j}\binom{n-k}{j}r^j$ be the general term of the triangle associated to
 the power sequence $n\to r^n$.
\begin{enumerate}
\item The sequence $T(2n,n)$ has ordinary generating function
 $\frac{1}{\sqrt{1-2(r+1)x+(r-1)^2x^2}}$, exponential generating function $e^{(r+1)x}I_0(2\sqrt{r}x)$,
 and corresponds to the coefficients of $x^n$ in $(1+(r+1)x+rx^2)^n$.
\item The numbers $T(2n+1,n)$ have generating function
$(\frac{1-(r-1)x}{\sqrt{1-2(r+1)x+(r-1)^2x^2}}-1)/(2x)$ and
exponential generating function
$e^{(r+1)x}(I_0(2\sqrt{r}x)+\sqrt{r}I_1(2\sqrt{r}x))$.
\item $T(2n-1,n-1)$
represents the coefficient of $x^n$ in $((1-(r-1)x)/(1-rx))^n$.
\item The
generalized Catalan numbers
$c(n;r^n)=T(2n,n)-T(2n,n-1)$ associated to
the triangle have ordinary generating function
$\frac{1-(r-1)x-\sqrt{1-2(r+1)x+(r-1)^2x^2}}{2x}$.
\item The sequence
$c(n+1;r^n)$ has exponential generating function
$\frac{1}{\sqrt{r}x}e^{(r+1)x}I_1(2\sqrt{r}x)$.
\item
The sequence $nc(n;r^n)=\sum_{k=0}^n\binom{n}{k}\binom{n+k}{k+1}\frac{r^{n-k}}{r+1}$ has exponential generating function
$$\frac{1}{\sqrt{r}x}e^{(r+1)x}I_1(2\sqrt{r}).$$
\item The sequence
$c(n;r^n)-0^n$ is expressible as $\sum_{k=0}^{\lfloor \frac{n-1}{2}
\rfloor}\binom{n-1}{2k}c(k)(r+1)^{n-2k-1}r^k$ and counts the number
of Motzkin paths of length n in which the level steps have $r+1$
colours and the up steps have $r$ colours. It is the series
reversion of $\frac{x}{1+(r+1)x+rx^2}$.
\end{enumerate}
\end{proposition}
Pascal's triangle can be generated by the well-know recurrence
\begin{displaymath} \binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}.\end{displaymath}
The following proposition gives the corresponding recurrence for the case of the triangle associated
to the sequence $n\to r^n$.
\begin{proposition} Let $T(n,k)=\sum_{k=0}^{n-k}\binom{k}{j}\binom{n-k}{j}r^j$. Then
\begin{displaymath}T(n,k)=T(n-1,k-1)+(r-1)T(n-2,k-1)+T(n-1,k).\end{displaymath}
\end{proposition}
\begin{proof} The triangle in question has Riordan array
representation
\begin{displaymath}\left(\frac{1}{1-x},\frac{x(1+(r-1)x)}{1-x}\right)\end{displaymath}
Thus the bivariate generating function of this triangle is given
by \begin{eqnarray*}
F(x,y)&amp;=&amp;\frac{1}{1-x}\frac{1}{1-y\frac{x(1+(r-1)x)}{1-x}}\\
&amp;=&amp;\frac{1}{1-x-xy-(r-1)x^2y}\end{eqnarray*}
\end{proof}
In this simple case, it is possible to characterize the inverse of
the triangle. We have
\begin{proposition} The inverse of the triangle associated to the sequence $n\to r^n$ is given by
the Riordan array $(1-u,u)$ where
\begin{displaymath}u=\frac{\sqrt{1+2(2r-1)x+x^2}-x-1}{2(r-1)}.\end{displaymath}
\end{proposition}
\begin{proof}
Let $(g^*,\bar{f})=(\frac{1}{1-x},\frac{x(1+(r-1)x)}{1-x})^{-1}$. Then
\begin{displaymath} \frac{\bar{f}(1+(r-1)\bar{f})}{1-\bar{f}}=x \Rightarrow \bar{f}=\frac{\sqrt{1+2(2r-1)x+x^2}-x-1}{2(r-1)}.\end{displaymath}
Since $g^*=\frac{1}{g\circ \bar{f}}=1-\bar{f}$ we obtain the result.
\end{proof}
\begin{corollary} The row sums of the inverse of the triangle associated with $n \to r^n$ are $1,0,0,0,\ldots$.
\end{corollary}
\begin{proof} The row sums of the inverse $(1-u,u)$ have generating function given by $\frac{1-u}{1-u}=1$.
In other words, the row sums of the inverse are $0^n=1,0,0,0,\ldots$.
\end{proof}

Other examples of these triangles are given by \seqnum{A081577}, \seqnum{A081578}, \seqnum{A081579}, and \seqnum{A081580}.
\section{The Jacobsthal and the Fibonacci cases}
We now look at the triangles generated by sequences whose elements
can be expressed in Binet form as a simple sum of powers. In the
first example of this section, the powers are of integers, while in
the second case (Fibonacci numbers) we indicate that the formalism
can be extended to non-integers under the appropriate conditions.
\begin{example} The Jacobsthal numbers $J(n+1)$, \seqnum{A001045}, have generating function $\frac{1}{1-x-2x^2}$ and general term
$J(n+1)=2.2^n/3+(-1)^n/3$. Using our previous examples, we see that the triangle defined by $J(n+1)$
\begin{displaymath}\left(\begin{array}{cccccccc} 1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp;\ldots \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp;\ldots \\ 1 &amp; 2 &amp; 1
&amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 3 &amp; 3 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp;\ldots \\ 1 &amp; 4 &amp;
8 &amp; 4 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots
\\1 &amp; 5 &amp; 16 &amp; 16 &amp; 5 &amp; 1 &amp; 0 &amp; \ldots\\1 &amp; 6 &amp; 27 &amp; 42 &amp; 27 &amp; 6 &amp; 1 &amp; \ldots \\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath} or \seqnum{A114202}, is the scaled sum
of the Riordan arrays discussed above, given by
\begin{displaymath} \frac{2}{3}(\frac{1}{1-x},\frac{x(1+x)}{1-x})+\frac{1}{3}(\frac{1}{1-x},\frac{x(1-2x)}{1-x}).\end{displaymath}
In particular, the $k$-th column of the triangle has generating function
\begin{eqnarray*}g_k(x)&amp;=&amp;\frac{x^k}{(1-x)^{k+1}}\left(\frac{2}{3}(1+x)^k+\frac{1}{3}(1-2x)^k\right)\\
&amp;=&amp;\frac{x^k}{(1-x)^{k+1}}\sum_{j=0}^k\binom{k}{j}\frac{1}{3}(2+(-2)^j)x^j.\end{eqnarray*}
We recognize in the sequence $\frac{1}{3}(2+(-2)^n)$ the inverse
binomial transform of $J(n+1)$.

Obviously, the inverse binomial transform of the row sums of the
matrix are given by
\begin{displaymath}\frac{2}{3}2^{\lfloor \frac{n}{2}\rfloor}+\frac{1}{3}(-1)^{\lfloor \frac{n}{2} \rfloor }\end{displaymath}
or $1,1,1,1,3,3,5,5,\ldots$, the doubled sequence of $J(n+1)$.

The terms $T(2n,n)$ for this triangle can be seen to have generating
function
$\frac{2}{3}\frac{1}{\sqrt{1-6x+x^2}}+\frac{1}{3}\frac{1}{\sqrt{1+4x^2}}$
and exponential generating function
$\frac{2}{3}e^{3x}I_0(2\sqrt{2}x)+\frac{1}{3}I_0(2\sqrt{-1}x)$.

The generalized Catalan numbers for this triangle are
$$1, 1, 4, 15, 60, 262, 1204, 5707, 27724, \ldots$$
whose generating function is
$\frac{3-\sqrt{1+4x^2}-2\sqrt{1-6x+x^2}}{6x}$.

To find the relationship between $T(n,k)$ and its `previous' elements, we proceed as follows,
where we write $T(n,k)=T(n,k;J(n+1))$ to indicate its dependence on $J(n+1)$.
\begin{eqnarray*}T(n,k;J(n+1))&amp;=&amp;\sum_{j=0}\binom{k}{j}\binom{n-k}{j}(\frac{2}{3}2^j+\frac{1}{3}(-1)^j)\\
&amp;=&amp;\frac{2}{3}\sum_{j=0}\binom{k}{j}\binom{n-k}{j}2^j+\frac{1}{3}\sum_{j=0}\binom{k}{j}\binom{n-k}{j}(-1)^j\\
&amp;=&amp;\frac{2}{3}T(n,k;2^n)+\frac{1}{3}T(n,k;(-1)^n)\\
&amp;=&amp;\frac{2}{3}(T(n-1,k-1;2^n)+T(n-2,k-1;2^n)+T(n-1,k;2^n))\\
&amp; &amp;+\frac{1}{3}(T(n-1,k-1;(-1)^n)-2T(n-2,k-1;(-1)^n)+T(n-1,k;(-1)^n))\\
&amp;=&amp;\frac{2}{3}T(n-1,k-1;2^n)+\frac{1}{3}T(n-1,k-1;(-1)^n)\\
&amp; &amp;+\frac{2}{3}(T(n-2,k-1;2^n)-T(n-2,k-1;(-1)^n))\\
&amp; &amp;+\frac{2}{3}T(n-1,k;2^n)+\frac{1}{3}T(n-1,k;(-1)^n)\\
&amp;=&amp;T(n-1,k-1;J(n+1))+2T(n-2,k-1;J(n))+T(n-1,k;J(n+1)).
\end{eqnarray*}
We see here the appearance of the non-invertible matrix based on $J(n)$. This begins as
\begin{displaymath}\left(\begin{array}{cccccccc} 0 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp;\ldots \\0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp;\ldots \\ 0 &amp; 1 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 2 &amp; 2 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp;\ldots \\ 0 &amp; 3 &amp;
5 &amp; 3 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots
\\0 &amp; 4 &amp; 9 &amp; 9 &amp; 4 &amp; 0 &amp; 0 &amp; \ldots\\0 &amp; 5 &amp; 14 &amp; 21 &amp; 14 &amp; 5 &amp; 0 &amp; \ldots \\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath}
\end{example}
\begin{example} We briefly look at the case of the Fibonacci sequence \begin{displaymath}
F(n+1)=\left((\frac{1+\sqrt{5}}{2})(\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})(\frac{1-\sqrt{5}}{2})^n\right)/\sqrt{5}.
\end{displaymath}
Again, we can display the associated triangle
\begin{displaymath}\left(\begin{array}{cccccccc} 1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp;\ldots \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp;\ldots \\ 1 &amp; 2 &amp; 1
&amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 3 &amp; 3 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp;\ldots \\ 1 &amp; 4 &amp;
7 &amp; 4 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots
\\1 &amp; 5 &amp; 13 &amp; 13 &amp; 5 &amp; 1 &amp; 0 &amp; \ldots\\1 &amp; 6 &amp; 21 &amp; 31 &amp; 21 &amp; 6 &amp; 1 &amp; \ldots \\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath} or \seqnum{A114197} as a sum of scaled `Riordan arrays' as follows:
\begin{displaymath}
\frac{1+\sqrt{5}}{2}(\frac{1}{1-x},\frac{x(1+(\frac{1+\sqrt{5}}{2}-1)x)}{1-x})-\frac{1-\sqrt{5}}{2}(\frac{1}{1-x},\frac{x(1+(\frac{1-\sqrt{5}}{2}-1)x)}{1-x}).
\end{displaymath}
Hence the $k$-th column of the associated triangle has generating function given by
\begin{displaymath}\frac{x^k}{(1-x)^{k+1}}(\frac{1+\sqrt{5}}{2}(1+(\frac{1+\sqrt{5}}{2}-1)x)^k+\frac{1-\sqrt{5}}{2}(1+(\frac{1-\sqrt{5}}{2}-1)x)^k).\end{displaymath}
Expanding, we find that the generating function of the $k$-th column of the triangle associated to $F(n+1)$ is
given by \begin{displaymath}\frac{x^k}{(1-x)^{k+1}}\sum_{j=0}^k\binom{k}{j}b_jx^j\end{displaymath}
where the sequence $b_n$ is the inverse binomial transform of $F(n+1)$. That is, we have
\begin{displaymath}b_n=\sum_{k=0}^n\binom{n}{k}(-1)^{n-k}F(k+1)=(\phi(\phi-1)^n+\frac{1}{\phi}(-\frac{1}{\phi}-1)^n)/\sqrt{5}\end{displaymath}
where $\phi=\frac{1+\sqrt{5}}{2}$.

Again, the inverse binomial transform of the row sums is given by $F({\lfloor \frac{n}{2} \rfloor}+1)$.

The term $T(2n,n)$ in this case is $\sum_{k=0}^n
\binom{n}{k}^2F(k+1)$, or $1,2,7,31,142,659,\ldots$ (\seqnum{A114198}). This has
ordinary generating function given by
\begin{displaymath}
\frac{\frac{1+\sqrt{5}}{2\sqrt{5}}}{\sqrt{1-2(\frac{1+\sqrt{5}}{2}+1)x+(\frac{1+\sqrt{5}}{2}-1)^2x^2}}-
\frac{\frac{1-\sqrt{5}}{2\sqrt{5}}}{\sqrt{1-2(\frac{1-\sqrt{5}}{2}+1)x+(\frac{1-\sqrt{5}}{2}-1)^2x^2}}
\end{displaymath}
and exponential generating function \begin{displaymath}
\frac{1+\sqrt{5}}{2\sqrt{5}}\exp(\frac{3+\sqrt{5}}{2}x)I_0(2\sqrt{\frac{1+\sqrt{5}}{2}}x)-
\frac{1-\sqrt{5}}{2\sqrt{5}}\exp(\frac{3-\sqrt{5}}{2}x)I_0(2\sqrt{\frac{1-\sqrt{5}}{2}}x).
\end{displaymath}

$T(n,k)$ satisfies the following recurrence
\begin{displaymath}T(n,k;F(n+1))=T(n-1,k-1;F(n+1))+T(n-2,k-1;F(n))+T(n-1,k;F(n+1))\end{displaymath}
where the triangle associated to $F(n)$ begins
\begin{displaymath}\left(\begin{array}{cccccccc} 0 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp;\ldots \\0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp;\ldots \\ 0 &amp; 1 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 2 &amp; 2 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp;\ldots \\ 0 &amp; 3 &amp;
5 &amp; 3 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots
\\0 &amp; 4 &amp; 9 &amp; 9 &amp; 4 &amp; 0 &amp; 0 &amp; \ldots\\0 &amp; 5 &amp; 14 &amp; 20 &amp; 14 &amp; 5 &amp; 0 &amp; \ldots \\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath}
We note that all Lucas sequences \cite{Lucas} can be treated in similar fashion.
\end{example}
\section{The general case}
\begin{proposition} Given an integer sequence $a_n$ with $a_0=1$, the centrally symmetric invertible triangle
associated to it by the above construction has the following generating function for its $k$-th column:
\begin{displaymath}\frac{x^k}{1-x}\sum_{j=0}^k\binom{k}{j}a_j\left(\frac{x}{1-x}\right)^j=\frac{x^k}{(1-x)^{k+1}}\sum_{j=0}^k\binom{k}{j}b_jx^j
\end{displaymath} where $b_n$ is the inverse binomial transform of $a_n$.\end{proposition}
\begin{proof}
We have
\begin{eqnarray*}[x^n]\frac{x^k}{1-x}\sum_{j=0}^k\binom{k}{j}a_j\left(\frac{x}{1-x}\right)^j&amp;=&amp;[x^{n-k}]\sum_{j=0}^k\binom{k}{j}a_j\frac{x^j}{(1-x)^{j+1}}\\
&amp;=&amp;\sum_{j}\binom{k}{j}a_j[x^{n-k-j}](1-x)^{-(j+1)}\\
&amp;=&amp;\sum_{j}\binom{k}{j}a_j[x^{n-k-j}]\sum_{i}\binom{j+i}{i}x^i\\
&amp;=&amp;\sum_{j}\binom{k}{j}a_j\binom{j+n-k-j}{n-k-j}\\
&amp;=&amp;\sum_{j}\binom{k}{j}\binom{n-k}{j}a_j\\
&amp;=&amp;T(n,k).\end{eqnarray*}
Similarly,
\begin{eqnarray*}[x^n]\frac{x^k}{(1-x)^{k+1}}\sum_{j=0}^k\binom{k}{j}b_jx^j&amp;=&amp;\sum_{j}\binom{k}{j}b_j[x^{n-k-j}](1-x)^{-(k+1)}\\
&amp;=&amp;\sum_{j}\binom{k}{j}b_j[x^{n-k-j}]\sum_{i}\binom{k+i}{i}x^i\\
&amp;=&amp;\sum_{j}\binom{k}{j}b_j\binom{k+n-k-j}{n-k-j}\\
&amp;=&amp;\sum_{j}\binom{k}{j}\binom{n-j}{k}b_j.
\end{eqnarray*}
Now \begin{eqnarray*}
\sum_{j=0}\binom{k}{j}\binom{n-k}{j}a_j&amp;=&amp;\sum_{j=0}\binom{k}{j}\binom{n-k}{j}\sum_{i=0}^j\binom{j}{i}b_i\\
&amp;=&amp;\sum_j\sum_i\binom{k}{j}\binom{n-k}{j}\binom{j}{i}b_i\\
&amp;=&amp;\sum_j\sum_i\binom{k}{j}\binom{j}{i}\binom{n-k}{j}b_i\\
&amp;=&amp;\sum_j\sum_i\binom{k}{i}\binom{k-i}{j-i}\binom{n-k}{j}b_i\\
&amp;=&amp;\sum_i\binom{k}{i}b_i\sum_j\binom{k-i}{k-j}\binom{n-k}{j}\\
&amp;=&amp;\sum_i\binom{k}{i}b_i\binom{n-i}{k}\\
&amp;=&amp;\sum_j\binom{k}{j}\binom{n-j}{k}b_j.\end{eqnarray*}
\end{proof}
\begin{corollary} The following relationship exists between a sequence $a_n$ and its inverse binomial transform $b_n$:
\begin{displaymath}\sum_j\binom{k}{j}\binom{n-k}{j}a_j=\sum_j\binom{k}{j}\binom{n-j}{k}b_j.\end{displaymath}
\end{corollary}
It is possible of course to reverse the above proposition to give us
the following:
\begin{proposition} Given a sequence $b_n$, the product of the
triangle whose $k$-th column has ordinary generating function $$
\frac{x^k}{(1-x)^{k+1}}\sum_{j=0}^k \binom{k}{j}b_jx^j$$ by the
binomial matrix is the centrally symmetric invertible triangle
associated to the binomial transform of $b_n$.\end{proposition}

\section{Exponential-factorial triangles}
In this section, we briefly describe an alternative method that produces
generalized Pascal matrices, based on suitably chosen sequences. For this,
we recall that the binomial matrix $\mathbf{B}$ may be represented as
\begin{displaymath}\mathbf{B}=\exp\left(\begin{array}{ccccccc} 0 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 2 &amp; 0 &amp;
0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; 3 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; 0
&amp; 4 &amp; 0 &amp; 0 &amp; \ldots
\\0 &amp; 0 &amp; 0  &amp; 0 &amp; 5 &amp; 0 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath} while if we
write $a(n)=n$ then the general term $\binom{n}{k}$ of this matrix
can be written as
$$\binom{n}{k}=\frac{\prod_{j=1}^ka(n-j+1)}{\prod_{j=1}^ka(j)}.$$
Furthermore,
$$\mathbf{B}=\sum_{k=0}\frac{\mathbf{M}^k}{\prod_{j=1}^ka(j)}$$
where $\mathbf{M}$ is the sub-diagonal matrix formed from the
elements of $a(n)$.

We shall see that by generalizing this construction to suitably
chosen sequences $a(n)$ where $a(0)=0$ and $a(1)=1$, we can obtain
generalized Pascal triangles, some of which are well documented in
the literature. Thus we let $T(n,k)$ denote the matrix with general
term
$$T(n,k)=\frac{\prod_{j=1}^ka(n-j+1)}{\prod_{j=1}^ka(j)}=\binom{n}{k}_{a(n)}.$$
\begin{proposition} $T(n,n-k)=T(n,k)$, $T(n,1)=a(n)$, $T(n+1,1)=T(n+1,n)=a(n+1)$ \end{proposition}
\begin{proof} To prove the first assertion, we assume first that
$k\le n-k$. Then
\begin{eqnarray*} T(n,k)&amp;=&amp;\frac{a(n)\ldots a(n-k+1)}{a(1)\dots
a(k)}\\
&amp;=&amp;\frac{a(n)\ldots a(n-k+1)}{a(1)\dots a(k)}\frac{a(n-k)\ldots
a(k+1)}{a(k+1)\ldots a(n-k)}\\
&amp;=&amp;T(n,n-k).\end{eqnarray*} Secondly, if $k&gt;n-k$, we have
\begin{eqnarray*} T(n,n-k)&amp;=&amp;\frac{a(n)\ldots a(k+1)}{a(1) \ldots
a(n-k)}\\
&amp;=&amp;\frac{a(n)\ldots a(k+1)}{a(1) \ldots a(n-k)}\frac{a(k) \ldots
a(n-k+1)} {a(n-k+1) \ldots a(k)}\\
&amp;=&amp; T(n,k).\end{eqnarray*} Next, we have \begin{eqnarray*} T(n,1)&amp;=&amp;
\frac{\prod_{j=1}^1 a(n-j+1)}{\prod_{j=1}^1 a(j)}\\
&amp;=&amp;\frac{a(n-1+1)}{a(1)}=a(n).\end{eqnarray*} since $a(1)=1$.
Similarly, \begin{eqnarray*} T(n+1,1)&amp;=&amp;
\frac{\prod_{j=1}^1 a(n+1-j+1)}{\prod_{j=1}^1 a(j)}\\
&amp;=&amp;\frac{a(n+1-1+1)}{a(1)}=a(n+1).\end{eqnarray*}
\end{proof}
Thus for those choices of the sequence $a(n)$ for which the values
of $T(n,k)$ are integers, $T(n,k)$ represents a generalized Pascal
triangle with $T(n,1)=a(n+1)$. We shall use the notation
$\mathbf{P}_{a(n)}$ to denote the triangle constructed as above.

We define the
 \emph{generalized Catalan sequence associated to
$a(n)$ by this construction} to be the sequence with general term
$$\frac{T(2n,n)}{a(n+1)}.$$
\begin{example} The Fibonacci numbers. The matrix
$\mathbf{P}_{F(n)}$ with general term
$$\frac{\prod_{j=1}^kF(n-j+1)}{\prod_{j=1}^kF(j)}$$ which can be expressed as $$\sum_{k=0}\frac{\mathbf{M}_F^k}{\prod_{j=1}^kF(j)}$$
where $\mathbf{M}_F$ is the sub-diagonal matrix generated by $F(n)$:
\begin{displaymath}\mathbf{M}_F=\left(\begin{array}{ccccccc} 0 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 1 &amp; 0 &amp;
0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; 2 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; 0 &amp;
3 &amp; 0 &amp; 0 &amp; \ldots
\\0 &amp; 0 &amp; 0  &amp; 0 &amp; 5 &amp; 0 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath} is the much
studied Fibonomial matrix, \seqnum{A010048}, \cite{Knuth},
\cite{Gessel}, \cite{Krot1}, \cite{Filbert}, \cite{FibCoeff}. For
instance, the generalized Catalan numbers associated to this
triangle are the Fibonomial Catalan numbers, \seqnum{A003150}.
\end{example}
\begin{example} Let $a(n)=\frac{2^n}{2}-\frac{0^n}{2}$. The matrix
$\mathbf{P}_{a(n)}$ with general term
$$\frac{\prod_{j=1}^ka(n-j+1)}{\prod_{j=1}^ka(j)}$$ which can be expressed as $$\sum_{k=0}\frac{\mathbf{M}^k}{\prod_{j=1}^ka(j)}$$
where $\mathbf{M}$ is the sub-diagonal matrix generated by $a(n)$:
\begin{displaymath}\mathbf{M}=\left(\begin{array}{ccccccc} 0 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 2 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; 4 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp;
0 &amp; 8 &amp; 0 &amp; 0 &amp; \ldots
\\0 &amp; 0 &amp; 0  &amp; 0 &amp; 16 &amp; 0 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath} is given by
\begin{displaymath}\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 2 &amp; 1
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 4 &amp; 4 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 8 &amp;
16 &amp; 8 &amp; 1 &amp; 0 &amp; \ldots
\\1 &amp; 16 &amp; 64  &amp; 64 &amp; 16 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath} This is
\seqnum{A117401}. For this matrix, we have $T(2n,n)=2^{n^2}$ and
$c(n;a(n))=2^{n(n-1)}$. This is easily generalized to the sequence
$n \to \frac{k^n}{k}-\frac{0^n}{k}$. For this sequence, we obtain
$T(2n,n)=k^{n^2}$ and $c(n)=k^{n(n-1)}$.
\end{example}
\begin{example} We take the case $a(n)={\lfloor \frac{n+1}{2}
\rfloor}$. In this case, we obtain the matrix
\begin{displaymath}\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 1 &amp; 1 &amp;
0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 2 &amp; 2 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 2 &amp; 4 &amp;
2 &amp; 1 &amp; 0 &amp; \ldots
\\1 &amp; 3 &amp; 6  &amp; 6 &amp; 3 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath}
\end{example} which has general term $$\binom{\lfloor \frac{n}{2}
\rfloor}{\lfloor \frac{k}{2} \rfloor}\binom{\lceil \frac{n}{2}
\rceil}{\lceil \frac{k}{2} \rceil}.$$ This triangle counts the
number of symmetric Dyck paths of semi-length $n$ with $k$ peaks
(\seqnum{A088855}). We note that for this triangle, $T(2n,n)$ is
$\binom{n}{\lfloor \frac{n}{2} \rfloor}^2$ while $T(2n,n)-T(2n,n-1)$
is the sequence $$1, 0, 2, 0, 12, 0, 100, 0, 980, 0, 10584\ldots$$
\begin{example} The Jacobsthal numbers. Let $a(n)=J(n)=\frac{2^n}{3}-\frac{(-1)^n}{3}$.
We form the matrix with general term
$$\frac{\prod_{j=1}^kJ(n-j+1)}{\prod_{j=1}^kJ(j)}$$ which can be expressed as $$\sum_{k=0}\frac{\mathbf{M}_J^k}{\prod_{j=1}^kJ(j)}$$
where $\mathbf{M}_J$ is the sub-diagonal matrix generated by $J(n)$:
\begin{displaymath}\mathbf{M}_J=\left(\begin{array}{ccccccc} 0 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 1 &amp; 0 &amp;
0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; 3 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; 0 &amp;
5 &amp; 0 &amp; 0 &amp; \ldots
\\0 &amp; 0 &amp; 0  &amp; 0 &amp; 11 &amp; 0 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath} We obtain the
matrix
\begin{displaymath}\mathbf{P}_{J(n)}=\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 1 &amp; 1 &amp;
0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 3 &amp; 3 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 5 &amp; 15 &amp;
5 &amp; 1 &amp; 0 &amp; \ldots
\\1 &amp; 11 &amp; 55  &amp; 55 &amp; 11 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath} We recognize
in this triangle the unsigned version of the $q$-binomial triangle
for $q=-2$, \seqnum{A015109}, whose $k$-th column has generating
function
$$x^k \frac{1}{\prod_{j=0}^k (1-(-2)^jx)}.$$ Using the above
notation, this latter signed triangle is therefore
$\mathbf{P}_{(-1)^nJ(n)}$. Note that
$$\frac{x}{(1-x)(1+2x)}=\frac{x}{1+x-2x^2}$$ is the generating
function for $(-1)^nJ(n)$.

The generating function of the $k$-th column of $\mathbf{P}_{J(n)}$
is given by
$$ x^k\prod_{j=0}^k\frac{1}{(1-(-1)^{(j+k \bmod 2)}2^jx)}.$$

The generalized Catalan numbers for $\mathbf{P}_{J(n)}$ are given by
$\frac{\mathbf{P}_{J(n)}(2n,n)}{J(n+1)}$. These are \seqnum{A015056}
$$1,1,5,77,5117,1291677,\ldots$$

We can generalize these results to the following:
\begin{proposition} Let $a(n)$ be the solution to the recurrence
$$ a(n)=(r-1)a(n-1)+r^2a(n-2), \qquad a(0)=0, \qquad a(1)=1.$$
Then $\mathbf{P}_{a(n)}$ is a generalized Pascal triangle whose
$k$-th column has generating function given by
$$ x^k\prod_{j=0}^k\frac{1}{(1-(-1)^{(j+k \bmod 2)}r^jx)}.$$
\end{proposition}
\end{example}
\begin{example} The Narayana and related triangles. The Narayana triangle
$\tilde{N}$ is a generalized Pascal triangle in the sense of this
section. It is known that the generating function of its $k$-th
column is given by
$$ x^k\frac{\sum_{j=0}^k N(k,j)x^j}{(1-x)^{2k+1}}.$$
Now $a(n)=\tilde{N}(n,1)=\binom{n+1}{2}$ satisfies $a(0)=0$,
$a(1)=1$. It is not difficult to see that, in fact,
$\mathbf{\tilde{N}}=\mathbf{P}_{\binom{n+1}{2}}$. See
\cite{Helms}. $T(2n,n)$ for this triangle is \seqnum{A000891},
with exponential generating function $I_0(2x)I_1(2x)/x$. We note
that is this case, the numbers generated by
$\tilde{N}(2n,n)/a(n+1)$ do not produce integers. However the
sequence $\tilde{N}(2n,n)-\tilde{N}(2n,n+1)$ turns out to be the
product of successive Catalan numbers $c(n)c(n+1)$. This is
\seqnum{A005568}.

The triangle $\mathbf{P}_{\binom{n+2}{3}}$ is \seqnum{A056939} with
matrix
\begin{displaymath}\mathbf{P}_{\binom{n+2}{3}}=\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 4 &amp; 1
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 10 &amp; 10 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 20
&amp; 50 &amp; 20 &amp; 1 &amp; 0 &amp; \ldots
\\1 &amp; 30 &amp; 175  &amp; 175 &amp; 30 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath} The $k$-th
column of this matrix has generating function
$$ x^k\frac{\sum_{j=0}^k N_3(k,j)x^j}{(1-x)^{3k+1}}$$
where $N_3(n,k)$ is the triangle of 3-Narayana numbers,
\cite{SulankeNara3},  \seqnum{A087647}.

$\mathbf{P}_{\binom{n+3}{4}}$ is the number triangle
\seqnum{A056940}.
\end{example}
\section{Acknowledgments}
The author wishes to thank Laura L. M. Yang for her careful
reading of this manuscript and her helpful comments. The author
also thanks an anonymous referee for their pertinent comments and
suggestions.
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\bigskip
\hrule
\bigskip
\noindent 2000 {\it Mathematics Subject Classification}: Primary
11B83; Secondary 05A19, 11B37, 11B65.

\noindent \emph{Keywords:} Pascal's triangle, Narayana numbers,
Catalan numbers, Schr\"oder numbers, Delannoy numbers, Fibonacci numbers, Jacobsthal numbers.

\bigskip
\hrule
\bigskip

(Concerned with sequences \seqnum{A000045}, \seqnum{A000108},
\seqnum{A000129}, \seqnum{A000891}, \seqnum{A000984},
\seqnum{A001003}, \seqnum{A001045}, \seqnum{A001263},
\seqnum{A001700}, \seqnum{A001850}, \seqnum{A002002},
\seqnum{A003150}, \seqnum{A005568}, \seqnum{A006318},
\seqnum{A007318}, \seqnum{A007564}, \seqnum{A008288},
\seqnum{A009545},
\seqnum{A010048},
\seqnum{A015056}, \seqnum{A015109},
\seqnum{A016116}, \seqnum{A047891}, \seqnum{A056939},
\seqnum{A056940}, \seqnum{A059231}, \seqnum{A060693},
\seqnum{A078009}, \seqnum{A078018}, \seqnum{A081178},
\seqnum{A081577}, \seqnum{A081578}, \seqnum{A081579},
\seqnum{A081580}, \seqnum{A082147}, \seqnum{A082148},
\seqnum{A082181}, \seqnum{A082201}, \seqnum{A082301},
\seqnum{A082302}, \seqnum{A082305}, 
\seqnum{A082366}, \seqnum{A082367}, \seqnum{A087647},
\seqnum{A088218}, \seqnum{A088617}, \seqnum{A088855},
\seqnum{A114197}, \seqnum{A114198},
\seqnum{A114202}, \seqnum{A117401}. )

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  December 7 2005;
revised version received  April 21 2006.
Published in {\it Journal of Integer Sequences},  May 19 2006.

\bigskip
\hrule
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