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\begin{center}
\vskip 1cm{\LARGE\bf On a Generalization of the Narayana Triangle} \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} 
By studying various ways of describing the Narayana
triangle, we provide a number of generalizations of this triangle and
study some of their properties. In some cases, the diagonal sums of
these triangles provide examples of Somos-$4$ sequences via their
Hankel transforms.
\end{abstract}

\section{Introduction}

In this note, we shall construct a generalization of the Narayana triangle by two means, and show that both constructions lead to the 
same result. The first is based on a closed expression for the general term of the triangle, inspired by a trinomial expression. The 
second construction is inspired by a certain bivariate continued fraction expression for the generating function of the Narayana 
triangle. This expression also allows us to easily investigate the diagonal sums of the the generalized Narayana triangles under study. 
In certain circumstances (but not all), the Hankel transforms of these diagonal sums turn out to be generalized Somos-$4$ sequences.

The Narayana triangle, and the polynomials for which it is the coefficient array, have many applications that span such areas as 
combinatorics \cite{Chapoton, Doslic, Fomin_Reading, Kostov, Sulanke_Nara1, SulankeNara3}, probability \cite{Narayana, Sulanke_Nara2} 
and statistics \cite{Narayana, Sulanke_Nara2}, random matrix theory \cite{Dumitriu, Dumitriu_Path, Marcenko} and wireless 
communications \cite{Barry_Mimo, Groh, Muller, Tulino} . We have previously studied \cite{Barry_Mimo} methods to describe this 
triangle, or more correctly, the three closely related triangles that are each commonly called the Narayana triangle. Thus we 
distinguish between the triangle $N_1(n,k)$ \seqnum{A131198} that has general term
$$N_1(n,k)=\frac{1}{k+1}\binom{n-1}{k}\binom{n}{k},$$ the triangle $N_2$ \seqnum{A090181} that has general term
\begin{eqnarray}N_2(n,k)&=&[k \le n]N_1(n,n-k)=0^{n+k}+\frac{1}{n+0^{nk}}\binom{n}{k}\binom{n}{k-1}\\
&=&\frac{1}{n-k+1}\binom{n-1}{n-k}\binom{n}{k},\end{eqnarray} (this is the reversal of $N_1$), and the triangle $N=N_3$
\seqnum{A001263} which has general term
$$N_3(n,k)=\frac{1}{k+1}\binom{n}{k}\binom{n+1}{k}.$$
Thus $N=N_3$ is the familiar ``Pascal-like'' triangle that begins
\begin{displaymath}\mathbf{N}_3=\left(\begin{array}{ccccccc}
1 & 0 & 0 & 0 & 0 & 0 & \ldots \\1 & 1 & 0 & 0 & 0 &
0 & \ldots \\ 1 & 3 & 1 & 0 & 0 & 0 & \ldots \\ 1 & 6 & 6 & 1
&
0 & 0 & \ldots \\ 1 & 10 & 20 & 10 & 1 & 0 & \ldots \\1 & 15 &
50 & 50 & 15
& 1 &\ldots\\ \vdots & \vdots & \vdots & \vdots & \vdots &
\vdots & \ddots\end{array}\right).\end{displaymath}
We clearly have $N(n,k)=N(n,n-k)$. A number triangle with this property will be called ``Pascal-like''.
Some generalizations of the Narayana numbers have been studied for instance in \cite{SulankeNara3}.

During this paper, we will use the following notational conventions. 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.
The element 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 power series $f(x)=\sum_{k=0}^{\infty}a_k 
\frac{x^k}{k!}$ is called the \emph{exponential generating function} or e.g.f. of the sequence. We use the notation
$0^n=[x^n]1$ for the sequence $1,0,0,0,\ldots,$ \seqnum{A000007}. Thus $0^n=[n=0]=\delta_{n,0}=\binom{0}{n}$.
Here,
we have used the Iverson bracket notation \cite{Concrete},
defined by $[\mathcal{P}]=1$ if the proposition $\mathcal{P}$
is true, and
$[\mathcal{P}]=0$ if $\mathcal{P}$ is false.

For a power series
$f(x)=\sum_{n=0}^{\infty}a_n x^n$ with $f(0)=0$ we define the reversion or compositional inverse of $f$ to be the
power series $\bar{f}(x)$ such that $f(\bar{f}(x))=x$. We sometimes write
$\bar{f}= \text{Rev}f$.

The Hankel transform \cite{CRI, Layman} of a sequence $a_n$ is the sequence
$h_n=|a_{i+j}|_{0\le i,j\le n}$. For certain sequences, this may be calculated by means of
continued fraction representations \cite{Kratt, Kratt1}. We shall use continued fractions \cite{Barry_CF, Wall} extensively in what 
follows. We also use some moment or integral representations of sequences, which derive for instance from the theory of orthogonal 
polynomials \cite{Chihara, Gautschi, Szego}. In particular we recall that
$$C_n=\frac{1}{2\pi} \int_0^4 x^n \frac{\sqrt{x(4-x)}}{x}\,dx$$ and
$$\binom{2n}{n}=\frac{1}{\pi}\int_0^4 x^n \frac{1}{\sqrt{x(4-x)}}\,dx.$$

We shall have occasion to use the language of Riordan arrays in this note.
The \emph{Riordan group} \cite{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 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)\cdot(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 $f$. Also called the reversion of $f$, we will use the notation
$\bar{f}=\text{Rev}(f)$ as well.

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

The row sums of the matrix $(g, f)$ have generating function
$$(g,f)\cdot \frac{1}{1-x}=\frac{g(x)}{1-f(x)}$$
 while the diagonal sums of $(g, f)$ have generating
function $g(x)/(1-xf(x))$.

In similar fashion we can also define exponential Riordan arrays $[g,f]$, using the e.g.f \cite{Barry_Exp, DeutschShap} instead of the
ordinary generating function. For instance, the identity matrix is given by $[1,x]$ and the
binomial matrix is given by $[e^x,x]$.

If $a_n$ is the sequence $a_0,a_1,a_2,\ldots$ with g.f. $f(x)$, then the sequence
$a_0,0,a_1,0,a_2,0,\ldots$ with g.f. $f(x^2)$ is called the \emph{aeration} of $a_n$.

Many interesting examples of number triangles, including Riordan arrays, can be found in Neil Sloane's On-Line
Encyclopedia of Integer Sequences \cite{SL1, SL2}. Sequences are frequently referred to by their
OEIS number. For instance, the matrix $\mathbf{B}$ is \seqnum{A007318}.
We observe that the Narayana triangles are not Riordan arrays.
\section{Trinomials and the Narayana triangle}
In order to explore how we might generalize the Narayana triangle $N$, we start with the following observation concerning powers of 
trinomials (the paper \cite{Noe} provides an introduction to the combinatorics of trinomials).
\begin{proposition} The number triangle with general term
$$[x^{2k}](1+\sqrt{\alpha}x+x^2)^n$$ is ``Pascal-like''.\end{proposition}
\begin{proof}
We first calculate $T(n,k)$. We have
\begin{eqnarray*}T(n,k)&=&[x^{2k}](1+\beta x+x^2)^n, \quad \beta=\sqrt{\alpha},\\
&=&[x^{2k}]\sum_{j=0}^n \binom{n}{j}x^{2j}(1+\beta x)^{n-j}\\
&=&[x^{2k}]\sum_{j=0}^n \binom{n}{j}x^{2j}\sum_{i=0}^{n-j}\binom{n-j}{i}\beta^i x^i\\
&=&[x^{2k}] \sum_{j=0}^n \sum_{i=0}^{n-j}\binom{n}{j}\binom{n-j}{i}\beta^i x^{i+2j}\\
&=&\sum_{j=0}^n \binom{n}{j}\binom{n-j}{2k-2j}\beta^{2k-2j}\\
&=&\sum_{j=0}^n \binom{n}{j}\binom{n-j}{2k-2j}\alpha^{k-j}.\end{eqnarray*}

\noindent We now wish to show that $T(n,k)=T(n,n-k)$, or that
$$\sum_{j=0}^n \binom{n}{j}\binom{n-j}{2k-2j}\alpha^{k-j}=\sum_{j=0}^n \binom{n}{j}\binom{n-j}{2(n-k)-2j}\alpha^{n-k-j}.$$ To do this, 
we will show that the coefficients of powers of $\alpha$ are equal on both sides of the sought identity. We have on the one hand that
\begin{eqnarray*}
\binom{n}{j}\binom{n-j}{2k-2j}\alpha^{k-j}&=&\binom{n}{n-j}\binom{n-j}{2k-2j}\alpha^{k-j}\\
&=&\binom{n}{2k-2j}\binom{n-2k+2j}{n-j-2k+2j}\alpha^{k-j}\\
&=&\binom{n}{2(k-j)}\binom{n-2(k-j)}{n+j-2k}\alpha^{k-j}\\
&=&\binom{n}{2u}\binom{n-2u}{n-k-u}\alpha^u,\quad u=k-j.\end{eqnarray*} On the other hand, we have
\begin{eqnarray*}
\binom{n}{j}\binom{n-j}{2(n-k)-2j}\alpha^{n-k-j}&=&\binom{n}{n-k-u}\binom{u+k}{2u}\alpha^u, \quad u=n-k-j\\
&=& \binom{n}{u+k}\binom{u+k}{2u}\alpha^u\\
&=& \binom{n}{2u}\binom{n-2u}{k-u}\alpha^u\\
&=& \binom{n}{2u}\binom{n-2u}{n-k-u}\alpha^u.\qedhere \end{eqnarray*}
\end{proof}
We are indebted to an anonymous referee  for the following simpler proof of the above result. 
Observe that $x^{2n}f_n(1/x)=f_n(x)$, where $f_n(x)=(1+\sqrt{\alpha}x+x^2)^n$. Hence we have 
\begin{eqnarray*} T(n,k)&=&[x^{2k}]f_n(x)\\
&=&[x^{2k}]x^{2n}f_n(1/x)\\
&=&[x^{2k-2n}]f_n(1/x)\\
&=&[x^{2n-2k}]f_n(x)=T(n,n-k).\end{eqnarray*}
\begin{example} \label{Pas}
\textbf{Pascal's triangle} The binomial triangle $\binom{n}{k}$ is given by
$$\binom{n}{k}=[x^{2k}](1+x^2)^n=\sum_{j=0}^n \binom{n}{j}\binom{n-j}{2k-2j}0^{k-j}.$$
\end{example}
\begin{example}
The triangle with general term
$$[x^{2k}](1+\sqrt{2}x+x^2)^n=\sum_{j=0}^n \binom{n}{j}\binom{n-j}{2k-2j}2^{k-j}$$ begins
 \begin{displaymath}\left(\begin{array}{cccccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & \ldots \\1 & 1 & 0 & 0 & 0 &
0 & 0 &\ldots \\ 1 & 4 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 9 & 9 & 1
&
0 & 0 & 0 &\ldots \\ 1 & 16 & 34 & 16 & 1 & 0 & 0 &\ldots \\1 & 25 &
90 & 90 & 25
& 1 & 0 &\ldots\\1 & 36 & 195 & 328 & 195 & 36 & 1 &\ldots \\  \vdots & \vdots & \vdots & \vdots & \vdots &
\vdots & \ddots\end{array}\right).\end{displaymath} This is \seqnum{A124216}.
\end{example}
We can easily characterize the central coefficients of the triangle with general term  $\sum_{j=0}^n 
\binom{n}{j}\binom{n-j}{2k-2j}r^{k-j}$, such as the numbers $1,4,34,328,\ldots$ for $r=2$ above.
\begin{proposition} The central coefficients $T(2n,n)$ of the triangle with general term $\sum_{j=0}^n 
\binom{n}{j}\binom{n-j}{2k-2j}r^{k-j}$ have moment representation
$$T(2n,n)=\frac{1}{\pi}\int_{\sqrt{r}-2}^{\sqrt{r}+2} \frac{x^{2n}}{\sqrt{4-(x-\sqrt{r})^2}}\,dx.$$
\end{proposition}
\begin{proof}
We have $$[x^n](1+rx+x^2)^n=[x^n]\frac{1}{\sqrt{(1-rx)^2-4x^2}},$$ 
and  $$[x^n]\frac{1}{\sqrt{(1-rx)^2-4x^2}}=\frac{1}{\pi}\int_{r-2}^{r+2}\frac{x^n}{\sqrt{4-(x-r)^2}}\,dx,$$ where this last equation 
can be established using Stieltjes-Perron techniques (\cite{Barry_Moment, Gautschi, Henrici, Wall}). 
Now $$T(2n,n)=[x^{2n}](1+\sqrt{r}x+x^2)^{2n},$$ whence the assertion.
\end{proof}

In order to relate the foregoing to the Narayana triangle, we use an ``umbral'' \cite{Roman} approach. We recall that the Catalan 
numbers
$$C_n=\frac{1}{n+1}\binom{2n}{n}$$ have the following moment representation:
$$C_n = \frac{1}{2\pi}\int_{0}^4 x^n \frac{\sqrt{x(4-x)}}{x} dx = \int_\mathbb{R} x^n d\mu_C(x).$$
We now define
$$T_C(n,k)=\int_\mathbb{R} [y^{2k}](1+\sqrt{x}y+y^2)^n d\mu_C(x).$$
\begin{lemma} $$T_C(n,k)=\sum_{j=0}^n \binom{n}{j}\binom{n-j}{2(k-j)}C_{k-j}.$$
\end{lemma}
\begin{proof}
\begin{eqnarray*} T_C(n,k)&=&\int_\mathbb{R} [y^{2k}](1+\sqrt{x}y+y^2)^n d\mu_C\\
&=&\frac{1}{2\pi}\int_0^4 [y^{2k}](1+\sqrt{x}y+y^2)^n \frac{\sqrt{x(4-x)}}{x} dx\\
&=&\frac{1}{2\pi}\int_0^4 \sum_{j=0}^n \binom{n}{j}\binom{n-j}{2(k-j)}x^{k-j}\frac{\sqrt{x(4-x)}}{x} dx\\
&=&\sum_{j=0}^n \binom{n}{j}\binom{n-j}{2(k-j)}\frac{1}{2\pi}\int_0^4 x^{k-j}\frac{\sqrt{x(4-x)}}{x} dx\\
&=&\sum_{j=0}^n \binom{n}{j}\binom{n-j}{2(k-j)}C_{k-j}. \qedhere \end{eqnarray*}
\end{proof}
\newpage
\begin{proposition}
$$T_C(n,k)=N(n,k).$$
\end{proposition}
\begin{proof}
We have
\begin{eqnarray*}
T_C(n,k)&=&\sum_{j=0}^n \binom{n}{j}\binom{n-j}{2(k-j)}C_{k-j}\\
&=&\sum_{j=0}^n \binom{n}{j}\binom{n-j}{2(k-j)}\binom{2(k-j)}{k-j}\frac{1}{k-j+1}\\
&=&\sum_{j=0}^n \binom{n}{j}\binom{n-j}{k-j}\binom{n-j-k+j}{2k-2j-k+j}\frac{1}{k-j+1}\\
&=&\sum_{j=0}^n \binom{n}{j}\binom{n-j}{k-j}\binom{n-k}{k-j}\frac{1}{k-j+1}\\
&=&\sum_{j=0}^n \binom{n}{n-j}\binom{n-j}{k-j}\binom{n-k}{k-j}\frac{1}{k-j+1}\\
&=&\sum_{j=0}^n \binom{n}{k-j}\binom{n-k+j}{n-j-k+j}\binom{n-k}{k-j}\frac{1}{k-j+1}\\
&=&\sum_{j=0}^n \binom{n}{n-k+j}\binom{n-k+j}{n-k}\binom{n-k}{k-j}\frac{1}{k-j+1}\\
&=&\sum_{j=0}^n \binom{n}{n-k}\binom{n-(n-k)}{n-k+j-(n-k)}\binom{n-k}{k-j}\frac{1}{k-j+1}\\
&=&\sum_{j=0}^n \binom{n}{k}\binom{k}{j}\binom{n-k}{k-j}\frac{1}{k-j+1}\\
&=&\frac{1}{k+1}\binom{n}{k}\sum_{j=0}^n \binom{k}{j}\frac{k+1}{k-j+1}\binom{n-k}{k-j}\\
&=&\frac{1}{k+1}\binom{n}{k}\sum_{j=0}^n \binom{k+1}{j}\binom{n-k}{k-j}\\
&=&\frac{1}{k+1}\binom{n}{k}\binom{n+1}{k},\end{eqnarray*} or
$$T_C(n,k)=\sum_{j=0}^n \binom{n}{j}\binom{n-j}{2(k-j)}C_{k-j}=\frac{1}{k+1}\binom{n}{k}\binom{n+1}{k}=N(n,k).\qedhere $$
\end{proof}
Many interesting results about the Narayana triangle may be deduced from the above. An example is the following integral representation 
of the central coefficients of the Narayana triangle, which is the sequence \seqnum{A000891} with general term
$N(2n,n)=\frac{1}{n+1}\binom{2n}{n}\binom{2n+1}{n}=\frac{1}{2n+1}\binom{2n+1}{n}^2$ and begins
$$1, 3, 20, 175, 1764, 19404, 226512, 2760615, 34763300, \ldots$$
\begin{proposition}
We have
\begin{equation}
N(2n,n)=\frac{1}{2\pi^2}\int_0^4 \int_{\sqrt{y}-2}^{\sqrt{y}+2}\frac{x^{2n}}{\sqrt{4-(x-\sqrt{y})^2}}
\frac{\sqrt{y(4-y)}}{y}\,dx\,dy\end{equation}
\end{proposition}

  We have thus established that the Narayana triangle is intimately connected to the sequence of Catalan numbers. We now generalize 
  this construction to general integer sequences $a_n$ with $a_0=1$. Thus if we are given an integer sequence $(a_n)_{n\ge 0}$  with 
  $a_0=1$, we define the number triangle with general term

 $$T_a(n,k)=\sum_{j=0}^n \binom{n}{j}\binom{n-j}{2(k-j)}a_{k-j}$$ to be the \emph{generalized Narayana triangle associated to} $a_n$.

\begin{example} $a_n=0^n$, with term $1,0,0,0,\ldots$. We have seen in example \ref{Pas} that the generalized Narayana triangle 
associated to $0^n$ is Pascal's triangle.
\end{example}
\begin{example} $a_n=2^n$. We have already met this example, where $$T_{2^n}(n,k)=[x^{2k}](1+\sqrt{2}x+x^2)^n.$$
\end{example}
\begin{example} We have $$T_{\binom{2n}{n}}(n,k)=\binom{n}{k}^2.$$
\begin{proof}
We have
\begin{eqnarray*} T_{\binom{2n}{n}}(n,k)&=&\sum_{j=0}^n \binom{n}{j}\binom{n-j}{2(k-j)}\binom{2(k-j)}{k-j}\\
&=& \quad \ldots \\
&=& \sum_{j=0}^n \binom{n}{k}\binom{k}{j}\binom{n-k}{k-j}\\
&=& \binom{n}{k}\sum_{j=0}^n \binom{k}{j}\binom{n-k}{k-j}\\
&=& \binom{n}{k}^2. \qedhere \end{eqnarray*}
\end{proof}
We can also deduce the following about the central terms $\binom{2n}{n}^2$ of this triangle:
\begin{eqnarray*}\binom{2n}{n}^2&=&\sum_{j=0}^n\binom{2n}{j}\binom{n}{j}\binom{2n-j}{n}\\
&=&\frac{1}{\pi^2}\int_0^4 \int_{\sqrt{y}-2}^{\sqrt{y}+2}\frac{x^{2n}}{\sqrt{4-(x-\sqrt{y})^2}} \frac{1}{\sqrt{y(4-y)}}\,dx\,dy.
\end{eqnarray*}
\end{example}
\begin{example} The generalized Catalan triangle associated to the Fibonacci numbers $F_{n+1}$ has general term
 $$\sum_{j=0}^n \binom{n}{j}\binom{n-j}{2(k-j)}F_{k-j+1}$$ and begins

 \begin{displaymath}\left(\begin{array}{cccccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & \ldots \\1 & 1 & 0 & 0 & 0 &
0 & 0 &\ldots \\ 1 & 3 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 6 & 6 & 1
&
0 & 0 & 0 &\ldots \\ 1 & 10 & 20 & 10 & 1 & 0 & 0 &\ldots \\1 & 15 &
50 & 50 & 15
& 1 & 0 &\ldots\\1 & 21 & 105 & 173 & 105 & 21 & 1 &\ldots \\  \vdots & \vdots & \vdots & \vdots & \vdots &
\vdots & \ddots\end{array}\right).\end{displaymath} Note the $N$ has row $[1,21,105,175,105,21,1]$ where this triangle has
$[1,21,105,173,105,21,1]$. This triangle is \seqnum{A162745}.
\end{example}
\section{Continued fractions}
It is known that $N$ has a generating function given by the continued fraction \cite{Barry_CF, Branden_Catalan, Zeilberger}
$$g_N(x,y)=\cfrac{1}{1-xy-
\cfrac{x}{1-
\cfrac{xy}{1-
\cfrac{x}{1-\cdots}}}}.$$ This can also be expressed as
\begin{equation}\label{Nar_gf} g_N(x,y)=\frac{1-x(1+y)-\sqrt{1-2x(1+y)+x^2(1-y)^2}}{2x^2y}.\end{equation} For our purposes, we find it 
easier to work with another continued fraction. Thus we have
\begin{proposition} The bivariate generating function of the Narayana triangle may be expressed in continued fraction form as:
\begin{equation}\label{Nar_CF}g_N(x,y)=
\cfrac{1}{1-x-xy-
\cfrac{x^2y}{1-x-xy-
\cfrac{x^2y}{1-x-xy-\ldots}}}.\end{equation}
\end{proposition}
\begin{proof} We may solve the equation
$$u(x,y)=\frac{1}{1-x-xy-x^2y u(x,y)}$$ to obtain the corresponding closed form. We find
$$u(x,y)=g_N(x,y).$$
\end{proof}
\noindent We recall that the Catalan numbers have generating function
$$g_C(x)=\frac{1-\sqrt{1-4x}}{2x}$$ which is also given by
$$g_C(x)=\cfrac{1}{1-\cfrac{x}{1-\cfrac{x}{1-\ldots}}}.$$ Thus
$$g_C(x)=\cfrac{1}{1-\cfrac{\alpha_1x}{1-\cfrac{\alpha_2 x}{1-\ldots}}},$$ where
in this case, $\alpha_n=1$ for $n \ge 1$. These coefficients $\alpha_n=1$ are reproduced in
equation (\ref{Nar_CF}) as the coefficients of $x^2y$.

This leads us to the following generalization: if $a_n$ is a sequence with generating function
expressible as
$$g_a(x)=\cfrac{1}{1-\cfrac{\alpha_1x}{1-\cfrac{\alpha_2 x}{1-\ldots}}},$$ then we define as the
\emph{generalized Narayana triangle associated to} $g_a$ the number triangle with bivariate generating function given by
\begin{equation}\label{Gen_CF}
\cfrac{1}{1-x-xy-
\cfrac{\alpha_1 x^2y}{1-x-xy-
\cfrac{\alpha_2 x^2y}{1-x-xy-\ldots}}}.\end{equation} We call sequences with generating functions of the above form $g_a(x)$
``Catalan-like''.

\noindent For such ``Catalan-like'' sequences, there is a binomial transform interpretation to this bivariate continued fraction. We 
can write it in the form
$$\cfrac{1}{1-x(1+y)-
\cfrac{\alpha_1 x^2(\sqrt{y})^2}{1-x(1+y)-
\cfrac{\alpha_2 x^2(\sqrt{y})^2}{1-x(1+y)-\ldots}}}$$ which can then \cite{Barry_CF} be regarded
as the $(1+y)$-th binomial transform of the aeration of the Catalan-like sequence with g.f.
$$\cfrac{1}{1-\cfrac{\alpha_1 \sqrt{y}x}{1-\cfrac{\alpha_2\sqrt{y} x}{1-\ldots}}}.$$
We can for instance easily characterize the row sums of these generalized Narayana triangles.
\begin{proposition} Let $N_a$ be the generalized Narayana triangle of the Catalan-like sequence $a_n$. Then the row sums of
$N_a$ are the second binomial transform of the aeration of $a_n$.
\end{proposition}
\begin{proof} The row sums have continued fraction generating function given by (set $y=1$):
$$\cfrac{1}{1-2x-
\cfrac{\alpha_1 x^2}{1-2x-
\cfrac{\alpha_2 x^2}{1-2x-\ldots}}}$$ The result follows immediately from \cite{Barry_CF}.
\end{proof}
Obviously, the row sums of the second inverse binomial transform of a generalized Narayana triangle $N_a$, that is, of the triangle
 $\mathbf{B}^{-2}N_a$, are equal to the aeration of $a_n$. We note that this triangle, up to signs, is also Pascal-like. The triangle
 $\mathbf{B}^{-1}N_a$ is of a specific shape, typified by $\mathbf{B}^{-1}N=\mathbf{B}^{-1}N_C$:
 \begin{displaymath}\left(\begin{array}{cccccccc}
\mathbf{1} & 0 & 0 & 0 & 0 & 0 & 0 & \ldots \\0 & 1 & 0 & 0 & 0 &
0 & 0 &\ldots \\ 0 & \mathbf{1} & 1 & 0 & 0 & 0 & 0 & \ldots \\ 0 & 0 & 3 & 1
&
0 & 0 & 0 &\ldots \\ 0 & 0 & \mathbf{2} & 6 & 1 & 0 & 0 &\ldots \\0 & 0 &
0 & 10 & 10
& 1 & 0 &\ldots\\0 & 0 & 0 & \mathbf{5} & 30 & 5 & 1 &\ldots \\  \vdots & \vdots & \vdots & \vdots & \vdots &
\vdots & \ddots\end{array}\right),\end{displaymath} where we see the aeration of the defining sequence appearing in the
$(2n,n)$-positions. It is clear that the row sums of this triangle will be the binomial transform of the aeration of the defining 
sequence (in this case, we get the Motzkin numbers \seqnum{A001006}). The triangle $\mathbf{B}^{-1}N_a$ has generating function given 
by the continued fraction
$$\cfrac{1}{1-xy-
\cfrac{\alpha_1 x^2y}{1-xy-
\cfrac{\alpha_2 x^2y}{1-xy-\ldots}}}.$$
\begin{example} The central binomial coefficients $\binom{2n}{n}$ have g.f. given by
$$\cfrac{1}{1-\cfrac{2x}{1-\cfrac{x}{1-\cfrac{x}{1-\ldots}}}}.$$ The corresponding generalized Narayana triangle thus has generating 
function
$$g(x,y)=\cfrac{1}{1-x-xy-
\cfrac{2 x^2y}{1-x-xy-
\cfrac{ x^2y}{1-x-xy-
\cfrac{ x^2y}{1-x-xy-\ldots}}}}.$$ This is in fact the triangle $\binom{n}{k}^2$.

We have the following:
\begin{proposition} \label{preLeg}
$$g(x,y)=\frac{1}{\sqrt{1-2x(1+y)+x^2(1-y)^2}}.$$
\end{proposition}
\begin{proof}
We obviously have
$$g(x,y)=\frac{1}{1-x-xy-2x^2y g_N(x,y)}.$$
Simplifying, we get the result.
\end{proof}
\begin{corollary}
$$g(x,y)=\sum_{n=0}^{\infty} P_n\left(\frac{1+y}{1-y}\right)x^n(1-y)^n,$$ where
$P_n$ is the $n$-th Legendre polynomial.
\end{corollary}
\begin{proof}
This follows since we have
$$\frac{1}{1-2ut+t^2}=\sum_{n=0}^{\infty} P_n(u)t^n.$$
\end{proof}
\end{example}
\begin{example}
The (large) Schr\"oder numbers \seqnum{A006318} have generating function given by
$$\cfrac{1}{1-\cfrac{2x}{1-\cfrac{x}{1-\cfrac{2x}{1-\cfrac{x}{1-\ldots}}}}},$$ with coefficient
pattern $(2,1,2,1,2,1,2,1,\ldots)$. The corresponding generalized Narayana triangle thus has generating function
$$g_S(x,y)=\cfrac{1}{1-x-xy-
\cfrac{2 x^2y}{1-x-xy-
\cfrac{ x^2y}{1-x-xy-
\cfrac{ 2x^2y}{1-x-xy-\ldots}}}}.$$ This triangle begins
 \begin{displaymath}\left(\begin{array}{cccccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & \ldots \\1 & 1 & 0 & 0 & 0 &
0 & 0 &\ldots \\ 1 & 4 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 9 & 9 & 1
&
0 & 0 & 0 &\ldots \\ 1 & 16 & 36 & 16 & 1 & 0 & 0 &\ldots \\1 & 25 &
100 & 100 & 25
& 1 & 0 &\ldots\\1 & 36 & 225 & 402 & 225 & 36 & 1 &\ldots \\  \vdots & \vdots & \vdots & \vdots & \vdots &
\vdots & \ddots\end{array}\right)\end{displaymath}
We can obtain $g_S(x,y)$ in closed form by solving the equation
$$g_S(x,y)=\cfrac{1}{1-x-xy-
\cfrac{2 x^2y}{1-x-xy-x^2y g_S(x,y)}}.$$
\end{example}
\begin{example} The double factorials $(2n-1)!!=\frac{(2n)!}{n!2^n}$, \seqnum{A001147}, have g.f. given by
$$\cfrac{1}{1-\cfrac{x}{1-\cfrac{2x}{1-\cfrac{3x}{1-\ldots}}}}.$$ The triangle associated to this generating function thus has 
bivariate generating function
$$\cfrac{1}{1-x-xy-
\cfrac{2 x^2y}{1-x-xy-
\cfrac{ 3x^2y}{1-x-xy-
\cfrac{ 4x^2y}{1-x-xy-\ldots}}}}.$$ This triangle begins
 \begin{displaymath}\left(\begin{array}{cccccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & \ldots \\1 & 1 & 0 & 0 & 0 &
0 & 0 &\ldots \\ 1 & 3 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 1 & 6 & 6 & 1
&
0 & 0 & 0 &\ldots \\ 1 & 10 & 21 & 10 & 1 & 0 & 0 &\ldots \\1 & 15 &
55 & 55 & 15
& 1 & 0 &\ldots\\1 & 21 & 120 & 215 & 120 & 21 & 1 &\ldots \\  \vdots & \vdots & \vdots & \vdots & \vdots &
\vdots & \ddots\end{array}\right).\end{displaymath}
The row sums of this triangle are \seqnum{A005425} with e.g.f. $e^{2x+x^2/2}$. The triangle has e.g.f. $e^{x+xy+yx^2/2}$. This is 
\seqnum{A100862}. It is the exponential Riordan array
$$\left[e^x, x(1+\frac{x}{2})\right],$$ with general term \cite{Barry_Exp}
$$\frac{n!}{k!}\sum_{j=0}^k \binom{n}{j}\frac{1}{(n-k-j)!}\left(\frac{1}{2}\right)^j.$$
\end{example}
\begin{example} The Fibonacci numbers $F(n+1)$ have generating function
$$\frac{1}{1-x-x^2}=\cfrac{1}{1-\cfrac{x}{1-\cfrac{x}{1+x}}}.$$ We thus associate to it the triangle with bivariate generating function 
$$\cfrac{1}{1-x-xy-
\cfrac{x^2y}{1-x-xy-
\cfrac{ x^2y}{1-x-xy+
\cfrac{ x^2y}{1-x-xy}}}}.$$ This can be expressed as
$$\frac{(1-x-xy)^3}{1-4x(1+y)+x^2(6+11y+6y^2)-2x^3(2+5y+5y^2+2y^3)+x^4(1+3y+3y^2+3y^3+y^4)}.$$
\end{example}
\noindent It is natural to ask if these two notions of generalized Narayana triangle coincide. That the answer is in the affirmative 
follows from the following theorem.
\begin{theorem} Let $a_n$ be a sequence whose generating function can be expressed as a continued fraction
$$g_a(x)=\cfrac{1}{1-\cfrac{\alpha_1 x}{1-\cfrac{\alpha_2 x}{1-\ldots}}},$$ which may be finite or infinite. Then the generalized 
Narayana triangle associated to $g_a$ with bivariate generating function
$$\cfrac{1}{1-x-xy-
\cfrac{\alpha_1 x^2y}{1-x-xy-
\cfrac{\alpha_2 x^2y}{1-x-xy-\ldots}}}$$
coincides with the generalized Narayana triangle associated to $a_n$, and thus has general term
$$\sum_{j=0}^n \binom{n}{j}\binom{n-j}{2(k-j)}a_{k-j}.$$
\end{theorem}
\begin{proof}
We have
$$g(x,y)=\cfrac{1}{1-x(1+y)-
\cfrac{\alpha_1 x^2y}{1-x(1+y)-
\cfrac{\alpha_2 x^2y}{1-x(1+y)-\ldots}}}$$ which is thus seen to be the g.f. of the
$(1+y)$-th binomial transform of $\sqrt{y}^n$ times the aeration of the sequence with g.f.
$$\cfrac{1}{1-
\cfrac{\alpha_1 x}{1-
\cfrac{\alpha_2 x}{1-\ldots}}}.$$ Thus the triangle in question is given by
$$[y^k]\sum_{j=0}^n \binom{n}{j}(1+y)^{n-j}y^{j/2}a_{j/2}\frac{1+(-1)^j}{2}.$$ 
\end{proof}
\begin{corollary}
The Narayana polynomials $N(x)=\sum_{k=0}^n N(n,k)x^k$ satisfy
$$N(x)=\sum_{k=0}^n \binom{n}{k}(1+x)^{n-k}x^{k/2}C_{k/2}\frac{1+(-1)^k}{2}.$$
\end{corollary}
\section{Diagonal sums of triangles}
The diagonal sums of the Narayana triangle, that is, the sequence with general term
$$r_n=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} N(n-k,k),$$ which begins
$$1, 1, 2, 4, 8, 17, 37, 82, 185, 423, 978, 2283, \ldots$$ (\seqnum{A004148}) have important applications in the area of secondary RNA 
structures \cite{Nkwanta}.
By our results above (see Equations (\ref{Nar_gf}) and (\ref{Nar_CF})), we see that this sequence has generating function
$$g_N(x,x)=\frac{1-x-x^2-\sqrt{1-2x-x^2-2x^3+x^4}}{2x^3}$$ which can be expressed in continued fraction form as
$$g_N(x,x)=\cfrac{1}{1-x-x^2-\cfrac{x^3}{1-x-x^2-\cfrac{x^3}{1-x-x^2-\ldots}}}.$$
The Hankel transform $(h_n)_{n \ge 0}$ of this sequence is the periodic sequence \seqnum{A046980} which begins
$$1, 1, 0, -1, -1, -1, 0, 1, 1, 1, 0, -1, -1, -1, 0, 1, 1, 1, 0, -1, -1,\ldots,$$ with
generating function $\frac{1+x-x^3}{1+x^4}$. This sequence, for those terms for which it makes sense, satisfies the
Somos-$4$ equation
$$h_n =\frac{h_{n-1}h_{n-3}-h_{n-2}^2}{h_{n-4}}.$$
\begin{example} We consider the generalized Narayana triangle associated to the central binomial coefficients $\binom{2n}{n}$. This 
triangle has general term $\binom{n}{k}^2$, and thus the diagonal sums $d_n$ are given by
$$d_n=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}\binom{n-k}{k}^2.$$ The generating function of this sequence \seqnum{A051286}, which 
begins
$$1, 1, 2, 5, 11, 26, 63, 153, 376, 931, 2317,\ldots,$$ is given by
$$\frac{1}{\sqrt{1-2x(1+x)+x^2(1-x)^2}}=\sum_{n=0}^{\infty}P_n\left(\frac{1+x}{1-x}\right)x^n(1-x)^n,$$ or in continued fraction form, 
as
$$\cfrac{1}{1-x-x^2-\cfrac{2x^3}{1-x-x^2-\cfrac{x^3}{1-x-x^2-\ldots}}}.$$ Its Hankel transform $h_n$ \seqnum{A174882} begins
$$1, 1, -2, -8, -16, -16, 32, 128, 256, 256, -512, \ldots$$ with g.f. $\frac{1+x-2x^2-8x^3}{1+16x^4}$. This sequence is a
$\left(\frac{3}{2},-1\right)$ Somos-$4$ sequence, in the sense that we have
$$h_n =\frac{\frac{3}{2}h_{n-1}h_{n-3}-h_{n-2}^2}{h_{n-4}}, \quad n\ge 4.$$
We note in passing that the sequence given by the logarithms (to base 2) of $|h_n|$ is the sequence (see \seqnum{A098181})
that begins
$$0,0,1, 3, 4, 4, 5, 7, 8, 8, 9, 11, 12,\ldots$$ with generating function $\frac{x^2(1+x)}{(1-x)^2(1+x^2)}$ and general term
$$\frac{1}{2}\left\{2n-1+\cos\left(\frac{\pi n}{2}\right)-\sin\left(\frac{\pi 
n}{2}\right)\right\}=\frac{1}{2}\left\{2n-1+(-1)^{\binom{n+1}{2}}\right\}.$$
Note that using the language of Riordan arrays, and the result of Proposition \ref{preLeg}, we have the following expressions for the 
generating function of the diagonal sums:
\begin{eqnarray*}\frac{1}{\sqrt{1-2x(1+x)+x^2(1-x)^2}}&=&\left(\frac{1}{1+x^2},\frac{x}{1+x^2}\right)\cdot 
\frac{1}{\sqrt{1-2x-3x^2}}\\
&=&\left(\frac{1}{1-x+x^2},\frac{x}{1-x+x^2}\right)\cdot\frac{1}{\sqrt{1-4x^2}} .\end{eqnarray*} In fact we have the following result.
\begin{proposition}
$$d_n=\left(\frac{1}{1+rx+x^2},\frac{x}{1+rx+x^2}\right)\cdot [u^n](1+(r+1)u+u^2)^n,$$ for any
integer $r$.
\end{proposition}
\end{example}
\begin{example} We consider the generalized Narayana triangle associated to $2^n C_n$,which has generating function
$$\cfrac{1}{1-x-xy-
\cfrac{2 x^2y}{1-x-xy-
\cfrac{2 x^2y}{1-x-xy-\ldots}}}$$ and general term
$$\sum_{j=0}^n \binom{n}{j}\binom{n-j}{2(k-j)}2^{k-j}C_{k-j}.$$
Then the diagonal sums of this triangle (\seqnum{A025245}) begin
$$1, 1, 2, 5, 11, 26, 65, 163, 416, 1081, 2837, \ldots$$ and have generating function
$$\cfrac{1}{1-x-x^2-
\cfrac{2 x^3}{1-x-x^2-
\cfrac{2 x^3}{1-x-x^2-\ldots}}}$$ or
$$\frac{1-x-x^2-\sqrt{1-2x-x^2-6x^3+x^4}}{4x^3}.$$ The Hankel transform $h_n$ of the diagonal sums begins
$$1, 1, -2, -12, -64, -64, 6656, 143360, 3342336, -105906176, -15837691904, \ldots.$$ It is a $(4,-4)$ Somos-$4$ sequence in the sense 
that
$$h_n =\frac{4h_{n-1}h_{n-3}-4h_{n-2}^2}{h_{n-4}}, \quad n\ge 4.$$
\end{example}
We can in fact conjecture the following:
\begin{conjecture} The Hankel transform of the sequence with generating function
$$g(x)=\frac{1}{1-\alpha x - \beta x^2 -\gamma x^3 g(x)}$$ is a
$(\gamma^2,-\beta \gamma^2)$ Somos-$4$ sequence.
\end{conjecture}
\begin{example} A Gaussian integer Somos-$4$ sequence. 
We note the following consequence of this conjecture: the (complex) sequence with generating function
$$\cfrac{1}{1-x-x^2-
\cfrac{i x^3}{1-x-x^2-
\cfrac{i x^3}{1-x-x^2-
\cfrac{i x^3}{1-x-x^2-\ldots}}}}, $$ where $i=\sqrt{-1}$, has a Hankel transform that is
a $(-1,1)$ Somos-$4$ sequence:
$$h_n =\frac{-h_{n-1}h_{n-3}+h_{n-2}^2}{h_{n-4}}, \quad n\ge 4.$$
This corresponds to the diagonal sums of the triangle with general term
$$\sum_{j=0}^n \binom{n}{j}\binom{n-j}{2(k-j)}i^{k-j}C_{k-j}.$$ This sequence begins
$$1, 1, 2, 3 + i, 5 + 3i, 8 + 9i, 11 + 22i, 11 + 51i, -6 + 111i, -75 + 228i, -291 + 439i,\ldots$$ with
Hankel transform beginning
$$1, 1, 1 + i, -i, 3i, 2 - 3i, -2 + 4i, 6 + 7i, -4 + 15i, -24 + 29i, -39 - 15i\ldots.$$
\end{example}
\begin{example} 
The sums of the diagonals of the Narayana triangle associated to the Schr\"oder numbers begin
$$1, 1, 2, 5, 11, 26, 63, 153, 376,933, 2331,\ldots$$ and have generating function
$$\frac{1-2x-x^2+x^3+x^4-\sqrt{1-4x+2x^2+2x^3+7x^4-2x^5-9x^6-2x^7+x^8}}{2x^3(1-x-x^2)}$$ corresponding to the
continued fraction
$$\cfrac{1}{1-x-x^2-\cfrac{2x^3}{1-x-x^2-\cfrac{x^3}{1-x-x^2-\cfrac{2x^3}{1-x-x^2-\ldots}}}}.$$
\noindent We note that the Hankel transform of this sequence is not a Somos-$4$ sequence.
\end{example}

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\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B83; Secondary 05A10, 05A15, 11B37, 11Y55, 15A15, 15A30, 30B70.

\noindent \emph{Keywords:} Narayana triangle, continued fraction,
Riordan array, Hankel determinant, Hankel transform, Somos-$4$.

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\noindent (Concerned with sequences
\seqnum{A000007},
\seqnum{A000045},
\seqnum{A000108},
\seqnum{A000891},
\seqnum{A001006},
\seqnum{A001147},
\seqnum{A001263},
\seqnum{A004148},
\seqnum{A005425},
\seqnum{A006318},
\seqnum{A007318},
\seqnum{A025245},
\seqnum{A025250},
\seqnum{A025253},
\seqnum{A046980},
\seqnum{A050512},
\seqnum{A051286},
\seqnum{A090181},
\seqnum{A098181},
\seqnum{A100862},
\seqnum{A124216},
\seqnum{A131198},
\seqnum{A162745}, and
\seqnum{A174882}.)

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\vspace*{+.1in}
\noindent
Received September 23 2010;
revised version received  March 22 2011.
Published in {\it Journal of Integer Sequences}, March 26 2011.

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