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\begin{center}
\vskip 1cm{\LARGE\bf A Note on Narayana Triangles and Related\\
\vskip .04in
Polynomials, Riordan Arrays, and MIMO \\
\vskip .12in
Capacity Calculations
} 
\vskip 1cm \large
Paul Barry\\
School of Science\\
Waterford Institute of Technology\\
Ireland\\
\href{mailto:pbarry@wit.ie}{\tt pbarry@wit.ie} \\
\ \\
Aoife Hennessy\\
Department of Computing, Mathematics and Physics\\
Waterford Institute of Technology\\
Ireland\\
\href{mailto:aoife.hennessy@gmail.com}{\tt aoife.hennessy@gmail.com} \\
\end{center}
\vskip .2 in

\begin{abstract} 
We study the Narayana triangles and related families
of polynomials. We link this study to Riordan arrays and Hankel
transforms arising from a special case of capacity calculation related
to MIMO communication systems. A link is established between a channel
capacity calculation and a series reversion.
\end{abstract}

\section{Introduction}
The Narayana numbers, which are closely related to the ubiquitous Catalan numbers, have an important and growing
literature.
Their applications are varied. In this note, we look at the mathematics of one application in the area of
MIMO (multiple input, multiple output) wireless communication. For our purposes, it is useful to distinguish between three
different ``Narayana triangles'' and their associated
``Narayana polynomials''. These triangles are documented separately in 
Sloane's {\it Encyclopedia} 
\cite{SL1} along with other variants. We will find it useful in this note to
use the language of Riordan arrays \cite{SGWW} for later sections. In the next section we provide a quick introduction to
the Riordan group. We also use the notion of ``Deleham array'' \cite{CF}, which is explained
in Appendix B. Our approach to Deleham arrays is based on continued fractions \cite{Wall}. We shall be interested in the
Hankel transform \cite{Layman, Kratt, RPB} of a number of integer
sequences that we shall encounter. We recall that if
$a_n$ is a given sequence, then the sequence with general term given by the determinant
$|a_{i+j}|_{0\le i,j \le n}$ is called the Hankel transform of $a_n$. We shall mention well-known orthogonal polynomials
in this note. General references are \cite{Chihara, wgautschi}. Links between orthogonal polynomials and Riordan arrays have been
studied in \cite{PB_Moment, PB_Meixner}. Techniques to calculate
Hankel transforms using associated orthogonal polynomials will follow methods to be found in \cite{CRI, RPB}.
\newline\newline
For an integer sequence $a_n$, that is, an element of $\mathbb{Z}^\mathbb{N}$, the power series
$f(x)=\sum_{k=0}^{\infty}a_k x^k$ is called the \emph{ordinary generating function} or g.f. of the sequence.
$a_n$ is thus the coefficient of $x^n$ in this series. We denote this by
$a_n=[x^n]f(x)$. For instance, $F_n=[x^n]\frac{x}{1-x-x^2}$ is the $n$-th Fibonacci number \seqnum{A000045}, while
$C_n=[x^n]\frac{1-\sqrt{1-4x}}{2x}$ is the $n$-th Catalan number \seqnum{A000108}. 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
$\bar{f}= \text{Rev}f$.
\section{Riordan arrays}
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.

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

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

Many interesting examples of 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}.
\section{The Narayana Triangles and their generating functions}\label{Reversion}
In this section, we define four separate though related ``Narayana triangles'', and we describe their (bi-variate)
generating functions.

The number triangle $\mathbf{N}_0$ with general term
\begin{equation}N_0(n,k)=\frac{1}{n+0^n}\binom{n}{k}\binom{n}{k+1}\end{equation} has \cite{Zeilberger} generating function
$\phi_0(x,y)$
which satisfies the equation
$$xy \phi_0^2+(x+xy-1)\phi_0+x=0.$$ Solving for $\phi_0(x,y)$ yields
\begin{equation}\phi_0(x,y)=\frac{1-x(1+y)-\sqrt{1-2x(1+y)+x^2(1-y)^2}}{2xy}.\end{equation}
This triangle begins
\begin{displaymath}\mathbf{N}_0=\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 \\ 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}
The triangle $\mathbf{N}_1$ with general term
\begin{equation}N_1(n,k)=0^{n+k}+\frac{1}{n+0^n}\binom{n}{k}\binom{n}{k+1}=\frac{1}{k+1}\binom{n-1}{k}\binom{n}{k}\end{equation} which
begins
\begin{displaymath}\mathbf{N}_1=\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}
clearly has generating function
\begin{equation}\phi_1(x,y)=1+\phi_0(x,y)=\frac{1-x(1-y)-\sqrt{1-2x(1+y)+x^2(1-y)^2}}{2xy}.\end{equation}
The triangle $\mathbf{N}_2$, which is the reversal of $\mathbf{N}_1$,  has general term
\begin{eqnarray}N_2(n,k)&amp;=&amp;[k \le n]N_1(n,n-k)=0^{n+k}+\frac{1}{n+0^{nk}}\binom{n}{k}\binom{n}{k-1}\\
&amp;=&amp;\frac{1}{n-k+1}\binom{n-1}{n-k}\binom{n}{k},\end{eqnarray}
 and begins
\begin{displaymath}\mathbf{N}_2=\left(\begin{array}{ccccccc} 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; 1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots
\\
0 &amp; 1 &amp; 3 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 1 &amp; 6 &amp; 6 &amp; 1 &amp; 0 &amp;
\ldots \\0 &amp; 1 &amp;
10 &amp; 20 &amp; 10 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\vdots &amp; \vdots &amp; \ddots\end{array}\right).\end{displaymath}
This triangle has generating function
\begin{equation} \phi_2(x,y)=1+y \phi_0(x,y)=\frac{1+x(1-y)-\sqrt{1-2x(1+y)+x^2(1-y)^2}}{2x}.\end{equation}
Finally the ``Pascal-like'' variant $\mathbf{N}_3$ with general term
\begin{equation}N_3(n,k)=N_0(n+1,k)=\frac{1}{n+1}\binom{n+1}{k}\binom{n+1}{k+1}\end{equation} which begins
\begin{displaymath}\mathbf{N}_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; 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} has generating function
\begin{equation}\phi_3(x,y)=\frac{\phi_0(x,y)}{x}=\frac{1-x(1+y)-\sqrt{1-2x(1+y)+x^2(1-y)^2}}{2x^2y}.\end{equation}
\begin{center}\begin{tabular}{|c|c|c|} \hline
$ Triangle $ &amp; A-number &amp; Generating function  \\ \hline
$\mathbf{N}_1$   &amp; \seqnum{A131198} &amp; $\phi_1(x,y)=\frac{1-x(1-y)-\sqrt{1-2x(1+y)+x^2(1-y)^2}}{2xy}$  \\ \hline
$\mathbf{N}_2$   &amp; \seqnum{A090181} &amp; $\phi_2(x,y)=\frac{1+x(1-y)-\sqrt{1-2x(1+y)+x^2(1-y)^2}}{2x}$ \\ \hline
$\mathbf{N}_3$   &amp; \seqnum{A001263} &amp; $\phi_3(x,y)=\frac{1-x(1+y)-\sqrt{1-2x(1+y)+x^2(1-y)^2}}{2x^2y}$ \\ \hline
\end{tabular}\end{center}
\section{Narayana Triangles and series reversion}
Using the generating functions obtained in the last section, we can relate the Narayana triangles to the process of
reverting sequences.
\begin{proposition} We have
\begin{equation}\phi_1(x,y)=\frac{1}{x}\text{Rev}_x \frac{x(1-xy)}{1-x(y-1)}.\end{equation}
\end{proposition}
\begin{proof}
We calculate the reversion of the expression
$$\frac{x(1-xy)}{1-x(y-1)},$$ considered as a function in $x$, with parameter $y$. This amounts to solving the equation
$$\frac{u(1-uy)}{1-u(y-1)}=x$$ for the unknown $u$.
We obtain $$u=x\phi_1(x,y).$$ Thus we have
$$\phi_1(x,y)=\frac{1}{x}\text{Rev}_x\frac{x(1-xy)}{1-x(y-1)},$$
as required.
\end{proof}

In like manner, we have
\begin{proposition}
\begin{equation}\phi_2(x,y)=\frac{1}{x}\text{Rev}_x\frac{x(1-x)}{1-x(1-y)}.\end{equation}
\end{proposition}
\begin{proof}
We calculate the $x$-reversion of the expression $\frac{x(1-x)}{1-x(1-y)}$. Thus we wish to solve for $u$, where
$$\frac{u(1-u)}{1-u(1-y)}=x.$$ We obtain
$$u=x \phi_2(x,y).$$
Thus
$$\phi_2(x,y)=\frac{1}{x}\text{Rev}_x\frac{x(1-x)}{1-x(1-y)}.$$
\end{proof}
In similar fashion, we can establish that
\begin{proposition}
\begin{equation}\phi_3(x,y)=\frac{1}{x}\text{Rev}_x\frac{x}{1+(1+y)x+yx^2}.\end{equation}
\end{proposition}
We note that the Narayana triangles are not Riordan arrays.
\section{The Narayana Triangles and continued fractions}\label{N_CF}
In this section, we develop continued fraction versions for each of the generating functions $\phi_1, \phi_2, \phi_3$.
In this case of $\phi_3$, we give two distinct (but equivalent) continued fraction expressions.
\begin{proposition}\cite[Section 3.5]{Branden_Catalan}  We have the following continued fraction expansion of $\phi_1(x,y)$:
\begin{equation}\phi_1(x,y)=\cfrac{1}{1-
\cfrac{x}{1-
\cfrac{xy}{1-
\cfrac{x}{1-
\cfrac{xy}{1-\cdots}}}}}.\end{equation}
\end{proposition}
\begin{proof}
It is easy to see that $\phi_1(x,y)$ obeys the equation \cite{Branden_Catalan}
\begin{equation}\label{phi1_eq} xy \phi_1^2-(xy-x+1)\phi_1+1=0.\end{equation} Thus
$$\phi_1(1-x-xy\phi_1)=1-xy\phi_1$$ and hence
\begin{eqnarray*}\phi_1 &amp;=&amp; \frac{1-xy\phi_1}{1-xy\phi_1-x}\\
&amp;=&amp;\frac{1}{1-\frac{x}{1-xy\phi_1}}.\end{eqnarray*}
We thus obtain the result that $\phi_1(x,y)$ can be expressed as the continued fraction
$$\phi_1(x,y)=\cfrac{1}{1-
\cfrac{x}{1-
\cfrac{xy}{1-
\cfrac{x}{1-
\cfrac{xy}{1-\cdots}}}}}.$$
\end{proof}
\begin{corollary}
We have
$$\mathbf{N}_1=[1,0,1,0,1,0,1,\ldots] \quad \Delta \quad
[0,1,0,1,0,1,\ldots].$$
\end{corollary}
\begin{proposition} We have the following continued fraction expansion of $\phi_2(x,y)$:
\begin{equation}\phi_2(x,y)=\cfrac{1}{1-
\cfrac{xy}{1-
\cfrac{x}{1-
\cfrac{xy}{1-
\cfrac{x}{1-\cdots}}}}}.\end{equation}
\end{proposition}
\begin{proof} It is easy to establish that
\begin{equation} x\phi_2^2-(1+x-xy)\phi_2+1=0\end{equation}
from which we deduce
$$\phi_2(1-x\phi_2-xy)=1-x\phi_2$$ and hence
\begin{eqnarray*}
\phi_2&amp;=&amp;\frac{1-x\phi_2}{1-x\phi_2-xy}\\
&amp;=&amp;\frac{1}{1-\frac{xy}{1-x\phi_2}}.\end{eqnarray*}
The result follows from this.
\end{proof}
\begin{corollary}
We have
$$\mathbf{N}_2=[0,1,0,1,0,1,\ldots] \quad \Delta \quad
[1,0,1,0,1,0,1,\ldots].$$
\end{corollary}
\noindent In order to find an expression for $\phi_3$, we first note that
$$\phi_3=\frac{\phi_1-1}{x} \Rightarrow \phi_1=1+x\phi_3.$$
Substituting into Eq.~(\ref{phi1_eq}) and simplifying, we find that
\begin{equation}\phi_3(1-xy-x^2y\phi_3)=1+x\phi_3\end{equation} and hence
\begin{eqnarray*}\phi_3&amp;=&amp;\frac{1+x\phi_3}{1-xy(1+x\phi_3)}\\
&amp;=&amp;\frac{1}{-xy+\frac{1}{1+x\phi_3}}\\
&amp;=&amp;\frac{1}{-xy+\frac{1}{\phi_1}}.\end{eqnarray*}
But
$$\frac{1}{\phi_1}=1-\cfrac{x}{1-
\cfrac{xy}{1-
\cfrac{x}{1-\cdots}}}.$$
Hence we obtain:
\begin{proposition} We have the following continued fraction expansion of $\phi_3(x,y)$:
\begin{equation} \phi_3(x,y)=\cfrac{1}{1-xy-
\cfrac{x}{1-
\cfrac{xy}{1-
\cfrac{x}{1-\cdots}}}}.\end{equation}
\end{proposition}
\begin{corollary}
We have
$$\mathbf{N}_3=[0,1,0,1,0,1,\ldots] \quad \Delta^{(1)} \quad
[1,0,1,0,1,0,1,\ldots].$$
\end{corollary}
We end this section by expressing the g.f. of $\mathbf{N}_3$ in another way.
\begin{proposition} The generating function of $\mathbf{N}_3$ has the following continued fraction expression
$$
\phi_3(x,y)=\cfrac{1}{1-x-xy-
\cfrac{x^2y}{1-x-xy-
\cfrac{x^2y}{1-x-xy-
\cfrac{x^2y}{1-\cdots}}}}.$$
\end{proposition} 
\begin{proof} This can be seen by solving the equation
$$u=\frac{1}{1-x-xy-x^2y u}$$ and comparing the solution
$u(x,y)$ with $\phi_3(x,y)$.
\end{proof}

\section{Narayana polynomials and moment sequences}\label{Moments}
To each of the above triangles, there is a family of ``Narayana'' polynomials \cite{Sulanke_Nara1, SulankeNara3}, where
the triangles take on the role of
coefficient arrays. Thus we get the polynomials
\begin{eqnarray*}
\mathcal{N}_{1,n}(y)&amp;=&amp;\sum_{k=0}^n N_1(n,k)y^k\\
\mathcal{N}_{2,n}(y)&amp;=&amp;\sum_{k=0}^n N_2(n,k)y^k\\
\mathcal{N}_{3,n}(y)&amp;=&amp;\sum_{k=0}^n N_3(n,k)y^k. \end{eqnarray*}
Note that since $\mathbf{N}_2$ is the reversal of $\mathbf{N}_1$, we have
$$\mathcal{N}_{2,n}(y)=\sum_{k=0}^n N_2(n,k)y^k=\sum_{k=0}^n N_1(n,k)y^{n-k}.$$
\begin{example}
The first terms of the sequence $(\mathcal{N}_{1,n}(y))_{n\ge 0}$ are:
$$1,1,1+y,1+3y+y^2,1+6y+6y^2+y^3,1+10y+20y^2+10y^3+y^4,\ldots$$
\end{example}
\noindent Using the results of section \textbf{\ref{Reversion}}, we see that
\begin{eqnarray*}
\mathcal{N}_{1,n}(y)&amp;=&amp;[x^{n+1}]\text{Rev}_x\frac{x(1-xy)}{1-(y-1)x}\\
\mathcal{N}_{2,n}(y)&amp;=&amp;[x^{n+1}]\text{Rev}_x\frac{x(1-x)}{1-(1-y)x}\\
\mathcal{N}_{3,n}(y)&amp;=&amp;[x^{n+1}]\text{Rev}_x\frac{x}{1+(1+y)x+yx^2}.\end{eqnarray*}
Values of these polynomials are often of significant combinatorial interest.
Sample values for these polynomials are tabulated below.
\begin{center}\begin{tabular}{|c|c|c|c|} \hline
$y$ &amp; $\mathcal{N}_{1,0}(y),\mathcal{N}_{1,1}(y),\mathcal{N}_{1,2}(y),\ldots$ &amp; A-number &amp; Name \\ \hline
$1$   &amp; $1,1,2,5,14,42,\ldots$ &amp; \seqnum{A000108} &amp; Catalan numbers\\ \hline
$2$   &amp; $1,1,3,11,45,197\ldots$ &amp; \seqnum{A001003} &amp; little Schr\"oder numbers \\ \hline
$3$   &amp; $1,1,4,19,100,562, \ldots$ &amp; \seqnum{A007564} &amp;  \\ \hline
$4$   &amp; $1,1,5,29,185,1257, \ldots$ &amp; \seqnum{A059231} &amp; \\ \hline
\end{tabular}\end{center}

\begin{center}\begin{tabular}{|c|c|c|c|} \hline
$y$ &amp; $\mathcal{N}_{2,0}(y),\mathcal{N}_{2,1}(y),\mathcal{N}_{2,2}(y),\ldots$ &amp; A-number &amp; Name \\ \hline
$1$   &amp; $1,1,2,5,14,42,\ldots$ &amp; \seqnum{A000108} &amp; Catalan numbers \\ \hline
$2$   &amp; $1,2,6,22,90,394,\ldots$ &amp; \seqnum{A006318} &amp; Large Schr\"oder numbers \\  \hline
$3$   &amp; $1,3,12,57,300,1686, \ldots$ &amp; \seqnum{A047891} &amp;\\ \hline
$4$   &amp; $1,4,20,116,740,5028,\ldots$ &amp; \seqnum{A082298} &amp; \\ \hline
\end{tabular}\end{center}

\begin{center}\begin{tabular}{|c|c|c|c|} \hline
$y$ &amp; $\mathcal{N}_{3,0}(y),\mathcal{N}_{3,1}(y),\mathcal{N}_{3,2}(y),\ldots$ &amp; A-number &amp; Name  \\ \hline
$1$   &amp; $1,2,5,14,42,132,\ldots$ &amp; \seqnum{A000108(n+1)} &amp; shifted Catalan numbers \\ \hline
$2$   &amp; $1,3,11,45,197,903,\ldots$ &amp; \seqnum{A001003(n+1)} &amp; shifted little Schr\"oder numbers \\ \hline
$3$   &amp; $1,4,19,100,562,3304, \ldots$ &amp; \seqnum{A007564(n+1)} &amp; \\ \hline
$4$   &amp; $1,5,29,185,1257, 8925,\ldots$ &amp; \seqnum{A059231(n+1)} &amp; \\ \hline
\end{tabular}\end{center}

We can derive a moment representation for these polynomials using the generating functions above and the Stieltjes
transform. We obtain the following\,:
\begin{proposition} The families of polynomials $(\mathcal{N}_{1,n}(y))_{n\ge 0}$, $(\mathcal{N}_{2,n}(y))_{n\ge 0}$,
$(\mathcal{N}_{3,n}(y))_{n\ge 0}$, are each a family of moments corresponding to an associated family of orthogonal functions.
\end{proposition}
\begin{proof} Using the established generating functions $\phi_1(x,y)$, $\phi_2(x,y)$ and $\phi_3(x,y)$, and the Stieltjes-Perron
transform (see Appendix C), we can establish the following moment representations, for the densities shown.
\begin{eqnarray*}
\mathcal{N}_{1,n}(y)&amp;=&amp;\frac{y-1}{y}0^n+\frac{1}{2\pi}\int_{y-2\sqrt{y}+1}^{y+2\sqrt{y}+1}x^n
\frac{\sqrt{-x^2+2x(1+y)-(1-y)^2}}{2y} dx,\\ \\
\mathcal{N}_{2,n}(y)&amp;=&amp;\frac{1}{2\pi}\int_{y-2\sqrt{y}+1}^{y+2\sqrt{y}+1}x^n \frac{\sqrt{-x^2+2x(1+y)-(1-y)^2}}{x} dx,\\
\\
\mathcal{N}_{3,n}(y)&amp;=&amp;\frac{1}{2\pi}\int_{y-2\sqrt{y}+1}^{y+2\sqrt{y}+1}x^n \frac{\sqrt{-x^2+2x(1+y)-(1-y)^2}}{y}
dx.\end{eqnarray*}
The associated orthogonal polynomials are determined by the densities shown.
\end{proof}
\noindent Using the theory developed in \cite{PB_Moment, PB_Meixner}, we can exhibit these families of polynomials as the first columns
of three related Riordan arrays. More precisely, we have
\begin{proposition}
The elements of the three families of polynomials $(\mathcal{N}_{1,n}(y))_{n\ge 0}$, $(\mathcal{N}_{2,n}(y))_{n\ge 0}$,
$(\mathcal{N}_{3,n}(y))_{n\ge 0}$ are given by the first column of the inverse Riordan arrays given by
$\left(\frac{1}{1+x},\frac{x}{(1+x)(1+yx)}\right)$, $\left(\frac{1}{1+yx},\frac{x}{(1+x)(1+yx)}\right)$, and
$\left(\frac{1}{(1+x)(1+yx)},\frac{x}{(1+x)(1+yx)}\right)$, respectively. These Riordan arrays are the coefficient arrays of the
corresponding families of orthogonal polynomials. Thus

\begin{eqnarray*}
\mathcal{N}_{1,n}(y)&amp;&amp; \textrm{is given by the first column of }\quad
\left(\frac{1}{1+x},\frac{x}{(1+x)(1+yx)}\right)^{-1},\\
\mathcal{N}_{2,n}(y)&amp;&amp; \textrm{is given by the first column of }\quad
\left(\frac{1}{1+yx},\frac{x}{(1+x)(1+yx)}\right)^{-1}\\
\mathcal{N}_{3,n}(y)&amp;&amp; \textrm{is given by the first column of }\quad
\left(\frac{1}{(1+x)(1+yx)},\frac{x}{(1+x)(1+yx)}\right)^{-1}.\end{eqnarray*}
\end{proposition}
\begin{proof} We look at the case of $\mathcal{N}_{1,n}$, as the other cases are proved in similar manner.
Thus we let
$$\left(\frac{1}{1+x},\frac{x}{(1+x)(1+yx)}\right)=(g,f).$$ We wish then to show that
$$\phi_1(x,y)=\frac{1}{g(\text{Rev}_x f(x,y))}.$$
 For $f(x,y)=\frac{x}{(1+x)(1+yx)}$, we find that
$$\text{Rev}_x f(x,y)=\frac{1-x(1+y)-\sqrt{1-2x(1+y)+x^2(1-y)^2}}{2xy}.$$
Then since $g(x)=\frac{1}{1+x}$, we find that
$$\frac{1}{g(\text{Rev}_x f(x,y))}=1+\text{Rev}_x f(x,y)=1+\frac{1-x(1+y)-\sqrt{1-2x(1+y)+x^2(1-y)^2}}{2xy}=\phi_1(x,y)$$ as required.
Now
\begin{eqnarray*}\left(\frac{1}{1+x},\frac{x}{(1+x)(1+yx)}\right)&amp;=&amp;\left(\frac{1+yx}{(1+x)(1+yx)},\frac{x}{(1+x)(1+yx)}\right)\\
&amp;=&amp;
\left(\frac{1+yx}{1+(1+y)x+yx^2},\frac{x}{1+(1+y)x+yx^2}\right),\end{eqnarray*} and hence
\cite{PB_Meixner} $\left(\frac{1}{1+x},\frac{x}{(1+x)(1+yx)}\right)$ is the coefficient array of a family of orthogonal polynomials.
\end{proof}
\section{An investigation inspired by a MIMO application of the Narayana numbers}
The role of the Catalan numbers and more recently the Narayana polynomials in the elucidation of the behaviour of certain
families of random matrices, along with applications to
areas such as MIMO wireless communication, is an active field of research. See for instance \cite{Dumitriu, Dumitriu_Path, Groh,
Muller, Muller_2, Silverstein_Bai, Tulino}. Other areas
where Narayana polynomials and their generalizations find applications include that of associahedra \cite{Chapoton, Fomin_Reading,
Postnikov} and secondary RNA structures \cite{Doslic_RNA}.

\noindent The investigations in this section and those that follow are inspired by MIMO (multiple input, multiple output) data-communication applications in  \cite{Muller} and \cite{Tulino}. The reader is referred to
Appendix A for the link with MIMO capacity calculations.
We let $$G_{\beta}(z)=-\frac{1}{2}+\frac{\beta-1}{2z}+\sqrt{\frac{(1-\beta)^2}{4z^2}+\frac{1}{4}-\frac{1+\beta}{2z}}.$$
In terms of wireless transmission,
$$\beta=\frac{T}{R}$$ where we have $T$ transmit antennas and $R$ receive antennas (see Appendix A). In this
section $\beta$ can be treated as a parameter.
Then the function $$g_{\beta}(x)=-\frac{1}{x}G_{\beta}(\frac{1}{x})$$ satisfies
$$g_{\beta}(x)=\frac{1+(1-\beta)x-\sqrt{1-2x(1+\beta)+(1-\beta)^2 x^2}}{2x}$$ and generates the sequence
$$1, \beta, \beta(\beta+1), \beta(\beta^2+3\beta+1), \beta(\beta^3+6\beta^2+6\beta+1),\ldots$$
In other words, $g_{\beta}(x)$ is the generating function of the sequence
$$a_n^{(\beta)}=\sum_{k=0}^n N_2(n,k)\beta^k =\mathcal{N}_{2,n}(\beta).$$
Thus $$g_{\beta}(x)=\phi_2(x,\beta).$$
We have the following moment representation:
\begin{eqnarray*}a_n^{(\beta)}&amp;=&amp;\frac{1}{2\pi}\int_{1+\beta-2\sqrt{\beta}}^{1+\beta+2\sqrt{\beta}}x^n\frac{\sqrt{-x^2+2x(1+\beta)-(1-\beta)^2}}{x}dx\\
&amp;=&amp;\frac{1}{2\pi}\int_{(1-\sqrt{\beta})^2}^{(1+\sqrt{\beta})^2}x^n\frac{\sqrt{((1-\sqrt{\beta})^2-x)(x-(1+\sqrt{\beta})^2)}}{x}dx\\
&amp;=&amp;\frac{\sqrt{\beta}}{\pi}\int_{(1-\sqrt{\beta})^2}^{(1+\sqrt{\beta})^2}x^n\frac{\sqrt{1-\left(\frac{1+\beta-x}{2\sqrt{\beta}}\right)^2}}{x}dx\\
&amp;=&amp;\frac{\sqrt{\beta}}{\pi}\int_{(1-\sqrt{\beta})^2}^{(1+\sqrt{\beta})^2}x^n \frac{w_U
\left(\frac{1+\beta-x}{2\sqrt{\beta}}\right)}{x}dx
\end{eqnarray*}
where $w_U(x)=\sqrt{1-x^2}$ is the weight function for the Chebyshev polynomials of the second kind.
This is an example of the well-known Mar\v{c}enko-Pastur \cite{Marcenko} distribution.
\section{Riordan arrays, orthogonal polynomials and $\mathbf{N}_2$}
We now note that $xg_{\beta}(x)$ is in fact the series reversion of the function
$$\frac{x(1-x)}{1+(\beta-1)x}.$$ This simple form leads us to investigate the nature of the coefficient array of
the orthogonal polynomials $P_n^{(\beta)}(x)$ associated to the weight function
$$w(x)=\frac{1}{2\pi}\frac{\sqrt{-x^2+2x(1+\beta)-(1-\beta)^2}}{x}=\frac{1}{2\pi}\frac{\sqrt{4\beta-(x-1-\beta)^2}}{x}dx$$
for which the elements
$$a_n^{(\beta)}=\sum_{k=0}^n N_2(n,k)\beta^k$$ are the moments. Put otherwise, these are the family of
orthogonal polynomials associated to the Narayana polynomials $\mathcal{N}_{2,n}$.
These polynomials can be expressed in terms of the Hankel determinants associated to the sequence $a_n^{(\beta)}$.
We find that the coefficient array of the polynomials $P_n^{(\beta)}(x)$  is given by the Riordan array
$$\left(\frac{1}{1+\beta x}, \frac{x}{1+(1+\beta )x+\beta x^2}\right)$$ whose inverse is given by
$$\mathbf{L}=\left(g_{\beta}(x), \frac{g_{\beta}(x)-1}{\beta}\right)=\left(\phi_2(x,\beta),\frac{\phi_2(x,\beta)-1}{\beta}\right).$$
The Jacobi-Stieltjes array \cite{ProdMat_0, ProdMat, PPWW}
for $\mathbf{L}$ is found \cite{PB_Moment, PB_Meixner} to
be \begin{displaymath}\left(\begin{array}{ccccccc} \beta &amp; 1 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\\beta &amp; \beta+1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; \beta &amp; \beta+1
&amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; \beta &amp; \beta+1 &amp; 1 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp;
0 &amp; \beta &amp; \beta+1 &amp; 1 &amp; \ldots
\\0 &amp; 0 &amp; 0 &amp; 0 &amp; \beta &amp; \beta+1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right)\end{displaymath}
indicating that the Hankel transform of the sequence $a_n^{(\beta)}$ is $\beta^{\binom{n+1}{2}}$, and that
$$g_{\beta}(x)=\cfrac{1}{1-\beta x -
\cfrac{\beta x^2}{1-(\beta+1)x-
\cfrac{\beta x^2} {1-(\beta +1)x-
\cfrac{\beta x^2}{1-\cdots}}}}.$$
We note that the coefficient array $\mathbf{L}^{-1}$ can be factorized as follows:
\begin{equation}\label{Fact}\mathbf{L}^{-1}=\left(\frac{1}{1+\beta x}, \frac{x}{1+(1+\beta )x+\beta
x^2}\right)=\left(1,\frac{x}{1+x}\right)\cdot \left(\frac{1-x}{1+(\beta-1)x},\frac{x(1-x)}{1+(\beta-1)x}\right).\end{equation}

\noindent Hence
\begin{eqnarray*}\mathbf{L}&amp;=&amp;\left(\frac{1-x}{1+(\beta-1)x},\frac{x(1-x)}{1+(\beta-1)x}\right)^{-1}\cdot \left(1,\frac{x}{1+x}\right)^{-1}\\
&amp;=&amp;(g_{\beta}(x),xg_{\beta}(x))\cdot
\left(1,\frac{x}{1-x}\right).\end{eqnarray*}

\noindent The general term of the matrix
$$\left(\frac{1-x}{1+(\beta-1)x},\frac{x(1-x)}{1+(\beta-1)x}\right)^{-1}=(g_{\beta}(x),xg_{\beta}(x))$$
is given by
$$\frac{k+1}{n+1}\sum_{j=0}^{n-k}\binom{n+1}{k+j+1}\binom{n+j}{j}(\beta-1)^{n-k-j}=\sum_{j=0}^{n-k}\frac{k+1}{k+j+1}\binom{n}{k+j}\binom{n+j}{j}(\beta-1)^{n-k-j}.$$
For instance, when $\beta=1$, which is the case of the matrix $(1-x,x(1-x))^{-1}$, we get the expression
$$\frac{k+1}{n+1}\binom{2n-k}{n-k}$$ for the general term. Now the general term of the matrix
$\left(1,\frac{1}{1-x}\right)$ is given by
$$\binom{n-1}{k-1}+0^n (-1)^k.$$
Hence the general term of $\mathbf{L}$ is given by
\begin{equation}\label{genTerm}\sum_{j=0}^n \sum_{i=0}^{n-j}
\frac{j+1}{i+j+1}\binom{n}{i+j}\binom{n+i}{i}(\beta-1)^{n-j-i}(\binom{j-1}{k-1}+0^j(-1)^k).\end{equation}

\noindent It is interesting to note that
\begin{equation}\mathbf{L}^{-1}=\left(\frac{1+x}{(1+x)(1+\beta x)},\frac{x}{(1+x)(1+\beta x)}\right).\end{equation}
We can use the factorization in Eq.~(\ref{Fact}) to express the orthogonal polynomials $P_n^{(\beta)}(x)$ in terms of the
Chebyshev polynomials of the second kind $U_n(x)$. Thus we recognize \cite{PB_Meixner} that the Riordan array
$$\left(\frac{1}{1+(1+\beta)x+\beta x^2},\frac{x}{1+(1+\beta)x+\beta x^2}\right)$$ is the
coefficient array of the modified Chebyshev polynomials of the second kind
$\beta^{\frac{n}{2}}U_n(\frac{x-(\beta+1)}{2\sqrt{\beta}})$. Hence by the factorization in Eq.~(\ref{Fact}) we obtain
\begin{equation}P_n^{(\beta)}(x)=\beta^{\frac{n}{2}}U_n\left(\frac{x-(\beta+1)}{2\sqrt{\beta}}\right)+\beta^{\frac{n-1}{2}}U_{n-1}\left(\frac{x-(\beta+1)}{2\sqrt{\beta}}\right).\end{equation}
We state this as a proposition.
\begin{proposition} The family $\{P_n(x)\}$  of orthogonal polynomials for which the Narayana polynomials $a_n^{(\beta)}=\mathcal{N}_{2,n}(x)$ are moments is given by
$$P_n^{(\beta)}(x)=\beta^{\frac{n}{2}}U_n\left(\frac{x-(\beta+1)}{2\sqrt{\beta}}\right)+\beta^{\frac{n-1}{2}}U_{n-1}\left(\frac{x-(\beta+1)}{2\sqrt{\beta}}\right).$$
\end{proposition}
\section{On the Hankel transform of the row sums of $\mathbf{L}$}
We recall that
$$\mathbf{L}=\left(g_{\beta}(x),\frac{g_{\beta}(x)-1}{\beta}\right)$$ is the matrix whose first column is given by
terms of the Narayana polynomial sequence $\mathcal{N}_{2,n}(\beta)$. We now wish to calculate the Hankel transform of the row sums
$s_n^{\beta}$ of the matrix $\mathbf{L}$.
The generating function of these row sums is given by
$$g_s(x)=\frac{(\beta+1)\sqrt{1-2x(\beta+1)+(1-\beta)^2 x^2}+(\beta-1)(x(\beta+1)+1)}{2(1-2x(1+\beta))}.$$
We infer from this (using Stieltjes-Perron) that the row sum elements $s_n^{(\beta)}$ are the moments for the following weight function
:
$$w_s(x)=\frac{1}{2\pi} \frac{\sqrt{-x^2+2x(1+\beta)-(\beta-1)^2}(\beta+1)}{x(2(1+\beta)-x)}$$ with
support on the interval $[1+\beta-2\sqrt{\beta},1+\beta+2\sqrt{\beta}]$.
From this we can determine that the Hankel transform of $s_n^{(\beta)}$ is given by
$$(\beta+1)^n \beta^{\binom{n}{2}}.$$

\noindent In fact, if we let
$$\mathbf{H}_s=\mathbf{L}_s \mathbf{D}_s \mathbf{L}_s^t$$ be the
$LDU$ decomposition \cite{PPWW} of the Hankel matrix associated with
$s_n^{(\beta)}$, then we have
$$\mathbf{L}_s=\left(g_s(x),
\frac{1-(1+\beta)x-\sqrt{1-2x(1+\beta)+(1-\beta)^2 x^2}}{2 \beta
x}\right).$$ Equivalently,
$$\mathbf{L}_s^{-1}=\left(\frac{1-x^2}{1+(1+\beta)x+\beta x^2},\frac{x}{1+(1+\beta)x+\beta
x^2}\right) $$ is the coefficient array of the orthogonal
polynomials associated to the sequence $s_n^{(\beta)}$. This is so
since the Stieltjes-Jacobi matrix associated to $s_n^{(\beta)}$ takes the form
\begin{displaymath}\left(\begin{array}{ccccccc} \beta+1 &amp; 1 &amp; 0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\\beta+1 &amp; \beta+1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\
0 &amp; \beta &amp; \beta+1 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; \beta &amp; \beta+1
&amp; 1 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; 0 &amp; \beta &amp; \beta+1 &amp; 1 &amp; \ldots
\\0 &amp; 0 &amp; 0 &amp; 0 &amp; \beta &amp; \beta+1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \ddots\end{array}\right).\end{displaymath}

\section{The Hankel transform of the row sums of
$(g_{\beta}(x),xg_{\beta}(x))$}

We have seen that
$$\mathbf{L}=(g_{\beta}(x),xg_{\beta}(x))\cdot \left(1,\frac{x}{1-x}\right).$$
In this section, we look at the Hankel transform of the row sums of
the factor $(g_{\beta}(x),xg_{\beta}(x))$ of $\mathbf{L}$. The row
sums in question have generating function
$$\frac{g_{\beta}(x)}{1-xg_{\beta}(x)}=\frac{1-(1+\beta)x-\sqrt{1-2(1+\beta)x+(1-\beta)^2 x^2}}{2 \beta x^2}.$$
From this we can infer that the row sums of
$(g_{\beta}(x),xg_{\beta}(x))$ are the moments for the weight
function
$$w(x)=\frac{1}{2\pi}\frac{\sqrt{-x^2+2(1+\beta)x-(1-\beta)^2}}{\beta}.$$
This then allows us to prove that the Hankel transform sought is
$\beta^{\binom{n+1}{2}}$.

\section{Formulas for the row sums of $(g_{\beta}(x),xg_{\beta}(x))$}
 We can
characterize the row sums of $(g_{\beta}(x),xg_{\beta}(x))$ by
observing that
$$\frac{xg_{\beta}(x)}{1-xg_{\beta}(x)}=\frac{1-(1+\beta)x-\sqrt{1-2(1+\beta)x+(1-\beta)^2 x^2}}{2 \beta
x}$$ is the series reversion of the function
$$\frac{x}{1+(1+\beta)x+\beta x^2}.$$ Hence the row sums are given
by the $(n+1)$-st term of the series reversion of
$\frac{x}{1+(1+\beta)x+\beta x^2}$. Thus the row sums of $(g_{\beta}(x),xg_{\beta}(x))$ are precisely
$$\mathcal{N}_{3,n}(\beta)=\sum_{k=0}^n N_3(n,k)\beta^k.$$ This may also be expressed  by
$$\sum_{k=0}^n \frac{k+1}{n+1}\sum_{j=0}^{n-k}
\binom{n+1}{k+j+1}\binom{n+j}{j}(\beta-1)^{n-k-j}$$ or by
$$\sum_{k=0}^n \sum_{j=0}^{n-k}
\frac{k+1}{k+j+1}\binom{n}{k+j}\binom{n+j}{j}(\beta-1)^{n-k-j}.$$

We have seen that the general term of the matrix $\mathbf{L}$ is given by Eq.~(\ref{genTerm})
and hence the general term $s_n^{(\beta)}$ of the row sums of
$\mathbf{L}$ is given by
$$s_n^{(\beta)}=\sum_{k=0}^n \sum_{j=0}^n \sum_{i=0}^{n-j}
\frac{j+1}{j+i+1}\binom{n}{i+j}\binom{n+i}{i}(\beta-1)^{n-j-i}(\binom{j-1}{k-1}+0^j
(-1)^k).$$

\section{Narayana polynomials and hypergeometric functions}
We recall that the hypergeometric function $_2F_1(\alpha, \beta;\gamma;x)$ is defined by
$$_2F_1(\alpha, \beta;\gamma;x)=\sum_{k=0}^{\infty} \frac{(\alpha)_k (\beta)_k }{(\gamma)_k}\frac{x^k}{k!},$$
where $$(\alpha)_k=\prod_{j=0}^{k-1}(\alpha+j).$$
\noindent For $n \in \mathbb{Z}$, we have $(n)_k =(-1)^k k! \binom{-n}{k}$.
Thus we have
\begin{eqnarray*}
_2F_1(-n,-n-1;2;x)&amp;=&amp;\sum_{k=0}^{\infty} \frac{(-n)_k (-n-1)_k}{(2)_k}\frac{x^k}{k!}\\
&amp;=&amp; \sum_{k=0}^{\infty} \frac{(-1)^k k!\binom{n}{k} (-1)^k k! \binom{n+1}{k}}{(n+1)!}\frac{x^k}{k!}\\
&amp;=&amp; \sum_{k=0}^n \frac{1}{k+1} \binom{n}{k}\binom{n+1}{k}x^k\\
&amp;=&amp; \sum_{k=0}^n N_3(n,k)x^k \\
&amp;=&amp; \mathcal{N}_{3,n}(x).\end{eqnarray*}
In two recent articles \cite{Branden_Stack, Kostov}, a link between $\mathcal{N}_{3,n}(x)$ and the Jacobi polynomials has
been established.
This is that
\begin{equation}\label{Hyper}\mathcal{N}_{3,n}(x)=\frac{1}{n+1}(1-x)^n P_n^{(1,1)}\left(\frac{1+x}{1-x}\right).\end{equation}
Now $$P_n^{\alpha,\beta}(x)=\binom{n+\alpha}{n} \,_2F_1(-n,n+\alpha+\beta+1;\alpha+1;\frac{1}{2}(1-x)),$$ and so
\begin{equation}\mathcal{N}_{3,n}(x)=(1-x)^n \,_2F_1\left(-n,n+3;2;\frac{x}{x-1}\right).\end{equation}
We can modify Eqn. (\ref{Hyper}) to obtain the following expression for $\mathcal{N}_{1,n}(x)$\,:
$$\mathcal{N}_{1,n}(x)=\frac{1-x\cdot0^n}{n+0^n}(1-x)^{n-1}P_{n+0^n-1}^{(1,1)}\left(\frac{1+x}{1-x}\right).$$
Straight-forward calculation also establishes that
$$N_2(n,k)=[x^{n-k}]\,_2F_1(k+1,k;2;x).$$

\noindent Note also that the triangle with general term
$$T(n,k)=[x^{n-k}]\,_2F_1(k+1,k;1;x)$$ is the triangle (see \seqnum{A103371}) with general term
$$T(n,k)=0^{n+k}+\binom{n-1}{k-1}\binom{n}{k},$$ which begins
\begin{displaymath}\left(\begin{array}{ccccccc} 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;
2 &amp; 1 &amp; 0 &amp; 0
&amp; 0 &amp; \ldots \\ 0 &amp; 3 &amp; 6 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 4 &amp; 18 &amp;
12 &amp; 1 &amp; 0 &amp; \ldots \\0 &amp; 5 &amp; 40 &amp; 60 &amp; 20 &amp; 1 &amp;\ldots\\
\vdots
&amp; \vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath}
This matrix and $\mathbf{N}_2$ are related as follows:
$$N_2(n,k)=\frac{T(n,k)}{(n-k+1)}=\frac{0^{n+k}+\binom{n-1}{k-1}\binom{n}{k}}{n-k+1}.$$
For the matrix $\mathbf{N}_1$, we have the following:
$$\mathcal{N}_{1,n}(x)=\,_2F_1(-n,-n+1,2,x).$$ Since $\mathbf{N}_2$ is the reversal of $\mathbf{N}_1$, we obtain
$$\mathcal{N}_{2,n}(x)=x^n \,_2F_1(-n,-n+1,2,1/x).$$
We can also note the following. We have seen (section \textbf{\ref{N_CF}}) that $\mathbf{N}_3$ has generating function
$$
\cfrac{1}{1-x-xy-
\cfrac{x^2y}{1-x-xy-
\cfrac{x^2y}{1-x-xy-
\cfrac{x^2y}{1-\cdots}}}}.$$
\noindent 
$\mathbf{N}_3$ is thus seen \cite{CF} to be the binomial transform of the array with generating function
$$
\cfrac{1}{1-xy-
\cfrac{x^2y}{1-xy-
\cfrac{x^2y}{1-xy-
\cfrac{x^2y}{1-\cdots}}}}.$$
This array begins
\begin{displaymath}\left(\begin{array}{ccccccc} 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;
1 &amp; 1 &amp; 0 &amp; 0
&amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; 3 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; 2 &amp;
6 &amp; 1 &amp; 0 &amp; \ldots \\0 &amp; 0 &amp; 0 &amp; 10 &amp; 10 &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{A107131}, which is the coefficient array of the polynomials given by
$$x^n\,_2F_1\left(\frac{1}{2}-\frac{n}{2},-\frac{n}{2};2;\frac{4}{x}\right).$$ It also counts the number of Motzkin paths
of
length $n$ with $k$ steps $U=(1,1)$ or $H=(1,0)$. The general term of this array is
$$[k \le n] \binom{n}{2n-2k}C_{n-k}.$$ Thus
\begin{equation}N_3(n,k)=\sum_{j=0}^n \binom{n}{j}\binom{j}{2(j-k)}C_{j-k}.\end{equation}
Since the row sums of $\mathbf{N}_3$ yield the shifted Catalan numbers, we arrive at the identity
\begin{equation}C_{n+1}=\sum_{k=0}^n \sum_{j=0}^n \binom{n}{j}\binom{j}{2(j-k)}C_{j-k}.\end{equation}


\section{Appendix A - calculation of MIMO capacity}

We follow \cite{Groh} to derive an expression for MIMO capacity in a
special case. This is a form of transmission technology which increases the transmission channel capacity by taking advantage of the
multipath nature of transmission when many antennas transmit to many receivers. Thus we assume that we have R receive antennas and T
transmit antennas, modeled by
$$\mathbf{r}=\mathbf{H}\mathbf{s}+\mathbf{n}$$ where $\mathbf{r}$ is the receive signal vector, $\mathbf{s}$ is the
source signal vector, $\mathbf{n}$ is an additive white Gaussian noise
(AWGN) vector, which is a realization of a complex normal
distribution $N(\mathbf{0}, \sigma^2 \mathbf{I}_R)$, and the channel
is represented by the complex matrix $\mathbf{H}\in \mathbb{C}^{R \times T}$.
We have the eigenvalue decomposition
$$\mathbf{H}^H \mathbf{H}=\frac{1}{T} \mathbf{Q}\mathbf{\Lambda}\mathbf{Q}^H.$$
We assume $T &lt; R$. Then the capacity of the uncorrelated MIMO
channels is given by \cite{Muller}

\begin{eqnarray*}C_{MIMO}&amp;=&amp;\frac{1}{R}\log_2
\det(\mathbf{I}_T+\mathbf{H}^H(\sigma^2\mathbf{I}_R)^{-1}\mathbf{H})\\
&amp;=&amp;\frac{1}{R}\log_2
\det(\mathbf{I}_T+\frac{1}{\sigma^2}\mathbf{H}^H\mathbf{H})\\
&amp;=&amp;\frac{1}{R}\log_2
\det(\mathbf{I}_T+\frac{1}{\sigma^2}\frac{1}{T}\mathbf{Q}\mathbf{\Lambda}\mathbf{Q}^H)\\
&amp;=&amp;\frac{1}{R}\log_2 \det(\mathbf{I}_T+\frac{1}{\sigma^2 T}\mathbf{\Lambda})\\
&amp;=&amp;\frac{T}{R}\frac{1}{T}\sum_{i=1}^T \log_2(1+\frac{1}{\sigma^2
T}\lambda_i)\\
&amp;=&amp;\frac{\beta}{\ln2}\frac{1}{T}\sum_{i=1}^T \ln(1+\frac{1}{\sigma^2
T}\lambda_i)\end{eqnarray*} where we have set
$$\beta=\frac{T}{R}.$$ Now
\begin{eqnarray*} \ln(1+x)&amp;=&amp;\ln(1+x_0)+\sum_{k=1}^N
(-1)^{k-1}\frac{(x-x_0)^k}{k(1+x_0)^k}, \qquad |x-x_0|&lt;1\\
&amp;=&amp;\ln(1+x_0)+\sum_{k=1}^N
\frac{(-1)^{k-1}}{k(1+x_0)^k}\sum_{j=0}^k\binom{k}{j}x^j(-1)^{k-j}x_0^{k-j}\\
&amp;=&amp;\ln(1+x_0)+\sum_{k=1}^n\sum_{j=0}^k\binom{k}{j}(-1)^{j-1}\frac{x_0^{k-j}}{k(1+x_0)^k}x^j\\
&amp;=&amp;\sum_{k=0}^N p_k x^k,\end{eqnarray*} where it is appropriate to
take $x_0=\frac{1}{\sigma^2}$. We thus obtain \begin{eqnarray*}
C_{MIMO}&amp;=&amp;\frac{\beta}{\ln2}\frac{1}{T}\sum_{i=1}^T \sum_{k=0}^N
p_k\left(\frac{\lambda_i}{\sigma^2 T}\right)^k\\
&amp;=&amp;\frac{\beta}{\ln2}\sum_{k=0}^N \frac{p_k}{(\sigma^2 T)^k}
\left(\frac{1}{T}\sum_{i=1}^T \lambda_i^k\right)\\
&amp;=&amp;\frac{\beta}{\ln2}\sum_{k=0}^N \frac{p_k}{(\sigma^2 T)^k} m_k\\
&amp;=&amp;\frac{\beta}{\ln2}\sum_{k=0}^N\frac{p_k}{(\sigma^2 T)^k}
\sum_{j=0}^k N_2(k,j)\beta^j.\end{eqnarray*} Here, we have replaced
the expression $m_k=\frac{1}{T}\sum_{i=1}^T \lambda_i^k$ by the $k$-th
moment of the limiting eigenvalue distribution function, which
following \cite{Silverstein_Bai} can to be shown to have Stieltjes
transform
$$G_{\beta}(z)=-\frac{1}{2}+\frac{\beta-1}{2z}+\sqrt{\frac{(1-\beta)^2}{4z^2}+\frac{1}{4}-\frac{1+\beta}{2z}}.$$
Thus by our preceding results (see Sections \textbf{\ref{Reversion}} and \textbf{\ref{Moments}}) we arrive at the new expression
\begin{equation}C_{MIMO}=\frac{\beta}{\ln2}\sum_{k=0}^N \frac{p_k}{(\sigma^2 T)^k}[x^{k+1}]\text{Rev}_x\left[
\frac{x(1-x)}{1-(1-\beta)x}\right].\end{equation}
\section{Appendix B - the Deleham construction}
For the purposes of this note, we define the \emph{Deleham construction} \cite{CF} as follows.
Given two sequences $r_n$ and $s_n$,we use the notation
$$ r \quad \Delta \quad s = [r_0,r_1,r_2,\ldots] \quad \Delta \quad [s_0,s_1,s_2,\ldots]$$ to denote the
number triangle whose bi-variate generating function is given by
$$\cfrac{1}{1-
\cfrac{(r_0x+s_0 xy)}{1-
\cfrac{(r_1x+s_1 xy)}{1-
\cfrac{(r_2x+s_2 xy)}{1-\cdots}}}}.$$ We furthermore define
$$ r \quad \Delta^{(1)} \quad s = [r_0,r_1,r_2,\ldots] \quad \Delta^{(1)} \quad [s_0,s_1,s_2,\ldots]$$ to denote
the number triangle whose bi-variate generating function is given by
$$\cfrac{1}{1-(r_0x+s_0 xy)-
\cfrac{(r_1x+s_1 xy)}{1-
\cfrac{(r_2x+s_2 xy)}{1-\cdots}}}.$$
\noindent See \seqnum{A084938} for the original definition.

\section{Appendix C - The Stieltjes transform of a measure} The
\emph{Stieltjes transform} of a measure $\mu$ on $\mathbb{R}$
is a function $G_{\mu}$
defined on $\mathbb{C}\setminus \mathbb{R}$ by
$$G_{\mu}(z)=\int_{\mathbb{R}}\frac{1}{z-t}\mu(t).$$ If $f$ is
a bounded continuous function on
$\mathbb{R}$, we have $$\int_{\mathbb{R}}f(x)\mu(x)=-\lim_{y
\to 0^{+}}\int_{\mathbb{R}}f(x)\Im G_{\mu}(x+iy)dx.$$ If $\mu$
has compact
support, then $G_{\mu}$ is holomorphic at infinity and for
large $z$, $$G_{\mu}(z)=\sum_{n=0}^{\infty}
\frac{a_n}{z^{n+1}},$$ where
$a_n=\int_{\mathbb{R}}t^n \mu(t)$ are the moments of the
measure. If $\mu(t)=d\psi(t)=\psi'(t)dt$ then (Stieltjes-Perron)
$$\psi(t)-\psi(t_0)=-\frac{1}{\pi}\lim_{y \to
0^{+}}\int_{t_0}^t \Im G_{\mu}(x+iy)dx.$$ If now $g(x)$ is the
generating function of a
sequence $a_n$, with $g(x)=\sum_{n=0}^{\infty}a_n x^n$, then
we
can define
$$G(z)=\frac{1}{z}g\left(\frac{1}{z}\right)=\sum_{n=0}^{\infty}\frac{a_n}{z^{n+1}}.$$
By this means, under the right circumstances we can
retrieve the density function for the measure that defines the
elements $a_n$ as moments.
\begin{example}
We let $g(z)=\frac{1-\sqrt{1-4z}}{2z}$ be the g.f. of the Catalan numbers. Then
$$G(z)=\frac{1}{z}g\left(\frac{1}{z}\right)=\frac{1}{2}\left(1-\sqrt{\frac{x-4}{x}}\right).$$
Then $$\Im G_{\mu}(x+iy)=-\frac{\sqrt{2}\sqrt{\sqrt{x^2+y^2}\sqrt{x^2-8x+y^2+16}-x^2+4x-y^2}}{4\sqrt{x^2+y^2}},$$
and so we obtain \begin{eqnarray*}\psi'(x)&amp;=&amp;-\frac{1}{\pi}\lim_{y \to
0^{+}}\left\{-\frac{\sqrt{2}\sqrt{\sqrt{x^2+y^2}\sqrt{x^2-8x+y^2+16}-x^2+4x-y^2}}{4\sqrt{x^2+y^2}}\right\}\\
&amp;=&amp; \frac{1}{2\pi}\frac{\sqrt{x(4-x)}}{x}.\end{eqnarray*}
\end{example}

\section{Acknowledgements} The second-named author was supported by a Council of Directors Strand I grant during this work.

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\bigskip
\hrule
\bigskip
\noindent 2000 {\it Mathematics Subject Classification}: Primary
11B83; Secondary 05A10, 05A19, 94A05, 94A11.


\noindent \emph{Keywords:} Integer sequence, Narayana triangle, Narayana polynomial, Riordan array, Hankel transform,
orthogonal polynomials, multiple-input multiple-output (MIMO) systems, Mar\v{c}enko-Pastur.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000007},
\seqnum{A000045},
\seqnum{A000108},
\seqnum{A001003},
\seqnum{A001263},
\seqnum{A006318},
\seqnum{A007318},
\seqnum{A007564},
\seqnum{A047891},
\seqnum{A059231},
\seqnum{A082298},
\seqnum{A084938},
\seqnum{A090181},
\seqnum{A103371},
\seqnum{A107131}, and
\seqnum{A131198}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received April 1 2009;
revised version received  June 30 2010; September 15 2010; March 10 2011.
Published in {\it Journal of Integer Sequences}, March 26 2011.

\bigskip
\hrule
\bigskip

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