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\begin{center}
\vskip 1cm{\LARGE\bf Generalized Narayana Polynomials, \\
\vskip .1in
Riordan Arrays, and Lattice Paths} \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} \\
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\vskip .2 in

\begin{abstract}
We study a family of polynomials in two variables,
identifying them as the moments of a two-parameter family of orthogonal
polynomials. The coefficient array of these orthogonal polynomials is
shown to be an ordinary Riordan array. We express the generating
function of the sequence of polynomials under study as a continued
fraction, and determine the corresponding Hankel transform. An
alternative characterization of the polynomials in terms of a related
Riordan array is also given. This Riordan array is associated with 
{\L}ukasiewicz paths. The special form of the production matrices is
exhibited in both cases. This allows us to produce a bijection from a
set of colored {\L}ukasiewicz paths to a set of colored Motzkin paths.
The polynomials studied generalize the notion of Narayana polynomial.
\end{abstract}

\section{Introduction}
In this note, we shall be concerned with the family of bivariate polynomials defined by
$$Q_n(r,s)=\sum_{k=0}^n \sum_{j=0}^k \binom{n}{j}\binom{n-j}{2(k-j)}a_{k-j}(s)r^k,$$
where
$$a_n(s)=\sum_{k=0}^n \binom{2n-k-1}{n-k}\frac{k+0^{n+k}}{n+0^{n k}} s^k.$$
Our main goal is to characterize $Q_n(r,s)$ as the moment sequence of a family of orthogonal polynomials, and in so doing also show that
the sequence $Q_n(r,s)$ has Hankel transform $h_n(r,s)=\det(Q_{i+j}(r,s))_{0 \le i,j \le n}$ \cite{Layman} given by
$$h_n(r,s)=s^n r^{\binom{n+1}{2}}.$$

Examples of these sequences include the shifted Catalan numbers $Q_n(1,1)$, the central binomial coefficients $Q_n(1,2)$, the central Delannoy numbers $Q_n(2,2)$, and many other sequences that are documented in the On Line Encyclopedia of Integer Sequences \cite{SL1, SL2}.

We use the production matrix of a particular Riordan array to establish the existence of a family of orthogonal polynomials for which the generalized Narayana polynomials are moments. We find it interesting to look at a closely related Riordan array, again with the generalized Narayana polynomials as elements of the first column, but where now the production matrix is not tri-diagonal. It does however have a structure that is interesting in its own right. The Hankel transform of the row sum sequence of this second Riordan array is also interesting to study.

\section{Preliminaries on integer sequences, Riordan arrays and Hankel transforms}
For an integer sequence $a_n$, that is, an element of $\mathbb{Z}^\mathbb{N}$, the power series
$f(x)=\sum_{n=0}^{\infty}a_n x^n$ 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 $f(\bar{f}(x))=x$.

For a lower triangular matrix $(a_{n,k})_{n,k \ge 0}$ the row sums give the sequence with general term
$\sum_{k=0}^n a_{n,k}$.


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
$g(x)=1+g_1x+g_2x^2+\cdots$ and $f(x)=f_1x+f_2x^2+\cdots$ where
$f_1\ne 0$ \cite{Spru}. We assume in addition that $f_1=1$ in what follows. 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 $g, f$ 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, 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
$\mathbf{a}=(a_0,a_1,\ldots)'$ is an integer sequence with ordinary
generating function $\cal{A}$ $(x)$, then the sequence
$\mathbf{M}\mathbf{a}$ has ordinary generating function
$g(x)$$\cal{A}$$(f(x))$. The (infinite) matrix $(g,f)$ can thus be considered to act on the ring of
integer sequences $\mathbb{Z}^\mathbb{N}$ by multiplication, where a sequence is regarded as a
(infinite) column vector. We can extend this action to the ring of power series
$\mathbb{Z}[[x]]$ by
$$(g,f):\cal{A}(\mathnormal{x}) \mapsto \mathnormal{(g,f)}\cdot
\cal{A}\mathnormal{(x)=g(x)}\cal{A}\mathnormal{(f(x))}.$$
\begin{example} The so-called \emph{binomial matrix} $\mathbf{B}$ is the element
$(\frac{1}{1-x},\frac{x}{1-x})$ of the Riordan group. It has general
element $\binom{n}{k}$, and hence as an array coincides with Pascal's triangle. More generally, $\mathbf{B}^m$ is the
element $(\frac{1}{1-m x},\frac{x}{1-mx})$ of the Riordan group,
with general term $\binom{n}{k}m^{n-k}$. It is easy to show that the
inverse $\mathbf{B}^{-m}$ of $\mathbf{B}^m$ is given by
$(\frac{1}{1+mx},\frac{x}{1+mx})$.
\end{example}

For an invertible matrix $L$, its production matrix (also called its Stieltjes matrix) \cite{ProdMat_0, ProdMat} is the matrix
$$P_L = L^{-1} \hat{L},$$ where $\hat{L}$ is the matrix $L$ with its first row removed.
A Riordan array $L$ is the inverse of the coefficient array of a family of orthogonal polynomials if and only if
$P_L$ is tri-diagonal \cite{Barry_Moment, Barry_Meixner}. Necessarily, the Jacobi coefficients of these orthogonal polynomials are then constant. Such types of polynomials have been studied in many contexts, including orthogonal polynomials \cite{Szego, Geronimus, Askey_Ismail_1, Askey_Ismail_2}, random walks \cite{Karlin, Grunbaum}, group theory \cite{Cohen}, non-commutative probability \cite{Saitoh, Ans, BB}, and the study of random matrices \cite{Collins}.

An important feature of Riordan arrays is that they have a number of sequence characterizations \cite{Cheon, He}. The simplest of
these
is as follows.
\begin{proposition} \label{Char} \cite[Theorem 2.1, Theorem 2.2]{He} Let $D=[d_{n,k}]$ be an infinite triangular matrix. Then $D$ is a Riordan array if and only if there
exist two sequences $A=[a_0,a_1,a_2,\ldots]$ and $Z=[z_0,z_1,z_2,\ldots]$ with $a_0 \neq 0$, $z_0 \neq 0$ such that
\begin{itemize}
\item $d_{n+1,k+1}=\sum_{j=0}^{\infty} a_j d_{n,k+j}, \quad (k,n=0,1,\ldots)$
\item $d_{n+1,0}=\sum_{j=0}^{\infty} z_j d_{n,j}, \quad (n=0,1,\ldots)$.
\end{itemize}
\end{proposition}
The coefficients $a_0,a_1,a_2,\ldots$ and $z_0,z_1,z_2,\ldots$ are called the $A$-sequence and the $Z$-sequence of the Riordan array
$L=(g(x),f(x))$, respectively.
Letting $A(x)$ be the generating function of the $A$-sequence and $Z(x)$ be the generating function of the $Z$-sequence, we have
\begin{equation}\label{AZ_eq} A(x)=\frac{x}{\bar{f}(x)}, \quad Z(x)=\frac{1}{\bar{f}(x)}\left(1-\frac{1}{g(\bar{f}(x))}\right).\end{equation}

The first column of $P_L$ is then generated by $Z(x)$, while the $k$-th column is generated by $x^{k-1}A(x)$ (taking the first column to be indexed by $0$).

For a given sequence $a_n$, the Hankel transform \cite{Layman} of $a_n$ is the sequence $\det(a_{i+j})_{0 \le i,j \le n}$. The Hankel transforms we shall be interested in will all be related to orthogonal polynomials, which in turn will be related to lattice paths \cite{Flaj, Viennot}.
A lattice path~\cite{Lehner} is a sequence  of vertices in the integer lattice $\mathbb{Z}^2$. A pair of consecutive vertices is called a step of the path. The height of a step is given by the $y$-value of the first vertex. A valuation is an integer function on the set of possible steps of  $\mathbb{Z}^2 \times \mathbb{Z}^2$. A valuation of a path is the product of the valuations of its steps. We concern ourselves with Motzkin paths, $k$-colored Motzkin paths~\cite{Barcucci} and \L ukasiewicz paths~\cite{Viennot}, which are defined below. A path of length $n$ begins at $(0, 0)$ and ends at $(n, 0)$. The valuations will be independent of the $x$-coordinates of the points, therefore the $x$-coordinates are redundant and we can represent a path $\pi$ by the sequence of its $y$-coordinates $\pi =(\pi(0) \cdots \pi(n)).$\\
\begin{definition}
A path $\pi =(\pi(0) \cdots \pi(n))$ is a Motzkin path~\cite{Lehner} if it satisfies the following conditions
\begin{enumerate}
\item The elementary steps can be north-east, east and south-east.
\item Steps never go below the $x$ axis.
\item $\pi(0) = (0,0)$ and $\pi(n) = (n,0)$
\end{enumerate}
If the east steps are labelled by $k$ colors we obtain the $k$-colored Motzkin paths.
\end{definition}
\begin{definition}
A path $\pi =(\pi(0) \cdots \pi(n))$ is a \L ukasiewicz path~\cite{Lehner} if it satisfies the following conditions
\begin{enumerate}
\item The elementary steps can be north - east and  east as those in Motzkin paths.
\item In addition south-east steps from level $k$ can fall to any level above or on the $x$ axis, and are denoted as $\lambda_{k,l}$ and referred to as \L ukasiewicz steps, where $k$ is the starting point of the south-east step and $l$ the level where the step ends.
\item Steps never go below the $x$ axis.
\end{enumerate}
\end{definition}
\begin{theorem}~\cite[Theorem 2.3] {Lehner}\label{path}
Let $$ \mu_n = \sum_{\pi \in \mathbf{M}} v(\pi)$$
where the sum is over the set of Motzkin paths $\pi =(\pi(0) \cdots \pi(n))$ of length $n$. Here $\pi(j)$ is the level after the $j^{th}$ step, and the valuation of a path is the product of the valuations of its steps $v(\pi) = \prod_{i = 1}^{n} v_i$ where
\begin{displaymath}
v_i = v(\pi(i-1),\pi(i)) = 
\begin{cases}
1, &amp; \text{if the $i^{th}$ step rises;} \\
\beta_{\pi(i-1)}, &amp; \text{if the $i^{th}$ step is horizontal;} \\
\alpha_{\pi(i-1)}, &amp; \text{if the $i^{th}$ step falls.}
\end{cases}
\end{displaymath}

\begin{center}
\begin{tikzpicture}
\draw [help lines] (-1,-1) grid (2,2);
\draw (-1,0) --  (0,1);
\draw (1,1) -- node[above=1pt] {$1$} (1,1);
\draw (-2,0) -- node[above=1pt] {\emph{level} $i$} (-2,0);
\draw (-1,0) -- (0,0);
\draw (1,0) -- node[above=1pt] {$\beta_{\pi(i-1)}$} (1,0);
\draw (-1,0) -- (0,-1);
\draw (1,-1) -- node[above=1pt] {$\alpha_{\pi(i-1)}$} (1,-1);
\end{tikzpicture}
\end{center}


Then the g.f. of the sequence $\mu_n$ is given by $$ \mathbf{M}(x) = \sum_{n=0}^{\infty} \mu_n x^n. $$ A continued fraction expansion of the g.f. is then $$ M(x) = \cfrac{1}{1- \beta_0 x - \cfrac{\alpha_1 x^2}{1 - \beta_1 x - \cfrac{\alpha_2 x^2}{1 - \beta_2 x - \cfrac{\alpha_3 x^2}{\ddots}}}}.$$
\end{theorem}
Now, from \cite{Kratt1} we have the following result,
\begin{theorem}\label{Kratt}~\cite[Theorem 11] {Kratt}
Let $(a_n)_{n \ge 0}$ be a sequence of numbers with g.f. $ \sum_{n=0}^{\infty} a_n x^n $ expressible in the form
\begin{displaymath}\label{cont}
\sum_{n=0}^{\infty} a_n x^n = \cfrac {a_0}{{1 - \beta_0 x - \cfrac{\alpha_1 x^2}{1 - \beta_1 x - \cfrac {\alpha_2 x^2}{\ddots}}}}.
\end{displaymath}
Then the Hankel determinant $\det_{0 \le i,j \le n-1}(a_{i+j})$ is given by
\begin{displaymath}
a_0^n \alpha_1^{n-1} \alpha_2^{n-2}\dots \alpha_{n-2}^2 \alpha_{n-1} = a_0^n \prod_{k=1}^{n-1} \alpha_k^{n-k},
\end{displaymath}
where the sequences $\{ \alpha_n \}_{n \geq 1 }$ and $\{ \beta_n \}_{n \geq 0}$ are the coefficients in the recurrence relation
\begin{displaymath}
P_n(x) = (x- \beta_n) P_{n-1} (x) - \alpha_n P_{n-2}(x), \quad n=1,2,3,4,\dots
\end{displaymath}
of the family of orthogonal polynomials $P_n$ for which $a_n$ forms the moment sequence.
\end{theorem}

\section{Generalized Narayana polynomials}

The sequence $a_n(s)=\sum_{k=0}^n \binom{2n-k-1}{n-k}\frac{k+0^{n+k}}{n+0^{n k}} s^k$ has generating function \cite{Barry_Catalan}
$$ g_s(x)=\frac{1}{1-s xc(x)}=(1,xc(x))\cdot \frac{1}{1-sx},$$  where $c(x)=\frac{1-\sqrt{1-4x}}{2x}$ is the generating function of the Catalan numbers, and $(1,xc(x))$ is the Riordan array \cite{SGWW} whose $k$-th column has generating function $(xc(x))^k$ \cite{Barry_Catalan}.  Since $c(x)$ has continued fraction expansion
$$c(x)=
\cfrac{1}{1-
\cfrac{x}{1-
\cfrac{x}{1-\cdots}}},$$ the generating function $g_s(x)$ has expansion
$$g_s(x)=
\cfrac{1}{1-
\cfrac{sx}{1-
\cfrac{x}{1-
\cfrac{x}{1-\cdots}}}} .$$
The matrix with general term
$$N_{n,k}(s)=\sum_{j=0}^k \binom{n}{j}\binom{n-j}{2(k-j)}a_{k-j}(s)$$ is the generalized Narayana triangle \cite{Barry_Nar} defined by the sequence $a_n(s)$, and the polynomials $Q_n(r,s)$ thus represent generalized Narayana polynomials. The Narayana triangle \seqnum{A001263} with general term
$$\frac{1}{k+1}\binom{n+1}{k}\binom{n}{k}$$ corresponds to $s=1$. For $s=2$ we get the number triangle with general term $\binom{n}{k}^2$ \seqnum{A008459}.

\begin{proposition}
 The generating function of the polynomials $Q_n(r,s)=\sum_{k=0}^n N_{n,k}(s)r^k$ is given by
$$g_Q(x)=\cfrac{1}{1-(r+1)x-
\cfrac{srx^2}{1-(r+1)x-
\cfrac{rx^2}{1-(r+1)x-
\cfrac{rx^2}{1-\cdots}}}}.$$
\end{proposition}
\begin{proof} See Theorem $21$, \cite{Barry_Nar}.
\end{proof}
\begin{corollary} The Hankel transform of $Q_n(r,s)$ is given by
$$h_n=s^n r^{\binom{n+1}{2}}.$$
\end{corollary}
\begin{proof} This follows from \cite{Kratt, Kratt1}.
\end{proof}
We know \cite{Barry_MIMO} that the Narayana polynomials $\sum_{k=0}^n \frac{1}{k+1} \binom{n}{k}\binom{n+1}{k}r^k$ can be characterized as being the first column elements of the inverse Riordan array $$\left(\frac{1}{1+(r+1)x+rx^2}, \frac{x}{1+(r+1)x+rx^2}\right)^{-1}.$$
The following proposition generalizes this result, by providing a characterization of the polynomials $Q_n(r,s)$ in terms of Riordan arrays.
\begin{proposition} The polynomials $Q_n(r,s)$ are given by the terms of the first column of the Riordan array
$$L=\left(\frac{1+r(1-s)x^2}{1+(r+1)x+rx^2}, \frac{x}{1+(r+1)x+rx^2}\right)^{-1}.$$
\end{proposition}
\begin{proof}
We can express $g_Q(x)$ in closed form by noticing that
$$g_Q(x)=\frac{1}{1-(r+1)x-sr x^2 v},$$ where $v$ is the solution of the equation
$$v=\frac{1}{1-(r+1)x-r x^2 v}.$$ (The function $v$ is the generating function of the Narayana triangle). We find that
$$g_Q(x)=\frac{2}{s\sqrt{x^2(r-1)^2-2x(r+1)+1}+(x(r+1)-1)(s-2)}.$$

Standard Riordan array techniques \cite{Barry_Moment} now show that this is equal to the generating function of the first column of the inverse array $\left(\frac{1+r(1-s)x^2}{1+(r+1)x+rx^2}, \frac{x}{1+(r+1)x+rx^2}\right)^{-1}$.
\end{proof}
Note that we can also express $g_Q(x)$ as follows.
$$g_Q(x)=\frac{s \sqrt{x^2(r-1)^2-2x(r+1)+1}+(2-s)(x(r+1)-1)}{2(x^2(r^2(s - 1) - r(s^2 - 2s + 2) + s - 1) + (s-1)(1-2x(r+1)))}.$$
We have
$$L=\left(g_Q(x), \frac{1-x(r+1)-\sqrt{1-2x(r+1)+x^2(r-1)^2}}{2rx}\right).$$
\begin{corollary} The polynomials $Q_n(r,s)$ are the moments of the family of orthogonal polynomials $P_n(x)$ that satisfy the recurrence
$$P_n(x)=(x-(r+1))P_{n-1}(x)-rP_{n-2}(x), \quad n &gt;2,$$
$P_0(x)=1$, $P_1(x)=x-(r+1)$, $P_2(x)=(x-(r+1))P_1(x)-srP_0(x)=x^2-2(r+1)x+r^2+r(2-s)+1$.
\end{corollary}
\begin{proof} We need to show \cite{Barry_Moment, Barry_Meixner} that the production matrix \cite{ProdMat_0, ProdMat} of
$$L=\left(\frac{1+r(1-s)x^2}{1+(r+1)x+rx^2}, \frac{x}{1+(r+1)x+rx^2}\right)^{-1}$$ is tri-diagonal. Standard Riordan array techniques show that this is so, with form
\begin{displaymath}P_L=\left(\begin{array}{ccccccc} r+1 &amp; 1 &amp;
0
&amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\sr &amp; r+1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\ 0 &amp; r
&amp; r+1 &amp; 1 &amp; 0 &amp;
0 &amp; \cdots \\ 0 &amp; 0 &amp; r &amp; r+1 &amp; 1 &amp; 0 &amp; \cdots \\ 0 &amp; 0 &amp; 0
&amp; r &amp; r+1 &amp; 1 &amp; \cdots \\0 &amp; 0 &amp; 0 &amp; 0 &amp; r &amp; r+1
&amp;\cdots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath}
The three-term recurrence coefficients for $P_n(x)$ can be read from $P_L$.
\end{proof}

\begin{corollary}
The Hankel transform of $Q_n(r,s)$ is given by
$$h_n(r,s)=s^n r^{\binom{n+1}{2}}.$$
\end{corollary}
\begin{proof}
This is a direct consequence of the form of $g_Q$; equivalently it is derived from the form of the production matrix above.
\end{proof}

We note that $Q_n(r,j/r)$ is an integer sequence for $0 \le j \le r-1$. For instance, $Q_n(2,1/2)$ is the sequence that begins
$$1, 3, 10, 36, 138, 558, 2362, 10398, 47326, 221562, \ldots$$ It has generating function
$$\cfrac{1}{1-3x-
\cfrac{x^2}{1-3x-
\cfrac{2x^2}{1-3x-
\cfrac{2x^2}{1-\cdots}}}}.$$ 
This means \cite{Barry_CF} that $Q_n(2,1/2)$ is the third binomial transform of the aeration of the sequence with generating function
$$\cfrac{1}{1-
\cfrac{x}{1-
\cfrac{2x}{1-
\cfrac{2x}{1-\cdots}}}}.$$
This latter sequence is the sequence of generalized Catalan numbers $C(2;n)$ \seqnum{A064062} with g.f.
$$\frac{1}{1-xc(2x)}.$$ Thus the g.f. of $Q_n(2,1/2)$ can be represented as
$$\frac{1}{1-3x} \frac{1}{1-\left(\frac{x}{1-3x}\right)^2 c\left(2\left(\frac{x}{1-3x}\right)^2\right)}.$$
The Hankel transform of $Q_n(2,1/2)$ is given by $h_n(2,1/2)=\left(\frac{1}{2}\right)^n 2^{\binom{n+1}{2}}=2^{\binom{n}{2}}$, \seqnum{A006125}, a sequence with many combinatorial interpretations. One such is that it represents the number of graphs on $n$ labeled nodes.



Similarly the sequence $Q_n(3,2/3)$ which begins
$$1, 4, 18, 88, 458, 2504, 14244, 83696, 505114, 3116968, \ldots,$$ with g.f.
$$\cfrac{1}{1-4x-
\cfrac{2x^2}{1-4x-
\cfrac{3x^2}{1-4x-
\cfrac{3x^2}{1-\cdots}}}},$$ can be seen to be the fourth binomial transform of the aeration of the sequence with g.f.
$$\frac{1}{1-2xc(3x)}.$$ 
The Hankel transform of $Q_n(3,2/3)$ is given by
$$h_n(3,2/3)=\left(\frac{2}{3}\right)^n 3^{\binom{n+1}{2}},$$ which begins
$$1, 2, 12, 216, 11664, 1889568, 918330048,\ldots.$$
This is \seqnum{A083667}, the number of antisymmetric binary relations on a set of $n$ labeled points.

We note that in general, we can interpret the continued fraction form of $g_Q(x)$ as saying that $Q_n(r,s)$ is the
$(r+1)$-st binomial transform of the aeration of the sequence with g.f.
$$\frac{1}{1-sr xc(rx)}.$$ Thus we may express $g_Q(x)$ as
$$ g_Q(x)=\frac{1}{1-(r+1)x} \frac{1}{1-sr\left(\frac{x}{1-(r+1)x}\right)^2 c\left(r\left(\frac{x}{1-(r+1)x}\right)^2\right)}.$$
Other interesting sequences are $Q_n(n,0)=(n+1)^n$, or \seqnum{A000169}, $Q_n(n,1)=\sum_{k=0}^n \frac{1}{k+1}\binom{n}{k}\binom{n+1}{k}n^k$, which is \seqnum{A099169}, and $Q_n(n,2)=\sum_{k=0}^n \binom{n}{k}^2 n^k$, which is \seqnum{A187021}.
\section{Moment representation}
We have the following moment representation of $Q_n(r,s)$.

$$Q_n(r,s)=
\begin{cases} 1, \quad \text{if } r=0;\\
(r+1)^n, \quad \text{if } s=0;\\ \\
\sum_{k=0}^n \frac{1}{k+1}\binom{n+1}{k}\binom{n}{k}r^k=\frac{1}{2\pi}\int_{r+1-2\sqrt{r}}^{r+1+2\sqrt{r}} x^n \frac{ \sqrt{-x^2+2x(r+1)-(r-1)^2}}{r}\,dx, \quad \text{if } s=1, r \ne 0;\\ \\
\begin{split}\frac{s}{2 \pi}\int_{r+1-2\sqrt{r}}^{r+1+2\sqrt{r}} x^n \frac{\sqrt{-x^2+2x(r+1)-(r-1)^2}}{1-s+r(s^2-2s+2)-r^2(s-1)+2x(r+1)(s-1)-x^2(s-1)}\,dx + \\ \frac{s-2}{2(s-1)}\left(\left(r+1-s\sqrt{\frac{r}{s-1}}\right)^n+\left(r+1+s\sqrt{\frac{r}{s-1}}\right)^n\right), \quad \text{otherwise}.\end{split}\end{cases}$$
This density function can be seen as a generalization of the Mar\v{c}enko-Pastur density \cite{Marcenko}.

\section{A product matrix}

The motivation for the work in this section comes from the observation that the Riordan array
$$\left(\frac{1+(c-a)x+(d-b)x^2}{1+ax+bx^2},\frac{x}{1+ax+bx^2}\right)^{-1}$$ has
production array
\begin{displaymath}\left(\begin{array}{ccccccc} c &amp; 1 &amp;
0
&amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\d &amp; a &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\ 0 &amp; b
&amp; a &amp; 1 &amp; 0 &amp;
0 &amp; \cdots \\ 0 &amp; 0 &amp; b &amp; a &amp; 1 &amp; 0 &amp; \cdots \\ 0 &amp; 0 &amp; 0
&amp; b &amp; a &amp; 1 &amp; \cdots \\0 &amp; 0 &amp; 0 &amp; 0 &amp; b &amp; a
&amp;\cdots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right),\end{displaymath} while the Riordan array given by the product
$$\left(\frac{1+(c-a)x+(d-b)x^2}{1+ax+bx^2},\frac{x}{1+ax+bx^2}\right)^{-1}\cdot \left(1,\frac{x}{1+tx}\right)$$ has
 production matrix
\begin{displaymath}\left(\begin{array}{ccccccc} c &amp; 1 &amp;
0
&amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\d &amp; a-t &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\ dt &amp; b
&amp; a-t &amp; 1 &amp; 0 &amp;
0 &amp; \cdots \\ dt^2 &amp; bt &amp; b &amp; a-t &amp; 1 &amp; 0 &amp; \cdots \\ dt^3 &amp; bt^2 &amp; bt
&amp; b &amp; a-t &amp; 1 &amp; \cdots \\dt^4 &amp; bt^3 &amp; bt^2 &amp; bt &amp; b &amp; a-t
&amp;\cdots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath} In other words, we can show that the product matrix has
$$Z(x)=\frac{c-(ct-d)x}{1-tx},\quad A(x)=\frac{1-(2t-a)x-(at-b-t^2)x^2}{1-tx}.$$
It is possible to display $Q_n(r,s)$ as the first column of a Riordan array in many ways.
The Riordan array in the last section was chosen because its inverse is the coefficient array of the associated family of
orthogonal polynomials. Another interesting Riordan array is the array
\begin{eqnarray*} \tilde{L}&amp;=&amp;\left(\left(1,\frac{x}{1-x}\right) \cdot \left(\frac{1+r(1-s)x^2}{1+(r+1)x+rx^2},\frac{x}{1+(r+1)x+rx^2}\right)\right)^{-1}\\
&amp;=&amp;\left(\frac{1-2x-(r(s-1)-1)x^2}{1+(r-1)x}, \frac{x(1-x)}{1+(r-1)x}\right)^{-1}\\
&amp;=&amp;\left(\frac{1+r(1-s)x^2}{1+(r+1)x+rx^2},\frac{x}{1+(r+1)x+rx^2}\right)^{-1}\cdot  \left(1,\frac{x}{1+x}\right)\\
&amp;=&amp; L \cdot \left(1,\frac{x}{1+x}\right). \end{eqnarray*}
We have
$$\tilde{L}=\left(g_Q(x), \frac{1-x(r-1)-\sqrt{1-2x(r+1)+x^2 (r-1)^2}}{2}\right).$$
The matrix $\tilde{L}$ then has the production matrix
\begin{displaymath}P_{\tilde{L}}=\left(\begin{array}{ccccccc} r+1 &amp; 1 &amp;
0
&amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\rs &amp; r &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\ rs &amp; r
&amp; r &amp; 1 &amp; 0 &amp;
0 &amp; \cdots \\ rs &amp; r &amp; r &amp; r &amp; 1 &amp; 0 &amp; \cdots \\ rs &amp; r &amp; r
&amp; r &amp; r &amp; 1 &amp; \cdots \\rs &amp; r &amp; r &amp; r &amp; r &amp; r
&amp;\cdots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath}
Such production matrices are linked to \L ukasiewicz paths \cite{Hennessy-Thesis}.
\begin{example} We take the case $r=2$, $s=\frac{1}{2}$. In this case we obtain the production matrix
\begin{displaymath}P_{\tilde{L}}=\left(\begin{array}{ccccccc} 3 &amp; 1 &amp;
0
&amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\1 &amp; 2 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\ 1 &amp; 2
&amp; 2 &amp; 1 &amp; 0 &amp;
0 &amp; \cdots \\ 1 &amp; 2 &amp; 2 &amp; 2 &amp; 1 &amp; 0 &amp; \cdots \\ 1 &amp; 2 &amp; 2
&amp; 2 &amp; 2 &amp; 1 &amp; \cdots \\1 &amp; 2 &amp; 2 &amp; 2 &amp; 2 &amp; 2
&amp;\cdots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath}
The associated Riordan array is
$$\left(\frac{1-2x+2x^2}{1+x},\frac{x(1-x)}{1+x}\right)^{-1}=\left(\frac{3-9x-\sqrt{1-6x+x^2}}{2(1-6x+10x^2)},\frac{1-x-\sqrt{1-6x+10x^2}}{2}\right).$$
This matrix begins
\begin{displaymath}\left(\begin{array}{ccccccc}
1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\
3 &amp; 1  &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\
10 &amp; 5 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\
36 &amp; 22 &amp; 7 &amp; 1 &amp; 0 &amp; 0 &amp; \cdots \\
138 &amp; 96 &amp; 38 &amp; 9 &amp; 1 &amp; 0 &amp; \cdots \\
588 &amp; 426 &amp; 192 &amp; 58 &amp; 11 &amp; 1 &amp;\cdots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right),\end{displaymath} and has row sums $u_n$ that begin
$$1, 4, 16, 66, 282, 1246, 5674, 26530, 126910, 619086, \ldots,$$ with g.f.
$$\frac{1-3x-2x^2-(1-2x)\sqrt{1-6x+x^2}}{2x(1-6x+10x^2)}.$$
We have the moment representation
$$u_n = \frac{1}{2 \pi} \int_{3-2\sqrt{2}}^{3+2\sqrt{2}} x^n \frac{(x-2)\sqrt{-x^2+6x-1}}{x^2-6x+10}\,dx.$$
Note also that the first column sequence $1,3,10,36,\dots$ has the moment representation
$$u_n = \frac{1}{2 \pi} \int_{3-2\sqrt{2}}^{3+2\sqrt{2}} x^n \frac{\sqrt{-x^2+6x-1}}{x^2-6x+10}\,dx.$$
The Hankel transform of the row sum sequence $u_n$ is equal to
$$2^{\binom{n}{2}} [x^n] \frac{1-x}{1-x+2x^2}.$$
\end{example}
In general, the rows sums of $\tilde{L}$ can be shown to have Hankel transform
$$r^{\binom{n}{2}} [x^n] \frac{1-x}{1-rs x + rx^2}.$$
\begin{example} We consider the case $r=1$, $s=3$. Then
$$\tilde{L}=(1-2x-x^2,x(1-x))^{-1}=\left(\frac{1}{1-2xc(x)-x^2c(x)^2},xc(x)\right).$$ 
The row sums $u_n$ of $\tilde{L}$  have g.f.
$$\frac{1}{(1-2xc(x)-x^2c(x)^2)(1-xc(x))}=\frac{1-5x-(1+x)\sqrt{1-4x}}{2x(x^2+8x-2)}.$$ We have the moment representation
$$u_n=\frac{1}{2\pi} \int_{0}^4 x^n \frac{(1+x)\sqrt{x(4-x)}}{1+8x-2x^2}\,dx+\frac{1}{8+4\sqrt{2}}\left(2-\frac{3}{\sqrt{2}}\right)^n
+\frac{1}{8-4\sqrt{2}}\left(2+\frac{3}{\sqrt{2}}\right)^n.$$
The Hankel transform of the row sums is equal to
$$[x^n]\frac{1-x}{1-3x +x^2},$$ 
or $F_{2n+1}$. The row sum sequence is \seqnum{A143464}, which begins
$$1, 3, 11, 42, 164, 649, 2591, 10408, \ldots.$$
\end{example}
We can also look at sequences with negative values of $r$ and $s$.
For instance, the sequence $Q_n(-1,-1)$ is associated with the production matrix
\begin{displaymath}P_{\tilde{L}}=\left(\begin{array}{ccccccc} 0 &amp; 1 &amp;
0
&amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\1 &amp; -1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\ 1 &amp; -1
&amp; -1 &amp; 1 &amp; 0 &amp;
0 &amp; \cdots \\ 1 &amp; -1 &amp; -1 &amp; -1 &amp; 1 &amp; 0 &amp; \cdots \\ 1 &amp; -1 &amp; -1
&amp; -1 &amp; -1 &amp; 1 &amp; \cdots \\1 &amp; -1 &amp; -1 &amp; -1 &amp; -1 &amp; -1
&amp;\cdots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath}
The sequence $Q_n(-1,-1)$ begins
$$1, 0, 1, 0, 0, 0, 1, 0, -2, 0, 6, 0, -18, 0, 57, 0, -186, 0, 622,\ldots,$$
and represents the aeration of $C(-1;n)$ \seqnum{A064310}. The sequence $Q_n(-1,-1)$ has
Hankel transform equal to $(-1)^n$.
\begin{example} The case $r=-1$, $s=1$ is also linked to the Fibonacci numbers. For this, we find that the row sums
are $$1, 1, -1, -2, 2, 5, -5, -14, 14, 42, -42, -132, 132, 429, \ldots,$$ a sequence of signed doubled Catalan numbers.
The Hankel transform in this case is $(-1)^{\binom{n+1}{2}}F_{n+2}$.
\end{example}
\begin{example} Similarly, the case $r=-1$, $s=-1$ is also linked to the Fibonacci numbers. The row sums are given by
$$1, 1, 1, 0, 0, 1, 1, -2, -2, 6, 6, -18, -18, 57, 57, -186, -186, 622, \ldots,$$ a sequence of signed doubled
generalized Catalan numbers $C(-1;n)$ (\seqnum{A064310}), given by $C(-1; \lfloor \frac{n+1}{2} \rfloor)$. The Hankel transform in this case is
$$1, 0, -1, -1, 2, 3, -5, -8, 13, 21, -34,\ldots,$$ 
or $(-1)^{\binom{n}{2}}F_{n-1}$.
The matrix in question is
$$\left(\frac{2}{3-\sqrt{1-4x^2}},\frac{1+2x-\sqrt{1-4x^2}}{2}\right)=\left(\frac{1-2x-x^2}{1-2x},\frac{x(1-x)}{1-2x}\right)^{-1}.$$
\end{example}
\section{A series reversion}
It is interesting to study the reversion of $xg_Q(x)$, that is, to study the appropriate solution of the equation
$$u g_Q(u) =x.$$
We obtain
$$\frac{u(x)}{x}=\frac{s \sqrt{4rx^2(s - 1) + 1} + 2x(1 - s)(r + 1) - s + 2}{2(1 - x(r + 1)(s - 2) + (1 - s + r(s^2 - 2s + 2) - r^2(s - 1))x^2)}.$$
This again is the generating function of a sequence of polynomials in $r$ and $s$, which begins
$$1, -(r + 1), r^2 + r(2 - s) + 1, - r^3 + r^2(2s - 3) + r(2s - 3) - 1, r^4 + r^3(4 - 3s) + r^2(2s^2 - 7s + 6) + r(4 - 3s) + 1,\ldots$$ As polynomials in the variable $r$, these polynomials have coefficient array given by
\begin{displaymath}\bar{N}(s)=\left(\begin{array}{ccccccc} 1 &amp; 0 &amp;
0
&amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\-1 &amp; -1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\ 1 &amp; 2-s
&amp; 1 &amp; 0 &amp; 0 &amp;
0 &amp; \cdots \\ -1 &amp; 2s-3 &amp; 2s-3 &amp; -1 &amp; 0 &amp; 0 &amp; \cdots \\ 1  &amp; 4-3s &amp; 2s^2-7s+6
&amp; 4-3s &amp; 1 &amp; 0 &amp; \cdots \\-1 &amp; 4s-5 &amp; 2(7s-5)-5s^2 &amp; 2(7s-5)-5s^2 &amp; 4s-5 &amp; -1
&amp;\cdots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right),\end{displaymath}
which apart from signs, is also Pascal-like. Examples for $s=0,\ldots, 5$ of these matrices $\bar{N}(s)$ are given at the end of the last section. The central coefficients $M(s)$
$$1, 2 - s, 2s^2 - 7s + 6, - 5s^3 + 24s^2 - 38s + 20, 14s^4 - 84s^3 + 188s^2 - 187s + 70,\ldots$$ of this triangle are of interest. As polynomials in $s$, they can be expressed as
$$M(s)=\sum_{k=0}^n \binom{2n}{n-k}\frac{2k+1}{n+k+1}(-1)^{n-k}(s-1)^{n-k}=\sum_{k=0}^n \binom{2n-k-1}{n-k}\frac{k+0^{n+k}}{n+0^{n k}}(1-s)^{n-k}(2-s)^k. $$
The g.f. of $M(s)$ is given by

$$g_M(s)=\frac{1}{1+(s-2)x c((1-s)x)}=\frac{s-(s-2)\sqrt{1-(1-s)4x}}{2(1-(s-2)^2 x)}.$$ They are the moments of the family of orthogonal polynomials $R_n(x)$ defined by
$$R_n(x)=(x-2(1-s))R_{n-1}(x)-(s-1)^2 R_{n-2}(x), \quad n &gt;2,$$ with $R_0(x)=1$, $R_1(x)=x-(s-2)$ and $R_2(x)=x^2+x(3s-4)+(s-1)(s-2)$.  The coefficient array of this family of polynomials is given by the Riordan array
$$\left(\frac{1-sx-(1-s)x^2}{1+2(1-s)x+(s-1)^2 x^2},\frac{x}{1+2(1-s)x+(s-1)^2 x^2}\right).$$ The moments $M(s)$ are then the elements of the first column of the inverse of this array. They can be represented as
$$M_n(s)=\frac{1}{\pi} \int_{4(1-s)}^0 x^n \frac{(s-2)\sqrt{x(4(1-s)-x)}}{2x(x-s^2-4(1-s))}\,dx$$ for
$s \ne 1,2$. The Hankel transform of $M_n(s)$ is given by
$$h_n(s)=(s-2)^n (s-1)^{n^2}.$$
\section{A bijection between paths related to $P_L$ and $P_{\tilde{L}}$}

Returning to the production matrix introduced above, \begin{displaymath}P_L=\left(\begin{array}{ccccccc} r+1 &amp; 1 &amp;
0
&amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\sr &amp; r+1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\ 0 &amp; r
&amp; r+1 &amp; 1 &amp; 0 &amp;
0 &amp; \cdots \\ 0 &amp; 0 &amp; r &amp; r+1 &amp; 1 &amp; 0 &amp; \cdots \\ 0 &amp; 0 &amp; 0
&amp; r &amp; r+1 &amp; 1 &amp; \cdots \\0 &amp; 0 &amp; 0 &amp; 0 &amp; r &amp; r+1
&amp;\cdots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right),\end{displaymath}
we are now interested in these matrix entries and their corresponding Motzkin paths. It is known \cite{Flaj, Hennessy-Thesis, Viennot} that the elements of the first column of the Riordan array corresponding to $P_L$ count Motzkin paths returning to the $x$ axis. Motzkin paths related to the matrix $P_L$ above are referred to as $(r+1)$-colored Motzkin paths, where we have a choice of $(r+1)$ colors for each east step. Similarly, each south-east step has a choice of  $r$ colors. We note that for the matrix above, the south-east step returning to the $x$ axis has a choice of $rs$ colors. For this reason, we will refer to such Motzkin weighted paths as \emph{($s$-$e_{x-axis}, e_{x-axis}; s$-$e, e)$-Motzkin paths}. Thus, $P_L$ corresponds to a $(rs, r+1; r, r+1)$-Motzkin path. Similarly, it can be shown \cite{Hennessy-Thesis} that the first column of the  Riordan array corresponding to the production matrix
\begin{displaymath}P_{\tilde{L}}=\left(\begin{array}{ccccccc} r+1 &amp; 1 &amp;
0
&amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\rs &amp; r &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\ rs &amp; r
&amp; r &amp; 1 &amp; 0 &amp;
0 &amp; \cdots \\ rs &amp; r &amp; r &amp; r &amp; 1 &amp; 0 &amp; \cdots \\ rs &amp; r &amp; r
&amp; r &amp; r &amp; 1 &amp; \cdots \\rs &amp; r &amp; r &amp; r &amp; r &amp; r
&amp;\cdots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right)\end{displaymath} counts \L ukasiewicz paths returning to the $x$ axis. Adopting similar notation as for the Motzkin paths above, $P_{\tilde{L}}$ corresponds with a $(rs, r+1; r, r)$-\L ukasiewicz path. Now, as the first column of the Riordan arrays corresponding to $P_L$ and $P_{\tilde{L}}$ both have generating function $g_Q(x),$ we introduce the following bijection between Motzkin and \L ukasiewicz paths.
\begin{proposition}\label{bijection}
A bijection exists between the set of $(rs, r+1; r, r+1)$-Motzkin paths of length $n$ and the set of  $(rs, r+1; r, r)$-\L ukasiewicz paths of length $n,$ for any $n &gt;0$.
\end{proposition}
We choose an ordering of colors of the east steps. In the proof below we refer to the east steps that are colored with the $(r+1)$-th color as $e_{r+1}$
\begin{proof}
We construct a map $\phi: \mathbb{M} \to \mathbb{L}.$ Given a $(rs, r+1; r, r+1)$-Motzkin path $P$ of length $n,$ we can obtain a path $\phi(P)$ as follows:
\begin{itemize}
\item $i$-colored east steps remain unchanged, for $i&lt;r+1$.
\item East steps $e_{r+1},$ become \L ukasiewicz steps, taking the color of its associated north-east, south-east pair.
\end{itemize}
Conversely, given a $(rs, r+1; r, r)$-\L ukasiewicz path we can obtain a $(rs, r+1; r, r+1)$-Motzkin path by
\begin{itemize}
\item $i$-colored east steps remain unchanged, for $i&lt;r+1$.
\item \L ukasiewicz steps form the $(r+1)^{th}$ colored east steps, $e_{r+1}$.
\end{itemize}
$$ \begin{tikzpicture}[domain=0:2]
\draw[thin,color=gray,step=.5cm,
dashed] (0,0) grid (15,1.5);
\draw[dashed, -&gt;] (-.5,0) -- (15.5,0)
node[below right] {$y$ $axis$};
\draw [-](0,0) -- (.5,0.5);
\draw [-](0.5,0.5) --  (1,0.5);
\draw [-](1,0.5) -- (1.5,0);
\draw [-](1.5,0) -- node[above = 10pt] {$ (r+1)rs $}(1.5,0);
%\draw[-&gt;] (0,-1) -- (0,1.5)
node[left] {$$};
\color{black}
\draw[&lt;-&gt;] (2.5,.5) --  (3,.5);
%\draw[thin,color=gray,step=.5cm,
%dashed] (3.5,0) grid (7,1);
\draw [-](3.5,0) -- (4,0.5);
\draw [-](4,0.5) -- (4.5,1);
\draw [-](4.5,1) -- node[above = 5pt] {$rs$} (5,0);
\draw [-](5.5,0) -- (6,0.5);
\draw [-](6,0.5) -- (6.5,0.5);
\draw [-](6.5,0.5) -- node[above = 5pt] {$r^2s$} (7,0);
%\draw[-&gt;] (0,-1) -- (0,1.5)
node[left] {$$};
\color{black}
%\draw[thin,color=gray,step=.5cm,
%dashed] (7.5,0) grid (13,1);
\draw [-](8,0.5) -- (8.5,1);
\draw [-](8.5,1) --  (9,1);
\draw [-](9,1) -- (9.5,0.5);
\draw [-](9.5,0.5) -- node[above = 10pt] {$ (r+1)r $}(9.5,0.5);
\draw[&lt;-&gt;] (10.5,1) --  (11,1);
\draw [-](11.5,.5) -- (12,1);
\draw [-](12,1) -- (12.5,1.5);
\draw [-](12.5,1.5) -- node[above = 5pt] {$r$} (13,0.5);
\draw [-](13.5,0.5) -- (14,1);
\draw [-](14,1) -- (14.5,1);
\draw [-](14.5,1) -- node[above = 5pt] {$r^2$} (15,0.5);
%\draw[-&gt;] (0,-1) -- (0,1.5)
node[left] {$$};
\draw plot[id=x] function{x*x};
\end{tikzpicture}$$

We illustrate the \L ukasiewicz path formed from the mapping $\phi,$ for a Motzkin path with all east steps $e_{r+1}$. We see that each north-east step is paired with its right horizontally visible south-east step. All east steps at the height of the south-east step belong to that north-east, south-east pair. For a north-east, south-east pair, with m level steps, at height $(y+1),$ the \L ukasiewicz path becomes a north-east path of $(m+1)$ steps with a \L ukasiewicz step $\lambda_{(y+m+1,y)}.$ The \L ukasiewicz step now takes the choice of colors of the original south-east step. North-east, south-east pairs with no east steps remain unchanged. We note that it is clear from above that the length is preserved.

$$\begin{tikzpicture}[domain=0:2]
\draw[thin,color=gray,step=.5cm,
dashed] (0,0) grid (7.5,1);
\color{red}
\draw [-](0,0)--(.5,.5);
\color{black}
\draw [-](0.5,0.5)--(1,.5);
\draw [-](1,0.5)--(1.5,.5);
\draw [-](1.5,0.5)--(2,.5);
\color{blue}
\draw [-](2,0.5)--(2.5,1);
\draw [-](2.5,1)--(3,.5);
\color{black}
\draw [-](3,.5)--(3.5,.5);
\color{orange}
\draw [-](3.5,.5)--(4,1);
\color{black}
\draw [-](4,1)--(4.5,1);
\draw [-](4.5,1)--(5,1);
\color{orange}
\draw [-](5,1)--(5.5,.5);
\color{black}
\draw [-](5.5,.5)--(6,0.5);
\draw [-](6,0.5) -- (6.5,0.5);
\draw [-](6.5,0.5) -- (7,0.5);
\color{red}
\draw [-](7,0.5) -- (7.5,0);
\color{black}
\draw[&lt;-&gt;] (7.5,.5) -- (8,.5);
\draw[thin,color=gray,step=.5cm,
dashed] (8,0) grid (15.5,4);
\color{red}
\draw [-](8,0)--(8.5,.5);
\color{black}
\draw [-](8.5,.5)--(9,1) ;
\draw [-](9,1)--(9.5,1.5);
\draw [-](9.5,1.5)--(10,2);
\color{blue}
\draw [-](10,2)--(10.5,2.5);
\draw [-](10.5,2.5)--(11,2);
\color{black}
\draw [-](11,2)--(11.5,2.5);
\color{orange}
\draw [-](11.5,2.5)--(12,3);
\color{black}
\draw [-](12,3)--(12.5,3.5);
\draw [-](12.5,3.5)--(13,4);
\color{orange}
\draw [-](13,4)--(13.5,2.5);
\color{black}
\draw [-](13.5,2.5)--(14,3);
\draw [-](14,3)--(14.5,3.5);
\draw [-](14.5,3.5)--(15,4);
\color{red}
\draw [-](15,4) -- (15.5,0);
\color{black}
\draw plot[id=x] function{x*x};
\end{tikzpicture} $$

Conversely, beginning with the \L ukasiewicz path above, again we pair north-east steps with their right horizontally visible step, be that a south-east or \L ukasiewicz step. The north-east step, whose left vertex corresponds to the right vertex of a \L ukasiewicz step, remains a north-east step. All other north-east steps who are right horizontally paired with a \L ukasiewicz step, form level steps colored $e_{r+1}.$ North-east steps corresponding to right horizontally visible south-east steps remain unchanged, as do south-east steps.\\

Again, looking at the path above, the \L ukasiewicz path corresponds to the Motzkin path with all east steps $e_{r+1}$. Considering all possible colorings of the Motzkin path, we have $(r+1)^9 r^2 rs$ Motzkin paths. Now, from $\phi,$ counting all corresponding \L ukasiewicz paths, with a choice of nine level steps and $e_{r+1}$ forming \L ukasiewicz steps, gives $\sum_{i=0}^{9}\binom{9}{i}r^i r^2 rs$ paths. For any path, considering a choice of all $(r+1)$ colors for $m$ east steps at level $y$, belonging to a north-east, south-east step pair(not returning to the $x$ axis) we have $ (r+1)^m r.$ Now, $ (r+1)^m r = \sum_{j=0}^{m}\binom{m}{j}r^j r$ which counts all combinations of \L ukasiewicz paths with $i$-colored east steps, $i&lt;r+1$. (We note that for any $m$ level steps belonging to a south-east step which returns to the $x$-axis, we have $ (r+1)^m rs = \sum_{j=1}^{m}\binom{m}{j}r^j rs$ which counts all combinations of \L ukasiewicz paths with east steps with a choice of $r$ colors.)

%\noindent Thus for a Motzkin path of length $n$, with $m$ north-east steps, $m-t$ south-east steps, $t$ south-east steps returning to the $x$ axis and $l$  level steps on the $x$ axis, we have
%$$ \mu_n = (r+1)^l r^{m-t} (r+1)^{n-l-2m} (rs)^t.$$
%\noindent We can re-write $\mu_n $ above as
%$$ (r+1)^l r^{m-t} (rs)^t \sum_{j=0}^{n-l-2m} \binom{n-l-2m}{j}r^{j} ,$$
%which counts the number of \L ukasiewicz paths of length $n$ with a choice of $r$ colors for each level step, except level steps on the $x$ axis %which have a choice of $r+1$ colors, and $r$ colors for all south-east steps except for south-east steps returning to the $x$ axis, which have $rs$ colors to choose.
%$$ \begin{tikzpicture}
%\draw (0:0.7cm) -- (0:1.5cm)
%arc (0:45:1.5cm) -- (45:0.7cm);
%\end{tikzpicture}$$
\end{proof}

\begin{example}
For $r,s = 1$ we have the $(1,2;1,2)$-Motzkin paths and the $(1,2;1,1)$-\L ukasiewicz paths, both counted by the shortened Catalan numbers \seqnum{A000108}. Let us look at the paths corresponding to one such Motzkin path of length $n=8$, letting the level step colors be red and yellow, and choosing yellow to be $e_{2}.$ As the \L ukasiewicz steps retain the same colors as the north-east steps, we will only show the colorings of the level steps.
\end{example}

\begin{tiny}

$$\begin{tikzpicture}[domain=0:2]
\draw[thin,color=gray,step=.5cm,
dashed] (0,0) grid (4,1);
\draw [-](0,0)--(.5,.5);
\color{yellow}
\draw [-](0.5,0.5)--(1,.5);
\draw [-](1,0.5)--(1.5,.5);
\color{black}
\draw [-](1.5,.5)--(2,1);
\color{yellow}
\draw [-](2,1)--(2.5,1);
\color{black}
\draw [-](2.5,1)--(3,.5);
\color{yellow}
\draw [-](3,.5)--(3.5,.5);
\color{black}
\draw [-](3.5,.5)--(4,0);
\draw[&lt;-&gt;] (4.1,.5) -- (4.4,.5);
\draw[thin,color=gray,step=.5cm,
dashed] (4.5,0) grid (8.5,2.5);
\draw [-](4.5,0)--(5,.5);
\draw [-](5,.5)--(5.5,1) ;
\draw [-](5.5,1)--(6,1.5);
\color{black}
\draw [-](6,1.5)--(6.5,2);
\draw [-](6.5,2)--(7,2.5);
\color{black}
\draw [-](7,2.5)--(7.5,1.5);
\draw [-](7.5,1.5)--(8,2);
\color{black}
\draw [-](8,2)--(8.5,0);
\draw plot[id=x] function{x*x};
\end{tikzpicture}\quad \begin{tikzpicture}[domain=0:2]
\draw[thin,color=gray,step=.5cm,
dashed] (0,0) grid (4,1);
\draw [-](0,0)--(.5,.5);
\color{red}
\draw [-](0.5,0.5)--(1,.5);
\draw [-](1,0.5)--(1.5,.5);
\color{black}
\draw [-](1.5,.5)--(2,1);
\color{red}
\draw [-](2,1)--(2.5,1);
\color{black}
\draw [-](2.5,1)--(3,.5);
\color{red}
\draw [-](3,.5)--(3.5,.5);
\color{black}
\draw [-](3.5,.5)--(4,0);
\draw[&lt;-&gt;] (4.1,.5) -- (4.4,.5);
\draw[thin,color=gray,step=.5cm,
dashed] (4.5,0) grid (8.5,1);
\draw [-](4.5,0)--(5,.5);
\color{red}
\draw [-](5,.5)--(5.5,.5) ;
\draw [-](5.5,.5)--(6,.5);
\color{black}
\draw [-](6,.5)--(6.5,1);
\color{red}
\draw [-](6.5,1)--(7,1);
\color{black}
\draw [-](7,1)--(7.5,.5);
\color{red}
\draw [-](7.5,.5)--(8,.5);
\color{black}
\draw [-](8,.5)--(8.5,0);
\draw plot[id=x] function{x*x};
\end{tikzpicture}$$
$$\begin{tikzpicture}[domain=0:2]
\draw[thin,color=gray,step=.5cm,
dashed] (0,0) grid (4,1);
\draw [-](0,0)--(.5,.5);
\color{yellow}
\draw [-](0.5,0.5)--(1,.5);
\color{red}
\draw [-](1,0.5)--(1.5,.5);
\color{black}
\draw [-](1.5,.5)--(2,1);
\color{red}
\draw [-](2,1)--(2.5,1);
\color{black}
\draw [-](2.5,1)--(3,.5);
\color{red}
\draw [-](3,.5)--(3.5,.5);
\color{black}
\draw [-](3.5,.5)--(4,0);
\draw[&lt;-&gt;] (4.1,.5) -- (4.4,.5);
\draw[thin,color=gray,step=.5cm,
dashed] (4.5,0) grid (8.5,1.5);
\draw [-](4.5,0)--(5,.5);
\draw [-](5,.5)--(5.5,1) ;
\color{red}
\draw [-](5.5,1)--(6,1);
\color{black}
\draw [-](6,1)--(6.5,1.5);
\color{red}
\draw [-](6.5,1.5)--(7,1.5);
\color{black}
\draw [-](7,1.5)--(7.5,1);
\color{red}
\draw [-](7.5,1)--(8,1);
\color{black}
\draw [-](8,1)--(8.5,0);
\draw plot[id=x] function{x*x};
\end{tikzpicture}\quad \begin{tikzpicture}[domain=0:2]
\draw[thin,color=gray,step=.5cm,
dashed] (0,0) grid (4,1);
\draw [-](0,0)--(.5,.5);
\color{red}
\draw [-](0.5,0.5)--(1,.5);
\color{yellow}
\draw [-](1,0.5)--(1.5,.5);
\color{black}
\draw [-](1.5,.5)--(2,1);
\color{red}
\draw [-](2,1)--(2.5,1);
\color{black}
\draw [-](2.5,1)--(3,.5);
\color{red}
\draw [-](3,.5)--(3.5,.5);
\color{black}
\draw [-](3.5,.5)--(4,0);
\draw[&lt;-&gt;] (4.1,.5) -- (4.4,.5);
\draw[thin,color=gray,step=.5cm,
dashed] (4.5,0) grid (8.5,1.5);
\draw [-](4.5,0)--(5,.5);
\color{red}
\draw [-](5,.5)--(5.5,.5) ;
\color{black}
\draw [-](5.5,.5)--(6,1);
\color{black}
\draw [-](6,1)--(6.5,1.5);
\color{red}
\draw [-](6.5,1.5)--(7,1.5);
\color{black}
\draw [-](7,1.5)--(7.5,1);
\color{red}
\draw [-](7.5,1)--(8,1);
\color{black}
\draw [-](8,1)--(8.5,0);
\draw plot[id=x] function{x*x};
\end{tikzpicture}$$
$$\begin{tikzpicture}[domain=0:2]
\draw[thin,color=gray,step=.5cm,
dashed] (0,0) grid (4,1);
\draw [-](0,0)--(.5,.5);
\color{red}
\draw [-](0.5,0.5)--(1,.5);
\draw [-](1,0.5)--(1.5,.5);
\color{black}
\draw [-](1.5,.5)--(2,1);
\color{yellow}
\draw [-](2,1)--(2.5,1);
\color{black}
\draw [-](2.5,1)--(3,.5);
\color{red}
\draw [-](3,.5)--(3.5,.5);
\color{black}
\draw [-](3.5,.5)--(4,0);
\draw[&lt;-&gt;] (4.1,.5) -- (4.4,.5);
\draw[thin,color=gray,step=.5cm,
dashed] (4.5,0) grid (8.5,1.5);
\draw [-](4.5,0)--(5,.5);
\color{red}
\draw [-](5,.5)--(5.5,.5) ;
\draw [-](5.5,.5)--(6,.5);
\color{black}
\draw [-](6,.5)--(6.5,1);
\draw [-](6.5,1)--(7,1.5);
\color{black}
\draw [-](7,1.5)--(7.5,.5);
\color{red}
\draw [-](7.5,.5)--(8,.5);
\color{black}
\draw [-](8,.5)--(8.5,0);
\draw plot[id=x] function{x*x};
\end{tikzpicture}\quad \begin{tikzpicture}[domain=0:2]
\draw[thin,color=gray,step=.5cm,
dashed] (0,0) grid (4,1);
\draw [-](0,0)--(.5,.5);
\color{red}
\draw [-](0.5,0.5)--(1,.5);
\draw [-](1,0.5)--(1.5,.5);
\color{black}
\draw [-](1.5,.5)--(2,1);
\color{red}
\draw [-](2,1)--(2.5,1);
\color{black}
\draw [-](2.5,1)--(3,.5);
\color{yellow}
\draw [-](3,.5)--(3.5,.5);
\color{black}
\draw [-](3.5,.5)--(4,0);
\draw[&lt;-&gt;] (4.1,.5) -- (4.4,.5);
\draw[thin,color=gray,step=.5cm,
dashed] (4.5,0) grid (8.5,1);
\draw [-](4.5,0)--(5,.5);
\color{red}
\draw [-](5,.5)--(5.5,.5) ;
\draw [-](5.5,.5)--(6,.5);
\color{black}
\draw [-](6,.5)--(6.5,1);
\color{red}
\draw [-](6.5,1)--(7,1);
\color{black}
\draw [-](7,1)--(7.5,.5);
\draw [-](7.5,.5)--(8,1);
\color{black}
\draw [-](8,1)--(8.5,0);
\draw plot[id=x] function{x*x};
\end{tikzpicture}$$
$$\begin{tikzpicture}[domain=0:2]
\draw[thin,color=gray,step=.5cm,
dashed] (0,0) grid (4,1);
\draw [-](0,0)--(.5,.5);
\color{yellow}
\draw [-](0.5,0.5)--(1,.5);
\draw [-](1,0.5)--(1.5,.5);
\color{black}
\draw [-](1.5,.5)--(2,1);
\color{red}
\draw [-](2,1)--(2.5,1);
\color{black}
\draw [-](2.5,1)--(3,.5);
\color{red}
\draw [-](3,.5)--(3.5,.5);
\color{black}
\draw [-](3.5,.5)--(4,0);
\draw[&lt;-&gt;] (4.1,.5) -- (4.4,.5);
\draw[thin,color=gray,step=.5cm,
dashed] (4.5,0) grid (8.5,2);
\draw [-](4.5,0)--(5,.5);
\draw [-](5,.5)--(5.5,1) ;
\draw [-](5.5,1)--(6,1.5);
\color{black}
\draw [-](6,1.5)--(6.5,2);
\color{red}
\draw [-](6.5,2)--(7,2);
\color{black}
\draw [-](7,2)--(7.5,1.5);
\color{red}
\draw [-](7.5,1.5)--(8,1.5);
\color{black}
\draw [-](8,1.5)--(8.5,0);
\draw plot[id=x] function{x*x};
\end{tikzpicture}\quad \begin{tikzpicture}[domain=0:2]
\draw[thin,color=gray,step=.5cm,
dashed] (0,0) grid (4,1);
\draw [-](0,0)--(.5,.5);
\color{yellow}
\draw [-](0.5,0.5)--(1,.5);
\color{red}
\draw [-](1,0.5)--(1.5,.5);
\color{black}
\draw [-](1.5,.5)--(2,1);
\color{yellow}
\draw [-](2,1)--(2.5,1);
\color{black}
\draw [-](2.5,1)--(3,.5);
\color{red}
\draw [-](3,.5)--(3.5,.5);
\color{black}
\draw [-](3.5,.5)--(4,0);
\draw[&lt;-&gt;] (4.1,.5) -- (4.4,.5);
\draw[thin,color=gray,step=.5cm,
dashed] (4.5,0) grid (8.5,2);
\draw [-](4.5,0)--(5,.5);
\draw [-](5,.5)--(5.5,1) ;
\color{red}
\draw [-](5.5,1)--(6,1);
\color{black}
\draw [-](6,1)--(6.5,1.5);
\draw [-](6.5,1.5)--(7,2);
\color{black}
\draw [-](7,2)--(7.5,1);
\color{red}
\draw [-](7.5,1)--(8,1);
\color{black}
\draw [-](8,1)--(8.5,0);
\draw plot[id=x] function{x*x};
\end{tikzpicture}$$
$$\begin{tikzpicture}[domain=0:2]
\draw[thin,color=gray,step=.5cm,
dashed] (0,0) grid (4,1);
\draw [-](0,0)--(.5,.5);
\color{yellow}
\draw [-](0.5,0.5)--(1,.5);
\color{red}
\draw [-](1,0.5)--(1.5,.5);
\color{black}
\draw [-](1.5,.5)--(2,1);
\color{red}
\draw [-](2,1)--(2.5,1);
\color{black}
\draw [-](2.5,1)--(3,.5);
\color{yellow}
\draw [-](3,.5)--(3.5,.5);
\color{black}
\draw [-](3.5,.5)--(4,0);
\draw[&lt;-&gt;] (4.1,.5) -- (4.4,.5);
\draw[thin,color=gray,step=.5cm,
dashed] (4.5,0) grid (8.5,1.5);
\draw [-](4.5,0)--(5,.5);
\draw [-](5,.5)--(5.5,1) ;
\color{red}
\draw [-](5.5,1)--(6,1);
\color{black}
\draw [-](6,1)--(6.5,1.5);
\color{red}
\draw [-](6.5,1.5)--(7,1.5);
\color{black}
\draw [-](7,1.5)--(7.5,1);
\draw [-](7.5,1)--(8,1.5);
\color{black}
\draw [-](8,1.5)--(8.5,0);
\draw plot[id=x] function{x*x};
\end{tikzpicture}\quad \begin{tikzpicture}[domain=0:2]
\draw[thin,color=gray,step=.5cm,
dashed] (0,0) grid (4,1);
\draw [-](0,0)--(.5,.5);
\color{red}
\draw [-](0.5,0.5)--(1,.5);
\color{yellow}
\draw [-](1,0.5)--(1.5,.5);
\color{black}
\draw [-](1.5,.5)--(2,1);
\color{yellow}
\draw [-](2,1)--(2.5,1);
\color{black}
\draw [-](2.5,1)--(3,.5);
\color{red}
\draw [-](3,.5)--(3.5,.5);
\color{black}
\draw [-](3.5,.5)--(4,0);
\draw[&lt;-&gt;] (4.1,.5) -- (4.4,.5);
\draw[thin,color=gray,step=.5cm,
dashed] (4.5,0) grid (8.5,2);
\draw [-](4.5,0)--(5,.5);
\color{red}
\draw [-](5,.5)--(5.5,.5) ;
\color{black}
\draw [-](5.5,.5)--(6,1);
\color{black}
\draw [-](6,1)--(6.5,1.5);
\draw [-](6.5,1.5)--(7,2);
\color{black}
\draw [-](7,2)--(7.5,1);
\color{red}
\draw [-](7.5,1)--(8,1);
\color{black}
\draw [-](8,1)--(8.5,0);
\draw plot[id=x] function{x*x};
\end{tikzpicture}$$
$$\begin{tikzpicture}[domain=0:2]
\draw[thin,color=gray,step=.5cm,
dashed] (0,0) grid (4,1);
\draw [-](0,0)--(.5,.5);
\color{red}
\draw [-](0.5,0.5)--(1,.5);
\color{yellow}
\draw [-](1,0.5)--(1.5,.5);
\color{black}
\draw [-](1.5,.5)--(2,1);
\color{red}
\draw [-](2,1)--(2.5,1);
\color{black}
\draw [-](2.5,1)--(3,.5);
\color{yellow}
\draw [-](3,.5)--(3.5,.5);
\color{black}
\draw [-](3.5,.5)--(4,0);
\draw[&lt;-&gt;] (4.1,.5) -- (4.4,.5);
\draw[thin,color=gray,step=.5cm,
dashed] (4.5,0) grid (8.5,1.5);
\draw [-](4.5,0)--(5,.5);
\color{red}
\draw [-](5,.5)--(5.5,.5) ;
\color{black}
\draw [-](5.5,.5)--(6,1);
\color{black}
\draw [-](6,1)--(6.5,1.5);
\color{red}
\draw [-](6.5,1.5)--(7,1.5);
\color{black}
\draw [-](7,1.5)--(7.5,1);
\draw [-](7.5,1)--(8,1.5);
\color{black}
\draw [-](8,1.5)--(8.5,0);
\draw plot[id=x] function{x*x};
\end{tikzpicture}\quad \begin{tikzpicture}[domain=0:2]
\draw[thin,color=gray,step=.5cm,
dashed] (0,0) grid (4,1);
\draw [-](0,0)--(.5,.5);
\color{red}
\draw [-](0.5,0.5)--(1,.5);
\draw [-](1,0.5)--(1.5,.5);
\color{black}
\draw [-](1.5,.5)--(2,1);
\color{yellow}
\draw [-](2,1)--(2.5,1);
\color{black}
\draw [-](2.5,1)--(3,.5);
\color{yellow}
\draw [-](3,.5)--(3.5,.5);
\color{black}
\draw [-](3.5,.5)--(4,0);
\draw[&lt;-&gt;] (4.1,.5) -- (4.4,.5);
\draw[thin,color=gray,step=.5cm,
dashed] (4.5,0) grid (8.5,1.5);
\draw [-](4.5,0)--(5,.5);
\color{red}
\draw [-](5,.5)--(5.5,.5) ;
\draw [-](5.5,.5)--(6,.5);
\color{black}
\draw [-](6,.5)--(6.5,1);
\draw [-](6.5,1)--(7,1.5);
\color{black}
\draw [-](7,1.5)--(7.5,.5);
\draw [-](7.5,.5)--(8,1);
\color{black}
\draw [-](8,1)--(8.5,0);
\draw plot[id=x] function{x*x};
\end{tikzpicture}$$
$$\begin{tikzpicture}[domain=0:2]
\draw[thin,color=gray,step=.5cm,
dashed] (0,0) grid (4,1);
\draw [-](0,0)--(.5,.5);
\color{red}
\draw [-](0.5,0.5)--(1,.5);
\color{yellow}
\draw [-](1,0.5)--(1.5,.5);
\color{black}
\draw [-](1.5,.5)--(2,1);
\color{yellow}
\draw [-](2,1)--(2.5,1);
\color{black}
\draw [-](2.5,1)--(3,.5);
\color{yellow}
\draw [-](3,.5)--(3.5,.5);
\color{black}
\draw [-](3.5,.5)--(4,0);
\draw[&lt;-&gt;] (4.1,.5) -- (4.4,.5);
\draw[thin,color=gray,step=.5cm,
dashed] (4.5,0) grid (8.5,2);
\draw [-](4.5,0)--(5,.5);
\color{red}
\draw [-](5,.5)--(5.5,.5) ;
\color{black}
\draw [-](5.5,.5)--(6,1);
\color{black}
\draw [-](6,1)--(6.5,1.5);
\draw [-](6.5,1.5)--(7,2);
\color{black}
\draw [-](7,2)--(7.5,1);
\draw [-](7.5,1)--(8,1.5);
\color{black}
\draw [-](8,1.5)--(8.5,0);
\draw plot[id=x] function{x*x};
\end{tikzpicture}\quad \begin{tikzpicture}[domain=0:2]
\draw[thin,color=gray,step=.5cm,
dashed] (0,0) grid (4,1);
\draw [-](0,0)--(.5,.5);
\color{yellow}
\draw [-](0.5,0.5)--(1,.5);
\color{red}
\draw [-](1,0.5)--(1.5,.5);
\color{black}
\draw [-](1.5,.5)--(2,1);
\color{yellow}
\draw [-](2,1)--(2.5,1);
\color{black}
\draw [-](2.5,1)--(3,.5);
\color{yellow}
\draw [-](3,.5)--(3.5,.5);
\color{black}
\draw [-](3.5,.5)--(4,0);
\draw[&lt;-&gt;] (4.1,.5) -- (4.4,.5);
\draw[thin,color=gray,step=.5cm,
dashed] (4.5,0) grid (8.5,2);
\draw [-](4.5,0)--(5,.5);
\draw [-](5,.5)--(5.5,1) ;
\color{red}
\draw [-](5.5,1)--(6,1);
\color{black}
\draw [-](6,1)--(6.5,1.5);
\draw [-](6.5,1.5)--(7,2);
\color{black}
\draw [-](7,2)--(7.5,1);
\draw [-](7.5,1)--(8,1.5);
\color{black}
\draw [-](8,1.5)--(8.5,0);
\draw plot[id=x] function{x*x};
\end{tikzpicture}$$
$$\begin{tikzpicture}[domain=0:2]
\draw[thin,color=gray,step=.5cm,
dashed] (0,0) grid (4,1);
\draw [-](0,0)--(.5,.5);
\color{yellow}
\draw [-](0.5,0.5)--(1,.5);
\draw [-](1,0.5)--(1.5,.5);
\color{black}
\draw [-](1.5,.5)--(2,1);
\color{yellow}
\draw [-](2,1)--(2.5,1);
\color{black}
\draw [-](2.5,1)--(3,.5);
\color{red}
\draw [-](3,.5)--(3.5,.5);
\color{black}
\draw [-](3.5,.5)--(4,0);
\draw[&lt;-&gt;] (4.1,.5) -- (4.4,.5);
\draw[thin,color=gray,step=.5cm,
dashed] (4.5,0) grid (8.5,2.5);
\draw [-](4.5,0)--(5,.5);
\draw [-](5,.5)--(5.5,1) ;
\draw [-](5.5,1)--(6,1.5);
\color{black}
\draw [-](6,1.5)--(6.5,2);
\draw [-](6.5,2)--(7,2.5);
\color{black}
\draw [-](7,2.5)--(7.5,1.5);
\color{red}
\draw [-](7.5,1.5)--(8,1.5);
\color{black}
\draw [-](8,1.5)--(8.5,0);
\draw plot[id=x] function{x*x};
\end{tikzpicture}\quad \begin{tikzpicture}[domain=0:2]
\draw[thin,color=gray,step=.5cm,
dashed] (0,0) grid (4,1);
\draw [-](0,0)--(.5,.5);
\color{yellow}
\draw [-](0.5,0.5)--(1,.5);
\draw [-](1,0.5)--(1.5,.5);
\color{black}
\draw [-](1.5,.5)--(2,1);
\color{red}
\draw [-](2,1)--(2.5,1);
\color{black}
\draw [-](2.5,1)--(3,.5);
\color{yellow}
\draw [-](3,.5)--(3.5,.5);
\color{black}
\draw [-](3.5,.5)--(4,0);
\draw[&lt;-&gt;] (4.1,.5) -- (4.4,.5);
\draw[thin,color=gray,step=.5cm,
dashed] (4.5,0) grid (8.5,2);
\draw [-](4.5,0)--(5,.5);
\draw [-](5,.5)--(5.5,1) ;
\draw [-](5.5,1)--(6,1.5);
\color{black}
\draw [-](6,1.5)--(6.5,2);
\color{red}
\draw [-](6.5,2)--(7,2);
\color{black}
\draw [-](7,2)--(7.5,1.5);
\draw [-](7.5,1.5)--(8,2);
\color{black}
\draw [-](8,2)--(8.5,0);
\draw plot[id=x] function{x*x};
\end{tikzpicture}$$
\end{tiny}


%\end{example}



\section{Examples}
The table below lists the first six generalized Narayana triangles, in the sense of this article.
For $s=0$, we have the usual Pascal's triangle, \seqnum{A007318}. For $s=1$, we have the Narayana triangle \seqnum{A001263}, while
for $s=2$ we have \seqnum{A008459}, which has general term $\binom{n}{k}^2$.

\begin{tabular} {|c|c|c|} \hline  $\begin{array}{cccccc} 1 &amp; 0 &amp;
0
&amp; 0 &amp; 0 &amp; 0 \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\ 1 &amp; 2
&amp; 1 &amp; 0 &amp; 0 &amp; 0  \\ 1 &amp; 3 &amp; 3 &amp; 1 &amp; 0 &amp; 0 \\ 1  &amp; 4 &amp; 6
&amp; 4 &amp; 1 &amp; 0  \\1 &amp; 5 &amp; 10 &amp; 10 &amp; 5 &amp; 1
\end{array}$&amp; $\begin{array}{cccccc} 1 &amp; 0 &amp;
0
&amp; 0 &amp; 0 &amp; 0 \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\ 1 &amp; 3
&amp; 1 &amp; 0 &amp; 0 &amp; 0  \\ 1 &amp; 6 &amp; 6 &amp; 1 &amp; 0 &amp; 0 \\ 1  &amp; 10 &amp; 20
&amp; 10 &amp; 1 &amp; 0  \\1 &amp; 15 &amp; 50 &amp; 50 &amp; 15 &amp; 1
\end{array}$&amp; $\begin{array}{cccccc} 1 &amp; 0 &amp;
0
&amp; 0 &amp; 0 &amp; 0 \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\ 1 &amp; 4
&amp; 1 &amp; 0 &amp; 0 &amp; 0  \\ 1 &amp; 9 &amp; 9 &amp; 1 &amp; 0 &amp; 0 \\ 1  &amp; 16 &amp; 36
&amp; 16 &amp; 1 &amp; 0  \\1 &amp; 25 &amp; 100 &amp; 100 &amp; 25 &amp; 1
\end{array}$\\
$s=0$ &amp; $s=1$ &amp; $s=2$ \\ \hline
$\begin{array}{cccccc} 1 &amp; 0 &amp;
0
&amp; 0 &amp; 0 &amp; 0 \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\ 1 &amp; 5
&amp; 1 &amp; 0 &amp; 0 &amp; 0  \\ 1 &amp; 12 &amp; 12 &amp; 1 &amp; 0 &amp; 0 \\ 1  &amp; 22 &amp; 54
&amp; 22 &amp; 1 &amp; 0  \\1 &amp; 35 &amp; 160 &amp; 160 &amp; 35 &amp; 1
\end{array}$ &amp; $\begin{array}{cccccc} 1 &amp; 0 &amp;
0
&amp; 0 &amp; 0 &amp; 0 \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\ 1 &amp; 6
&amp; 1 &amp; 0 &amp; 0 &amp; 0  \\ 1 &amp; 15 &amp; 15 &amp; 1 &amp; 0 &amp; 0 \\ 1  &amp; 28 &amp; 74
&amp; 28 &amp; 1 &amp; 0  \\1 &amp; 45 &amp; 230 &amp; 230 &amp; 45 &amp; 1
\end{array}$&amp; $\begin{array}{cccccc} 1 &amp; 0 &amp;
0
&amp; 0 &amp; 0 &amp; 0 \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\ 1 &amp; 7
&amp; 1 &amp; 0 &amp; 0 &amp; 0  \\ 1 &amp; 18 &amp; 18 &amp; 1 &amp; 0 &amp; 0 \\ 1  &amp; 34 &amp; 96
&amp; 34 &amp; 1 &amp; 0  \\1 &amp; 55 &amp; 310 &amp; 310 &amp; 55 &amp; 1
\end{array}$\\
$s=3$ &amp; $s=4$ &amp; $s=5$ \\ \hline \end{tabular}
\newline\newline
\bigskip
\ \\
\bigskip
The next table lists some examples of the sequences $Q_n(r,s)$ for the values of
$r$ and $s$ shown.
\begin{center}\begin{tabular}{|c|c|c|}
\hline $(r,s)$
&amp; $Q_n(r,s)$ &amp; OEIS number   \\ \hline (0,1) &amp;  $1,1,1,1,1,1,\ldots$ &amp;
\seqnum{A000012} \\ \hline (1,1) &amp;
$1,2,5,14,42,132,\ldots$ &amp;
\seqnum{A000108}$(n+1)$  \\ \hline
(2,1) &amp;
$1,3,11,45,197,903,\ldots$ &amp; \seqnum{A001003}$(n+1)$  \\ \hline (3,1) &amp;  $1,4,19,100,562,3304,\ldots$ &amp; \seqnum{A007564}$(n+1)$  \\
\hline (4,1) &amp;  $1,5,29,185,1257,8925,\ldots$ &amp; \seqnum{A059231}$(n+1)$  \\
\hline (0,2) &amp;  $1,1,1,1,1,1,\ldots$ &amp;
\seqnum{A000012} \\ \hline (1,2) &amp;
$1,2,6,20,70,252,\ldots$ &amp;
\seqnum{A000984}  \\ \hline
(2,2) &amp;
$1,3,13,63,321,1683,\ldots$ &amp; \seqnum{A001850}  \\ \hline (3,2) &amp;  $1,4,22,136,886,5944,\ldots$ &amp; \seqnum{A069835}  \\
\hline (4,2) &amp;  $1,5,33,245,1921,15525,\ldots$ &amp; \seqnum{A084771} \\
\hline (1,3) &amp;  $1,2,7,26,100,392,\ldots$ &amp; \seqnum{A101850} \\
\hline (1,4) &amp;  $1,2,8,32,132,552,\ldots$ &amp; \seqnum{A155084} \\
\hline \end{tabular}\end{center}
\ \\
\newline
\ \\
\bigskip
Finally, we document the start of the matrices $\bar{N}(s)$ for $s=0,\ldots, 5$.

\bigskip

\begin{tabular} {|c|c|}
\hline  
$\begin{array}{cccccc} 
1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
-1 &amp; -1 &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\ 
1 &amp; 2 &amp; 1 &amp; 0 &amp; 0 &amp; 0  \\
-1 &amp; -3 &amp; -3 &amp; 1 &amp; 0 &amp; 0 \\ 
1  &amp; 4 &amp; 6 &amp; 4 &amp; 1 &amp; 0  \\
-1 &amp; -5 &amp; -10 &amp; -10 &amp; -5 &amp; -1
\end{array}$&amp;
$\begin{array}{cccccc} 
1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
-1 &amp; -1 &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\ 
1 &amp; 1 &amp; 1 &amp; 0 &amp; 0 &amp; 0  \\ 
-1 &amp; -1 &amp; -1 &amp; -1 &amp; 0 &amp; 0 \\ 
1  &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 0  \\
-1 &amp; -1 &amp; -1 &amp; -1 &amp; -1 &amp; -1
\end{array}$\\
$s=0$ &amp; $s=1$   \\ \hline
$\begin{array}{cccccc} 
1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
-1 &amp; -1 &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\ 
1 &amp; 0 &amp; 1 &amp; 0 &amp; 0 &amp; 0  \\ 
-1 &amp; 1 &amp; 1 &amp; -1 &amp; 0 &amp; 0 \\ 
1  &amp; -2 &amp; 0 &amp; -2 &amp; 1 &amp; 0  \\
-1 &amp; 3 &amp; -2 &amp; -2 &amp; 3 &amp; -1
\end{array}$&amp;
$\begin{array}{cccccc}
1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
-1 &amp; -1 &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\ 
1 &amp; -1 &amp; 1 &amp; 0 &amp; 0 &amp; 0  \\ 
-1 &amp; 3 &amp; 3 &amp; -1 &amp; 0 &amp; 0 \\ 
1  &amp; -5 &amp; 3 &amp; -5 &amp; 1 &amp; 0  \\
-1 &amp; 7 &amp; -13 &amp; -13 &amp; 7 &amp; -1
\end{array}$ \\
$s=2$ &amp; $s=3$  \\ \hline 
$\begin{array}{cccccc} 
1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
-1 &amp; -1 &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\ 1 &amp; -2
&amp; 1 &amp; 0 &amp; 0 &amp; 0  \\ -1 &amp; 5 &amp; 5 &amp; -1 &amp; 0 &amp; 0 \\ 1  &amp; -8 &amp; 10
&amp; -8 &amp; 1 &amp; 0  \\-1 &amp; 11 &amp; -34 &amp; -34 &amp; 11 &amp; -1
\end{array}$&amp; $\begin{array}{cccccc} 1 &amp; 0 &amp;
0
&amp; 0 &amp; 0 &amp; 0 \\-1 &amp; -1 &amp; 0 &amp; 0 &amp; 0 &amp; 0  \\ 1 &amp; -3
&amp; 1 &amp; 0 &amp; 0 &amp; 0  \\ -1 &amp; 7 &amp; 7 &amp; -1 &amp; 0 &amp; 0 \\ 1  &amp; -11 &amp; 21
&amp; -11 &amp; 1 &amp; 0  \\-1 &amp; 15 &amp; -65 &amp; -65 &amp; 15 &amp; -1
\end{array}$\\
$s=4$ &amp; $s=5$ \\ \hline \end{tabular}

\section{Acknowledgements} The author would like to  thank an anonymous referee for their informative remarks and bibliographical additions. 
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\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B83; Secondary
15B36, 42C05, 11C20, 15B05.

\noindent \emph{Keywords: } 
Narayana polynomial, Riordan array, production matrix, orthogonal
polynomial, Hankel transform.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000007},
\seqnum{A000012},
\seqnum{A000045},
\seqnum{A000108},
\seqnum{A000169},
\seqnum{A000984},
\seqnum{A001003},
\seqnum{A001263},
\seqnum{A001850},
\seqnum{A006125},
\seqnum{A007318},
\seqnum{A007564},
\seqnum{A008459},
\seqnum{A059231},
\seqnum{A064062},
\seqnum{A064310},
\seqnum{A069835},
\seqnum{A083667},
\seqnum{A084771},
\seqnum{A099169},
\seqnum{A101850},
\seqnum{A143464},
\seqnum{A155084},
\seqnum{A187021}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  November 9 2011;
revised version received  April 17 2012.
Published in {\it Journal of Integer Sequences}, April 20 2012.

\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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\end{document}

                                                                                

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