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\begin{center}
\vskip 1cm{\LARGE\bf Comparing Two Matrices of Generalized \\
\vskip .02in
Moments Defined by Continued Fraction 
\vskip .12in
Expansions} \vskip 1cm \large
Paul Barry\\
School of Science\\
Waterford Institute of Technology\\
Ireland\\
\href{mailto:pbarry@wit.ie}{\tt pbarry@wit.ie} \\

\end{center}
\vskip .2 in

\begin{abstract}
We study two matrices $N$ and $M$ defined by the parameters of
equivalent $S$- and $J$-continued fraction expansions, and compare them
by examining the product $N^{-1}M$. Using examples based on the Catalan
numbers, the little Schr\"oder numbers and powers of $q$, we indicate
that this matrix product is an object worthy of study. In the case of
the little Schr\"oder numbers, we find that the matrix $N$ has an
interleaved structure based on two Riordan arrays.
\end{abstract}

\section{Introduction}

In this note, we study two matrices whose elements may be considered to be generalized moments. The matrices are defined using the coefficients of simple Jacobi and Stieltjes continued fractions.

In familiar cases, these matrices are well-known, though this study examines them from a fresh perspective. It will be assumed that the reader is familiar with the basics of orthogonal polynomials \cite{Chihara, Gautschi, Szego}, Riordan arrays \cite{SGWW}, production matrices \cite{Prod1, Prod2, PW}, continued fractions \cite{Wall} and the interplay between these areas \cite{Barry_Meixner, Barry_Moment}.

Our point of departure is a sequence $a_n$, with $a_1=1$, whose elements are either integers or polynomials with integer coefficients.

We will use these numbers to define two lower-triangular matrices, which we then compare.

In both cases, the elements of the first column will be the sequence $\mu_n$ generated by the continued fraction

$$\cfrac{1}{1-\cfrac{a_1 x}{1-\cfrac{a_2x}{1-\cfrac{a_3 x}{1-\cdots}}}}.$$


We require that this sequence be Catalan-like, in the sense that we require all the Hankel determinants $|\mu_{i+j}|_{0 \le i,j \le n}$ to be non-zero.

An equivalence transformation ensures that the sequence $\mu_n$ is the same as that generated by
$$\cfrac{1}{1-a_1 x- \cfrac{a_1 a_2 x^2}{1-(a_2+a_3)x-\cfrac{a_3 a_4 x^2}{1-(a_4+a_5)x-\cdots}}}.$$
This exhibits $\mu_n$ as the moment sequence of the family of orthogonal polynomials $P_n(x)$ that satisfy
$$P_n(x)=(x-(a_{2n-2}+a_{2n-1}))P_{n-1}(x)-a_{2n-3}a_{2n-2}P_{n-2}(x),$$ with $P_0(x)=1$, $P_1(x)=x-a_1$.

The first matrix $M$ that we shall be interested in is the inverse of the matrix of coefficients of these polynomials.
The theory of production matrices and orthogonal polynomials tells us that this production matrix is given by

\begin{displaymath}\left(\begin{array}{ccccccc} a_1 & 1 &
0
& 0 & 0 & 0 & \ldots \\a_1 a_2 & a_2+a_3 & 1 & 0 & 0 & 0 & \ldots \\ 0 & a_3 a_4
& a_4+a_5 & 1 & 0 &
0 & \ldots \\ 0 & 0 & a_5 a_6 & a_6+a_7 & 1 & 0 & \ldots \\ 0 & 0 & 0
& a_7 a_8 & a_8+ a_9 & 1 & \ldots \\0 & 0 & 0 & 0 & a_9 a_{10} & a_{10}+a_{11}
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).\end{displaymath}
 The form of this production matrix ensures that the matrix $M$ generated by it will be lower-triangular with $1$'s on the diagonal. We obtain a matrix which begins

\begin{tiny}\begin{displaymath}\left(\begin{array}{cccccc}
1 & 0 & 0 & 0 & 0 & \ldots \\
\alpha & 1 & 0 & 0 & 0  & \ldots \\
\alpha(\alpha+\beta) & \alpha+\beta+\gamma & 1 & 0  & 0 & \ldots \\
\alpha((\alpha+\beta)^2+\beta \gamma) & (\alpha+\beta)^2+\beta \gamma+\gamma(\alpha+\beta+\gamma+\delta) & \alpha+\beta+\gamma+\delta+\epsilon  & 1 & 0 & \ldots \\

\alpha((\alpha+\beta)^3+\beta \gamma (2\alpha+2\beta+\gamma+\delta) & (\alpha+\beta+\gamma+\delta+\epsilon)^2-\alpha(\gamma+\delta+\epsilon)-\beta(\delta+\epsilon)-\epsilon(\gamma-\phi) & \cdots
& \alpha+\beta+\gamma+\delta+\epsilon+\phi+\psi & 1  & \ldots \\
\vdots &
\vdots & \vdots &  \vdots & \vdots &
\ddots\end{array}\right), \end{displaymath} \end{tiny}
where we have written $\alpha=a_1$, $\beta=a_2$, and so on.

In order to define the second matrix $N$, which again will have $\mu_n$ in the first column, we also use a production matrix.
To construct this production matrix, we have two alternative routes. The first one proceeds as follows; we take the inverse of the matrix
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0 & \ldots \\-a_1 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 0 & -a_2
& 1 & 0 & 0 &
0 & \ldots \\ 0 & 0 & -a_3 & 1 & 0 & 0 & \ldots \\ 0 & 0 & 0
& -a_4 & 1 & 0 & \ldots \\0 & 0 & 0 & 0 & -a_5 & 1
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right) \end{displaymath} to obtain the matrix

\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0 & \ldots \\a_1 & 1 & 0 & 0 & 0 & 0 & \ldots \\ a_1 a_2 & a_2
& 1 & 0 & 0 &
0 & \ldots \\ a_1 a_2 a_3 & a_2 a_3 & a_3 & 1 & 0 & 0 & \ldots \\ a_1 a_2 a_3 a_4 & a_2 a_3 a_4 & a_3 a_4
& a_4 & 1 & 0 & \ldots \\a_1 a_2 a_3 a_4 a_5 & a_2 a_3 a_4 a_5 & a_3 a_4 a_5 & a_4 a_5 & a_5 & 1
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right). \end{displaymath}
We now behead this matrix (we remove the first row) to obtain the following production matrix.
\begin{displaymath}\left(\begin{array}{ccccccc} a_1 & 1 & 0 & 0 & 0 & 0 & \ldots \\ a_1 a_2 & a_2
& 1 & 0 & 0 &
0 & \ldots \\ a_1 a_2 a_3 & a_2 a_3 & a_3 & 1 & 0 & 0 & \ldots \\ a_1 a_2 a_3 a_4 & a_2 a_3 a_4 & a_3 a_4
& a_4 & 1 & 0 & \ldots \\a_1 a_2 a_3 a_4 a_5 & a_2 a_3 a_4 a_5 & a_3 a_4 a_5 & a_4 a_5 & a_5 & 1
&\ldots\\
a_1 a_2 a_3 a_4 a_5 a_6 & a_2 a_3 a_4 a_5 a_6 &
a_3 a_4 a_5 a_6
& a_4 a_5 a_6 & a_5 a_6 & a_6 & \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right). \end{displaymath}
The form of this production matrix ensures that the matrix $N$ that it generates will be lower-triangular with $1$'s on the diagonal. The matrix $N$ that we seek then begins

\begin{displaymath}\scriptsize\left(\begin{array}{cccccc} 1 & 0 &
0
& 0 & 0 & \ldots \\ \alpha & 1 & 0 & 0 & 0  & \ldots \\ \alpha(\alpha+\beta) & \alpha+\beta
& 1 & 0  &
0 & \ldots \\ \alpha((\alpha+\beta)^2+\beta \gamma) & (\alpha+\beta)^2+\beta \gamma & \alpha+\beta+\gamma & 1 & 0 & \ldots \\

\alpha((\alpha+\beta)^3+\beta \gamma (2\alpha+2\beta+\gamma+\delta) & (\alpha+\beta)^3+\beta \gamma (2\alpha+2\beta+\gamma+\delta & (\alpha+\beta)^2+\gamma(\alpha+\delta)+(\beta+\gamma)^2
& \alpha+\beta+\gamma+\delta & 1  & \ldots \\
\vdots &
\vdots & \vdots &  \vdots & \vdots &
\ddots\end{array}\right), \end{displaymath} where we have used $\alpha=a_1$, $\beta=a_2$, and so on.

There is an alternative production matrix approach to the construction of $N$. Multiplying the $(n,k)$-th element of $N$ by
$$\prod_{j=1}^k a_j$$ produces a lower triangular matrix whose first column is the same as that of $N$, and whose production matrix takes the simple form of
\begin{displaymath}\left(\begin{array}{ccccccc} a_1 & a_1 &
0
& 0 & 0 & 0 & \ldots \\a_2 & a_2 & a_2 & 0 & 0 & 0 & \ldots \\ a_3 & a_3
& a_3 & a_3 & 0 &
0 & \ldots \\ a_4 & a_4 & a_4 & a_4 & a_4 & 0 & \ldots \\ a_5 & a_5 & a_5
& a_5 & a_5 & a_5 & \ldots \\a_6 & a_6 & a_6 & a_6 & a_6 & a_6
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).  \end{displaymath}
We can clearly reverse this process, starting with the sequence $a_n$, to produce $N$.

In order to compare the two matrices $M$ and $N$, it is natural to examine the product $N^{-1}M$.

\begin{example} \textbf{The Catalan matrices}. We let $a_n=1$. Thus we are interested in the sequence
generated by the continued fraction
$$\cfrac{1}{1-\cfrac{ x}{1-\cfrac{x}{1-\cfrac{ x}{1-\cdots}}}}.$$

This is the sequence of Catalan numbers $C_n = \frac{1}{n+1}\binom{2n}{n}$, \seqnum{A000108}. In this instance,
the production matrix of  $N$ for both methods of generation is given by
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 1 &
0
& 0 & 0 & 0 & \ldots \\1 & 1 & 1 & 0 & 0 & 0 & \ldots \\ 1 & 1
& 1 & 1 & 0 &
0 & \ldots \\ 1 & 1 & 1 & 1 & 1 & 0 & \ldots \\ 1 & 1 & 1
& 1 & 1 & 1 & \ldots \\1 & 1 & 1 & 1 & 1 & 1
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right),  \end{displaymath} and $N$ is the Riordan array
$$N=(c(x), xc(x))=(1-x, x(1-x))^{-1} \quad\quad \seqnum{A033184}.$$

The matrix $M$ is given by the Riordan array
$$M=(c(x), xc(x)^2)=\left(\frac{1}{1+x}, \frac{x}{(1+x)^2}\right)^{-1} \quad\quad \seqnum{A039599},$$ and the associated orthogonal polynomials are the
Chebyshev polynomials $U_n(\frac{x}{2})$. The production matrix of $(c(x), xc(x)^2)$ is given by
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 1 &
0
& 0 & 0 & 0 & \ldots \\1 & 2 & 1 & 0 & 0 & 0 & \ldots \\ 0 & 1
& 2 & 1 & 0 &
0 & \ldots \\ 0 & 0 & 1 & 2 & 1 & 0 & \ldots \\ 0 & 0 & 0
& 1 & 2 & 1 & \ldots \\0 & 0 & 0 & 0 & 1 & 2
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right),  \end{displaymath} corresponding to the generating function
$$\cfrac{1}{1-x-\cfrac{x^2}{1-2x-\cfrac{x^2}{1-2x-\cdots}}}$$ of $C_n$.

A straight-forward Riordan array calculation now shows that in this case,
$$N^{-1}\cdot M = (c(x), xc(x))^{-1} \cdot (c(x), xc(x)^2)= \left(1, \frac{x}{1-x}\right),$$ which is the shifted binomial matrix
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0 & \ldots \\0 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 0 & 1
& 1 & 0 & 0 &
0 & \ldots \\ 0 & 1 & 2 & 1 & 0 & 0 & \ldots \\ 0 & 1 & 3
& 3 & 1 & 0 & \ldots \\0 & 1 & 4 & 6 & 4 & 1
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).  \end{displaymath}


\end{example}
\begin{example} \textbf{The $q$-case}. We take the example of
$$a_n = \frac{q^n}{q}+\frac{(q-1)0^n}{q},$$ so that 
$a_1=1$, $a_2=q$, $a_3=q^2$, and so on. 


Starting with the matrix
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0 & \ldots \\-1 & 1 & 0 & 0 & 0 & 0 & \ldots \\ 0 & -q
& 1 & 0 & 0 &
0 & \ldots \\ 0 & 0 & -q^2 & 1 & 0 & 0 & \ldots \\ 0 & 0 & 0
& -q^3 & 1 & 0 & \ldots \\0 & 0 & 0 & 0 & -q^4 & 1
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right), \end{displaymath} we invert it and behead the resulting matrix to get the production matrix
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 1 &
0
& 0 & 0 & 0 & \ldots \\q & q & 1 & 0 & 0 & 0 & \ldots \\ q^3 & q^3
& q^2 & 1 & 0 &
0 & \ldots \\ q^6 & q^6 & q^5 & q^3 & 1 & 0 & \ldots \\ q^{10} & q^{10} & q^9
& q^7 & q^4 & 1 & \ldots \\q^{15} & q^{15} & q^{14} & q^{12} & q^9 & q^5
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right),  \end{displaymath} which we use to generate the matrix N:
\begin{displaymath}\scriptsize\left(\begin{array}{cccccc} 1 & 0 &
0
& 0 & 0 & \ldots \\1 & 1 & 0 & 0 & 0 & \ldots \\ q+1 & q+1
& 1 & 0 & 0 &
 \ldots \\ q^3+q^2+2q+1 & q^3+q^2+2q+1 & q^2+q+1 & 1 & 0 &  \ldots \\ q^6+q^5+2q^4+3q^3+3q^2+3q+1 & q^6+q^5+2q^4+3q^3+3q^2+3q+1 & q^5+q^4+2q^3+2q^2+2q+1
& q^3+q^2+q+1 & 1 &  \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right). \end{displaymath}

In the left column we recognize the $q$-Catalan sequence $\mu_n$ with generating function
$$\cfrac{1}{1-\cfrac{ x}{1-\cfrac{qx}{1-\cfrac{q^2 x}{1-\cdots}}}}.$$

Alternatively we may begin with the production matrix
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 1 &
0
& 0 & 0 & 0 & \ldots \\q & q & q & 0 & 0 & 0 & \ldots \\ q^2 & q^2
& q^2 & q^2 & 0 &
0 & \ldots \\ q^3 & q^3 & q^3 & q^3 & q^3 & 0 & \ldots \\ q^4 & q^4 & q^4
& q^4 & q^4 & q^4 & \ldots \\q^5 & q^5 & q^5 & q^5 & q^5 & q^5
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right).  \end{displaymath} Let $\tilde{N}$ be the matrix generated by this production matrix. Dividing column $k$ of the $\tilde{M}$ by
$$\prod_{i=1}^k a_i =\prod_{i=1} q^i=q^{\binom{k}{2}},$$ we recover the matrix $N$.

The generating function $$\cfrac{1}{1-\cfrac{ x}{1-\cfrac{qx}{1-\cfrac{q^2 x}{1-\cdots}}}}$$ is equivalent to
$$\cfrac{1}{1-x-\cfrac{qx^2}{1-(q+q^2)x-\cfrac{q^5 x^2}{1-(q^3+q^4)x-\cfrac{q^9x^2}{1-(q^5+q^6)x-\cdots}}}}.$$

This leads to the production matrix
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 1 &
0
& 0 & 0 & 0 & \ldots \\q & q+q^2 & 1 & 0 & 0 & 0 & \ldots \\ 0 & q^5
& q^3+q^4 & 1 & 0 &
0 & \ldots \\ 0 & 0 & q^9 & q^5+q^6 & 1 & 0 & \ldots \\ 0 & 0 & 0
& q^{13} & q^7+q^8 & 1 & \ldots \\0 & 0 & 0 & 0 & q^{17} & q^9+q^{10}
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right), \end{displaymath} which generates the matrix $M$, with first column equal to $\mu_n$.


The inverse of the matrix $M$ is the coefficient array of the orthogonal polynomials defined by
$$P_n(x)=(x-q^{2n-3}(1+q))P_{n-1}(x)-q^{4n-7}P_{n-2}(x),$$ where $P_0(x)=1$ and $P_1(x)=x-1$.

For $N^{-1} \cdot M$, we obtain the matrix that begins

\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0& \ldots \\0 & 1 & 0 & 0 & 0 & 0& \ldots \\ 0 & q^2
& 1 & 0 & 0 &
 0& \ldots \\ 0 & q^5 & q^3+q^4 & 1 & 0 & 0& \ldots \\ 0 & q^9 & q^7+q^8+q^9
& q^4+q^5+q^6& 1 & 0&  \ldots \\
0 & q^{14} & q^{12}+q^{13}+q^{14}+q^{15}
& q^9+q^{10}+q^{11}+q^{12}+q^{13}& q^5+q^6+q^7+q^8 & 1&  \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right). \end{displaymath}


Dividing each element $M_{n,k}$ of $M$ by
$$q^{\binom{n-k+2}{2}-1},$$ we obtain the matrix
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0& \ldots \\0 & 1 & 0 & 0 & 0 & 0& \ldots \\ 0 & 1
& 1 & 0 & 0 &
 0& \ldots \\ 0 & 1 & q(1+q) & 1 & 0 & 0& \ldots \\ 0 & 1 & q^2(1+q+q^2)
& q^2(1+q+q^2) & 1 & 0&  \ldots \\
0 & 1 & q^3(1+q+q^2+q^3)
& q^4(1+q+2q^2+q^3+q^4) & q^3(1+q+q^2+q^3) & 1&  \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right), \end{displaymath} which is the Hadamard product of the matrices
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0& \ldots \\0 & 1 & 0 & 0 & 0 & 0& \ldots \\ 0 & 1
& 1 & 0 & 0 &
 0& \ldots \\ 0 & 1 & 1+q & 1 & 0 & 0& \ldots \\ 0 & 1 & 1+q+q^2
& 1+q+q^2 & 1 & 0&  \ldots \\
0 & 1 & 1+q+q^2+q^3
& 1+q+2q^2+q^3+q^4 & 1+q+q^2+q^3 & 1&  \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right), \end{displaymath} and
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0& \ldots \\0 & 1 & 0 & 0 & 0 & 0& \ldots \\ 0 & 1
& 1 & 0 & 0 &
 0& \ldots \\ 0 & 1 & q & 1 & 0 & 0& \ldots \\ 0 & 1 & q^2
& q^2 & 1 & 0&  \ldots \\
0 & 1 & q^3
& q^4 & q^3 & 1&  \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right), \end{displaymath} where the first matrix is the
$q$-Riordan array ${n-1 \brack n-k}_q$ \cite{qAnalogue}, and the second matrix is a shifted version of the matrix
$q^{k(n-k)}$.

The production matrix of $N^{-1}\cdot M$ begins
\begin{displaymath}\left(\begin{array}{ccccccc}
0 & 1 & 0 & 0 & 0 & 0 & \ldots \\
0 & q^2 & 1 & 0 & 0 & 0 & \ldots \\
0 & q^5(q-1) & q^2(q^2+q-1) & 1 & 0 &
0 & \ldots \\ 0 & 0 & q^7(q^2-1) & q^3(q^3+q^2-1) & 1 & 0 & \ldots \\ 0 & 0 & 0
& q^{10}(q^3-1) & q^4(q^4+q^3-1) & 1 & \ldots \\0 & 0 & 0 & 0 & q^{13}(q^4-1) & q^5(q^5+q^4-1)
&\ldots\\
\vdots &
\vdots & \vdots & \vdots & \vdots & \vdots &
\ddots\end{array}\right), \end{displaymath} indicating that in this case, the inverse matrix $(N^{-1}\cdot M)^{-1}=M^{-1}\cdot N$ is the coefficient array of a family of orthogonal polynomials whose parameters are given by the production matrix above.

We look more closely at the case of $q=2$. We find that
\begin{displaymath}M=\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0& \ldots \\1 & 1 & 0 & 0 & 0 & 0& \ldots \\ 3 & 7
& 1 & 0 & 0 &
 0& \ldots \\ 17 & 77 & 31 & 1 & 0 & 0& \ldots \\ 171 & 1471 & 1333
& 127 & 1 & 0&  \ldots \\
3113 & 51653 & 98487
& 21717 & 511 & 1&  \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right), \end{displaymath} while
\begin{displaymath}N=\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0& \ldots \\1 & 1 & 0 & 0 & 0 & 0& \ldots \\ 3 & 3
& 1 & 0 & 0 &
 0& \ldots \\ 17 & 17 & 7 & 1 & 0 & 0& \ldots \\ 171 & 171 & 77
& 51 & 1 & 0&  \ldots \\
3113 & 3113 & 1471
& 325 & 31 & 1&  \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right). \end{displaymath}

Then
\begin{displaymath}N^{-1}\cdot M=\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0& \ldots \\0 & 1 & 0 & 0 & 0 & 0& \ldots \\ 0 & 4
& 1 & 0 & 0 &
 0& \ldots \\ 0 & 32 & 24 & 1 & 0 & 0& \ldots \\ 0 & 512 & 896
& 112 & 1 & 0&  \ldots \\
0 & 16384 & 61440
& 17920 & 480 & 1&  \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right). \end{displaymath}
\noindent Looking at the reduced matrix
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0& \ldots  \\  4
& 1 & 0 & 0 &
 0& 0& \ldots \\ 32 & 24 & 1 & 0 & 0& 0&\ldots \\ 512 & 896
& 112 & 1 & 0& 0& \ldots \\
16384 & 61440
& 17920 & 480 & 1& 0& \ldots \\
1048576 & 8126464
& 5079040 & 317440 & 1984& 1& \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right) \end{displaymath} we see that it is the moment array of the family of orthogonal polynomials whose parameters are given in the production matrix
\begin{displaymath}\left(\begin{array}{ccccccc} 4 & 1 &
0
& 0 & 0 & 0& \ldots \\16 & 20 & 1 & 0 & 0 & 0& \ldots \\ 0 & 384
& 88 & 1 & 0 &
 0& \ldots \\ 0 & 0 & 7168 & 368 & 1 & 0& \ldots \\ 0 & 0 & 0
& 122880 & 1504 & 1 &  \ldots \\
0 & 0 & 0
& 0 & 2031616 & 6080&  \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right). \end{displaymath}  

We deduce that the sequence
$1,4,32,512,16384,\ldots$ or
$2^{n(n+3)/2}$ \seqnum{A036442} has a generating function given by
$$\cfrac{1}{1-4x-\cfrac{16x^2}{1-20x-\cfrac{384x^2}{1-88x-\cfrac{7168x^2}{1-368x-\cdots}}}},$$ or equivalently,
$$\cfrac{1}{1-\cfrac{4x}{1-\cfrac{4x}{1-\cfrac{16x}{1-\cfrac{24x}{1-\cfrac{64x}{1-\cdots}}}}}}.$$


In this latter expression, the coefficients are given by the sequence
$$b(n)=2^{n+2}-2^{(n+1)/2}(1-(-1)^n).$$
The Hankel transform of $2^{n(n+3)/2}$ is then given by \cite{Kratt1, Kratt2, Layman}
$$h_n=\prod_{k=0}^{n-1} (b(2k+1)b(2k+2))^{n-k}.$$
A similar analysis can be carried out for $q^{n(n+3)/2}$.


\end{example}

\begin{example} \textbf{The little Schr\"oder numbers}. In this example, we take a base sequence $a_n$ given by
$$1,1,2,1,2,1,2,1,2,1,2,1,2,1,2,\cdots.$$ The sequence with generating function
$$\cfrac{1}{1-\cfrac{a_1x}{1-\cfrac{a_2x}{1-\cfrac{a_3x}{1-\cdots}}}}=\cfrac{1}{1-\cfrac{x}{1-\cfrac{2x}{1-\cfrac{x}{1-\cdots}}}}$$ is the sequence of little Schr\"oder numbers \seqnum{A001003}
$$1, 1, 3, 11, 45, 197, 903,\ldots.$$ 

These numbers are also generated by
$$\cfrac{1}{1-x-\cfrac{2x^2}{1-3x-\cfrac{2x^2}{1-3x-\cfrac{2x^2}{1-3x-\cdots}}}}.$$ \noindent
We have
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0& \ldots  \\  -1
& 1 & 0 & 0 &
 0& 0& \ldots \\ 0 & -2 & 1 & 0 & 0& 0&\ldots \\ 0 & 0
& -1 & 1 & 0& 0& \ldots \\
0 & 0
& 0 & -2 & 1& 0& \ldots \\
 & 0
& 0 & 0 & -1 & 1& \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right)^{-1}=\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0& \ldots  \\  1
& 1 & 0 & 0 &
 0& 0& \ldots \\ 2 & 2 & 1 & 0 & 0& 0&\ldots \\ 2 & 2
& 1 & 1 & 0& 0& \ldots \\
4 & 4
& 2 & 2 & 1& 0& \ldots \\
4 & 4
& 2 & 2 & 1 & 1& \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right), \end{displaymath} so that the matrix $N$ in this case begins
\begin{displaymath}\left(\begin{array}{ccccccc} \color{blue}{1} & 0 &
0
& 0 & 0 & 0& \ldots  \\  \color{blue}{1}
& \color{red}{1} & 0 & 0 &
 0& 0& \ldots \\ \color{blue}{3} & \color{red}{3} & \color{blue}{1} & 0 & 0& 0&\ldots \\ \color{blue}{11} & \color{red}{11}
& \color{blue}{4} & \color{red}{1} & 0& 0& \ldots \\
\color{blue}{45} & \color{red}{45}
& \color{blue}{17} & \color{red}{6} & \color{blue}{1} & 0& \ldots \\
\color{blue}{197} & \color{red}{197}
& \color{blue}{76} & \color{red}{31} & \color{blue}{7} & \color{red}{1}& \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right) \end{displaymath} with production matrix
\begin{displaymath}\left(\begin{array}{ccccccc} \color{blue}{1} & \color{red}{1} &
0
& 0 & 0 & 0& \ldots \\\color{blue}{2} & \color{red}{2} & \color{blue}{1} & 0 & 0 & 0& \ldots \\
\color{blue}{2} & \color{red}{2} & \color{blue}{1} & \color{red}{1} & 0 &
 0& \ldots \\
 \color{blue}{4} & \color{red}{4} & \color{blue}{2} & \color{red}{2} & \color{blue}{1} & 0& \ldots \\
 \color{blue}{4} & \color{red}{4} & \color{blue}{2} & \color{red}{2} & \color{blue}{1} & \color{red}{1} &  \ldots \\
\color{blue}{8} & \color{red}{8} & \color{blue}{4}
& \color{red}{4} & \color{blue}{2} & \color{red}{2} &  \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right). \end{displaymath} 


For instance, we have
$$\color{blue}{17}=\color{blue}{1}\cdot\color{red}{11}\color{black}{+}\color{blue}{1}\cdot\color{blue}{4}\color{black}{+}\color{blue}{2}\cdot\color{red}{1}\color{black}{+}\color{blue}{2}\cdot0\color{black}{+}\cdots,$$
and
$$\color{red}{45}=\color{red}{1}\cdot \color{blue}{11}\color{black}{+}\color{red}{2}\cdot \color{red}{11}\color{black}{+}\color{red}{2}\cdot\color{blue}{4}\color{black}{+}\color{red}{4}\cdot\color{red}{1}\color{black}{+}\cdots.$$

The matrix $M$ is given by
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0& \ldots  \\  1
& 1 & 0 & 0 &
 0& 0& \ldots \\ 3 & 4 & 1 & 0 & 0& 0&\ldots \\ 11 & 17
& 7 & 1 & 0& 0& \ldots \\
45 & 76
& 40 & 10 & 1& 0& \ldots \\
197 & 353
& 216 & 72 & 13 & 1& \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right). \end{displaymath} \noindent This is the Riordan array \seqnum{A172094}
$$\left(\frac{1+x-\sqrt{1-6x+x^2}}{4x}, \frac{1-3x-\sqrt{1-4x+x^2}}{4x}\right)=\left(\frac{1}{1+x}, \frac{x}{1+3x+2x^2}\right)^{-1},$$ with production matrix
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 1 &
0
& 0 & 0 & 0& \ldots  \\  2
&3 & 1 & 0 &
 0& 0& \ldots \\ 0 & 2 & 3 & 1 & 0& 0&\ldots \\ 0 & 0
& 2 & 3 & 1& 0& \ldots \\
0 & 0
& 0 & 2 & 3& 1& \ldots \\
0 & 0
& 0 & 0 & 2 & 3& \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right). \end{displaymath}

The matrix $N$ is a ``mixture" (in left to right interleaved fashion) \cite{Davenport} of this Riordan array and the related Riordan array
$$\left(1, \frac{1-3x-\sqrt{1-6x+x^2}}{4x}\right)=\left(1, \frac{x}{1+3x+2x^2}\right)^{-1},$$
 which has production matrix
\begin{displaymath}\left(\begin{array}{ccccccc} 0 & 1 &
0
& 0 & 0 & 0& \ldots  \\  0
&3 & 1 & 0 &
 0& 0& \ldots \\ 0 & 2 & 3 & 1 & 0& 0&\ldots \\ 0 & 0
& 2 & 3 & 1& 0& \ldots \\
0 & 0
& 0 & 2 & 3& 1& \ldots \\
0 & 0
& 0 & 0 & 2 & 3& \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right). \end{displaymath}
We have
\begin{displaymath}
N_{n,k}=\begin{cases}
[x^n] \frac{1+x-\sqrt{1-6x+x^2}}{4x}\left(\frac{1-3x-\sqrt{1-6x+x^2}}{4}\right)^{k/2},\text{if $k$ is even}\\
[x^n] x^k \left(\frac{1-3x-\sqrt{1-6x+x^2}}{4x^2}\right)^{(k+1)/2}, \text{if $k$ is odd}.
\end{cases} \end{displaymath}
We note that in like fashion, the matrix $N^{-1}$, which begins
\begin{displaymath}\left(\begin{array}{ccccccc} \color{blue}{1} & 0 &
0
& 0 & 0 & 0& \ldots  \\  \color{red}{-1}
& \color{red}{1} & 0 & 0 &
 0& 0& \ldots \\ 0 & \color{blue}{-3} & \color{blue}{1} & 0 & 0& 0&\ldots \\ 0 & \color{red}{1}
& \color{red}{-4} & \color{red}{1} & 0& 0& \ldots \\
0 & 0
& \color{blue}{7} & \color{blue}{-6} & \color{blue}{1}& 0& \ldots \\
0 & 0
& \color{red}{-1} & \color{red}{11} & \color{red}{-7} & \color{red}{1} & \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right), \end{displaymath} is a ``mixture" (in shifted alternate row fashion) of the two matrices
$$\left(\frac{1}{1+3x+2x^2},\frac{x}{1+3x+2x^2}\right)\quad \text{and}\quad \left(\frac{1}{1+x}, \frac{x}{1+3x+2x^2}\right).$$


For instance, the array $\left(\frac{1}{1+3x+2x^2},\frac{x}{1+3x+2x^2}\right)$ begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0& \ldots  \\  -3
& 1 & 0 & 0 &
 0& 0& \ldots \\ 7 & -6 & 1 & 0 & 0 & 0&\ldots \\ -15 & 23
& -9 & 1 & 0& 0& \ldots \\
31 & -72
& 48 & -12 & 1& 0& \ldots \\
-63 & 201
& -198 & 82 & -15 & 1& \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right), \end{displaymath}
 while the array $\left(\frac{1}{1+x}, \frac{x}{1+3x+2x^2}\right)$ begins

\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0& \ldots  \\  -1
& 1 & 0 & 0 &
 0& 0& \ldots \\ 1 & -4 & 1 & 0 & 0& 0&\ldots \\ -1 & 11
& -7 & 1 & 0& 0& \ldots \\
1 & -26
& 30 & -10 & 1& 0& \ldots \\
-1 & 57
& -102 & 58 & -13 & 1& \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right). \end{displaymath}


We have
\begin{displaymath}N^{-1}\cdot M=\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0& \ldots  \\  0
& 1 & 0 & 0 &
 0& 0& \ldots \\ 0 & 1 & 1 & 0 & 0& 0&\ldots \\ 0 & 2
& 3 & 1 & 0& 0& \ldots \\
0 & 2
& 5 & 4 & 1& 0& \ldots \\
0 & 4
& 12 & 13 & 6 & 1& \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right). \end{displaymath}
We find that the production matrix of the inverse of this matrix is given by
\begin{displaymath}\left(\begin{array}{ccccccc} 0 & 1 &
0
& 0 & 0 & 0& \ldots \\0 & -1 & 1 & 0 & 0 & 0& \ldots \\ 0 & 0
& -2 & 1 & 0 &
 0& \ldots \\ 0 & 0 & 0 & -1 & 1 & 0& \ldots \\ 0 & 0 & 0
& 0 & -2 & 1 &  \ldots \\
0 & 0 & 0
& 0 & 0 & -1 &  \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right). \end{displaymath} 

This is the beheading of the inverse of the matrix
\begin{displaymath}\left(\begin{array}{ccccccc} 1 & 0 &
0
& 0 & 0 & 0& \ldots  \\  0
& 1 & 0 & 0 &
 0& 0& \ldots \\ 0 & 1 & 1 & 0 & 0& 0&\ldots \\ 0 & 2
& 2 & 1 & 0& 0& \ldots \\
0 & 2
& 2 & 1 & 1& 0& \ldots \\
0 & 4
& 4 & 2 & 2 & 1& \ldots \\
\vdots &
\vdots & \vdots & \vdots & \vdots  & \vdots &
\ddots\end{array}\right). \end{displaymath}

The form of the production matrix of the inverse is reflected in the structure of $N^{-1}\cdot M$ as follows: the internal elements of each even row satisfy $$t_{i,j}=1\cdot t_{1-1,j-1}+1\cdot t_{i-1,j},$$ while for odd rows we have
$$t_{i,j}=1\cdot t_{i-1,j-1}+2\cdot t_{i-1,j}.$$
\noindent We remark that it is clear that the interleaved structure of $N$, based on two Riordan arrays, will be replicated in the case of any sequence $a_n$ of the form $1,1,r,1,r,1,r,1,\ldots$.

\end{example}
\section{Conclusion} Using the parameters of equivalent Stieltjes and Jacobi continued fractions, we have defined two matrices $N$ and $M$, and we have studied the product $N^{-1}M$ in three specific cases. In each case, some noteworthy results have emerged. We conclude that the matrix $N^{-1}M$ is worthy of further study.
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\bigskip
\hrule
\bigskip
\noindent 2010 {\it Mathematics Subject Classification}: Primary
15B36; Secondary 11B83, 15A09, 30B70, 42C05.


\noindent \emph{Keywords:} Matrix, Stieltjes continued fraction, Jacobi continued fraction, orthogonal polynomials, production matrix, Riordan array, Hankel transform

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000108},
\seqnum{A001003},
\seqnum{A033184},
\seqnum{A036442},
\seqnum{A039599},
\seqnum{A172094}.)

\bigskip
\hrule
\bigskip


\vspace*{+.1in}
\noindent
Received 
revised version received  
Published in {\it Journal of Integer Sequences},

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

                                                                                

