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\begin{center}
\vskip 1cm{\LARGE\bf  On the Hurwitz Transform of Sequences}  \vskip 1cm
\large
Paul Barry\\
School of Science\\
Waterford Institute of Technology\\
Ireland\\
\href{mailto:pbarry@wit.ie}{\tt pbarry@wit.ie}
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\begin{abstract} 
Based on classical concepts, we introduce and study the Hurwitz transform of sequences, relating this transform to the Hankel transform of sequences. We also define and study associated polynomials, including links to related families of orthogonal polynomials. Examples of these associated polynomials are given within the context of Riordan arrays.
\end{abstract}

\section{Introduction} Given a sequence $a_n$, we denote by $h_n$ the general term of the sequence
with $h_n=|a_{i+j}|_{0 \le i,j \le n}$. The sequence $h_n$ is called the Hankel transform of $a_n$ \cite{Kratt_1, Kratt_2, Layman}. This sequence of Hankel determinants has attracted much attention of late amongst those working in the area of integer and polynomial sequences in particular \cite{Chammam, Ismail, RadEff, Vein_Dale}. In this note we shall introduce the notion of a related Hurwitz transform, and we shall study some of its properties. As with the Hankel transform, this transform is based on classical results which have a rich literature. Part of this literature is captured in the review article by Holtz and Tyaglov \cite{Holtz}, which forms a good background to this note. Our Hurwitz transform will give rise to a sequence of determinant values, which can be related to the Hankel transform.

In the sequel, we shall be mainly concerned with integer sequences. Known integer sequences are often referred to by their OEIS number \cite{SL1, SL2}. For instance, the sequence of Catalan numbers $C_n=\frac{1}{n+1} \binom{2n}{n}$ is \seqnum{A000108}. Its generating function, defined by $\sum_{n=0}^{\infty} C_n x^n$, is equal to $c(x)=\frac{1-\sqrt{1-4x}}{2x}$. Its first elements are
$$1,1,2,5,14,42,132,\ldots.$$ This sequence finds many applications in combinatorics \cite{Stanley1, Stanley2}. It is the unique sequence whose Hankel transform, along with that of its first shift $C_{n+1}$, is the all $1$'s sequence \cite{Benjamin, Mays}. We use it in many of our examples, partly because of these properties.

We recall the following notational elements.
For an integer sequence $a_n$, that is, an element of $\mathbb{Z}^\mathbb{N}$, the power series
$f(x)=\sum_{k=0}^{\infty}a_n x^n$ 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
$a_n=[x^n]f(x)$ \cite{Merlini_MC}. For instance, $F_n=[x^n]\frac{x}{1-x-x^2}$ is the $n$-th Fibonacci number \seqnum{A000045}, while
$C_n=[x^n]\frac{1-\sqrt{1-4x}}{2x}$.  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 $f(\bar{f}(x))=x$.



\section{Definition of the Hurwitz transform}
We consider two sequences $a_n$ and $b_n$, and define the Hurwitz matrix of order $n$ defined by these sequences as follows. If $n$ is even, $n=2m$, then the Hurwitz matrix of order $n$ is defined to be the matrix
\begin{displaymath}\mathcal{H}_n=\left(\begin{array}{ccccccc} a_0 &amp; a_1 &amp; a_2 &amp; \cdots &amp; a_m &amp; \cdots &amp; a_{2m}  \\ b_0 &amp; b_1 &amp; b_2 &amp; \cdots &amp; b_m &amp; \cdots &amp; b_{2m}  \\ 0 &amp; a_0 &amp; a_1 &amp; \cdots &amp; a_{m-1} &amp;
\cdots &amp; a_{2m-1}  \\ 0 &amp; b_0 &amp; b_1 &amp; \cdots &amp; b_{m-1} &amp; \cdots &amp; b_{2m-1}  \\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots  \\ 0 &amp; 0 &amp; 0 &amp; \cdots &amp; a_0 &amp; \cdots &amp; a_{m}  \end{array}\right).\end{displaymath} If $n$ is odd, $n=2m+1$, then the Hurwitz matrix of order $n$ is defined to be the matrix
\begin{displaymath}\mathcal{H}_n=\left(\begin{array}{ccccccc} a_0 &amp; a_1 &amp; a_2 &amp; \cdots &amp; a_m &amp; \cdots &amp; a_{2m+1}  \\ b_0 &amp; b_1 &amp; b_2 &amp; \cdots &amp; b_m &amp; \cdots &amp; b_{2m+1}  \\ 0 &amp; a_0 &amp; a_1 &amp; \cdots &amp; a_{m-1} &amp;
\cdots &amp; a_{2m}  \\ 0 &amp; b_0 &amp; b_1 &amp; \cdots &amp; b_{m-1} &amp; \cdots &amp; b_{2m}  \\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots  \\ 0 &amp; 0 &amp; 0 &amp; \cdots &amp; a_0 &amp; \cdots &amp; a_{m+1}  \\0 &amp; 0  &amp; 0 &amp; \cdots &amp; b_0 &amp; \cdots &amp; b_{m+1} \end{array}\right).\end{displaymath} We shall call the sequence of determinants $\mathbb{H}_n=|\mathcal{H}_n|$ the \emph{Hurwitz transform} of the sequences $a_n$ and $b_n$ (in that order). We shall sometimes write $\mathbb{H}_n(a_n,b_n)$ or $\mathbb{H}_n(a,b)$ for the transform of $a_n$ and $b_n$, to make the dependence on $a_n$ and $b_n$ more explicit. By the definition, it is clear that if $a_n$ and $b_n$ are integer sequences, then $\mathbb{H}_n$ is an integer sequence. We have
$$\mathbb{H}_n(a,a)=a_0 0^n=\begin{cases} a_0, &amp; \text{if $n=0$} ; \\
0, &amp; n&gt;0.
\end{cases}$$

Note that we can express the general term $\mathcal{H}_{i,j}$ of the Hurwitz matrix in the following manner.

\begin{equation} \mathcal{H}_{i,j}= \begin{cases} 0, &amp; \text{if  $2j+2 \le i$}; \\ a_{j-\frac{i}{2}}, &amp; \text{if $2|i$}; \\ b_{j-\frac{i-1}{2}}, &amp; \text{otherwise}. \end{cases}
\end{equation}

\noindent We can associate a sequence $s_n$ with the two sequences $a_n$ and $b_n$ in the following manner. We can define $s_n$ implicity by the relations
$$a_n=\sum_{k=0}^n s_k b_{n-k},$$ which is a convolution equation for $s_n$. If $b_0 \ne 0$ (which we will assume henceforth), we have
$$s_n = [x^n] \frac{\sum_{j=0}^{\infty} a_j x^j}{\sum_{j=0}^{\infty} b_j x^j}.$$ That is, $s_n$ is the sequence whose generating function is the quotient of the generating function $f(x)=\sum_{n=0}^{\infty} a_n x^n$ of $a_n$ and of the generating function $g(x)=\sum_{n=0}^{\infty} b_n x^n$ of $b_n$.
Using generating functions allows us to express the elements of the Hurwitz matrix as follows.
\begin{equation} \mathcal{H}_{j,i}=\begin{cases} [x^i] x^{\frac{j}{2}}f(x), &amp; \text{if $2|j$}; \\ \\
[x^i] x^{\frac{j-1}{2}} g(x), &amp; \text{otherwise}.
\end{cases}
\end{equation}

We let $h_n$ denote the Hankel transform of $s_n$ and we let $h_n^*$ denote the Hankel transform of the shifted sequence $s_n^*=s_{n+1}$. Then we have the following proposition characterizing $\mathbb{H}_n$.
\begin{proposition} We have
$$\mathbb{H}_{2n}=b_0^{2n+1} h_n,\quad\quad \mathbb{H}_{2n+1}=(-1)^{n+1}b_0^{2n+2} h_n^*.$$
\end{proposition}
\begin{proof}
We take the case $n=2m$. Beginning with the original matrix, we carry out the following steps.
\begin{enumerate}
\item Factor $b_0$ out of column $0$.
\item Subtract $b_j$ times column $0$ from column $j$ for $1 \le j \le 2m$.
\item Factor $b_0$ out of column $1$.
\item Subtract $b_{j-1}$ times column $1$ from column $j$ for $2 \le j \le 2m$.
\item Factor $b_0$ out of column $2$.
\item Subtract $b_{j-2}$ times column $2$ from column $j$ for $3 \le j \le 2m$.
\item etc.
\end{enumerate}
One has now factored $b_0$ out $2m+1$ times, and the resulting matrix is
 \begin{displaymath}\left(\begin{array}{cccccccc}
 s_0 &amp; s_1 &amp; s_2 &amp; \cdots &amp; s_{m-1} &amp; s_m &amp; \cdots &amp; s_{2m} \\
 1 &amp; 0 &amp; 0 &amp; \cdots &amp; 0 &amp; 0 &amp; \cdots &amp; 0 \\
 0 &amp; s_0 &amp; s_1 &amp; \cdots &amp; s_{m-2} &amp; s_{m-1} &amp; \cdots &amp; s_{2m-1} \\
 0  &amp; 1 &amp; 0 &amp; \cdots &amp; 0 &amp;0&amp; \cdots  &amp; 0 \\
 0 &amp; 0 &amp; s_0 &amp; \cdots &amp; s_{m-3} &amp; s_{m-2} &amp; \cdots &amp; s_{2m-2} \\
 0 &amp; 0  &amp; 1 &amp; \cdots &amp; 0&amp; 0 &amp; \cdots &amp; 0\\
 \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots \\
  0 &amp; 0 &amp; 0 &amp; \cdots &amp; 1&amp; 0 &amp; \cdots &amp; 0 \\
 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 0&amp; s_0 &amp; \cdots &amp; s_{m}
 \end{array}\right).\end{displaymath}
Expanding iteratively along the
rows with a single 1 we get an ``upside-down" Hankel determinant
\begin{displaymath} \left|\begin{array}{ccc}
s_m &amp; \cdots &amp; s_{2m} \\
s_{m-1}&amp;  \cdots &amp; s_{2m-1} \\
\vdots &amp; \vdots &amp; \vdots \\
s_0 &amp; \cdots &amp; s_m \end{array}\right|. \end{displaymath}
Now looking at the case $n=2m+1$, we get the matrix
 \begin{displaymath}\left(\begin{array}{ccccccccc}
 s_0 &amp; s_1 &amp; s_2 &amp; \cdots &amp; s_{m-1} &amp; s_m &amp; s_{m+1} &amp; \cdots &amp; s_{2m+1} \\
 1 &amp; 0 &amp; 0 &amp; \cdots &amp; 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 0 \\
 0 &amp; s_0 &amp; s_1 &amp; \cdots &amp; s_{m-2} &amp; s_{m-1} &amp; s_m &amp;  \cdots &amp; s_{2m} \\
 0  &amp; 1 &amp; 0 &amp; \cdots &amp; 0 &amp;0&amp; 0 &amp; \cdots  &amp; 0 \\
 0 &amp; 0 &amp; s_0 &amp; \cdots &amp; s_{m-3} &amp; s_{m-2} &amp; s_{m-1} &amp; \cdots &amp; s_{2m-1} \\
 0 &amp; 0  &amp; 1 &amp; \cdots &amp; 0&amp; 0 &amp; 0 &amp; \cdots &amp; 0\\
 \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots \\

 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 0&amp; s_0 &amp; s_1 &amp; \cdots &amp; s_{m+1} \\
 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 0&amp; 1 &amp; 0 &amp; \cdots &amp; 0 \\
 \end{array}\right).\end{displaymath}
Again,  expanding iteratively along the
rows with a single 1 we get an ``upside-down" Hankel determinant
\begin{displaymath} \left|\begin{array}{ccc}
s_{m+1} &amp; \cdots &amp; s_{2m+1} \\
s_m &amp;  \cdots &amp; s_{2m} \\
\vdots &amp; \vdots &amp; \vdots \\
s_1 &amp; \cdots &amp; s_{m+1} \end{array}\right|. \end{displaymath}
Keeping track of signs now yields the result.



By looking at the transposed matrix, $\left(\mathcal{H}_{i,j}\right)^T$, we can interpret the above operations as follows, where we use the notation of Riordan arrays (see the section on Riordan arrays). We have
\begin{eqnarray*}
(g(x), x)^{-1}\cdot \left(\mathcal{H}_{i,j}\right)^T&amp;=&amp;\left(\frac{1}{g(x)}, x\right)\cdot \left(\mathcal{H}_{j,i}\right)\\
&amp;=&amp; \left(\frac{1}{g(x)}, x\right)\cdot \left(\begin{cases} [x^i] x^{\frac{j}{2}}f(x), &amp; \text{if $2|j$}; \\ \\
[x^i] x^{\frac{j-1}{2}} g(x), &amp; \text{otherwise}. \end{cases} \right)\\
&amp;=&amp; \left(\begin{cases} [x^i] x^{\frac{j}{2}}\frac{f(x)}{g(x)}, &amp; \text{if $2|j$}; \\ \\
[x^i] x^{\frac{j-1}{2}} \frac{g(x)}{g(x)}, &amp; \text{otherwise}.
\end{cases} \right)\\
&amp;=&amp; \left(\begin{cases} [x^i] x^{\frac{j}{2}} \frac{f(x)}{g(x)}, &amp; \text{if $2|j$}; \\ \\
[x^i] x^{\frac{j-1}{2}}, &amp; \text{otherwise}.
\end{cases} \right)\\
&amp;=&amp; \left(\begin{cases} [x^i] x^{\frac{j}{2}} s(x), &amp; \text{if $2|j$}; \\ \\
[x^i] x^{\frac{j-1}{2}}, &amp; \text{otherwise}.
\end{cases} \right).
\end{eqnarray*}
Transposing and taking the first $n+1$ rows and columns (for $n=2m$ and $n=2m+1$) brings us back to the above cases.
Note that the determinant of $(g(x),x)_n^{-1}$ is $\frac{1}{b_0^{n+1}}$.

\end{proof}

We thus have

\begin{displaymath} \left(\begin{array}{ccccc}
b_0 &amp; &amp; &amp; &amp;  \cdots \\
b_1 &amp; b_0 &amp; &amp; &amp; \cdots \\
b_2 &amp; b_1 &amp; b_0 &amp; &amp; \cdots \\
b_3 &amp; b_2 &amp; b_1 &amp; b_0 &amp; \cdots \\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots \end{array}\right)^{-1} \cdot \left(\begin{array}{ccccc}
a_0 &amp; b_0 &amp; 0&amp; 0&amp;  \cdots \\
a_1 &amp; b_1 &amp; a_0&amp; b_0&amp; \cdots \\
a_2 &amp; b_2 &amp; a_1 &amp; b_1&amp; \cdots \\
a_3 &amp; b_3 &amp; a_2 &amp; b_2 &amp; \cdots \\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots \end{array}\right)=\left(\begin{array}{ccccc}
s_0 &amp; 1&amp; 0&amp; 0&amp;  \cdots \\
s_1 &amp; 0 &amp; s_0 &amp; 1&amp; \cdots \\
s_2 &amp; 0 &amp; s_1 &amp;0 &amp; \cdots \\
s_3 &amp; 0 &amp; s_2 &amp; 0 &amp; \cdots \\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots \end{array}\right) \end{displaymath} and hence
\begin{displaymath}
\left(\begin{array}{ccccc}
a_0 &amp; b_0 &amp; 0&amp; 0&amp;  \cdots \\
a_1 &amp; b_1 &amp; a_0&amp; b_0&amp; \cdots \\
a_2 &amp; b_2 &amp; a_1 &amp; b_1&amp; \cdots \\
a_3 &amp; b_3 &amp; a_2 &amp; b_2 &amp; \cdots \\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots \end{array}\right)=\left(\begin{array}{ccccc}
b_0 &amp; &amp; &amp; &amp;  \cdots \\
b_1 &amp; b_0 &amp; &amp; &amp; \cdots \\
b_2 &amp; b_1 &amp; b_0 &amp; &amp; \cdots \\
b_3 &amp; b_2 &amp; b_1 &amp; b_0 &amp; \cdots \\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots \end{array}\right) \cdot \left(\begin{array}{ccccc}
s_0 &amp; 1&amp; 0&amp; 0&amp;  \cdots \\
s_1 &amp; 0 &amp; s_0 &amp; 1&amp; \cdots \\
s_2 &amp; 0 &amp; s_1 &amp;0 &amp; \cdots \\
s_3 &amp; 0 &amp; s_2 &amp; 0 &amp; \cdots \\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots \end{array}\right). \end{displaymath}
\begin{example} It is possible to gain further insight into this result by using Gaussian elimination in the following way.
Let $s(x)$ be the g.f. of $s_n$. Then we have
$$s(x)=\frac{f(x)}{g(x)} \Rightarrow f(x)=s(x)g(x).$$ That is
$$\sum_{n=0}^{\infty} a_n x^n = \left(\sum_{n=0}^{\infty}s_n x^n\right)\left(\sum_{n=0}^{\infty}b_n x^n\right)=\sum_{n=0}^{\infty} \sum_{k=0}^n s_k b_{n-k} x^n,$$ or
$$a_n=\sum_{k=0}^n s_k b_{n-k}.$$
We now substitute for $a_n$ in the definitions of $\mathcal{H}_n$. For instance, we get
 \begin{displaymath}\mathcal{H}_n=\left(\begin{array}{ccccccc} s_0 b_0 &amp; s_0 b_1+s_1 b_0 &amp; s_0 b_2+ s_1 b_1+ s_2 b_0 &amp; \cdots &amp; s_0 b_m+\ldots &amp; \cdots &amp; s_0 b_{2m}+\ldots  \\ b_0 &amp; b_1 &amp; b_2 &amp; \cdots &amp; b_m &amp; \cdots &amp; b_{2m}  \\ 0 &amp; s_0 b_0 &amp; s_0 b_1+s_1 b_0 &amp; \cdots &amp; s_0 b_{m-1}+\ldots &amp;
\cdots &amp; s_0 b_{2m-1}+\ldots  \\ 0 &amp; b_0 &amp; b_1 &amp; \cdots &amp; b_{m-1} &amp; \cdots &amp; b_{2m-1}  \\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots  \\ 0 &amp; 0 &amp; 0 &amp; \cdots &amp; s_0 b_0 &amp; \cdots &amp; s_0 b_{m+1}+\ldots  \end{array}\right)\end{displaymath} for $n=2m$.

\noindent For instance, assuming that $s_0 \ne 0$, $s_1 \ne 0$, we have
$$\mathbb{H}_2 = \left|\begin{array}{ccc}
s_0 b_0 &amp; s_0 b_1+s_1 b_0 &amp; s_0 b_2+s_1 b_1+s_2 b_0 \\
b_0 &amp; b_1 &amp; b_2 \\
0 &amp; s_0 b_0 &amp; s_0 b_1+s_1 b_0 \end{array}\right| $$

$$=
\frac{1}{s_0} \left|\begin{array}{ccc}
s_0b_0 &amp; s_0 b_1+s_1 b_0 &amp; s_0 b_2+s_1 b_1+s_2 b_0 \\
s_0 b_0 &amp; s_0 b_1 &amp; s_0 b_2 \\
0 &amp; s_0 b_0 &amp; s_0 b_1+s_1 b_0 \end{array}\right| $$

$$=
\frac{1}{s_0} \left|\begin{array}{ccc}
s_0 b_0 &amp; s_0 b_1+s_1 b_0 &amp; s_0 b_2+s_1 b_1+s_2 b_0 \\
0 &amp; -s_1 b_0 &amp; -s_1 b_1-s_2 b_0 \\
0 &amp; s_0 b_0 &amp; s_0 b_1+s_1 b_0 \end{array}\right| $$

$$= \frac{1}{s_0^2 s_1} \left|\begin{array}{ccc}
s_0 b_0 &amp; s_0 b_1+s_1 b_0 &amp; s_0 b_2+s_1 b_1+s_2 b_0 \\
0 &amp; -s_0 s_1 b_0 &amp; -s_0 s_1 b_1-s_0 s_2 b_0 \\
0 &amp; 0  &amp; (s_1^2-s_0 s_2)b_0 \end{array}\right| $$

$$= b_0^3 (s_0 s_2-s_1^2) $$

$$= b_0^3 \left|\begin{array}{cc} s_0 &amp; s_1 \\
s_1 &amp; s_2 \end{array}\right|.$$

\noindent Similarly, we obtain

$$\mathbb{H}_3=\frac{1}{s_0^4s_1^2(s_1^2-s_0s_2)}\left|\begin{array}{cccc}
s_0 b_0 &amp; \ldots &amp; \ldots &amp; \ldots \\
0 &amp; -s_0s_1 b_0 &amp; \ldots &amp; \ldots \\
0 &amp; 0 &amp; s_0s_1(s_1^2-s_0s_2)b_0 &amp; \ldots \\
0 &amp; 0 &amp; 0 &amp; s_0 s_1 (s_2^2-s_1 s_3) b_0 \end{array}\right|,$$
and hence we have
$$\mathbb{H}_3= b_0^4 (s^2_2-s_1 s_3)=b_0^4 \left| \begin{array}{cc} s_1 &amp; s_2 \\ s_2 &amp; s_3\end{array}\right|.$$
\end{example}

\noindent Finally we note that

$$\mathbb{H}_n(\alpha a_n, \beta b_n)=\alpha^{\lfloor \frac{n+2}{2} \rfloor}\beta^{\lfloor \frac{n+1}{2} \rfloor} \mathbb{H}_n(a_n, b_n).$$

\begin{example} It is well-known that the Hankel transform of the Catalan numbers $C_n$, along with that of the shifted sequence $C_{n+1}$, is given by the all $1$'s sequence.
We thus turn to the Catalan numbers to provide an example of a pair of sequences whose Hurwitz transform is the all $1$'s sequence.
We recall that the sequence $C_n$ has generating function
$$c(x)=\frac{1-\sqrt{1-4x}}{2x}.$$

\noindent We define $C_{-1}=0$. Then the Hurwitz transform of the pair
$$a_n=(-1)^n (C_n+C_{n-1}), \quad \quad b_n=(-1)^n \binom{1}{n}$$ is such that $$\mathbb{H}_n=1 \quad \text{for all $n$.}$$
\noindent This follows since in this case,
$$f(x)=(1-x)c(-x), \quad \quad g(x)=1-x.$$ Then
$$\frac{f(x)}{g(x)}=\frac{(1-x)c(-x)}{1-x}=c(-x),$$ which is the generating function of $(-1)^n C_n$. The Hankel transform of $(-1)^n C_n$ is $1,1,1,\ldots$ while that of $(-1)^{n+1} C_{n+1}$ is $(-1)^{n+1}$, hence the result.

It is clear that any pair of sequences $a_n$, $b_n$ such that $\frac{f(x)}{g(x)}=c(-x)$ will furnish a Hurwitz transform consisting of the all $1$'s sequence. Thus, as with the Hankel transform, the Hurwitz transform is not injective.
\end{example}
\begin{example}
We now look at an example where $b_0 \ne 1$. Thus we take
$$a_n=C_n, \quad b_n=C_n+C_{n+1}, $$ with
$$f(x)=c(x),\quad g(x)=c(x)+c(x)^2, \quad s(x)=\frac{1}{1+c(x)}=\frac{1+2x+\sqrt{1-4x}}{2(x+2)}.$$
\noindent We have $b_0=2$. In this case, we find that
$$s_n=\frac{1}{2^{n+1}} \sum_{k=0}^{n-1} \frac{n-k}{n} \binom{n+k-1}{k}2^k,$$ which begins
$$\frac{1}{2}, -\frac{1}{4}, -\frac{3}{8}, -\frac{13}{16}, -\frac{67}{32}, -\frac{381}{64},\ldots.$$
We find that
$$2^{2n+1} h_n=(-1)^n (n+1),\quad 2^{2n+2}(-1)^{n+1}h_n^*=1,$$ and so the Hurwitz transform $\mathbb{H}_n(C_n, C_n+C_{n+1})$ is given by
$$1, 1, -2, 1, 3, 1, -4, 1, 5, 1, -6,\ldots.$$
\end{example}
\begin{example} This example  uses the Motzkin numbers \seqnum{A001006}
$$M_n=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} C_k,$$ with generating function
$$\frac{1-x-\sqrt{1-2x-3x^2}}{2x^2}.$$
We let $$a_n=M_n,\quad b_n=M_n+ M_{n+1}.$$
We find that $s_n$ has the generating function
$$\frac{1+3x+\sqrt{1-2x-3x^2}}{6x+4},$$ and begins
$$\frac{1}{2}, -\frac{1}{4}, -\frac{1}{8}, -\frac{5}{16}, -\frac{17}{32}, -\frac{77}{64},\ldots.$$
We find that $2^{2n+1} h_n$ is the periodic sequence that begins
$$1, -1, 0, 1, -1, 0,1, -1, 0, \ldots,$$ while $2^{2n+2}(-1)^{n+1}h_n^*$ is the all $1$'s sequence.
Thus we find that $\mathbb{H}_n(M_n, M_n+M_{n+1})$ is the periodic sequence
$$1, 1, -1, 1, 0, 1, 1, 1, -1, 1, 0, 1, 1, 1, -1, 1, 0, 1, 1, 1, -1,\ldots,$$ with generating function
$$\frac{1+x-x^2+x^3+x^5}{1-x^6}.$$
\end{example}
\begin{example} We define the Hurwitz transform of a single sequence $a_n$ to be the Hurwitz transform of the pair $(a_n, 0^n)$. In this example, we take $a_n$ to be the sequence \seqnum{A025262}$(n+1)$, which begins
$$ 1, 1, 3, 8, 23, 68, 207, 644, 2040, 6558, 21343, \ldots.$$ This sequence has generating function
$$f(x)=\frac{1-2x-\sqrt{1-4x+4x^3}}{2x^2}.$$
Its Hankel transform $h_n$ is an example of a Somos-$4$ \cite{Chang, Gale, Xin} sequence. This means that it satisfies the recurrence
$$h_{n-1}h_{n-3}+h_{n-2}^2= h_n h_{n-4}, \quad n \ge 3.$$ In this case, $h_n$ begins
$$1, 2, 3, 7, 23, 59, 314, 1529, 8209, 83313, 620297,\ldots.$$ This is \seqnum{A006720}$(n+3)$.

More generally, we say that a sequence $e_n$ is a $(\alpha, \beta)$ Somos-$4$ sequence if we have
$$\alpha h_{n-1}h_{n-3}+\beta h_{n-2}^2= h_n h_{n-4}, \quad n \ge 4.$$

\noindent Now $h_n^*$ begins $$1, -1, -5, -4, 29, 129, -65, -3689, -16264, 113689, 2382785,\ldots,$$  and hence
$\mathbb{H}_n$ begins
$$1, -1, 2, -1, 3, 5, 7, -4, 23, -29, 59, 129, 314, \ldots.$$

\noindent Numerical evidence suggests that $\mathbb{H}_n$ is then a $(-1,1)$ Somos-$4$ sequence.

\end{example}
\begin{example} We let $a_n$ be the sequence \seqnum{A160702}$(n+1)$. This sequence begins
$$1, 1, 5, 19, 79, 333, 1441, 6351, 28451, 129185,\ldots,$$ and its Hankel transform $h_n$  is a $(4,24)$ Somos-$4$ sequence, as is the Hankel transform $h_n^*$ of $a_{n+1}$. We can then conjecture that the Hurwitz transform of $a_n$ is a $(-2,2)$ Somos-$4$ sequence. $\mathbb{H}_n$ begins
$$1, -1, 4, -6, 20, 88, 464, 512, 17024, -173568, 1632256,\ldots,$$ and our claim is that
$$(-2)\mathbb{H}_{n-1}\mathbb{H}_{n-3}+2 \mathbb{H}_{n-2}^2= \mathbb{H}_n \mathbb{H}_{n-4}, \quad n \ge 4.$$
We are not at present able to prove this assertion.
\end{example}
\begin{example} We finish this section with an example which recalls the use of the Hurwitz matrix to determine if a polynomial is stable. We let
$e_n=\binom{n}{\frac{n}{2}}$, and we set
$$a_n=e_{2n+1}=\binom{2n+1}{n+1},\quad b_n=e_{2n}=\binom{2n}{n}.$$
We find that the Hurwitz transform $\mathbb{H}_n(e_{2n+1},e_{2n})$ in this case is given by
$$1, -1, 1, 1, 1, -1, 1, 1, 1, -1, 1,\ldots.$$
\end{example}


\section{Hurwitz associated polynomials}
One important application of Hankel determinants is in the construction of orthogonal polynomials \cite{Chihara, Szego}, where the Hankel determinants in question have elements that are the moments of the density associated with the orthogonal polynomials. We now use the Hurwitz matrix to construct families of polynomials, which we then relate to the polynomials defined by $s_n$ and $s_n^*$ by the Hankel construction.

We let

$$P_n^{(s)} (x)=\Delta_n^{(s)}(1,x,x^2,\cdots,x^n)=\left|\begin{array}{ccccc} s_0 &amp; s_1&amp;s_2 &amp;\cdots &amp; s_n \\
s_1 &amp; s_2&amp;s_3 &amp;\cdots &amp; s_{n+1} \\
\vdots &amp; \vdots &amp; \vdots &amp;\cdots &amp; \vdots \\
s_{n-1} &amp; s_{n-2}&amp;s_{n-3} &amp;\cdots &amp; s_{2n-1} \\
1 &amp; x&amp;x^2 &amp;\cdots &amp; x^n\end{array}\right|,$$ respectively
 $$P_n^{(s^*)} (x)=\Delta_n^{(s^*)}(1,x,x^2,\cdots,x^n)=\left|\begin{array}{ccccc} s_1 &amp; s_2&amp;s_3 &amp;\cdots &amp; s_{n+1} \\
s_2 &amp; s_3&amp;s_4 &amp;\cdots &amp; s_{n+2} \\
\vdots &amp; \vdots &amp; \vdots &amp;\cdots &amp; \vdots \\
s_n &amp; s_{n-1}&amp;s_{n-2} &amp;\cdots &amp; s_{2n} \\
1 &amp; x&amp;x^2 &amp;\cdots &amp; x^n\end{array}\right|,$$and let $L^{(s)}$ (respectively $L^{(s^*)}$) be the coefficient array of the family of orthogonal polynomials $P_n^{(s)} (x)$ (respectively $P_n^{(s^*)} (x))$.

If $n$ is even, $n=2m$, we set
\begin{displaymath}\mathcal{D}_n(1,x,\ldots, x^m)=\left|\begin{array}{ccccccc} a_0 &amp; a_1 &amp; a_2 &amp; \cdots &amp; a_m &amp; \cdots &amp; a_{2m}  \\ b_0 &amp; b_1 &amp; b_2 &amp; \cdots &amp; b_m &amp; \cdots &amp; b_{2m}  \\ 0 &amp; a_0 &amp; a_1 &amp; \cdots &amp; a_{m-1} &amp;
\cdots &amp; a_{2m-1}  \\ 0 &amp; b_0 &amp; b_1 &amp; \cdots &amp; b_{m-1} &amp; \cdots &amp; b_{2m-1}  \\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots  \\ 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 1 &amp; \cdots &amp; x^m  \end{array}\right|.\end{displaymath}
If $n$ is odd, $n=2m+1$, then we let
\begin{displaymath}\mathcal{D}_n(1,x,\ldots,x^{m+1})=\left|\begin{array}{ccccccc} a_0 &amp; a_1 &amp; a_2 &amp; \cdots &amp; a_m &amp; \cdots &amp; a_{2m+1}  \\ b_0 &amp; b_1 &amp; b_2 &amp; \cdots &amp; b_m &amp; \cdots &amp; b_{2m+1}  \\ 0 &amp; a_0 &amp; a_1 &amp; \cdots &amp; a_{m-1} &amp;
\cdots &amp; a_{2m}  \\ 0 &amp; b_0 &amp; b_1 &amp; \cdots &amp; b_{m-1} &amp; \cdots &amp; b_{2m}  \\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots  \\ 0 &amp; 0 &amp; 0 &amp; \cdots &amp; a_0 &amp; \cdots &amp; a_{m+1}  \\0 &amp; 0  &amp; 0 &amp; \cdots &amp; 1 &amp; \cdots &amp; x^{m+1} \end{array}\right|.\end{displaymath}
We can gain insight into this construction again by looking at the transpose of the underlying matrix. Using an obvious notation \cite{Merlini_MC}, we have

\begin{equation} \mathcal{H}_{j,i}(x)=\begin{cases} [t^i] t^{\frac{j}{2}}f(t), &amp; \text{if $2|j$ and $j&lt;n$}; \\ \\
[t^i] t^{\frac{j-1}{2}} g(t), &amp; \text{if $2 \nmid j$  and  $j&lt;n$};\\ \\
[t^i] \frac{t^{\lfloor \frac{j}{2} \rfloor}}{1-xt}, &amp; \text{if  $j=n$}.
\end{cases}
\end{equation}


\noindent Finally we let $M=(b_{i-j}[j \le i])$ be the sequence (or renewal) array of the sequence $b_n$, and $M_n$ represent the matrix composed of the first $n+1$ rows and columns of $M$ (where we use a similar subscript notation to denote the first
$n+1$ rows and columns of other matrices). We have $M=(g(x),x)$ as a Riordan array.
\begin{proposition}
We have $$\mathcal{D}_{2n}(1,x,\ldots, x^n)=  (-1)^n b_0^{2n+1}\Delta_n^{(s^*)}(1,\phi_1(x),\phi_2(x),\ldots,\phi_n(x))$$ where
$$(1,\phi_1(x),\phi_2(x),\ldots,\phi_n(x))^t=M_n^{-1}(1,x,x^2,\ldots,x^n)^t,$$
and we have $$\mathcal{D}_{2n+1}(1,x,\ldots, x^{n+1})=  b_0^{2n+2} \Delta_{n+1}^{(s)}(1,\phi_1(x),\phi_2(x),\ldots,\phi_n(x),\phi_{n+1}(x))$$ where
$$(1,\phi_1(x),\phi_2(x),\ldots,\phi_n(x),\phi_{n+1}(x))^t=M_{n+1}^{-1}(1,x,x^2,\ldots,x^n)^t.$$
Equivalently, if $\tilde{L}$ is the coefficient array of the family of polynomials $Q_n(x)=\mathcal{D}_{2n}(1,x,\ldots, x^n)$ (respectively $\tilde{L}^*$ is the coefficient array of the polynomials $R_{n+1}(x)=\mathcal{D}_{2n+1}(1,x,\ldots, x^{n+1})$, $R_0(x)=1$) then we have
$$\tilde{L}= S\cdot L^{(s)}\cdot M^{-1}$$
(respectively
$$\tilde{L}^*=L^{(s^*)} \cdot M^{-1}),$$ 
where $S=\text{diag}(1,-1,1,-1,\ldots)$.
\end{proposition}
\begin{proof} Carrying out the eliminations on $\mathcal{D}_{2m}$ that we have used in the first proposition, we get, for $n=2m$,
 \begin{displaymath}\left(\begin{array}{cccccccc}
 s_0 &amp; s_1 &amp; s_2 &amp; \cdots &amp; s_{m-1} &amp; s_m &amp; \cdots &amp; s_{2m} \\
 1 &amp; 0 &amp; 0 &amp; \cdots &amp; 0&amp; 0 &amp; \cdots &amp; 0 \\
 0 &amp; s_0 &amp; s_1 &amp; \cdots &amp; s_{m-2} &amp; s_{m-1} &amp; \cdots &amp; s_{2m-1} \\
 0  &amp; 1 &amp; 0 &amp; \cdots &amp; 0&amp; 0 &amp; \cdots &amp; 0 \\
 0 &amp; 0 &amp; s_0 &amp; \cdots  &amp;s_{m-3} &amp; s_{m-2} &amp; \cdots &amp; s_{2m-2} \\
 0 &amp; 0  &amp; 1 &amp; \cdots &amp; 0&amp;  0 &amp; \cdots &amp; 0 \\
 \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots \\
 0 &amp; 0 &amp; 0 &amp; \cdots &amp; s_0&amp; s_1 &amp; \cdots &amp; s_{m+1}\\
 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 1&amp; 0 &amp; \cdots &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; \cdots &amp; 0&amp; 1 &amp; \cdots &amp; \phi_m(x)
\end{array}\right).\end{displaymath}
Expanding iteratively along the
rows with a single 1 we get an ``upside-down" augmented Hankel determinant
\begin{displaymath} \left|\begin{array}{ccc}
s_{m+1} &amp; \cdots &amp; s_{2m} \\

\vdots &amp; \vdots &amp; \vdots \\
s_1 &amp; \cdots &amp; s_{m+1}\\
1 &amp; \cdots &amp;\phi_m(x)  \end{array}\right|. \end{displaymath}
Similarly for $n=2m+1$, we obtain in $\mathcal{D}_{2m+1}$ upon elimination the following matrix.
 \begin{displaymath}\left(\begin{array}{cccccccc}
 s_0 &amp; s_1 &amp; s_2 &amp; \cdots &amp; s_{m-1} &amp; s_m &amp; \cdots &amp; s_{2m+1} \\
 1 &amp; 0 &amp; 0 &amp; \cdots &amp; 0&amp; 0 &amp; \cdots &amp; 0 \\
 0 &amp; s_0 &amp; s_1 &amp; \cdots &amp; s_{m-2} &amp; s_{m-1} &amp; \cdots &amp; s_{2m} \\
 0  &amp; 1 &amp; 0 &amp; \cdots &amp; 0&amp; 0 &amp; \cdots &amp; 0 \\
 0 &amp; 0 &amp; s_0 &amp; \cdots  &amp;s_{m-3} &amp; s_{m-2} &amp; \cdots &amp; s_{2m-1} \\
 0 &amp; 0  &amp; 1 &amp; \cdots &amp; 0&amp;  0 &amp; \cdots &amp; 0 \\
 \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots \\
 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 1&amp; 0 &amp; \cdots &amp; 0\\
 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 0&amp; s_0 &amp; \cdots &amp; s_{m+1}\\

0 &amp; 0 &amp; 0 &amp; \cdots &amp; 0&amp; 1 &amp; \cdots &amp; \phi_{m+1}(x)
\end{array}\right).\end{displaymath}
Expanding iteratively along the
rows with a single 1 we get an ``upside-down" augmented Hankel determinant
\begin{displaymath} \left|\begin{array}{ccc}
s_{m} &amp; \cdots &amp; s_{2m+1} \\
\vdots &amp; \vdots &amp; \vdots \\
s_0 &amp; \cdots &amp; s_{m+1}\\
1 &amp; \cdots &amp;\phi_{m+1}(x)  \end{array}\right|. \end{displaymath}

\noindent The result follows from this.
Note that in terms of the matrix elements $\mathcal{H}_{i,j}(x)$, we can interpret the above as
\begin{eqnarray*}
M_n^{-1}\cdot (\mathcal{H}_{j,i}(x))_n&amp;=&amp; \left(\frac{1}{g(t)}, t\right)_n \cdot (\mathcal{H}_{j,i}(x))_n\\
&amp;=&amp; \left(\begin{cases} [t^i] t^{\frac{j}{2}}f(t)/g(t), &amp; \text{if $2|j$ and  $j&lt;n$}; \\ \\
[t^i] t^{\frac{j-1}{2}} g(t)/g(t), &amp; \text{if $2 \nmid j$  and  $j&lt;n$};\\ \\
[t^i] t^{\lfloor \frac{j}{2} \rfloor}\frac{1}{g(t)}\frac{1}{1-xt},&amp; \text{if $j=n$}.
\end{cases} \right) \\
&amp;=&amp;\left(\begin{cases} [t^i] t^{\frac{j}{2}} s(t), &amp; \text{if $2|j$ and $j&lt;n$}; \\ \\
[t^i] t^{\frac{j-1}{2}},   &amp; \text{if $2 \nmid j$ and  $j&lt;n$};\\ \\
[t^i] t^{\lfloor \frac{j}{2} \rfloor}\frac{1}{g(t)}\frac{1}{1-xt}, &amp; \text{if $j=n$}.
\end{cases} \right), \end{eqnarray*}
where $$\phi_n(x)=[t^n] \frac{1}{g(t)}\frac{1}{1-xt}.$$

\end{proof}
\noindent Note that we have
\begin{displaymath}
\left(\begin{array} {cccccc}
b_0 &amp; &amp; &amp; &amp; &amp;  \\
b_1 &amp; b_0 &amp; &amp; &amp; &amp;  \\
\vdots &amp; \vdots &amp; \vdots &amp; &amp; &amp; \\
b_m &amp; b_{m-1} &amp; b_{m-2} &amp; \vdots &amp; &amp;  \\
b_{m+1} &amp; b_{m} &amp; b_{m-1} &amp; \vdots &amp; &amp;  \\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;  \\
b_{2m} &amp; b_{2m-1} &amp; b_{2m-2} &amp; b_{2m-3} &amp; \cdots &amp; b_0 \end{array} \right)^{-1} \cdot
\left(\begin{array} {cccccc}
a_0 &amp; b_0 &amp; 0 &amp; 0 &amp; \cdots &amp; 0 \\
a_1 &amp; b_1 &amp; a_0&amp; b_0&amp; \cdots &amp; 0 \\
\vdots &amp; \vdots &amp; \vdots &amp;\vdots &amp;\cdots &amp;\vdots \\
a_m &amp; b_m &amp; a_{m-1} &amp; b_{m-1} &amp;\cdots &amp; 1 \\
a_{m+1} &amp; b_{m+1} &amp; a_{m} &amp; b_m &amp; \cdots &amp; x \\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots \\
a_{2m} &amp; b_{2m} &amp; a_{2m-1} &amp; b_{2m-1} &amp; \cdots &amp; x^m \end{array} \right)
\end{displaymath}
\begin{displaymath}
=\left(\begin{array} {cccccc}
s_0 &amp; 1 &amp; 0 &amp; 0 &amp; \cdots &amp; 0 \\
s_1 &amp; 0 &amp; s_0&amp; 1&amp; \cdots &amp; 0 \\
\vdots &amp; \vdots &amp; \vdots &amp;\vdots &amp;\cdots &amp;\vdots \\
s_m &amp; 0 &amp; s_{m-1} &amp; 0 &amp;\cdots &amp; 1 \\
s_{m+1} &amp; 0 &amp; s_{m} &amp; 0 &amp; \cdots &amp; \phi_1(x) \\
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots \\
s_{2m} &amp; 0 &amp; s_{2m-1} &amp; 0 &amp; \cdots &amp; \phi_m(x) \end{array} \right)
\end{displaymath}
with a corresponding matrix equation for the case $n=2m+1$.
\section{The Riordan array case}

We recall elements of the theory of Riordan arrays before giving examples of Hurwitz associated polynomials, where the relevant coefficient matrices happen to be Riordan arrays.
\noindent 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 $f(\bar{f}(x))=x$.
\newline\newline
For a lower triangular matrix $(a_{n,k})_{n,k \ge 0}$ the row sums give the sequence with general term
$\sum_{k=0}^n a_{n,k}$. More generally, a lower triangular matrix $(a_{n,k})_{n,k \ge 0}$ is the coefficient array of the polynomials
$$P_n(x)=\sum_{k=0}^n a_{n,k}x^k.$$

\noindent 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 often require in addition that $f_1=1$, but this is not the case in this note. 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 $f$.

Elements of the form $(g(x), x)$ form a subgroup called the \emph{Appell subgroup}.
\newline\newline If $\mathbf{M}$ is the matrix $(g,f)$, and
$\mathbf{a}=(a_0,a_1,\ldots)^t$ is an integer sequence (expressed as an infinite column vector) 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
$$(g,f):\cal{A}(\mathnormal{x}) \mapsto \mathnormal{(g,f)}\cdot
\cal{A}\mathnormal{(x)=g(x)}\cal{A}\mathnormal{(f(x))}.$$
\noindent This action is often referred to as the fundamental theorem of Riordan arrays.
\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}

\begin{example} The Riordan array $\left(\frac{1}{1+x}, \frac{x}{(1+x)^2}\right)$ has inverse $(c(x), xc(x)^2)$. It is the coefficient array of the unique family of orthogonal polynomials
$$P_n(x)=(x-2)P_{n-1}(x)-P_{n-2}(x)$$ for which the Catalan numbers $C_n$ are the moments.
\end{example}

\noindent 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 polynomial sequence
$P_n(t)$ for which $(g,f)$ is the coefficient array will have g.f.
$$\frac{g(x)}{1-t f(x)}.$$
\noindent
For an invertible matrix $M$, its production matrix (also called its Stieltjes matrix) \cite{ProdMat_0, ProdMat_1} is the matrix
$$P_M = M^{-1} \hat{M},$$ where $\hat{M}$ is the matrix $M$ with its first row removed.
A Riordan array $M$ is the inverse of the coefficient array of a family of orthogonal polynomials \cite{Chihara, Szego} if and only if
$P_M$ is tri-diagonal \cite{Barry_Meixner, Barry_Roh}. Necessarily, the Jacobi coefficients (i.e., the coefficients of the three-term recurrence \cite{Chihara} that defines the polynomials) of these orthogonal polynomials are then constant.

\begin{example} The production matrix of $(c(x), xc(x)^2)$ is given by
 \begin{displaymath}\left(\begin{array}{ccccccc} 1 &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 \\ 0 &amp; 1 &amp; 2 &amp;
1 &amp; 0 &amp; 0 &amp;
\cdots \\ 0  &amp; 0 &amp; 1 &amp; 2 &amp; 1 &amp; 0 &amp; \cdots \\ 0 &amp; 0 &amp; 0 &amp; 1 &amp;
2 &amp; 1 &amp; \cdots \\0 &amp; 0  &amp; 0 &amp; 0 &amp; 1 &amp; 2 &amp;\cdots\\ \vdots
&amp; \vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right). \end{displaymath}
\end{example}
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 $A=[\alpha_0,\alpha_1,\alpha_2,\ldots]$ and $Z=[z_0,z_1,z_2,\ldots]$ with $\alpha_0 \neq 0$, $z_0 \neq 0$ such that
\begin{itemize}
\item $d_{n+1,k+1}=\sum_{j=0}^{\infty} \alpha_j d_{n,k+j}, \quad (k,n=0,1,\ldots)$
\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 $\alpha_0,\alpha_1,\alpha_2,\ldots$ and $z_0,z_1,z_2,\ldots$ are called the $A$-sequence and the $Z$-sequence of the Riordan array
$M=(g(x),f(x))$, 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{d_{0,0}}{g(\bar{f}(x))}\right).\end{equation}

The first column of $P_M$ 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 $0$).
There is a close link between orthogonal polynomials whose defining three term recurrences have constant coefficients and Riordan arrays whose inverses have tri-diagonal production matrices \cite{Barry_Meixner, Barry_Roh}. We devote this section to examples where the
 coefficient array $L^{(s)}$ is a Riordan array
$$L^{(s)}=(u(x), v(x)).$$ We shall also assume that $a_0=b_0=1$ for the rest of this section. Note that we have, in this case,
$${L^{(s)}}^{-1}=(u(x), v(x))^{-1}=\left(\frac{1}{u(\bar{v}(x))},\bar{v}(x)\right)=(s(x), \bar{v}(x)).$$ In particular,
$$s(x)=\frac{1}{u(\bar{v}(x))} \Rightarrow u(x)=\frac{1}{s(v(x))}.$$
We have $$L^{(s)} \cdot M^{-1}=L^{(s)}\cdot (g(x),x)^{-1}=L^{(s)}\cdot \left(\frac{1}{g(x)},x\right)=\left(\frac{u(x)}{g(x)},v(x)\right).$$ \noindent Looking at inverses, we get
$$ (L^{(s)} \cdot M^{-1})^{-1}=M \cdot (L^{(s)})^{-1}=\left(\frac{g(x)}{u(\bar{v}(x))}, \bar{v}(x)\right).$$

We wish to compare the production array of the matrix $(L^{(s)})^{-1}$ with that of the matrix $(L^{(s)} \cdot M^{-1})^{-1}$. For the matrix
$(L^{(s)})^{-1}=(u, v)^{-1}=\left(\frac{1}{u(\bar{v}(x))}, \bar{v}(x)\right)$, we have
\begin{equation} A(x)=\frac{x}{v(x)},\quad Z(x)=\frac{1}{v}(1-u(x)),\end{equation} since by assumption $d_{0,0}=1$.
For $(L^{(s)} \cdot M^{-1})^{-1}=\left(\frac{g(x)}{u(\bar{v}(x))}, \bar{v}(x)\right)$ we find that
\begin{equation}\tilde{A}(x)=\frac{x}{v(x)},\quad \tilde{Z}(x)=\frac{1}{v(x)}\left(1-\frac{u(x)}{g(v(x))}\right).\end{equation}
In our case (that of $L^{(s)}$ being the coefficient array of a family of orthogonal polynomials), we have
$$L^{(s)}=\left(\frac{1+\alpha' x+ \beta' x^2}{1+\alpha x+\beta x^2},\frac{x}{1+\alpha x+\beta x^2}\right),$$ for suitable values of $\alpha, \beta, \alpha', \beta'$.
Then
$$A(x)=\tilde{A}(x)=1+\alpha x+\beta x^2,$$ and
$$Z(x)=\frac{1+\alpha x+\beta x^2}{x}\left(1-\frac{1+\alpha' x+ \beta' x^2}{1+\alpha x+\beta x^2}\right)=(\alpha-\alpha')+(\beta-\beta')x,$$ along with
$$\tilde{Z}(x)=\frac{1}{x}\left(1+\alpha x+\beta x^2-(1+\alpha' x+\beta' x^2)\cdot \frac{1}{g\left(\frac{x}{1+\alpha x+\beta x^2}\right)}\right).$$
Thus as expected, $(L^{(s)})^{-1}$ has a tri-diagonal production matrix, which begins
 \begin{displaymath}\left(\begin{array}{ccccccc} \alpha-\alpha' &amp; 1 &amp; 0 &amp; 0
&amp;
0 &amp; 0 &amp; \cdots \\ \beta-\beta' &amp; \alpha &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\ 0 &amp; \beta &amp; \alpha &amp;
1 &amp; 0 &amp; 0 &amp;
\cdots \\ 0  &amp; 0 &amp; \beta &amp; \alpha &amp; 1 &amp; 0 &amp; \cdots \\ 0 &amp; 0 &amp; 0 &amp; \beta &amp;
\alpha &amp; 1 &amp; \cdots \\0 &amp; 0  &amp; 0 &amp; 0 &amp; \beta &amp; \alpha &amp;\cdots\\ \vdots
&amp; \vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath} However, given the form of $\tilde{Z}(x)$, only in the exceptional case when $$\tilde{Z}(x)=\frac{1}{x}\left(1+\alpha x+\beta x^2- \frac{1+\alpha' x+\beta' x^2}{g\left(\frac{x}{1+\alpha x+\beta x^2}\right)}\right)$$ is of the form $\gamma + \delta x$ will the production matrix of $(L^{(s)} \cdot M^{-1})^{-1}$ be tri-diagonal. Only in this case are the polynomials $Q_n(x)$ orthogonal.

In the case when
$$L^{(s^*)}=(u^*(x), v^*(x))=(s^*(x), \bar{v^*}(x))^{-1}$$ is a Riordan array, a similar analysis is valid.

\begin{example} We look at the case where $a_n=\sum_{k=0}^n \binom{2k}{k} C_{n-k}=\binom{2n+1}{n+1}$ and $b_n=\binom{2n}{n}$. In this case, we have
$$f(x)=\frac{c(x)}{\sqrt{1-4x}},\quad g(x)=\frac{1}{\sqrt{1-4x}}, \quad s(x)=\frac{f(x)}{g(x)}=c(x),$$ and hence $s_n=C_n$.

\noindent We find that the coefficient array $\tilde{L}$ of the polynomials $Q_n(x)$ is the array
$$\left(\frac{1-x}{(1+x)^2},\frac{x}{(1+x)^2}\right),$$ which 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 \\ 5 &amp; -5 &amp; 1 &amp;
0 &amp; 0 &amp; 0 &amp;
\cdots \\ -7 &amp; 14 &amp; -7 &amp; 1 &amp; 0 &amp; 0 &amp; \cdots \\ 9 &amp; -30 &amp; 27 &amp; -9 &amp;
1 &amp; 0 &amp; \cdots \\-11 &amp; 55  &amp; -77 &amp; 44 &amp; -11 &amp; 1 &amp;\cdots\\ \vdots
&amp; \vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath} We have
$$Q_n(x)=\sum_{k=0}^n \frac{2n+1}{2k+1}\binom{n+k}{2k}(-1)^{n-k} x^k.$$ The inverse array $\tilde{L}^{-1}$ 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 \\ 35 &amp; 21 &amp; 7 &amp; 1 &amp; 0 &amp; 0 &amp; \cdots \\ 126 &amp; 84 &amp; 36 &amp; 9 &amp;
1 &amp; 0 &amp; \cdots \\462 &amp; 330  &amp; 165 &amp; 55 &amp; 11 &amp; 1 &amp;\cdots\\ \vdots
&amp; \vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right) \quad (\seqnum{A111418}).\end{displaymath}
By its form, we see that $\left(\frac{1-x}{(1+x)^2},\frac{x}{(1+x)^2}\right)$ is in this case the coefficient array of a family of orthogonal polynomials. This is verified by noting that
$$\tilde{Z}(x)=\frac{(1+x)^2}{x} \left(1-\frac{\frac{1}{1+x}}{\frac{1}{\sqrt{1-\frac{4x}{(1+x)^2}}}}\right)=3+x.$$ In fact, the production matrix of $\left(\frac{1-x}{(1+x)^2},\frac{x}{(1+x)^2}\right)^{-1}$ is given by
 \begin{displaymath}\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 \\ 0 &amp; 1 &amp; 2 &amp;
1 &amp; 0 &amp; 0 &amp;
\cdots \\ 0  &amp; 0 &amp; 1 &amp; 2 &amp; 1 &amp; 0 &amp; \cdots \\ 0 &amp; 0 &amp; 0 &amp; 1 &amp;
2 &amp; 1 &amp; \cdots \\0 &amp; 0  &amp; 0 &amp; 0 &amp; 1 &amp; 2 &amp;\cdots\\ \vdots
&amp; \vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right), \end{displaymath}
\noindent and so we have
$$Q_n(x)=(x-2)Q_{n-1}(x)-Q_{n-2}(x), \quad Q_0(x)=1,\, Q_1(x)=x-3.$$ \noindent
Looking at $M^{-1} \tilde{L}^{-1}$, we have
\begin{displaymath}\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0 &amp; 0
&amp;
0 &amp; 0 &amp; \cdots \\2 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\ 6 &amp; 2 &amp; 1 &amp;
0 &amp; 0 &amp; 0 &amp;
\cdots \\ 20 &amp; 6 &amp; 2 &amp; 1 &amp; 0 &amp; 0 &amp; \cdots \\ 70 &amp; 20 &amp; 6 &amp; 2 &amp;
1 &amp; 0 &amp; \cdots \\252 &amp; 70  &amp; 20 &amp; 6 &amp; 2 &amp; 1 &amp;\cdots\\ \vdots
&amp; \vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right)^{-1} \cdot \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 \\ 35 &amp; 21 &amp; 7 &amp; 1 &amp; 0 &amp; 0 &amp; \cdots \\ 126 &amp; 84 &amp; 36 &amp; 9 &amp;
1 &amp; 0 &amp; \cdots \\462 &amp; 330  &amp; 165 &amp; 55 &amp; 11 &amp; 1 &amp;\cdots\\ \vdots
&amp; \vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right)\end{displaymath}
\begin{displaymath}=\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 \\ 2 &amp; 3 &amp; 1 &amp;
0 &amp; 0 &amp; 0 &amp;
\cdots \\ 5 &amp; 9 &amp; 5 &amp; 1 &amp; 0 &amp; 0 &amp; \cdots \\ 14 &amp; 28 &amp; 20 &amp; 7 &amp;
1 &amp; 0 &amp; \cdots \\42 &amp; 90  &amp; 75 &amp; 35 &amp; 9 &amp; 1 &amp;\cdots\\ \vdots
&amp; \vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right) \quad (\seqnum{A039599}).\end{displaymath} This latter array is the inverse of $\left(\frac{1}{1+x},\frac{x}{(1+x)^2}\right)^{-1}$.  The array $\left(\frac{1}{1+x},\frac{x}{(1+x)^2}\right)$ is the coefficient array of the orthogonal polynomials whose moments are the Catalan numbers $C_n$.

Turning now to $s_n^*$, we have $s_n^*=C_{n+1}$. In this case, we find that
$$\tilde{L}^*=L^{(s^*)}\cdot M^{-1}=(u^*(x), v^*(x))=\left(\frac{1-x}{(1+x)^3},\frac{x}{(1+x)^2}\right).$$ This array represents the coefficients of a family of polynomials $R_n(x)$ that are ``almost orthogonal'', in the sense that the production matrix of the inverse of this matrix is of the form
\begin{displaymath}\left(\begin{array}{ccccccc} 4 &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 \\ 2 &amp; 1 &amp; 2 &amp;
1 &amp; 0 &amp; 0 &amp;
\cdots \\ -2 &amp; 0 &amp; 1 &amp; 2 &amp; 1 &amp; 0 &amp; \cdots \\ 2 &amp; 0 &amp; 0 &amp; 1 &amp;
2 &amp; 1 &amp; \cdots \\-2 &amp; 0  &amp; 0 &amp; 0 &amp; 1 &amp; 2 &amp;\cdots\\ \vdots
&amp; \vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath}
\noindent Thus we have
$$R_n(x)=(x-2) R_{n-1}(x)-R_{n-2}(x)-\tilde{z}_{n-1}, \quad R_0(x)=1,\, R_1(x)=x-4, $$
where $\tilde{z}_n$ is the sequence $0,0, 2, -2, 2, -2, 2,\ldots$.

The inverse matrix $(u^*(x), v^*(x))^{-1}$, which begins
\begin{displaymath}\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0 &amp; 0
&amp;
0 &amp; 0 &amp; \cdots \\4 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\ 15 &amp; 6 &amp; 1 &amp;
0 &amp; 0 &amp; 0 &amp;
\cdots \\ 56 &amp; 28 &amp; 8 &amp; 1 &amp; 0 &amp; 0 &amp; \cdots \\ 210 &amp; 120 &amp; 45 &amp; 10 &amp;
1 &amp; 0 &amp; \cdots \\792 &amp; 495  &amp; 220 &amp; 66 &amp; 12 &amp; 1 &amp;\cdots\\ \vdots
&amp; \vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right),\end{displaymath} has general term $\binom{2n+2}{n+k+2}$, and
satisfies
\begin{displaymath}\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0 &amp; 0
&amp;
0 &amp; 0 &amp; \cdots \\2 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\ 6 &amp; 2 &amp; 1 &amp;
0 &amp; 0 &amp; 0 &amp;
\cdots \\ 20 &amp; 6 &amp; 2 &amp; 1 &amp; 0 &amp; 0 &amp; \cdots \\ 70 &amp; 20 &amp; 6 &amp; 2 &amp;
1 &amp; 0 &amp; \cdots \\252 &amp; 70  &amp; 20 &amp; 6 &amp; 2 &amp; 1 &amp;\cdots\\ \vdots
&amp; \vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right)^{-1} \cdot \left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0 &amp; 0
&amp;
0 &amp; 0 &amp; \cdots \\4 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\ 15 &amp; 6 &amp; 1 &amp;
0 &amp; 0 &amp; 0 &amp;
\cdots \\ 56 &amp; 28 &amp; 8 &amp; 1 &amp; 0 &amp; 0 &amp; \cdots \\ 210 &amp; 120 &amp; 45 &amp; 10 &amp;
1 &amp; 0 &amp; \cdots \\792 &amp; 495  &amp; 220 &amp; 66 &amp; 12 &amp; 1 &amp;\cdots\\ \vdots
&amp; \vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right)\end{displaymath}
\begin{displaymath}=\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0 &amp; 0
&amp;
0 &amp; 0 &amp; \cdots \\2 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\ 5 &amp; 4 &amp; 1 &amp;
0 &amp; 0 &amp; 0 &amp;
\cdots \\ 14 &amp; 14 &amp; 6 &amp; 1 &amp; 0 &amp; 0 &amp; \cdots \\ 42 &amp; 48 &amp; 27 &amp; 8 &amp;
1 &amp; 0 &amp; \cdots \\132 &amp; 165  &amp; 110 &amp; 44 &amp; 10 &amp; 1 &amp;\cdots\\ \vdots
&amp; \vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right)=\left(c(x)^2,xc(x)^2\right),\end{displaymath} where the first column of this last matrix (which is \seqnum{A039598}) is given by
$s_n^*=C_{n+1}$. Note that the first column elements of $(u^*(x), v^*(x))^{-1}$ are given by  $$\sum_{k=0}^n \binom{2k}{k}C_{n-k+1}=\sum_{k=0}^n \binom{2n-2k}{n-k}C_{k+1}=\binom{2n+2}{n+2}, \quad (\seqnum{A001791}).$$ The inverse of this last matrix is the coefficient array
$$L^{(s^*)}=\left(\frac{1}{(1+x)^2},\frac{x}{(1+x)^2}\right)=(c(x)^2, xc(x)^2)^{-1}$$ for the family of orthogonal polynomials
$$\tilde{R}_n(x)=(x-2)\tilde{R}_{n-1}(x)-\tilde{R}_{n-2}(x), \quad \tilde{R}_0(x)=1,\, \tilde{R}_1(x)=x-2.$$
These polynomials have moments given by $C_{n+1}$.
We note that we have
$$\tilde{Z}^*(x)=\frac{(1+x)^2}{x} \left(1-\frac{\frac{1}{(1+x)^2}}{\frac{1}{\sqrt{1-\frac{4x}{(1+x)^2}}}}\right)=\frac{4+3x+x^2}{1+x},$$ where
$\frac{4+3x+x^2}{1+x}$ expands to give
$4, -1, 2, -2, 2, -2, 2,\ldots$.

\end{example}
\begin{example} In the last example, it happened that the family of polynomials $Q_n(x)$  constituted a family of orthogonal polynomials. This is not true in general. In this example, we let $$a_n=\sum_{k=0}^n F_{k+1} C_{n-k}, \quad b_n=F_{k+1},$$ where $F_n$ \seqnum{A000045} denotes the $n$-th Fibonacci number. Then we have
$$f(x)=\frac{c(x)}{1-x-x^2},\quad g(x)=\frac{1}{1-x-x^2}, \quad s(x)=\frac{f(x)}{g(x)}$$ and $s_n=C_n$. Thus again, $L^{(s)}=\left(\frac{1}{1+x},\frac{x}{(1+x)^2}\right)$ and hence
$$\tilde{L}=\left(\frac{1}{1+x},\frac{x}{(1+x)^2}\right) \cdot \left(\frac{1}{1-x-x^2},x\right)^{-1}=\left(\frac{1}{1+x},\frac{x}{(1+x)^2}\right) \cdot (1-x-x^2,x).$$
Thus
$$\tilde{L}=\left(\frac{1+3x+3x^2+3x^3+x^4}{(1+x)^5}, \frac{x}{(1+x)^2}\right),$$ whose inverse has production matrix \begin{displaymath}\left(\begin{array}{ccccccc} 2 &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 \\ -2 &amp; 1 &amp; 2 &amp;
1 &amp; 0 &amp; 0 &amp;
\cdots \\ 5 &amp; 0 &amp; 1 &amp; 2 &amp; 1 &amp; 0 &amp; \cdots \\ -9 &amp; 0 &amp; 0 &amp; 1 &amp;
2 &amp; 1 &amp; \cdots \\ 14 &amp; 0  &amp; 0 &amp; 0 &amp; 1 &amp; 2 &amp;\cdots\\ \vdots
&amp; \vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath}
\noindent This follows since
$$\tilde{Z}(x)=\frac{1}{x}\left((1+x)^2-\frac{1+x}{g\left(\frac{x}{(1+x)^2}\right)}\right)=\frac{2+7x+7x^2+4x^3+x^4}{(1+x)^3},$$ which expands to
$$2, 1, -2, 5, -9, 14, -20, 27, -35,\ldots.$$
This means that $\tilde{L}$ is the coefficient array of the ``almost-orthogonal" polynomials $Q_n(x)$ that satisfy
$$Q_n(x)=(x-2)Q_{n-1}-Q_{n-2}(x)-\tilde{z}_{n-1},$$
where $\tilde{z}_n$ is the sequence $0,0,-2,5,-9, 14, -20, 27, -35,\ldots$.

\noindent Looking now at $s^*$, we have
$$\tilde{L}^* =\left(\frac{1}{(1+x)^2},\frac{x}{(1+x)^2}\right)\cdot (1-x-x^2,x)=\left(\frac{1+3x+3x^2+3x^3+x^4}{(1+x)^6},\frac{x}{(1+x)^2}\right).$$
The inverse of $\tilde{L}^*$ then has production array
\begin{displaymath}\left(\begin{array}{ccccccc} 3 &amp; 1 &amp; 0 &amp; 0
&amp;
0 &amp; 0 &amp; \cdots \\0 &amp; 2 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots \\ -1 &amp; 1 &amp; 2 &amp;
1 &amp; 0 &amp; 0 &amp;
\cdots \\ 6 &amp; 0 &amp; 1 &amp; 2 &amp; 1 &amp; 0 &amp; \cdots \\ -15 &amp; 0 &amp; 0 &amp; 1 &amp;
2 &amp; 1 &amp; \cdots \\29 &amp; 0  &amp; 0 &amp; 0 &amp; 1 &amp; 2 &amp;\cdots\\ \vdots
&amp; \vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath} We have
$$\tilde{Z}^*(x)=\frac{1}{x}\left((1+x)^2-\frac{1}{g\left(\frac{x}{(1+x)^2}\right)}\right)=\frac{3+12x+17x^2+14x^3+6x^4+x^5}{(1+x)^3},$$
which expands to $3,0,-1,6,-15,29,\ldots$.
Thus in this case we have
$$R_n(x)=(x-2)R_{n-1}(x)-R_{n-2}(x)-\tilde{z}_{n-1}^*, \quad R_0(x)=1,\, R_1(x)=x-3.$$
\end{example}
\begin{example} For any element $T=(u(x),x)$ of the Appell subgroup of Riordan arrays \cite{SGWW}, it is clear that
$$ \mathbb{H}_n (Ta, Tb)=\mathbb{H}_n(a,b).$$ This is so because
for the pair $Ta$, $Tb$, we have
$$s(x) = \frac{u(x) f(x)}{u(x) g(x)}= \frac{f(x)}{g(x)},$$ by the fundamental theorem of Riordan arrays.
Thus for instance we have
$$\mathbb{H}_n(a_n, b_n)=\mathbb{H}_n(\sum_{k=0}^n a_k, \sum_{k=0}^n b_k),$$ since the partial sum operator is equal to the Riordan array
$\left(\frac{1}{1-x}, x\right)$.

Similarly we have
$$\mathbb{H}_n(a_n, b_n)=\mathbb{H}_n(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} a_{n-2k}, \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} b_{n-2k}),$$ where in this case the Riordan array is $\left(\frac{1}{1-x^2}, x\right)$.
\end{example}
\section{Further examples}

We list below a table showing a small sample of Hurwitz transforms for the pairs of sequences shown.

\begin{center}\begin{tabular}{|c|c|c|c|} \hline $a_n$
&amp; $b_n$ &amp; $s(x)$ &amp; Hurwitz transform \\ \hline

 $(-1)^nC_{n+1}$
&amp; $(-1)^nC_n$ &amp;$c(-x)$ &amp;
$1,1,1,1,1,1,\ldots$ \\ \hline
$C_n$ &amp; $C_{n+1}$ &amp; $\frac{1}{c(x)}$ &amp;
$1,1,-2,1,3,1,-4,1,5,1,\ldots$ \\ \hline
$C_{n+1}$ &amp; $C_n$
&amp; $c(x)$ &amp; $1,-1,1,1,1,-1,1,1,1,-1,1,\ldots$ \\ \hline
$(-1)^n\binom{1}{n}$ &amp; $(-1)^n(C_n+C_{n-1})$ &amp; $\frac{1}{c(-x)}$ &amp;
$1,-1,-2,1,3,-1,-4,1,5,\ldots$
\\ \hline $C_n$ &amp; $0^n$ &amp;
$c(x)$ &amp; $1,-1,1,1,1,-1,1,1,1,-1,1,\ldots$ \\ \hline
$C_n$ &amp; $(-1)^n \binom{1}{n}$ &amp;
$\frac{c(x)}{1-x}$ &amp; $1,-2,0,2,-1,1,-1,-5,0,5,1,\ldots$ \\ \hline
$C_n$ &amp; $1$ &amp;
$(1-x)c(x)$ &amp; $1,0,1,-1,-1,1,-4,0,-4,-1,1,\ldots$ \\ \hline
$C_n$ &amp; $\frac{1}{1-2n}\binom{2n}{n}$ &amp;
$\frac{c(x)}{\sqrt{1-4x}}$ &amp; $1,-3,1,5,1,-7,1,9,1,-11,1,\ldots$ \\ \hline
$C_n+C_{n+1}$ &amp; $C_n$ &amp; $1+c(x)$ &amp; $2, -1, 3, 1, 4, -1, 5, 1, 6, -1, 7,\dots$ \\ \hline
$C_\frac{n}{2}\frac{1+(-1)^n}{2}$ &amp; $2^n$ &amp;
$(1-2x)c(x^2)$ &amp; $1,2,-3,3,-3,4,5,5,5,6,-7,\ldots$ \\ \hline
$T_n=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}\binom{n}{2k}\binom{2k}{k}$ &amp; $T_{n+1}$ &amp; $\frac{\sqrt{1-2x-3x^2}+3x-1}{x(1-3x)}$ &amp; $1, 2, -2, 4, 0, 8, 8, 16, -16, 32, 0,\ldots$ \\ \hline
\end{tabular}\end{center}


\section{Conclusion} Since the notion of Hurwitz transform proposed here has been linked to the Hankel transform in an easily understood way, it may be said that this notion does not add much to what is already known. What it does add, however, is a fresh perspective, both on the Hankel transform, and on the applications of the Hurwitz matrix. Some natural questions arise in this context. What known sequences are the Hurwitz transforms of pairs of sequences? Are there sequences which cannot be the Hurwitz transform of a sequence? Given that many pairs of sequences may have the same Hurwitz transform, what notion of inverse transform can we formulate?

With regard to new perspectives, the Hurwitz transform makes us look at the pair $(h_n, h_n^*)$ whenever we wish to study $h_n$. An interesting example of this is the case of the Narayana polynomials,
$$s_n=\sum_{k=0}^n \frac{1}{k+1} \binom{n+1}{k}\binom{n}{k} x^k.$$
It is well known that in this case, we have
$$ h_n=x^{\binom{n+1}{2}}.$$ In looking at the pair $(h_n, h_n^*)$, we discover that
$$ h_n^*=x^{\binom{n+1}{2}}\frac{1-x^{n+2}}{1-x}.$$

\noindent Other examples of the pairing $(h_n,h_n^*)$ have been studied in different contexts \cite{Barry, Boj_1, Boj_2, Kratt_2}.

\section{Acknowledgement} The author is indebted to an anonymous reviewer whose careful reading and perceptive remarks have hopefully contributed to this version being more complete and more readable. The author is particularly grateful for the reviewer's insight into the proof of the propositions in the paper.

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\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11C20; Secondary 11B83, 15A15, 15B05, 15B36.

\noindent \emph{Keywords:}
Hurwitz matrix, Hankel matrix, Hankel transform,
integer sequence, Riordan array.

\bigskip
\hrule
\bigskip


\noindent (Concerned with sequences
\seqnum{A000007},
\seqnum{A000045},
\seqnum{A000108},
\seqnum{A001006},
\seqnum{A001791},
\seqnum{A006720},
\seqnum{A025262},
\seqnum{A039598},
\seqnum{A039599},
\seqnum{A111418}, and
\seqnum{A160702}.)


\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  July 26 2012;
revised version received  October 9 2012.
Published in {\it Journal of Integer Sequences}, October 10 2012.

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