\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \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} \end{center} \vskip .2 in \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 & a_1 & a_2 & \cdots & a_m & \cdots & a_{2m} \\ b_0 & b_1 & b_2 & \cdots & b_m & \cdots & b_{2m} \\ 0 & a_0 & a_1 & \cdots & a_{m-1} & \cdots & a_{2m-1} \\ 0 & b_0 & b_1 & \cdots & b_{m-1} & \cdots & b_{2m-1} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & a_0 & \cdots & 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 & a_1 & a_2 & \cdots & a_m & \cdots & a_{2m+1} \\ b_0 & b_1 & b_2 & \cdots & b_m & \cdots & b_{2m+1} \\ 0 & a_0 & a_1 & \cdots & a_{m-1} & \cdots & a_{2m} \\ 0 & b_0 & b_1 & \cdots & b_{m-1} & \cdots & b_{2m} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & a_0 & \cdots & a_{m+1} \\0 & 0 & 0 & \cdots & b_0 & \cdots & 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 $$\mathbb{H}_n(a,a)=a_0 0^n=\begin{cases} a_0, & \text{if $n=0$} ; \\ 0, & n>0. \end{cases}$$ 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, & \text{if $2j+2 \le i$}; \\ a_{j-\frac{i}{2}}, & \text{if $2|i$}; \\ b_{j-\frac{i-1}{2}}, & \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), & \text{if $2|j$}; \\ \\ [x^i] x^{\frac{j-1}{2}} g(x), & \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 & s_1 & s_2 & \cdots & s_{m-1} & s_m & \cdots & s_{2m} \\ 1 & 0 & 0 & \cdots & 0 & 0 & \cdots & 0 \\ 0 & s_0 & s_1 & \cdots & s_{m-2} & s_{m-1} & \cdots & s_{2m-1} \\ 0 & 1 & 0 & \cdots & 0 &0& \cdots & 0 \\ 0 & 0 & s_0 & \cdots & s_{m-3} & s_{m-2} & \cdots & s_{2m-2} \\ 0 & 0 & 1 & \cdots & 0& 0 & \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1& 0 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0& s_0 & \cdots & 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 & \cdots & s_{2m} \\ s_{m-1}& \cdots & s_{2m-1} \\ \vdots & \vdots & \vdots \\ s_0 & \cdots & 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 & s_1 & s_2 & \cdots & s_{m-1} & s_m & s_{m+1} & \cdots & s_{2m+1} \\ 1 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 \\ 0 & s_0 & s_1 & \cdots & s_{m-2} & s_{m-1} & s_m & \cdots & s_{2m} \\ 0 & 1 & 0 & \cdots & 0 &0& 0 & \cdots & 0 \\ 0 & 0 & s_0 & \cdots & s_{m-3} & s_{m-2} & s_{m-1} & \cdots & s_{2m-1} \\ 0 & 0 & 1 & \cdots & 0& 0 & 0 & \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0& s_0 & s_1 & \cdots & s_{m+1} \\ 0 & 0 & 0 & \cdots & 0& 1 & 0 & \cdots & 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} & \cdots & s_{2m+1} \\ s_m & \cdots & s_{2m} \\ \vdots & \vdots & \vdots \\ s_1 & \cdots & 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&=&\left(\frac{1}{g(x)}, x\right)\cdot \left(\mathcal{H}_{j,i}\right)\\ &=& \left(\frac{1}{g(x)}, x\right)\cdot \left(\begin{cases} [x^i] x^{\frac{j}{2}}f(x), & \text{if $2|j$}; \\ \\ [x^i] x^{\frac{j-1}{2}} g(x), & \text{otherwise}. \end{cases} \right)\\ &=& \left(\begin{cases} [x^i] x^{\frac{j}{2}}\frac{f(x)}{g(x)}, & \text{if $2|j$}; \\ \\ [x^i] x^{\frac{j-1}{2}} \frac{g(x)}{g(x)}, & \text{otherwise}. \end{cases} \right)\\ &=& \left(\begin{cases} [x^i] x^{\frac{j}{2}} \frac{f(x)}{g(x)}, & \text{if $2|j$}; \\ \\ [x^i] x^{\frac{j-1}{2}}, & \text{otherwise}. \end{cases} \right)\\ &=& \left(\begin{cases} [x^i] x^{\frac{j}{2}} s(x), & \text{if $2|j$}; \\ \\ [x^i] x^{\frac{j-1}{2}}, & \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 & & & & \cdots \\ b_1 & b_0 & & & \cdots \\ b_2 & b_1 & b_0 & & \cdots \\ b_3 & b_2 & b_1 & b_0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array}\right)^{-1} \cdot \left(\begin{array}{ccccc} a_0 & b_0 & 0& 0& \cdots \\ a_1 & b_1 & a_0& b_0& \cdots \\ a_2 & b_2 & a_1 & b_1& \cdots \\ a_3 & b_3 & a_2 & b_2 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array}\right)=\left(\begin{array}{ccccc} s_0 & 1& 0& 0& \cdots \\ s_1 & 0 & s_0 & 1& \cdots \\ s_2 & 0 & s_1 &0 & \cdots \\ s_3 & 0 & s_2 & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array}\right) \end{displaymath} and hence \begin{displaymath} \left(\begin{array}{ccccc} a_0 & b_0 & 0& 0& \cdots \\ a_1 & b_1 & a_0& b_0& \cdots \\ a_2 & b_2 & a_1 & b_1& \cdots \\ a_3 & b_3 & a_2 & b_2 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array}\right)=\left(\begin{array}{ccccc} b_0 & & & & \cdots \\ b_1 & b_0 & & & \cdots \\ b_2 & b_1 & b_0 & & \cdots \\ b_3 & b_2 & b_1 & b_0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array}\right) \cdot \left(\begin{array}{ccccc} s_0 & 1& 0& 0& \cdots \\ s_1 & 0 & s_0 & 1& \cdots \\ s_2 & 0 & s_1 &0 & \cdots \\ s_3 & 0 & s_2 & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \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 & s_0 b_1+s_1 b_0 & s_0 b_2+ s_1 b_1+ s_2 b_0 & \cdots & s_0 b_m+\ldots & \cdots & s_0 b_{2m}+\ldots \\ b_0 & b_1 & b_2 & \cdots & b_m & \cdots & b_{2m} \\ 0 & s_0 b_0 & s_0 b_1+s_1 b_0 & \cdots & s_0 b_{m-1}+\ldots & \cdots & s_0 b_{2m-1}+\ldots \\ 0 & b_0 & b_1 & \cdots & b_{m-1} & \cdots & b_{2m-1} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & s_0 b_0 & \cdots & 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 $$\mathbb{H}_2 = \left|\begin{array}{ccc} s_0 b_0 & s_0 b_1+s_1 b_0 & s_0 b_2+s_1 b_1+s_2 b_0 \\ b_0 & b_1 & b_2 \\ 0 & s_0 b_0 & s_0 b_1+s_1 b_0 \end{array}\right| $$ $$= \frac{1}{s_0} \left|\begin{array}{ccc} s_0b_0 & s_0 b_1+s_1 b_0 & s_0 b_2+s_1 b_1+s_2 b_0 \\ s_0 b_0 & s_0 b_1 & s_0 b_2 \\ 0 & s_0 b_0 & s_0 b_1+s_1 b_0 \end{array}\right| $$ $$= \frac{1}{s_0} \left|\begin{array}{ccc} s_0 b_0 & s_0 b_1+s_1 b_0 & s_0 b_2+s_1 b_1+s_2 b_0 \\ 0 & -s_1 b_0 & -s_1 b_1-s_2 b_0 \\ 0 & s_0 b_0 & s_0 b_1+s_1 b_0 \end{array}\right| $$ $$= \frac{1}{s_0^2 s_1} \left|\begin{array}{ccc} s_0 b_0 & s_0 b_1+s_1 b_0 & s_0 b_2+s_1 b_1+s_2 b_0 \\ 0 & -s_0 s_1 b_0 & -s_0 s_1 b_1-s_0 s_2 b_0 \\ 0 & 0 & (s_1^2-s_0 s_2)b_0 \end{array}\right| $$ $$= b_0^3 (s_0 s_2-s_1^2) $$ $$= b_0^3 \left|\begin{array}{cc} s_0 & s_1 \\ s_1 & s_2 \end{array}\right|.$$ \noindent Similarly, we obtain $$\mathbb{H}_3=\frac{1}{s_0^4s_1^2(s_1^2-s_0s_2)}\left|\begin{array}{cccc} s_0 b_0 & \ldots & \ldots & \ldots \\ 0 & -s_0s_1 b_0 & \ldots & \ldots \\ 0 & 0 & s_0s_1(s_1^2-s_0s_2)b_0 & \ldots \\ 0 & 0 & 0 & s_0 s_1 (s_2^2-s_1 s_3) b_0 \end{array}\right|,$$ and hence we have $$\mathbb{H}_3= b_0^4 (s^2_2-s_1 s_3)=b_0^4 \left| \begin{array}{cc} s_1 & s_2 \\ s_2 & s_3\end{array}\right|.$$ \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 $$P_n^{(s)} (x)=\Delta_n^{(s)}(1,x,x^2,\cdots,x^n)=\left|\begin{array}{ccccc} s_0 & s_1&s_2 &\cdots & s_n \\ s_1 & s_2&s_3 &\cdots & s_{n+1} \\ \vdots & \vdots & \vdots &\cdots & \vdots \\ s_{n-1} & s_{n-2}&s_{n-3} &\cdots & s_{2n-1} \\ 1 & x&x^2 &\cdots & x^n\end{array}\right|,$$ respectively $$P_n^{(s^*)} (x)=\Delta_n^{(s^*)}(1,x,x^2,\cdots,x^n)=\left|\begin{array}{ccccc} s_1 & s_2&s_3 &\cdots & s_{n+1} \\ s_2 & s_3&s_4 &\cdots & s_{n+2} \\ \vdots & \vdots & \vdots &\cdots & \vdots \\ s_n & s_{n-1}&s_{n-2} &\cdots & s_{2n} \\ 1 & x&x^2 &\cdots & x^n\end{array}\right|,$$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 & a_1 & a_2 & \cdots & a_m & \cdots & a_{2m} \\ b_0 & b_1 & b_2 & \cdots & b_m & \cdots & b_{2m} \\ 0 & a_0 & a_1 & \cdots & a_{m-1} & \cdots & a_{2m-1} \\ 0 & b_0 & b_1 & \cdots & b_{m-1} & \cdots & b_{2m-1} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1 & \cdots & 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 & a_1 & a_2 & \cdots & a_m & \cdots & a_{2m+1} \\ b_0 & b_1 & b_2 & \cdots & b_m & \cdots & b_{2m+1} \\ 0 & a_0 & a_1 & \cdots & a_{m-1} & \cdots & a_{2m} \\ 0 & b_0 & b_1 & \cdots & b_{m-1} & \cdots & b_{2m} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & a_0 & \cdots & a_{m+1} \\0 & 0 & 0 & \cdots & 1 & \cdots & 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), & \text{if $2|j$ and $j