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\begin{center}
\vskip 1cm{\LARGE\bf The Euler-Seidel Matrix, Hankel Matrices \\
\vskip .1in
and Moment Sequences} \vskip 1cm \large
Paul Barry\\
School of Science\\
Waterford Institute of Technology\\
Ireland\\
\href{mailto:pbarry@wit.ie}{\tt pbarry@wit.ie} \\
\ \\
Aoife Hennessy\\
Department of Computing, Mathematics and Physics\\
Waterford Institute of Technology\\
Ireland\\
\href{mailto:aoife.hennessy@gmail.com}{\tt aoife.hennessy@gmail.com} \\
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\vskip .2 in


\begin{abstract} We study the Euler-Seidel matrix of certain integer sequences, using the binomial transform and Hankel matrices. For moment sequences, we give an integral representation of the Euler-Seidel matrix. Links are drawn to Riordan arrays, orthogonal polynomials, and Christoffel-Darboux expressions.
\end{abstract}
\section{Introduction}

The purpose of this note is to investigate the close relationship that exists between the Euler-Seidel matrix \cite{Dil_Kurt_2009, Dil_Kurt, Dil_3, Dumont, Mezo_Dil} of an integer sequence, and the Hankel matrix \cite{Layman} of that sequence. We do so in the context of sequences that have integral moment representations, though many of the results are valid in a more general context. While partly expository in nature, the note assumes a certain familiarity with generating functions, both ordinary and exponential, orthogonal polynomials \cite{Chihara, wgautschi, Szego} and Riordan arrays \cite{SGWW, Spru} (again, both ordinary, where we use the notation $(g,f)$, and exponential, where we use the notation $[g,f]$). Many interesting examples of sequences and Riordan arrays can be found in Neil Sloane's {\it On-Line
Encyclopedia of Integer Sequences},
\cite{SL1,SL2}. Sequences are frequently referred to by their
OEIS number. For instance, the binomial matrix $\mathbf{B}$ is \seqnum{A007318}.

The \emph{Euler-Seidel matrix} of a sequence $(a_n)_{n \ge 0}$, which we will denote by $\mathbf{E}=\mathbf{E}_a$, is defined to be the rectangular array $(a_{n,k})_{n,k \ge 0}$ determined by the recurrence $a_{0,k}=a_k$ ($k\ge 0$) and
\begin{equation}\label{EqBin} a_{n,k}=a_{n,k-1}+a_{n+1,k-1} \quad (n \ge 0, k \ge 1). \end{equation}
The sequence $(a_{0,k})$, the first row of the matrix, is usually called the \emph{initial sequence}, while the sequence $(a_{n,0})$, first column of the matrix, is called the \emph{final sequence}. They are related by the binomial transform (or Euler transform, after Euler, who first proved this \cite{Euler}). We recall that the binomial transform of a sequence $a_n$ has general term $b_n=\sum_{k=0}^n \binom{n}{k}a_k$. Thus the first row and column of the matrix are determined from Eq. (\ref{EqBin}) as follows:
\begin{eqnarray} a_{n,0}&amp;=&amp;\sum_{k=0}^n \binom{n}{k} a_{0,k}, \\a_{0,n}&amp;=&amp;\sum_{k=0}^n \binom{n}{k}(-1)^{n-k}a_{k,0}.\end{eqnarray}
In general, we have
\begin{equation}\label{EqMat}
a_{n,k}=\sum_{i=0}^n \binom{n}{i}a_{i+k,0}=\sum_{i=0}^n \binom{n}{i}a_{i+k}. \end{equation}

\begin{example}
We take $a_{0,n}=a_n=C_n=\frac{1}{n+1}\binom{2n}{n}$, the Catalan numbers. Thus the initial sequence, or first row, is the Catalan numbers, while the final sequence, or first column, will be the binomial transform of the Catalan numbers. We obtain the following matrix:
\begin{displaymath}\left(\begin{array}{ccccccc}
1 &amp; 1 &amp; 2 &amp; 5 &amp; 14 &amp; 42 &amp; \ldots \\
2 &amp; 3 &amp; 7 &amp; 19 &amp; 56 &amp; 174 &amp; \ldots \\
5 &amp; 10 &amp; 26 &amp; 75 &amp; 230 &amp; 735 &amp; \ldots \\
15 &amp; 36 &amp; 101 &amp; 305 &amp; 965 &amp; 3155 &amp; \ldots \\
51 &amp; 137 &amp; 406 &amp; 1270 &amp; 4120 &amp; 13726 &amp; \ldots \\
188 &amp; 543 &amp; 1676 &amp; 5390 &amp; 17846 &amp; 60398
&amp;\ldots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath}
We now remark that the Catalan numbers $C_n$ \seqnum{A000108} have the following moment representation:
\begin{equation}\label{IntCat} C_n=\frac{1}{2 \pi} \int_0^4 x^n \frac{\sqrt{x(4-x)}}{x} \,dx.\end{equation}
\end{example}
\noindent For the rest of this note, we shall assume that all sequences discussed have a moment representation of the form
$$a_n= \int_{\mathbf{R}} x^n d\mu_a $$ for a suitable measure $d\mu_a$. The reader is directed to the Appendix for the link between the generating function of a sequence and the corresponding measure, when it exits.

We give some examples of such sequences.
\begin{example} \textbf{The aerated Catalan numbers}.

We have seen (see Eq. (\ref{IntCat})) that the Catalan numbers are a moment sequence. Similarly, the
aerated Catalan numbers
$$1,0,1,0,2,0,5,0,14,\ldots$$ can by represented by
$$C_{\frac{n}{2}}\frac{1+(-1)^n}{2}=\frac{1}{2\pi} \int_{-2}^{2} x^n \sqrt{4-x^2} \,dx.$$
This is the famous semi-circle distribution, of importance in random matrix theory.
\end{example}
\begin{example} \textbf{The factorial numbers $n!$}.

We have the well-known integral representation of $n!$ \seqnum{A000142}
$$n! = \int_{0}^{\infty} x^n e^{-x}\, dx.$$
\end{example}
\begin{example} \textbf{The aerated double factorials}.

We recall that the double factorials \seqnum{A001147} are given by
$$(2n-1)!!=\prod_{k=1}^n (2k-1) =\frac{(2n)!}{n!\,2^n}.$$
The aerated double factorials, which begin
$$1,0,3,0,5,0,15, 0,105,0, 945, \ldots $$ have integral representation
$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} x^n\, e^{-\frac{x^2}{2}}\,dx.$$
The aerated double factorial numbers have exponential generating function $e^{\frac{x^2}{2}}$.
(Note that the numbers $2^n\cdot n!$ \seqnum{A000165} are also called double factorials).
\end{example}
\section{The Euler-Seidel and Hankel matrix for moment sequences}
We recall that for a sequence $(a_n)_{n\ge 0}$, its Hankel matrix is the matrix $\mathbf{H}=\mathbf{H}_a$ with general term $a_{n+k}$. Note that if $a_n$ has o.g.f. $A(x)$ then the bivariate generating function of $\mathbf{H}_a$ is given by
$$ \frac{xA(x)-yA(y)}{x-y}.$$

If $a_n$ has an exponential generating function $G(x)$, then the $n$-th row (and $n$-th column) of $\mathbf{H}_a$ has exponential generating function given by
$$\frac{d^n}{dx^n}G(x).$$
\begin{example}
We have seen that the aerated double factorial numbers have e.g.f. $e^{\frac{x^2}{2}}.$
Thus the $n$-th row of the Hankel matrix associated to them has e.g.f.
$$\frac{d^n}{dx^n} e^{\frac{x^2}{2}}=e^{\frac{x^2}{2}} \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2}(2k-1)!! x^{n-2k}.$$
(We note that written as
$$\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2}(2k-1)!!=e^{-\frac{x^2}{x}}\frac{d^n}{dx^n} e^{\frac{x^2}{x}},$$ this is a statement about scaled Hermite polynomials.)
\end{example}
\noindent  Note that if
$$a_n = \int x^n d\mu_a$$ then
$$a_{n+k}=\int x^{n+k} d\mu_a = \int x^nx^k d\mu_a.$$ The binomial matrix is the matrix $\mathbf{B}$ with general term $\binom{n}{k}$. The binomial transform of a sequence $a_n$ is the sequence with general term
$$b_n=\sum_{k=0}^n \binom{n}{k}a_k.$$
In this case, the sequence $b_n$ has o.g.f. given by
$$\frac{1}{1-x} A\left(\frac{x}{1-x}\right).$$ The sequence $(b_n)_{n\ge 0}$ can be viewed as
$$\mathbf{B}\cdot (a_n)^t.$$
Note that we have
\begin{eqnarray*} b_n&amp;=&amp;\sum_{k=0}^n \binom{n}{k}a_k\\
&amp;=&amp;\sum_{k=0}^n \binom{n}{k}\int x^k \,d\mu_a\\
&amp;=&amp;\int \sum_{k=0}^n \binom{n}{k} x^k \,d\mu_a\\
&amp;=&amp; \int (1+x)^n \, d\mu_a.\end{eqnarray*}
In similar fashion, we have
$$a_n = \int (x-1)^n \,d\mu_b.$$
\begin{proposition} We have
\begin{equation}\label{EqBH}
\mathbf{E}_a=\mathbf{B}\mathbf{H}_a.\end{equation} \end{proposition}
\begin{proof}
We have
$$\mathbf{B}\mathbf{H}_a=\left(\binom{n}{k}\right)\cdot \left(a_{n+k}\right).$$
The result follows from Eq. (\ref{EqMat}).
\end{proof}
\noindent We now let
$$b_n=\sum_{k=0}^n\binom{n}{k} a_k,$$ the binomial transform of $a_n$.
In the sequel, we shall be interested in the product $\mathbf{B}^{-1}\mathbf{H}_b$.

\begin{example} Taking $b_n=\sum_{k=0}^n \binom{n}{k}C_k$ \seqnum{A0007317}, the binomial transform of the Catalan numbers $C_n$, we obtain
\begin{displaymath}\mathbf{H}_b=\left(\begin{array}{ccccccc}
1 &amp; 2 &amp; 5 &amp; 15 &amp; 51 &amp; 188 &amp; \ldots \\
2 &amp; 5 &amp; 15 &amp; 51 &amp; 188 &amp; 731 &amp; \ldots \\
5 &amp; 15 &amp; 51 &amp; 188 &amp; 731 &amp; 2950 &amp; \ldots \\
15 &amp; 51 &amp; 188 &amp; 731 &amp; 2950 &amp; 12235 &amp; \ldots \\
51 &amp; 188 &amp; 731 &amp; 2950 &amp; 12235 &amp; 51822 &amp; \ldots \\
188 &amp; 731 &amp; 2950 &amp; 12235 &amp; 51822 &amp; 223191
&amp;\ldots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right)\end{displaymath} Multiplying by $\mathbf{B}^{-1}$, we obtain
\begin{displaymath}\mathbf{B}^{-1}\mathbf{H}_b=\left(\begin{array}{ccccccc} 1 &amp; 2 &amp;
5
&amp; 15 &amp; 51 &amp; 188 &amp; \ldots \\1 &amp; 3 &amp; 10 &amp; 36 &amp; 137 &amp; 543 &amp; \ldots \\ 2 &amp; 7
&amp; 26 &amp; 101 &amp; 406 &amp;
1676 &amp; \ldots \\ 5 &amp; 19 &amp; 75 &amp; 305 &amp; 1270 &amp; 5390 &amp; \ldots \\ 14 &amp; 56 &amp; 230
&amp; 965 &amp; 4120 &amp; 17846 &amp; \ldots \\42 &amp; 174 &amp; 735 &amp; 3155 &amp; 13726 &amp; 60398
&amp;\ldots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right),\end{displaymath}
which is the transpose of the Euler-Seidel matrix for $C_n$. \end{example}
This result is general. In order to prove this, we will use the follow lemma.
\begin{lemma}
\begin{equation} \label{Eqkn}
x^n(1+x)^k=\sum_{i=0}^k \binom{k}{i}x^{i+n}=\sum_{j=0}^n (-1)^{n-j}\binom{n}{j}\sum_{i=0}^{j+k} \binom{j+k}{i}x^i.\end{equation}
\end{lemma}
\begin{proof}

Since $(1+x)^k=\sum_{i=0}^k \binom{k}{i}x^i$ by the binomial theorem, we immediately have
$$x^n(1+x)^k=x^n\sum_{i=0}^k \binom{k}{i}x^i=\sum_{i=0}^k \binom{k}{i}x^{i+n}.$$
But also, we have
\begin{eqnarray*}
\sum_{j=0}^n (-1)^{n-j}\binom{n}{j}\sum_{i=0}^{j+k} \binom{j+k}{i}x^i&amp;=&amp;\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}(1+x)^{j+k}\\
&amp;=&amp;(1+x)^k \sum_{j=0}^n(-1)^{n-j}\binom{n}{j}(1+x)^j\\
&amp;=&amp;(1+x)^k x^n.\end{eqnarray*}

\end{proof}



\begin{proposition}
The Euler-Seidel matrix of the sequence $(a_n)_{n\ge 0}$ is equal to the transpose of the matrix given by
$\mathbf{B}^{-1}\mathbf{H}_b$,
where $\mathbf{H}_b$ is the Hankel matrix of the binomial transform $$b_n=\sum_{k=0}^n \binom{n}{k}a_k$$ of the initial sequence $a_n$. That is,
\begin{equation}\label{BiH} \mathbf{E}_a^t=\mathbf{B}^{-1}\mathbf{H}_b.\end{equation}
\end{proposition}
\begin{proof}
The general element of
$$\mathbf{B}^{-1}\mathbf{H}=\left((-1)^{n-k}\binom{n}{k}\right)\cdot \left (b_{n+k}\right)$$ is given by
$$\sum_{j=0}^n (-1)^{n-j}\binom{n}{j}b_{j+k}.$$ Now
\begin{eqnarray*}\sum_{j=0}^n (-1)^{n-j}\binom{n}{j}b_{j+k}&amp;=&amp;\sum_{j=0}^n (-1)^{n-j}\binom{n}{j}\sum_{i=0}^{j+k}\binom{j+k}{i}a_i.\end{eqnarray*}
We thus wish to show that
$$\sum_{j=0}^n (-1)^{n-j}\binom{n}{j}\sum_{i=0}^{j+k}\binom{j+k}{i}a_i=\sum_{i=0}^k \binom{k}{i}a_{i+n}.$$ To this end, we assume that
$$a_n = \int x^n d\mu_a, $$ and observe that
\begin{eqnarray*}
\sum_{i=0}^k \binom{k}{i}a_{i+n}&amp;=&amp;\sum_{i=0}^k \binom{k}{i}\int x^{i+n}d\mu_a\\
&amp;=&amp;\int \sum_{i=0}^k \binom{k}{i}x^{i+n}\,d\mu_a \\
&amp;=&amp; \int \sum_{j=0}^n (-1)^{n-j}\binom{n}{j}\sum_{i=0}^{j+k} \binom{j+k}{i}x^i \,d\mu_a\\
&amp;=&amp;\sum_{j=0}^n (-1)^{n-j}\binom{n}{j}\sum_{i=0}^{j+k} \binom{j+k}{i}a_i. \end{eqnarray*}

The result now follows from Eq. (\ref{EqMat}).
\end{proof}
\begin{corollary}
\begin{equation}\label{Enk} \mathbf{E}_a = \left(\int x^k(1+x)^n d\mu_a\right).\end{equation}
\end{corollary}
We are interested in characterising the main diagonal of the Euler-Seidel matrix of $a_n$, which by the above is the same as the main diagonal of $\mathbf{B}^{-1}\mathbf{H}_b$, where $\mathbf{H}$ is the Hankel matrix of $b_n$, the binomial transform of $a_n$.

Note that the diagonal is given by
$$a_{n,n}=\sum_{i=0}^n \binom{n}{i}a_{n+i}.$$

\begin{example} We have seen that the diagonal of the Euler-Seidel matrix for the Catalan numbers $C_n$ begins
$$1, 3, 26, 305, 4120, 60398, 934064,  \ldots$$ By the above, the general term of this sequence is
$$d_n=\sum_{i=0}^n \binom{n}{i}C_{n+i}.$$
Now consider the moment representation of the Catalan numbers given by
$$C_n= \int x^n d\mu = \frac{1}{2\pi}\int_{0}^4 x^n \frac{\sqrt{x(4-x)}}{x} dx.$$
We claim that
$$d_n =\int (x(1+x))^n d\mu = \frac{1}{2\pi}\int_{0}^4 (x(1+x))^n  \frac{\sqrt{x(4-x)}}{x} dx.$$ This follows from the result above, or directly, since
\begin{eqnarray*}
\int (x(1+x))^n d\mu&amp;=&amp;=\int (x+x^2)^n d\mu\\
&amp;=&amp;\int \sum_{i=0}^n \binom{n}{i}x^{2i}x^{n-i} d\mu \\
&amp;=&amp; \sum_{i=0}^n \binom{n}{i} \int x^{n+i} d\mu \\
&amp;=&amp; \sum_{i=0}^n \binom{n}{i} C_{n+i}.\end{eqnarray*}
\end{example}
\noindent Note that by the change of variable $y=x(1+x)$ we obtain in this case the alternative moment representation for $d_n$ given by 
$$d_n=\frac{1}{2\pi} \int_{0}^{20} y^n \frac{\sqrt{2}(1+\sqrt{1+4y})\sqrt{5\sqrt{1+4y}-2y-5}}{4y\sqrt{1+4y}}\,dy.$$
The above method of proof is easily generalised. Thus we have
\begin{proposition}
Let $a_n$ be a sequence which can be represented as the sequence of moments of a measure:
$$a_n=\int x^n d\mu_a.$$
Then the elements $d_n$ of the main diagonal of the Euler-Seidel matrix have moment representation given by
$$d_n = \int (x(1+x))^n d\mu_a.$$
\end{proposition}
\section{Examples: the Fibonacci and Jacobsthal cases}
\begin{example} We first of all look at the Fibonacci numbers, $F_{n+1}=F(n+1)$ \seqnum{A000045}.  It is well known that the binomial transform of $F(n+1)$ is $F(2n+1)$ \seqnum{A122367}:
$$F(2n+1)=\sum_{k=0}^n \binom{n}{k}F(k+1).$$
Letting $\phi=\frac{1+\sqrt{5}}{2}$ and $\bar{\phi}=\frac{1}{\phi}=\frac{1-\sqrt{5}}{2}$, we have
\begin{eqnarray*}F_{n+1}&amp;=&amp;
\frac{1}{\sqrt{5}}(\phi^{n+1}-\bar{\phi}^{n+1})\\
&amp;=&amp;
\frac{1}{\sqrt{5}}(\phi\phi^{n}-\bar{\phi}\bar{\phi}^{n})\\
&amp;=&amp;\frac{1}{\sqrt{5}}\int_{\mathbf{R}} x^n (\phi \delta_{\phi}-\bar{\phi}\delta_{\bar{\phi}}) \,dx.\end{eqnarray*}
Then the general element of the Euler-Seidel matrix for $F_{n+1}$ is given by
\begin{eqnarray*}\frac{1}{\sqrt{5}}\int_{\mathbf{R}} x^k (1+x)^n (\phi \delta_{\phi}-\bar{\phi}\delta_{\bar{\phi}}) \,dx&amp;=&amp;\frac{\phi \cdot \phi^k(1+\phi)^n-\bar{\phi}\cdot\bar{\phi}^k(1+\bar{\phi})^n}{\sqrt{5}}\\
&amp;=&amp;\frac{ \phi^{k+1}(1+\phi)^n-\bar{\phi}^{k+1}(1+\bar{\phi})^n}{\sqrt{5}},\end{eqnarray*} where
$\delta_a$ represents the Dirac delta ``function'' at $a$:
$$\int_{\mathbf{R}} f(x)\delta_a dx = &lt;\delta_a,f&gt;=f(a).$$
This gives us
\begin{displaymath}\mathbf{E}_{F_{n+1}}=\left(\begin{array}{ccccccc}
1 &amp; 1 &amp; 2 &amp; 3 &amp; 5 &amp; 8 &amp; \ldots \\
2 &amp; 3 &amp; 5 &amp; 8 &amp; 13 &amp; 21 &amp; \ldots \\
5 &amp; 8 &amp; 13 &amp; 21 &amp; 34 &amp; 55 &amp; \ldots \\
13 &amp; 21 &amp; 34 &amp; 55 &amp; 89 &amp; 144 &amp; \ldots \\
34 &amp; 55 &amp; 89 &amp; 144 &amp; 233 &amp; 377 &amp; \ldots \\
89 &amp; 144 &amp; 233 &amp; 377 &amp; 610 &amp; 987
&amp;\ldots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath}
The diagonal of this matrix begins
$$1,3,13,55,233,\ldots$$
This is $F_{3n+1}$ \seqnum{A033887}, which by the above can be written
$$F_{3n+1}=\frac{ \phi^{n+1}(1+\phi)^n-\bar{\phi}^{n+1}(1+\bar{\phi})^n}{\sqrt{5}}.$$ \end{example}
\begin{example}
Our next example is based on the Jacobsthal numbers $J_n$ \seqnum{A001045}, where
$$ J_n= \frac{2^n}{3}-\frac{(-1)^n}{3}.$$ We have
$$J_{n+1}=\frac{2\cdot2^n}{3}+\frac{(-1)^n}{3}=\frac{1}{3}\int_{\mathbf{R}}x^n \, (2\delta_2+\delta_{-1})\,dx .$$ Then the Euler-Seidel matrix for $J_{n+1}$ has general element given by
\begin{eqnarray*} \int x^k(1+x)^n \,(2\delta_2+\delta_{-1})\,dx&amp;=&amp;\frac{2\cdot 2^k (1+2)^n + (-1)^k (1+(-1))^n}{3}\\
&amp;=&amp;\frac{2^{k+1}3^n+(-1)^k\, 0^n}{3}.\end{eqnarray*}
We have
\begin{displaymath}\mathbf{E}_{J_{n+1}}=\left(\begin{array}{ccccccc}
1 &amp; 1 &amp; 3 &amp; 5 &amp; 11 &amp; 21 &amp; \ldots \\
2 &amp; 4 &amp; 8 &amp; 16 &amp; 32 &amp; 64 &amp; \ldots \\
6 &amp; 12 &amp; 24 &amp; 48 &amp; 96 &amp; 192 &amp; \ldots \\
18 &amp; 36 &amp; 72 &amp; 144 &amp; 288 &amp; 576 &amp; \ldots \\
54 &amp; 108 &amp; 216 &amp; 432 &amp; 864 &amp; 1728 &amp; \ldots \\
162 &amp; 324 &amp; 648 &amp; 1296 &amp; 2592 &amp; 5184
&amp;\ldots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath}
The diagonal sums, which begin
$$1,4,24,144,864,\ldots$$ are
\seqnum{A067411}, with general term
$$\frac{2^{n+1}\,3^n+(-1)^n\,0^n}{3}=\frac{2\cdot6^n+0^n}{3}.$$
\end{example}
\section{Hankel matrices, Riordan arrays and orthogonal polynomials}
From the last section, we have
$$\mathbf{E}_a = \mathbf{B}\mathbf{H}_a$$ and
$$\mathbf{E}_a^t=\mathbf{B}^{-1}\mathbf{H}_b$$ where $b_n$ is the binomial transform of $a_n$. The second equation shows us that
\begin{eqnarray*}\mathbf{E}_a&amp;=&amp;(\mathbf{B}^{-1}\mathbf{H}_b)^t\\
&amp;=&amp;\mathbf{H}_b^t (\mathbf{B}^{-1})^t\\
&amp;=&amp;\mathbf{H}_b (\mathbf{B}^t)^{-1},\end{eqnarray*} since
$\mathbf{H}_b$ is symmetric.
Thus we obtain
$$\mathbf{B}\mathbf{H}_a=\mathbf{H}_b (\mathbf{B}^t)^{-1},$$ which implies that
\begin{equation} \label{BHB} \mathbf{H}_b=\mathbf{B}\mathbf{H}_a\mathbf{B}^t. \end{equation}
Since $\text{det}(\mathbf{B})=1$, we deduce the well-known result that the Hankel transform of $b_n$ is equal to that of $a_n$. We can also use this result to relate the $LDU$ decomposition of $\mathbf{H}_b$ \cite{P_W, Woan} to that of $\mathbf{H}_a$. Thus we have
\begin{eqnarray*} \mathbf{H}_b&amp;=&amp;\mathbf{B}\mathbf{H}_a \mathbf{B}^t\\
&amp;=&amp; \mathbf{B}\cdot\mathbf{L}_a \mathbf{D}_a \mathbf{L}_a^t \cdot \mathbf{B}^t\\
&amp;=&amp; (\mathbf{B}\mathbf{L}_a) \mathbf{D}_a (\mathbf{B}\mathbf{L}_a)^t.\end{eqnarray*}

One consequence of this is that the coefficient triangle of the polynomials orthogonal with respect to $d\mu_b$ is given by
$$\mathbf{L}_a^{-1}\mathbf{B}^{-1},$$ where $\mathbf{L}_a^{-1}$ is the coefficient array of the polynomials orthogonal with respect to $d\mu_a$.
\begin{example}
We take the example of the Catalan numbers $a_n=C_n$ and their binomial transform $b_n=\sum_{k=0}^n C_k$. It is well known that the Hankel transform of $C_n$ is the all $1$'s sequence, which implies that $\mathbf{D}_a$ is the identity matrix. Thus in this case,
$$\mathbf{H}_a=\mathbf{L_a}\mathbf{L_a}^t$$ where
$$\mathbf{L}_a=\mathbf{L}_{C_n}=(c(x), xc(x)^2) \quad\quad (\text{\seqnum{A039599}}) $$ with
$$\mathbf{L}_a^{-1}=\left(\frac{1}{1+x},\frac{x}{(1+x)^2}\right) \quad\quad (\text{\seqnum{A129818}}) $$ with general term $(-1)^{n-k}\binom{n+k}{2k}$, where we have used the notation of Riordan arrays.
The polynomials
$$P_n(x)=\sum_{k=0}^n (-1)^{n-k}\binom{n+k}{2k}x^k$$ are thus a  family of polynomials orthogonal on $[0,4]$ with respect to the density function $\frac{1}{2\pi} \frac{\sqrt{x(4-x)}}{x}$ \cite{Morgan-Voyce}.
It is known that the bi-variate generating function of the inverse of the $n$-th principal minor of $\mathbf{H}_a$ is given by the Christoffel-Darboux quotient
$$\frac{P_{n+1}(x)P_n(y)-P_{n+1}(y)P_n(x)}{x-y}.$$
We deduce that the orthogonal polynomials defined by $b_n$ have coefficient matrix
\begin{eqnarray*}\mathbf{L}_b^{-1}&amp;=&amp;\mathbf{L}_a^{-1}\mathbf{B}^{-1}\\
&amp;=&amp;\left(\frac{1}{1+x},\frac{x}{(1+x)^2}\right) \cdot \left(\frac{1}{1+x}, \frac{x}{1+x}\right) \\
&amp;=&amp; \left(\frac{1+x}{1+3x+x^2}, \frac{x}{1+3x+x^2}\right).\end{eqnarray*}
It turns out that these polynomials $Q_n(x)$ are given simply by
$$Q_n(x)=P_n(x-1).$$
Thus $\mathbf{H}_b^{-1}$ has $n$-th principal minor generated by
$$\frac{Q_{n+1}(x)Q_n(y)-Q_{n+1}(y)Q_n(x)}{x-y}.$$
In similar manner, we can deduce that the Euler-Seidel matrix $\mathbf{E}_a=\mathbf{E}_{C_n}$ is such that the $n$-th principal minor of $\mathbf{E}_a^{-1}$ is generated by
$$\frac{P_{n+1}(x)P_n(y-1)-P_{n+1}(y-1)P_n(x)}{x-y}=\frac{P_{n+1}(x)Q_n(y)-Q_{n+1}(y)P_n(x)}{x-y}.$$
\end{example}
\begin{example}
For the aerated double factorials, we have
$$\mathbf{H}_a = \mathbf{L}_a \mathbf{D}_a \mathbf{L}_a^t$$
where $$\mathbf{L}_a=[e^{\frac{x^2}{2}},x], \quad \mathbf{D}_a=\text{diag}(n!).$$
The associated orthogonal polynomials (which are scaled Hermite polynomials) have coefficient matrix
$$\mathbf{L}_a^{-1}=[e^{-\frac{x^2}{2}},x],$$ and we have
$$P_n(x)=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k}(2k-1)!!(-1)^k x^{n-2k},$$ or equivalently,
$$P_n(x)=\sum_{k=0}^n Bessel^*\left(\frac{n+k}{2},k\right)(-1)^{\frac{n-k}{2}}\frac{1+(-1)^{n-k}}{2}x^k,$$
where $$Bessel^*(n,k)=\frac{(2n-k)!}{k!(n-k)!2^{n-k}}=\binom{n+k}{2k}(2k-1)!!,$$ (see \cite{BesRef}).
With $$Q_n(x)=P_n(x-1)$$ we again have that
the Euler-Seidel matrix $\mathbf{E}_a$ is such the $n$-th principal minor of $\mathbf{E}_a^{-1}$ is generated by
$$\frac{P_{n+1}(x)P_n(y-1)-P_{n+1}(y-1)P_n(x)}{x-y}=\frac{P_{n+1}(x)Q_n(y)-Q_{n+1}(y)P_n(x)}{x-y}.$$
\end{example}
\section{Appendix: The Stieltjes transform of a measure} The
\emph{Stieltjes transform} of a measure $\mu$ on $\mathbb{R}$
is a function $G_{\mu}$
defined on $\mathbb{C}\setminus \mathbb{R}$ by
$$G_{\mu}(z)=\int_{\mathbb{R}}\frac{1}{z-t}\mu(t).$$ If $f$ is
a bounded continuous function on
$\mathbb{R}$, we have $$\int_{\mathbb{R}}f(x)\mu(x)=-\lim_{y
\to 0^{+}}\int_{\mathbb{R}}f(x)\Im G_{\mu}(x+iy)dx.$$ If $\mu$
has compact
support, then $G_{\mu}$ is holomorphic at infinity and for
large $z$, $$G_{\mu}(z)=\sum_{n=0}^{\infty}
\frac{a_n}{z^{n+1}},$$ where
$a_n=\int_{\mathbb{R}}t^n \mu(t)$ are the moments of the
measure. If $\mu(t)=d\psi(t)=\psi'(t)dt$ then
$$\psi(t)-\psi(t_0)=-\frac{1}{\pi}\lim_{y \to
0^{+}}\int_{t_0}^t \Im G_{\mu}(x+iy)dx.$$ If now $g(x)$ is the
generating function of a
sequence $a_n$, with $g(x)=\sum_{n=0}^{\infty}a_n x^n$, then
we
can define
$$G(z)=\frac{1}{z}g\left(\frac{1}{z}\right)=\sum_{n=0}^{\infty}\frac{a_n}{z^{n+1}}.$$
By this means, under the right circumstances we can
retrieve the density function for the measure that defines the
elements $a_n$ as moments.
\begin{example}
We let $g(z)=\frac{1-\sqrt{1-4z}}{2z}$ be the g.f. of the Catalan numbers. Then
$$G(z)=\frac{1}{z}g\left(\frac{1}{z}\right)=\frac{1}{2}\left(1-\sqrt{\frac{x-4}{x}}\right).$$
Then $$\Im G_{\mu}(x+iy)=-\frac{\sqrt{2}\sqrt{\sqrt{x^2+y^2}\sqrt{x^2-8x+y^2+16}-x^2+4x-y^2}}{4\sqrt{x^2+y^2}},$$
and so we obtain \begin{eqnarray*}\psi'(x)&amp;=&amp;-\frac{1}{\pi}\lim_{y \to
0^{+}}\left\{-\frac{\sqrt{2}\sqrt{\sqrt{x^2+y^2}\sqrt{x^2-8x+y^2+16}-x^2+4x-y^2}}{4\sqrt{x^2+y^2}}\right\}\\
&amp;=&amp; \frac{1}{2\pi}\frac{\sqrt{x(4-x)}}{x}.\end{eqnarray*}
\end{example}
\section{Acknowledgements} The author would like to thank the anonymous referees for their careful reading and cogent suggestions which have hopefully
led to clearer paper.
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\end{thebibliography}

\bigskip
\hrule
\bigskip
\noindent 2010 {\it Mathematics Subject Classification}:
Primary
11B83; Secondary  11B39, 11C20, 15B05, 15B36, 42C05.

\noindent \emph{Keywords:} Euler-Seidel matrix, Hankel matrix, integer sequence, moment sequence, orthogonal polynomial,
Riordan array.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000045},
\seqnum{A000108},
\seqnum{A000142},
\seqnum{A000165},
\seqnum{A000731},
\seqnum{A001045},
\seqnum{A001147},
\seqnum{A007317},
\seqnum{A007318},
\seqnum{A033887},
\seqnum{A039599},
\seqnum{A067411},
\seqnum{A122367}, and
\seqnum{A129818}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received April 27 2010;
revised version received   July 21 2010.
Published in {\it Journal of Integer Sequences}, August 2 2010.

\bigskip
\hrule
\bigskip

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