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\begin{center}
\vskip 1cm{\LARGE\bf The Restricted Toda Chain, Exponential \\
\vskip .1in
Riordan Arrays, and Hankel Transforms} \vskip 1cm \large
Paul Barry\\
School of Science\\
Waterford Institute of Technology\\
Ireland\\
\href{mailto:pbarry@wit.ie}{\tt pbarry@wit.ie} \\

\end{center}
\vskip .2 in


\begin{abstract} We re-interpret results on the classification
of Toda chain solutions given by Sheffer class orthogonal
polynomials in terms of exponential Riordan arrays. We also
examine associated Hankel transforms.
\end{abstract}
\section{Introduction}


The restricted Toda chain equation \cite{Nakamura, Vinet} is
simply described by
\begin{equation}\label{Toda} \dot{u}_n=u_n(b_n-b_{n-1}),\quad
n=1,2,\ldots; \quad \dot{b}_n =u_{n+1}-u_{n}, \quad
n=0,1,\ldots\end{equation} with $u_0=0$, where the dot
indicates differentiation with respect to $t$. In this note, we
shall show how solutions to this equation can be formulated in
the context of exponential Riordan arrays. The Riordan arrays
we shall consider may be considered as parameterized (or
``time''-dependent) Riordan arrays. We have already considered
parameterized Riordan arrays \cite{PB_Moments}, exploring the links between these Riordan
arrays and orthogonal polynomials.

The restricted Toda chain equation is closely related to
orthogonal polynomials, since the functions $u_n$ and $b_n$ can
be considered as the coefficients in the usual three-term
recurrence \cite{Chihara, wgautschi, Szego} satisfied by orthogonal polynomials:
\begin{equation}\label{TT} P_{n+1}(x)+b_n P_n(x)+u_n
P_{n-1}(x)=xP_n(x), \quad n=1,2,\ldots \end{equation}
with initial conditions $P_0(x)=1$ and $P_1(x)=x-b_0$.

Nakamura and Zhedanov \cite{Nakamura, Nakamura_Special} give solutions of the
restricted Toda chain equation in terms of Sheffer polynomials. Given
the close links between Sheffer class polynomials and Riordan
arrays \cite{DR, He_Sheffer}, we find it instructive to
re-interpret these results in terms of Riordan arrays. Thus in the cases discussed below, we find that the coefficient arrays of the orthogonal polynomials linked to solutions of the restricted Toda chain are exponential Riordan arrays. We believe that this result concerning an application of exponential Riordan arrays to the Toda chain equation is new. In particular, the parameters of the solution are obtained by calculating the terms of the corresponding production matrices, using the procedures of Deutsch, Ferrari, and Rinaldi \cite{ProdMat_0, ProdMat}.

The reader is referred to \cite{Chihara, wgautschi, Szego} for
basic information on orthogonal polynomials (see also
\cite{Roman}), to \cite{PasTri, Lah, DeutschShap, SGWW} for
details of Riordan arrays,  to \cite{ProdMat_0, ProdMat} for
information on production (Stieltjes) matrices (see also
\cite{P_W}), and to \cite{CRI, Layman, Woan} for examples of
Hankel transforms. Many interesting examples of Riordan arrays
(both ordinary and exponential), and Hankel transforms, can be
found in Neil Sloane's On-Line
Encyclopedia of Integer Sequences (OEIS) \cite{SL1, SL2}. Integer equences are frequently referred to by their OEIS number. For instance, the binomial matrix $\mathbf{B}$ (``Pascal's triangle'')
is \seqnum{A007318}.

\section{Integer
sequences, Hankel transforms, exponential Riordan arrays,
orthogonal polynomials} In this section, we recall known
results on integer
sequences, Hankel transforms, exponential Riordan arrays and
orthogonal polynomials that will be useful for the sequel.

For an integer sequence $a_n$, that is, an element of
$\mathbb{Z}^\mathbb{N}$, the power series
$f_o(x)=\sum_{k=0}^{\infty}a_k x^k$ is called the
\emph{ordinary generating function} or
g.f.
of the sequence, while
$f_e(x)=\sum_{k=0}^{\infty}\frac{a_k}{k!} x^k$ is called the
\emph{exponential generating function} or e.g.f.
of
the sequence. $a_n$ is thus the coefficient of $x^n$ in
$f_o(x)$. We denote this by $a_n=[x^n]f_o(x)$. Similarly,
$a_n=n![x^n]f_e(x)$.
For instance, $F_n=[x^n]\frac{x}{1-x-x^2}$ is the $n$-th
Fibonacci number \seqnum{A000045}, while
$n!=n![x^n]\frac{1}{1-x}$, which
says
that $\frac{1}{1-x}$ is the e.g.f. of $n!$ \seqnum{A000142}.
For a power series $f(x)=\sum_{n=0}^{\infty}a_n x^n$ with
$f(0)=0$ and
$f'(0)\neq 0$ we
define the reversion or compositional inverse of $f$ to be the
power series $\bar{f}(x)=f^{[-1]}(x)$ such that
$f(\bar{f}(x))=x$. We
sometimes write $\bar{f}= \text{Rev}f$.

The {\it
Hankel
transform} \cite{Layman} of a given sequence
$A=(a_n)_{n \ge 0}$ is the
sequence of Hankel determinants $\{h_0, h_1, h_2,\dots \}$
where
$h_{n}=|a_{i+j}|_{i,j=0}^{n}$, i.e

\begin{center} \begin{equation}
 \label{gen1}
 A=(a_n)_{n\in\mathbb N_0}\quad \rightarrow \quad
 h=(h_n)_{n\in\mathbb N_0}:\quad
h_n=\left| \begin{array}{ccccc}
 a_0\ &amp; a_1\  &amp; \cdots &amp; a_n  &amp;  \\
 a_1\ &amp; a_2\  &amp;        &amp; a_{n+1}  \\
\vdots &amp;      &amp; \ddots &amp;          \\
 a_n\ &amp; a_{n+1}\ &amp;    &amp; a_{2n}
\end{array} \right|. \end{equation} \end{center} The Hankel
transform of a sequence $a_n$ and that of its binomial transform are
equal.

In the case that $a_n$ has g.f. $g(x)$ expressible in the form
$$g(x)=\cfrac{a_0}{1-\alpha_0 x-
\cfrac{\beta_1 x^2}{1-\alpha_1 x-
\cfrac{\beta_2 x^2}{1-\alpha_2 x-
\cfrac{\beta_3 x^2}{1-\alpha_3 x-\cdots}}}},$$ (with $\beta_i
\neq 0$ for all $i$) then
we have \cite{Kratt, Kratt1, Wall}
\begin{equation}h_n = a_0^n \beta_1^{n-1}\beta_2^{n-2}\cdots
\beta_{n-1}^2\beta_n=a_0^n\prod_{k=1}^n
\beta_k^{n-k+1}.\end{equation}
Note that this is independent of $\alpha_n$. In general
$\alpha_n$ and $\beta_n$ are not integers.
Such a continued fraction is associated to a monic family of
orthogonal polynomials which obey the three term recurrence
$$p_{n+1}(x)=(x-\alpha_n)p_n(x)-\beta_n p_{n-1}(x), \qquad
p_0(x)=1,\qquad p_1(x)=x-\alpha_0.$$
\noindent The terms appearing in the first column of the
inverse of the coefficient array of these polynomials are the
moments of
family.

The \emph{exponential Riordan group} \cite
{PasTri,DeutschShap,ProdMat}, is a set of infinite
lower-triangular integer matrices, where each matrix is defined
by a pair of generating functions $g(x)=g_0+g_1x+g_2x^2+\ldots$
and
$f(x)=f_1x+f_2x^2+\ldots$ where $f_1\ne 0$. The associated
matrix is the matrix whose $i$-th column has exponential
generating
function
$g(x)f(x)^i/i!$ (the first column being indexed by 0). The
matrix corresponding to the pair $f, g$ is denoted by $[g, f]$.
It is
\emph{monic} if $g_0=1$. If in addition $f_1=1$, then it is \emph{proper}. The group law is  given by
\begin{displaymath} [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 $f$. We use the notation
$\mathit{e}\mathcal{R}$ to denote this group. If $\mathbf{M}$
is the matrix $[g,f]$,
and $\mathbf{u}=(u_n)_{n \ge 0}$ is an integer sequence with
exponential generating function $\mathcal{U}$ $(x)$, then the
sequence
$\mathbf{M}\mathbf{u}^T$ has exponential generating function
$g(x)\mathcal{U}(f(x))$. Thus the row sums of the array $[g,f]$
are given
by
$g(x)e^{f(x)}$ since the sequence $1,1,1,\ldots$ has
exponential generating function $e^x$. \begin{example} The
\emph{binomial matrix}
is the matrix with general term $\binom{n}{k}$. It is realized
by Pascal's triangle. As an exponential Riordan array, it is
given by
$[e^x,x]$. We further have $$([e^x,x])^m=[e^{mx},x].$$
\end{example}



An important concept for the sequel is that of production
matrix. The concept of a \emph{production matrix}
\cite{ProdMat_0, ProdMat} is a general one, but for this note
we find it convenient to review it in the context of Riordan
arrays.
Thus
let $P$ be an infinite matrix (most often it will have integer
entries). Letting $\mathbf{r}_0$ be the row vector
$$\mathbf{r}_0=(1,0,0,0,\ldots),$$ we define
$\mathbf{r}_i=\mathbf{r}_{i-1}P$, $i \ge 1$. Stacking these
rows leads to another
infinite
matrix which we denote by $A_P$. Then $P$ is said to be the
\emph{production matrix} for $A_P$. \noindent If we let
$$u^T=(1,0,0,0,\ldots,0,\ldots)$$ then we have
$$A_P=\left(\begin{array}{c}
u^T\\u^TP\\u^TP^2\\\vdots\end{array}\right)$$ and
$$DA_P=A_PP$$ where $D=(\delta_{i,j+1})_{i,j \ge 0}$ (where
$\delta$ is the usual Kronecker symbol). Note that  \cite{P_W} $P$ is
also called the Stieltjes matrix associated to $A_P$. The
 following result \cite{ProdMat} concerning
matrices that are
production matrices for exponential Riordan arrays is important for the sequel.
\begin{proposition} Let $A=\left(a_{n,k}\right)_{n,k \ge
0}=[g(x),f(x)]$ be an
exponential Riordan array and let
\begin{equation}\label{seq_def} c(y)=c_0 + c_1 y +c_2 y^2 +
\ldots, \qquad r(y)=r_0 + r_1 y + r_2
y^2
+ \ldots\end{equation} be two formal power series that that
\begin{eqnarray}\label{r_def} r(f(x))&amp;=&amp;f'(x) \\ \label{c_def}
c(f(x))&amp;=&amp;\frac{g'(x)}{g(x)}. \end{eqnarray} Then
\begin{eqnarray} (i)\qquad a_{n+1,0}&amp;=&amp;\sum_{i} i! c_i a_{n,i}
\\ (ii)\qquad
a_{n+1,k}&amp;=&amp; r_0 a_{n,k-1}+\frac{1}{k!} \sum_{i\ge
k}i!(c_{i-k}+k r_{i-k+1})a_{n,i} \end{eqnarray}  or, defining
$c_{-1}=0$,
\begin{equation}\label{array_def}
a_{n+1,k}=\frac{1}{k!}\sum_{i\ge k-1} i!(c_{i-k}+k
r_{i-k+1})a_{n,i}.\end{equation} Conversely,
starting from the sequences defined by (\ref{seq_def}), the
infinite array $\left(a_{n,k}\right)_{n,k\ge 0}$ defined by
(\ref{array_def}) is an exponential Riordan array.
\end{proposition} \noindent A consequence of this proposition
is that
$P=\left(p_{i,j}\right)_{i,j\ge 0}$ where
$$p_{i,j}=\frac{i!}{j!}(c_{i-j}+jr_{r-j+1})  \qquad
(c_{-1}=0).$$ Furthermore, the bivariate
exponential generating function $$\phi_P(t,z)=\sum_{n,k}
p_{n,k}t^k \frac{z^n}{n!}$$ of the matrix $P$ is given by
$$\phi_P(t,z) =
e^{tz}(c(z)+t r(z)).$$ Note in particular that we have
$$r(x)=f'(\bar{f}(x))$$ and
$$c(x)=\frac{g'(\bar{f}(x))}{g(\bar{f}(x))}.$$
\begin{example} We consider the exponential Riordan array
$[\frac{1}{1-x},x]$, \seqnum{A094587}. This array \cite{Lah}
has elements
\begin{displaymath}\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0 &amp; 0 &amp;
0 &amp; 0 &amp; \ldots \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 2 &amp; 2 &amp; 1 &amp;
0 &amp; 0 &amp;
0 &amp; \ldots \\ 6 &amp; 6 &amp; 3 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 24 &amp; 24 &amp; 12 &amp;
4 &amp; 1 &amp; 0 &amp; \ldots \\120 &amp; 120  &amp; 60 &amp; 20 &amp; 5 &amp; 1 &amp;\ldots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right)\end{displaymath} and general term $[k
\le n] \frac{n!}{k!}$ with
inverse \begin{displaymath}\left(\begin{array}{ccccccc} 1 &amp; 0 &amp;
0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\-1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp;
-2 &amp; 1
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; -3 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp;
0 &amp; 0&amp; -4 &amp; 1 &amp; 0 &amp; \ldots \\0 &amp; 0  &amp; 0 &amp; 0 &amp; -5 &amp; 1 &amp;\ldots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right),\end{displaymath} which is the array
$[1-x,x]$. In particular, we
note that the row sums of the inverse, which begin
$1,0,-1,-2,-3,\ldots$ (that is, $1-n$), have e.g.f.
$(1-x)\exp(x)$. This sequence
is
thus the binomial transform of the sequence with e.g.f. $(1-x)$
(which is the sequence starting $1,-1,0,0,0,\ldots$). In order
to
calculate the production matrix $P$ of
$[\frac{1}{1-x},x]$ we note that $f(x)=x$, and hence we have
$f'(x)=1$ so
$f'(\bar{f}(x))=1$. Also $g(x)=\frac{1}{1-x}$ leads to
$g'(x)=\frac{1}{(1-x)^2}$, and so, since $\bar{f}({x})=x$, we
get
$$\frac{g'(\bar{f}(x))}{g(\bar{f}(x))}=\frac{1}{1-x}.$$ Thus
the generating function for $P$ is
$$e^{tz}\left(\frac{1}{1-z}+t\right).$$ Thus $P$ is
the matrix $[\frac{1}{1-x},x]$ with its first row removed.
\end{example}
\begin{example} We consider the exponential Riordan array $[1,
\frac{x}{1-x}]$. The general term of this matrix \cite{Lah} may
be
calculated as
follows:
\begin{eqnarray*}T_{n,k}&amp;=&amp;\frac{n!}{k!}[x^n]\frac{x^k}{(1-x)^k}\\
&amp;=&amp;\frac{n!}{k!}[x^{n-k}](1-x)^{-k}\\
&amp;=&amp;\frac{n!}{k!}[x^{n-k}]\sum_{j=0}^{\infty}\binom{-k}{j}(-1)^jx^j\\
&amp;=&amp;\frac{n!}{k!}[x^{n-k}]\sum_{j=0}^{\infty}\binom{k+j-1}{j}x^j\\
&amp;=&amp;\frac{n!}{k!}\binom{k+n-k-1}{n-k}\\
&amp;=&amp;\frac{n!}{k!}\binom{n-1}{n-k}.\end{eqnarray*} Thus its row
sums, which have e.g.f. $\exp
\left(\frac{x}{1-x}\right)$, have general term $\sum_{k=0}^n
\frac{n!}{k!}\binom{n-1}{n-k}$. This is \seqnum{A000262}, the
`number of
``sets of lists": the number of partitions of $\{1,..,n\}$ into
any number of lists, where a list means an ordered subset'.
Its
general
term is equal to $(n-1)!L_{n-1}(1,-1)$. The inverse of
$\left[1, \frac{x}{1-x}\right]$ is the exponential Riordan
array
$\left[1,\frac{x}{1+x}\right]$, \seqnum{A111596}. The row sums
of this sequence have e.g.f. $\exp\left(\frac{x}{1+x}\right)$,
and
start
$1, 1, -1, 1, 1, -19, 151, \ldots$. This is \seqnum{A111884}.
To calculate the production matrix of
$\left[1,\frac{x}{1+x}\right]$ we
note that $g'(x)=0$, while $\bar{f}(x)=\frac{x}{1+x}$ with
$f'(x)=\frac{1}{(1+x)^2}$. Thus $$f'(\bar{f}(x))=(1+x)^2,$$ and
so the
generating function of the production matrix is given by $$
e^{tz}t(1+z)^2.$$ \noindent The production matrix of the
inverse
begins \begin{displaymath}\left(\begin{array}{ccccccc} 0 &amp; 1 &amp;
0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\0 &amp; 2 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp;
2 &amp; 4 &amp;
1
&amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; 6 &amp; 6 &amp; 1 &amp; 0 &amp; \ldots \\0 &amp; 0 &amp; 0
&amp; 12 &amp; 8 &amp; 1 &amp; \ldots \\0 &amp; 0 &amp; 0 &amp; 0 &amp; 20 &amp; 10 &amp;\ldots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath} \end{example}
\begin{example} The exponential
Riordan array $\mathbf{A}=
\left[\frac{1}{1-x},\frac{x}{1-x}\right]$, or
\begin{displaymath}\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp; \ldots \\1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 2 &amp; 4 &amp; 1
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 6 &amp; 18 &amp; 9 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 24 &amp;
96 &amp; 72
&amp; 16 &amp; 1 &amp; 0 &amp; \ldots \\120 &amp; 600  &amp; 600 &amp; 200 &amp; 25 &amp; 1
&amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp;
\ddots\end{array}\right),\end{displaymath} has general term $$
T_{n,k}=\frac{n!}{k!}\binom{n}{k}.$$  Its inverse is
$\left[\frac{1}{1+x},\frac{x}{1+x}\right]$ with general term
$(-1)^{n-k}\frac{n!}{k!}\binom{n}{k}$. This is
\seqnum{A021009}, the triangle of coefficients of the Laguerre
polynomials $L_n(x)$.
The
production matrix of $\left[\frac{1}{1-x},\frac{x}{1-x}\right]$
is given by \begin{displaymath}\left(\begin{array}{ccccccc} 1 &amp;
1 &amp; 0 &amp; 0
&amp; 0 &amp; 0 &amp;
\ldots \\1 &amp; 3 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 4 &amp; 5 &amp; 1 &amp; 0 &amp;
0 &amp; \ldots \\ 0 &amp; 0 &amp; 9 &amp; 7 &amp; 1 &amp; 0 &amp; \ldots \\0 &amp; 0 &amp; 0 &amp; 16 &amp;
9 &amp; 1 &amp;
\ldots \\0 &amp; 0 &amp; 0 &amp; 0 &amp; 25 &amp; 11 &amp;\ldots\\ \vdots &amp; \vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath} \end{example}
\begin{example} The exponential Riordan array
$\left[e^x,\ln\left(\frac{1}{1-x}\right)\right]$, or
\begin{displaymath}\left(\begin{array}{ccccccc} 1 &amp; 0 &amp; 0 &amp; 0 &amp;
0 &amp; 0 &amp; \ldots
\\1
&amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 3 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots
\\ 1 &amp; 8 &amp; 6 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 1 &amp; 24 &amp; 29 &amp; 10 &amp; 1 &amp; 0 &amp;
\ldots
\\1 &amp; 89 &amp; 145 &amp; 75 &amp; 15 &amp; 1 &amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots
&amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right),\end{displaymath}
is the coefficient array for the polynomials $$_2F_0(-n,x;-1)$$
which are an unsigned version of the Charlier polynomials (of
order
$0$) \cite{wgautschi, Roman, Szego}. This is \seqnum{A094816}.
It is equal to $$[e^x,x]\left[ 1,
\ln\left(\frac{1}{1-x}\right)\right],$$ or the product of the
binomial array $\mathbf{B}$ and the array of (unsigned)
Stirling numbers
of the first kind. The production matrix of the inverse of this
matrix is given by
\begin{displaymath}\left(\begin{array}{ccccccc} -1
&amp;
1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\1 &amp; -2 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\
0 &amp; 2 &amp; -3 &amp; 1 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; 3 &amp; -4 &amp; 1 &amp; 0 &amp;
\ldots \\0
&amp; 0 &amp; 0 &amp; 4 &amp; -5 &amp; 1 &amp; \ldots \\0 &amp; 0 &amp; 0 &amp; 0 &amp; 5 &amp; -6
&amp;\ldots\\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots
&amp;
\ddots\end{array}\right)\end{displaymath} which indicates the
orthogonal nature of these polynomials. We can prove this as
follows. We
have $$\left[e^x,
\ln\left(\frac{1}{1-x}\right)\right]^{-1}=\left[e^{-(1-e^{-x})},1-e^{-x}\right].$$
Hence $g(x)=e^{-(1-e^{-x})}$ and
$f(x)=1-e^{-x}$. We are thus led to the equations
\begin{eqnarray*} r(1-e^{-x})&amp;=&amp;\,e^{-x},\\
c(1-e^{-x})&amp;=&amp;-e^{-x},\end{eqnarray*}
with solutions $r(x)=1-x$, $c(x)=x-1$. Thus the bivariate
generating function for the production matrix of the inverse
array is
$$e^{tz}(z-1+t(1-z)),$$ which is what is required.
\end{example}

\section{Hermite polynomials and the Toda chain}



We recall that the Hermite polynomials may be defined as
$$H_n(x)=n!\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}\frac{(-1)^k
(2x)^{n-2k}}{k!(n-2k)!}.$$
The generating function for $H_n(x)$ is given by
$$e^{2xt-t^2}=\sum_{n=0}^{\infty} H_n(x)\frac{t^n}{n!}.$$
The following result was proved \cite{PB_Moments}:
\begin{proposition} The proper exponential Riordan array
$$\mathbf{L}=\left[e^{2rx-x^2},x\right]$$ has as first column
the Hermite polynomials $H_n(r)$. This array has a tri-diagonal
production array.
\end{proposition}
\begin{proof} The first column of $\mathbf{L}$ has generating
function $e^{2rx-x^2}$, from which the first assertion follows.
Standard Riordan array techniques show us that the production
array $P$ of $\mathbf{L}$ is indeed tri-diagonal, beginning
with
\begin{displaymath}\left(\begin{array}{ccccccc} 2r &amp; 1 &amp;
0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\-2 &amp; 2r &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp;
-4
&amp; 2r &amp; 1 &amp; 0 &amp;
0 &amp; \ldots \\ 0 &amp; 0 &amp; -6 &amp; 2r &amp; 1 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; 0
&amp; -8 &amp; 2r &amp; 1 &amp; \ldots \\0 &amp; 0 &amp; 0 &amp; 0 &amp; -10 &amp; 2r
&amp;\ldots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath} This is so since
$f(x)=x$ gives us $\bar{f}(x)=x$ and $f'(x)=1$, and thus
$$r(x)=f'(\bar{f}(x))=f'(x)=1.$$ Similarly,
$$c(x)=\frac{g'(\bar{f}(x))}{g(\bar{f}(x))}=2(r-x),$$ and thus
the bivariate generating function $\phi_P(y,w)$ of $P$ is given
by
$$\phi_P(y,z)=e^{yz}(2(r-z)+y),$$ as required.
\end{proof}

\noindent We note that $\mathbf{L}$ starts with
\begin{displaymath}\left(\begin{array}{ccccccc} 1 &amp; 0 &amp;
0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\2r &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\
2(2r^2-1) &amp; 4r
&amp; 1 &amp; 0 &amp; 0 &amp;
0 &amp; \ldots \\ 4r(2r^2-3) &amp; 6(2r^2-1) &amp; 6r &amp; 1 &amp; 0 &amp; 0 &amp; \ldots
\\ 4(4r^3-12r^2+3)  &amp; 16r(2r^2-3) &amp; 12(2r^2-1)
&amp; 8r &amp; 1 &amp; 0 &amp; \ldots \\8r(4r^4-20r^2+15) &amp; 20(4r^4-12r^2+3) &amp;
40r(2r^2-3) &amp; 20(2r^2-1) &amp; 10r &amp; 1
&amp;\ldots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath}
Thus $$\mathbf{L}^{-1}=\left[e^{-2rx+x^2},x\right]$$ is the
coefficient array of a set of orthogonal polynomials which have
as moments the Hermite polynomials. These new orthogonal
polynomials satisfy the three-term recurrence
$$\mathfrak{H}_{n+1}(x)=(x-2r)\mathfrak{H}_n(x)+2n
\mathfrak{H}_{n-1}(x),$$ with
$\mathfrak{H}_0=1$, $\mathfrak{H}_1=x-2r$.
\newline\newline
We can now modify this result to give us our first Toda chain
result.
\begin{proposition} The exponential Riordan array
$$\left[e^{-2(z-t)x+x^2},x\right]$$ is the coefficient array of
a family of orthogonal polynomials $P_n(x)$ with
$$P_{n+1}(x)+b_n P_n(x)+u_n P_{n-1}(x)=xP_n(x),$$ where
$(u_n, b_n)$ is a solution to the restricted Toda chain.
\end{proposition}
\begin{proof}
We easily determine that the inverse matrix
$$\left[e^{2(z-t)x-x^2},x\right]$$ has the production matrix $P$ given by
\begin{displaymath}\left(\begin{array}{ccccccc} 2(z-t) &amp; 1 &amp;
0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\-2 &amp; 2(z-t) &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\
0 &amp; -4
&amp; 2(z-t) &amp; 1 &amp; 0 &amp;
0 &amp; \ldots \\ 0 &amp; 0 &amp; -6 &amp; 2(z-t) &amp; 1 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp;
0
&amp; -8 &amp; 2(z-t) &amp; 1 &amp; \ldots \\0 &amp; 0 &amp; 0 &amp; 0 &amp; -10 &amp; 2(z-t)
&amp;\ldots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath} This is so since
$f(x)=x$ gives us $\bar{f}(x)=x$ and $f'(x)=1$, and thus
$$r(x)=f'(\bar{f}(x))=f'(x)=1.$$ Similarly,
$$c(x)=\frac{g'(\bar{f}(x))}{g(\bar{f}(x))}=-2(x-z+t),$$ and
thus the bivariate generating function $\phi_P(y,w)$ of $P$ is
given by
$$\phi_P(y,w)=e^{yw}(-2(w-z+t)+y),$$ as required.

This verifies that $P_n(x)$ is indeed a family of orthogonal
polynomials, for which
$$u_n(t)=-2n, \quad b_n(t)=2(z-t).$$
It is immediate that these satisfy Eq. (\ref{Toda}).
\end{proof}
\noindent We now note that the moments of this polynomial
family (the first column of the inverse matrix) $m_n$ satisfy
the following relation:
\begin{eqnarray*}
m_n &amp;=&amp; n![x^n]e^{2(z-t)x-x^2}\\
&amp;=&amp;\frac{1}{e^{-t^2+2tz}}\frac{d^n}{dt^n}e^{-t^2+2tz}\\
&amp;=&amp; n!\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}
\binom{n-k}{k}(-1)^k \frac{(2(z-t))^{n-2k}}{(n-k)!}\\
&amp;=&amp;\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}
\binom{n}{k}\binom{n-k}{k}k!(2(z-t))^{n-2k}\\
&amp;=&amp;H_n(z-t).\end{eqnarray*}

From the form of the production matrix $P$, we infer that the generating function of $m_n$ may be expressed as
$$\cfrac{1}{1-2(z-t)x+
\cfrac{2x^2}{1-2(z-t)x+
\cfrac{4x^2}{1-2(z-t)x+
\cfrac{6x^2}{1-2(z-t)x+\cdots}}}}. $$
\noindent The Hankel transform of $m_n$, defined as the
sequence $h_n$ where
$$h_n = |m_{i+j}|_{0 \le i,j \le n}$$ is then given by
$$h_n=(-2)^{\binom{n+1}{2}} \prod_{k=0}^n k!.$$


\section{Charlier polynomials and the Toda chain}
\begin{proposition} The exponential Riordan array
$$\left[e^{-xe^t},\ln(1+x)\right]$$ is the coefficient array of
a family of orthogonal polynomials $P_n(x)$ with
$$P_{n+1}(x)+b_n P_n(x)+u_n P_{n-1}(x)=xP_n(x),$$ where
$(u_n, b_n)$ is a solution to the restricted Toda chain.
\end{proposition}
\begin{proof}
We determine that the inverse matrix
$$\left[e^{e^{t+x}-e^t},e^x-1\right]$$ has the production
matrix
\begin{displaymath}\left(\begin{array}{ccccccc} e^t &amp; 1 &amp;
0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\e^t &amp; e^t+1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\
0 &amp; 2e^t
&amp; e^t+2 &amp; 1 &amp; 0 &amp;
0 &amp; \ldots \\ 0 &amp; 0 &amp; 3e^t &amp; e^t+3 &amp; 1 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp;
0
&amp; 4e^t &amp; e^t+4 &amp; 1 &amp; \ldots \\0 &amp; 0 &amp; 0 &amp; 0 &amp; 5e^t &amp; e^t+5
&amp;\ldots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right) \end{displaymath} as follows. We have
$\bar{f}(x)=\ln(1+x)$ and $f'(x)=e^x$, and thus
$$r(x)=f'(\bar{f}(x))=1+x.$$ Similarly,
$$c(x)=\frac{g'(\bar{f}(x))}{g(\bar{f}(x))}=\frac{e^{xe^t+t}(1+x)}{e^{xe^t}}=e^t(1+x).$$
Thus the bivariate generating function for the production matrix
$P$ is given by $$\phi_P(y,z)=e^{yz}(e^t(1+z)+y(1+z)),$$ as
required. This verifies that $P_n(x)$ is indeed a family of
orthogonal polynomials, for which
$$u_n(t)=ne^t, \quad b_n(t)=n+e^t.$$
It is easy now to verify that, with these values, $(u_n,b_n)$
satisfies the Toda chain equations Eq. (\ref{Toda}).
\end{proof}

\noindent The moments $m_n$ of this family of orthogonal
polynomials may be expressed as:
\begin{eqnarray*} m_n&amp;=&amp;n![x^n]e^{e^{t+x}-e^t}\\
&amp;=&amp;\frac{1}{e^{e^t-1}}\frac{d^n}{dt^n}e^{e^t-1}\\
&amp;=&amp; \sum_{k=0}^n {n \brace k} e^{kt},
\end{eqnarray*} where
$${n \brace k}=\frac{1}{k!} \sum_{j=0}^k (-1)^{k-j}
\binom{k}{j}j^n$$ are the Stirling numbers of the second kind.
From the form of the production matrix $P$, we infer that the generating function of $m_n$ may be expressed as
$$\cfrac{1}{1-e^tx-
\cfrac{e^t}{1-(e^t+1)x-
\cfrac{2e^t}{1-(e^t+2)x-
\cfrac{3e^t}{1-(e^t+3)x-\cdots}}}}. $$
\noindent The Hankel transform of $m_n$ is then given by
$$h_n=(e^t)^{\binom{n+1}{2}}\prod_{k=0}^n k!$$

\noindent In particular we have
$$m_n(0)=\sum_{k=0}^n {n \brace k}=\text{Bell}(n),$$ where
$\text{Bell}(n)$ denotes the $n$-th Bell number, with Hankel
transform
$$h_n=\prod_{k=0}^n k!.$$

\section{Laguerre polynomials and the Toda chain}
We have seen that the exponential Riordan array
$$\left[\frac{1}{1-x},\frac{x}{1-x}\right]$$ is closely related
to the Laguerre polynomials. We introduce a ``time'' parameter
$t$ as follows:
$$\left[\frac{1}{1-\frac{x}{1+t}},\frac{x}{1-\frac{x}{1+t}}\right]=\left[\left(1-\frac{x}{1+t}\right)^{-1},\frac{x}{1-\frac{x}{1+t}}\right],$$
and further generalize this array by introducing a general
power factor $\alpha$, to get the array
$$\left[\left(1-\frac{x}{1+t}\right)^\alpha,\frac{x}{1-\frac{x}{1+t}}\right].$$
We then have the following proposition.
\begin{proposition} The exponential Riordan array
$$\left[\left(1-\frac{x}{1+t}\right)^\alpha,\frac{x}{1-\frac{x}{1+t}}\right]$$
is the coefficient array of a family of orthogonal polynomials
$P_n(x)$ with
$$P_{n+1}(x)+b_n P_n(x)+u_n P_{n-1}(x)=xP_n(x),$$ where
$(u_n, b_n)$ is a solution to the restricted Toda chain.
\end{proposition}
\begin{proof}
 The inverse matrix
 $$\left[\left(\frac{1+t+x}{1+t}\right)^{\alpha},\frac{(1+t)x}{1+t+x}\right]$$
 has the production matrix
\begin{displaymath}\left(\begin{array}{ccccccc}
\frac{\alpha}{1+t} &amp; 1 &amp;
0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\\frac{-\alpha}{(1+t)^2} &amp;
\frac{\alpha-2}{1+t} &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp;
\frac{2(1-\alpha)}{(1+t)^2}
&amp; \frac{\alpha-4}{1+t} &amp; 1 &amp; 0 &amp;
0 &amp; \ldots \\ 0 &amp; 0 &amp; \frac{3(2-\alpha)}{(1+t)^2} &amp;
\frac{\alpha-6}{1+t} &amp; 1 &amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; 0
&amp; \frac{4(3-\alpha)}{(1+t)^2} &amp; \frac{\alpha-8}{1+t} &amp; 1 &amp;
\ldots \\0 &amp; 0 &amp; 0 &amp; 0 &amp; \frac{5(4-\alpha)}{(1+t)^2} &amp;
\frac{\alpha-10}{1+t}
&amp;\ldots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath} This verifies that
$P_n(x)$ is indeed a family of orthogonal polynomials, for
which
$$u_n(t)=\frac{n(n-\alpha-1)}{(1+t)^2}, \quad
b_n(t)=\frac{\alpha-2n}{1+t}.$$
It is easy now to verify that, with these values, $(u_n,b_n)$
satisfies the Toda chain equations Eq. (\ref{Toda}).
\end{proof}

\noindent For this family of orthogonal polynomials, the
moments $m_n$ may be expressed as:
\begin{equation}\label{Laguerre_moment}
m_n=n![x^n]\left(1+\frac{x}{1+t}\right)^{\alpha}=\frac{1}{(1+t)^{\alpha}}\frac{d^n}{dt^n}(1+t)^{\alpha}=\frac{(\alpha)_n}{(1+t)^n},\end{equation}
where
$$(\alpha)_n=\alpha(\alpha-1)\cdots (\alpha-n+1).$$
From the form of the production matrix, we infer that the generating function of $m_n$ may be expressed as
$$\cfrac{1}{1-\frac{\alpha}{1+t}x-
\cfrac{\frac{0-\alpha}{(1+t)^2}x^2}{1-\frac{\alpha-2}{1+t}x-
\cfrac{\frac{2(1-\alpha)}{(1+t)^2}x^2}{1-\frac{\alpha-4}{1+t}x-
\cfrac{\frac{3(2-\alpha)}{(1+t)^2}x^2}{1-\frac{\alpha-6}{1+t}x-\cdots}}}}.$$
The Hankel transform of $m_n$ is then given by
$$h_n=\frac{(-1)^{\binom{n+1}{2}}}{(1+t)^{n(n+1)}}\prod_{k=0}^n
k!(\alpha-k)^{n-k}.$$

\noindent Noticing that 
$$\lim_{c \to 1} \frac{c(1+t)}{c-1}\ln\left(\frac{1-\frac{x}{c(1+t)}}{1-\frac{x}{1+t}}\right)=\frac{x}{1-\frac{x}{1+t}},$$ we can 'embed' this previous result in the following:
\begin{proposition} The exponential Riordan array
$$\left[\left(1-\frac{x}{1+t}\right)^\alpha,\frac{c(1+t)}{c-1}\ln\left(\frac{1-\frac{x}{c(1+t)}}{1-\frac{x}{1+t}}\right)\right]$$
is the coefficient array of a family of orthogonal polynomials
$P_n(x)$ with
$$P_{n+1}(x)+b_n P_n(x)+u_n P_{n-1}(x)=xP_n(x),$$ where for $c=1$, 
$(u_n, b_n)$ is a solution to the restricted Toda chain.
\end{proposition}
\begin{proof} We can show that the inverse array has a tri-diagonal production matrix with 
$$b(n)=\frac{\alpha c - n(c+1)}{c(1+t)},$$ and 
$$u(n)=\frac{n(n-1-\alpha)}{c(1+t)^2}.$$ 
Letting $c \to 1$ now gives us the result.
\end{proof}
\noindent Another direction of generalization is given be looking at the related exponential Riordan array 
$$\left[e^{\alpha x},\frac{x}{1+\frac{x}{1+t}}\right],$$ which has inverse
$$\left[e^{\frac{\alpha x}{1-\frac{x}{1+t}}},\frac{x}{1-\frac{x}{1+t}}\right].$$ For this, we have the following proposition.
\begin{proposition} The exponential Riordan array
$$\left[e^{\frac{\alpha x}{1-\frac{x}{1+t}}},\frac{x}{1-\frac{x}{1+t}}\right]$$
is the coefficient array of a family of orthogonal polynomials
$P_n(x)$ with
$$P_{n+1}(x)+b_n P_n(x)+u_n P_{n-1}(x)=xP_n(x),$$ where
$(u_n, b_n)$ is a solution to the restricted Toda chain.
\end{proposition}
\begin{proof}
We find that
$$u_n(t)=\frac{n(n-1)}{(1+t)^2}$$ and
$$b_n(t)=-\frac{\alpha(1+t)+2n}{1+t}.$$
\end{proof}
\noindent In this case, the moments are simply
$$m_n=(-\alpha)^n.$$
\section{Meixner polynomials and the Toda chain}
\begin{proposition} The exponential Riordan array
$$\left[\frac{1}{\sqrt{1-2x \tanh(t)-x^2 \mathrm{sech}(t)^2
}},\ln\left(\sqrt{\frac{1+xe^{-t}\mathrm{sech}(t)}{1-xe^t
\mathrm{sech}(t)}}\right)\right]$$ is the coefficient array of
a family of orthogonal polynomials $P_n(x)$ with
$$P_{n+1}(x)+b_n P_n(x)+u_n P_{n-1}(x)=xP_n(x),$$ where
$(u_n, b_n)$ is a solution to the restricted Toda chain.
\end{proposition}
\begin{proof}
 The inverse matrix
 $$\left[\frac{\textrm{sech}(x+t)}{\textrm{sech}(t)},\sinh(x)\frac{\textrm{sech}(x+t)}{\textrm{sech}(t)}\right]$$
 has the production matrix
\begin{displaymath}\left(\begin{array}{ccccccc} -\tanh(t) &amp; 1
&amp;
0
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ -\textrm{sech}^2(t) &amp; -3\,\tanh(t) &amp; 1
&amp; 0 &amp; 0 &amp; 0 &amp; \ldots \\ 0 &amp; -4\,\textrm{sech}^2(t)
&amp; -5\,\tanh(t) &amp; 1 &amp; 0 &amp;
0 &amp; \ldots \\ 0 &amp; 0 &amp; -9\,\textrm{sech}^2(t) &amp; -7\,\tanh(t) &amp; 1
&amp; 0 &amp; \ldots \\ 0 &amp; 0 &amp; 0
&amp; -16\,\textrm{sech}^2(t) &amp; -9\,\tanh(t) &amp; 1 &amp; \ldots \\0 &amp; 0 &amp;
0 &amp; 0 &amp; -25\,\textrm{sech}^2(t) &amp; -11\tanh(t)
&amp;\ldots\\
\vdots &amp;
\vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp;
\ddots\end{array}\right).\end{displaymath} This verifies that
$P_n(x)$ is indeed a family of orthogonal polynomials, for
which
$$u_n(t)=-n^2 \textrm{sech}^2(t), \quad
b_n(t)=-(2n+1)\tanh(t).$$
It is easy now to verify that, with these values, $(u_n,b_n)$
satisfies the Toda chain equations Eq. (\ref{Toda}).
\end{proof}
\noindent We may describe the moments $m_n$ of this family of
polynomials by
\begin{equation}\label{Meixner_moment} m_n=n![x^n]
\frac{\textrm{sech}(x+t)}{\textrm{sech}(t)}=\frac{1}{\textrm{sech}(t)}\frac{d^n}{dt^n}\textrm{sech}(t).\end{equation}

\noindent The Hankel transform of $m_n$ is then given by
$$h_n =
(-1)^{\binom{n+1}{2}}\textrm{sech}(t)^{n(n+1)}\prod_{k=0}^n
(k!)^2.$$

\section{Acknowledgements}

The author would like to express his gratitude for the careful reading
and helpful remarks of an anonymous reviewer, which have resulted in
what is hopefully a clearer paper.


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\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 42C05; Secondary
11B83,  11C20, 15B05, 15B36, 33C45.

\noindent \emph{Keywords:} Toda chain, orthogonal polynomials,
moments, Hermite polynomial, Charlier polynomial, Meixner
polynomial,
Riordan array, Hankel determinant, Hankel transform.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences  
\seqnum{A000045},
\seqnum{A000142},
\seqnum{A000262},
\seqnum{A007318},
\seqnum{A021009},
\seqnum{A094587},
\seqnum{A094816},
\seqnum{A111596}, and
\seqnum{A111884}.)

\bigskip
\hrule
\bigskip

\noindent
Received May 27 2010;
revised version received   August 3 2010.
Published in {\it Journal of Integer Sequences},
August 4 2010.

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