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\begin{center}
\vskip 1cm{\LARGE\bf Riordan Arrays, Sheffer Sequences and \\
\vskip .15in
``Orthogonal'' Polynomials} \vskip 1cm \large
\vskip .1in Giacomo Della Riccia \\
Dept.\ of Math.\ and Comp.\ Science - Research Center \textit{Norbert Wiener}\\
University of Udine \\
Via delle Scienze 206 \\
33100 Udine \\
Italy \\
\href{mailto:dlrca@uniud.it}{\tt dlrca@uniud.it}
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\vskip .2 in
\begin{abstract}
Riordan group concepts are combined with the basic properties of
convolution families of polynomials and Sheffer
sequences, to establish a duality law, canonical
forms $\rho(n,m)={n\choose m}c^mF_{n-m}(m),\ c\neq0,$ and
extensions
 $\rho(x,x-k)=(-1)^kx^{\underline{k+1}}c^{x-k}F_k(x)$, where the
$F_k(x)$ are polynomials in $x$, holding for each $\rho(n,m)$ in a
Riordan array. Examples $\rho(n,m)={n\choose m}S_k(x)$ are given, in
which the $S_k(x)$ are ``orthogonal'' polynomials currently found in
mathematical physics and combinatorial analysis.
\end{abstract}

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\section{Introduction}
We derive from basic principles in the theory of \textit{convolution
families} \cite{Knuth1992b} and \textit{Sheffer sequences}
\cite{Weisstein-a} of polynomials, \textit{canonical} forms
$\rho(n,m)={n\choose m}c^m\textrm{F}_{n-m}(m)$, $c\neq0$, and
\textit{extensions}
 $\rho(x,x-k)=x^{\underline{k}}c^{x-k}\textrm{F}_k(x-k)/k!=
(-1)^kx^{\underline{k+1}}c^{x-k}\rho_k(x)$, holding for all elements
$\rho$ in the Riordan group. We show in Section 3 that the
\textit{extensions} are of \textit{polynomial} type when $c=1$. In
Section 2, we define \textit{transformation rules} and a
\textit{duality law}, that will greatly simplify algebraic
manipulations. In Section 4, we give examples $\rho(n,m)={n\choose
m}\textrm{S}_{n-m}(m)$, in which $\textrm{S}_k(x)$ are ``orthogonal''
polynomials currently found in mathematical physics and
combinatorial analysis. The concluding remarks are in Section 5.

For easy reference, we recall some definitions. The Riordan group is
a set of invertible infinite lower triangular matrices
$M=\{\rho(n,m)\}_{n,m\geq0}$, called \textit{Riordan matrices}, with
entries:
\begin{eqnarray}
&&\rho(n,m)=\left[\frac{u^n}{n!}\right]g(u)\frac{f(u)^m}{m!}=
\frac{n!}{m!}[u^{n-m}]g(u)\left(\frac{f(u)}{u}\right)^m,\ \rho(0,0)=1,\label{gen}\\
&&\mbox{equivalently,}\quad\sum_{n\geq
 m\geq0}\rho(n,m)\frac{u^n}{n!}=g(u)\frac{f(u)^m}{m!}\quad
\mbox{(``exponential'' Riordan group)},\nonumber
\end{eqnarray}
where $g(u)=1+g_1u+g_2u^2+\ldots$ is invertible, with
$\widehat{g}(u)=1/g(u)=\sum_{n\geq0}\widehat{g}_nu^n$, and
$f(u)=f_1u+f_2u^2+\ldots$ is a \textit{delta series} with
compositional inverse $\overline{f}(u)$,
$f(\overline{f}(u))=\overline{f}(f(u))=u$. A group element is
denoted by $\rho=(g(u),\ f(u))$ and the group law $\star$ is
$(l(u),\ h(u))\star(g(u),\ f(u))=(l(u)g(h(u)),\ f(h(u))$, with
identity $I=(1,u)$ and group inverse
$\rho^{-1}=(1/g(\overline{f}(u)),\ \overline{f}(u))$. A
\textit{group representation} 
$\rho=\rho_1\star\rho_2$, in matrix notation, is
$\rho(n,m)=\sum_i\rho_1(n,i)\rho_2(i,m)$. As we know,
group representations are extensively used in the study of identities
\cite{Shapiro1991,ZhaoWang2003,Shapiro2003,Sprugnoli1994}
and they will play an essential role in this
work. A sequence $a_k(x)$, given by $\sum_ka_k(x)u^k=a(u)^x$, is a
\textit{convolution family} of polynomials iff $a(0)=1$
\cite{Knuth1992b}. In a convolution family the polynomials
$a_k(x)$ are multiples of $x$ for $k>0$, having degree $\leq k$, and
$xa_0(x)\equiv1$. A Sheffer sequence $\mathcal{S}_k(x)$ for $(g(u),\
f(u))$ \cite{Weisstein-a} is given by
$\sum_{k\geq0}\mathcal{S}_k(x)u^k/k!=(1/g(\overline{f}(u)))e^{x\overline{f}(u)}$,
where $(g(u),\ f(u))$ are as in Riordan group theory; the Sheffer
sequence for $f=(1,\ f(u))$ is the \textit{associated} sequence for
$f(u)$, and the Sheffer sequence for $(g(u),\ u)$ is the
\textit{Appell} sequence for $g(u)$.
\section{Transformation Rules and the Duality Law}
We first define some useful \textit{transformation rules}, noting
that transformation rules and Barry's ``transforms'' 
\cite{Barry2007b} are different concepts.

1) The \textit{duality rule} ``$\sim$'' for a function $f(n,m),\
n,m\ \mbox{integers}$, is $\widetilde{f}(n,m)=f(-m,-n)$ and for a
function $f_k(x)$, the rule is $\widetilde{f}_k(x)=f_k(k-x)$. When
$x=n$ and $k=n-m$, the two rules coincide. The following special
cases will be used in the sequel, without reference.
\begin{eqnarray*}
&&\widetilde{{n\choose m}}=\widetilde{{n\choose n-m}}=
{-m\choose n-m}=(-1)^{n-m}{n-1\choose m-1},\quad n\geq m\geq0,\\
&&\widetilde{\frac{n!}{m!}}=\widetilde{n^{\underline{n-m}}}=
(-m)^{\underline{n-m}}=(-1)^{n-m}(n-1)^{\underline{n-m}}=
(-1)^{n-m}\frac{n!}{m!}\frac{m}{n},\\
&&\widetilde{\frac{n!}{(m-1)!}}=
\widetilde{n^{\underline{n-m+1}}}=(-1)^{n-m+1}\frac{n!}{(m-1)!}.
\end{eqnarray*}
The dual $\widetilde{\rho}(n,m)$ of $\rho(n,m)$ and the dual of a
group representation $\rho=\rho_1\star\rho_2$ are
\begin{eqnarray*}
&&\widetilde{\rho}(n,m)=\rho(-m,-n)=
(-1)^{n-m}\frac{n!}{m!}\frac{m}{n}[u^{n-m}]
g(u)\left(\frac{f(u)}{u}\right)^{-n}\quad\mbox{from (\ref{gen})},\\
&&\widetilde{\rho}(n,m)= \sum_i\rho_1(-m,-i)\rho_2(-i,-n)=
\sum_i\widetilde{\rho_2}(n,i)\widetilde{\rho_1}(i,m),\quad
\widetilde{\rho}=\widetilde{\rho_2}\star\widetilde{\rho_1},
\end{eqnarray*}
after a change $i\rightarrow -i$; this change is legal because we
are allowed to use summation $\sum_i$ over an unbounded range.

2) For any number $\mu$, the \textit{scaling rule} ``$(\mu)$'' is a
group automorphism:
\begin{eqnarray*}
&&(\mu)\rho=(1,\ \mu u)\star\rho\star\left(1,\
\frac{u}{\mu}\right)=\left(g(\mu u),\ \frac{f(\mu
u)}{\mu}\right),\quad\lim_{\mu\rightarrow0}(\mu)(g(u),\ f(u))=(1,\ u)=I,\\
&&((\mu)\rho)(n,m)=\mu^{n-m}\rho(n,m),\quad((\mu)\rho)^{-1}=
(\mu)\rho^{-1},\quad(\mu)(\rho_1\star\rho_2)=(\mu)\rho_1\star(\mu)\rho_2.
\end{eqnarray*}

3) The \textit{negation rule} ``(-)'' is the special
\textit{scaling} $\mu=-1$:
$$(-)\rho=(g(-u),\ -f(-u)):\quad((-)\rho)(n,m)=(-1)^{n-m}\rho(n,m).$$
\textit{Duality} and \textit{negation} applied to (\ref{gen}) and
Lagrange's inversion formula yield, respectively,
\begin{eqnarray*}
(-1)^{n-m}\widetilde{f}(n,m)&=&(-1)^{n-m}f(-m,-n)=
\frac{n!}{m!}\frac{m}{n}[u^{n-m}]\left(\frac{f(u)}{u}\right)^{-n},\\
f^{-1}(n,m)&=&\frac{n!}{m!}[u^n]\overline{f}(u)^m=
\frac{n!}{m!}\frac{m}{n}[u^{n-m}]\left(\frac{f(u)}{u}\right)^{-n}.
\end{eqnarray*}
Combining the two formulas, we find a \emph{duality law} in the
\emph{associated} subgroup $\{(1,\ f(u))\}$:
\begin{equation}\label{dualinv}
f^{-1}(n,m)=(-1)^{n-m}\widetilde{f}(n,m),\quad
 f^{-1}=(-)\widetilde{f},
\end{equation}
For $g$ in the \emph{Appell} subgroup $\{(g(u),\ u)\}$ and
$\rho^{-1}=f^{-1}\star g^{-1}=(-)\widetilde{f}\star g^{-1}$,
(\ref{gen}) yields:
\begin{eqnarray*}
&&g(n,m)=\frac{n!}{m!}g_{n-m}={n\choose m}(n-m)!g_{n-m},\quad
g^{-1}(n,m)={n\choose m}(n-m)!\widehat{g}_{n-m},\\
&&\rho^{-1}(n,m)=\sum^{n-m}_{i=0}(-1)^i
\widetilde{f}(n,n-i)(n-i)^{\underline{n-m-i}}\widehat{g}_{n-m-i}.
\end{eqnarray*}
\section{Canonical Forms and Extensions}
Let us write the \textit{delta series} $f(u)$ in $f=(1,\ f(u))$ and
the defining relation (\ref{gen}) as:
\begin{eqnarray}
&&f(u)=uca(u)=uc(1+a_1u+\cdots),\quad c\neq0,\nonumber\\
&&\rho(n,m)={n\choose m}c^m\left[\frac{u^{n-m}}{(n-m)!}\right]
g(u)\left(\frac{f(u)}{cu}\right)^m={n\choose
m}c^mF_{n-m}(m),\label{can}
\end{eqnarray}
and call $c\neq0$ the \textit{weight} of $f$, $\{ca_n\}$ the
\textit{f-reference} sequence (\textit{f}-refseq) and ${n\choose
m}c^mF_{n-m}(m)$ the \textit{canonical} form of $\rho(n,m)$. The
remarkable structure of \textit{canonical} forms, namely, a binomial
coefficient multiplied by a factor $c^mF_{n-m}(m)$, implies the
following.

\begin{proposition}
A numerical array with entries $\rho(n,m)={n\choose m}c^mF_{n-m}(m)$
is a Riordan matrix iff the $F_k(x)$ are polynomials forming a
Sheffer sequence.
\end{proposition}

\begin{proof}
If the $\rho(n,m)$ are numbers related to an element $\rho=(g(u),\
f(u))$, then according to (\ref{can}) we can write
$$\sum_{n-m}\frac{F_{n-m}(m)}{(n-m)!}u^{n-m}=g(u)e^{m\ln\left(\frac{f(u)}{cu}\right)}=
g(u)\left(\frac{f(u)}{cu}\right)^m,$$ thus the $F_k(x)$ form a
Sheffer sequence given by
$\sum_kF_k(x)u^k/k!=g(u)e^{x\ln(f(u)/cu)}$. Conversely, if the above
Sheffer sequence is given, then, by letting $x=m$, we obtain
(\ref{can}), proving that the $\rho(n,m)$ define a Riordan matrix.
\end{proof}

\begin{corollary}\label{cor1}
Each $\rho(n,m)$ in a Riordan matrix has an extension
\begin{equation}\label{polext}
\rho(x,x-k)=(-1)^kx^{\underline{k+1}}\rho_k(x),\quad\rho_k(x)=
c^{x-k}\frac{(-1)^k}{k!}\frac{\textrm{F}_k(x-k)}{x-k},
\end{equation}
which is of \textit{polynomial} type when $c=1$.
\end{corollary}
\begin{proof}
Since $F_{n-m}(m)$ is a polynomial in $m$, we can put in the
canonical form $n-m=k$, $n=x$ and $m=x-k$, thus obtaining
immediately well-defined \textit{extensions} (\ref{polext}), which
are  clearly of \textit{polynomial} type when $c=1$.
\end{proof}
Since $f(u)/cu=a(u)$ is invertible and $a(0)=1$, the
coefficients $(-1)^kxf_k(x+k)$ in the power series expansion of
$(f(u)/cu)^x$, written as $(f(u)/cu)^x=x\sum_k(-1)^kf_k(x+k)u^k$,
form a \textit{convolution family}. Similarly, (\ref{gen}) written
for $f^{-1}(n,m)=(-1)^{n-m}\widetilde{f}(n,m)$, with $u\rightarrow
cu$ and $m=-x$, implies:
\begin{eqnarray*}
\left(\frac{c\overline{f}(u)}{u}\right)^m&=&
-m\sum_{n-m\geq0}(-1)^{n-m}f_{n-m}(-m)\frac{u^{n-m}}{c^{n-m}},\\
\left(\frac{\overline{f}(cu)}{u}\right)^{-x}&=&\left(\frac
{\overline{f(u)/c}}{u}\right)^{-x}=
x\sum_{k\geq0}(-1)^kf_k(x)u^k,\quad
c\overline{f}\left(f\left(\frac{u}{c}\right)\right)=u,
\end{eqnarray*}
showing that the $(-1)^{k+1}xf_k(-x)$ form a \textit{convolution
family} for $(\overline{f(u)/c}/u)^x$. Moreover,
\begin{eqnarray*}
f(n,m)&=&(-1)^{n-m}\frac{n!}{(m-1)!}c^mf_{n-m}(n),\\
f^{-1}(n,m)&=&(-1)^{n-m}\widetilde{f}(n,m)=
(-1)^{n-m+1}\frac{n!}{(m-1)!}c^{-n}f_{n-m}(-m)\\
f(x,x-k)&=&c^{x-k}(-1)^kx^{\underline{k+1}}f_k(x),\quad
 f^{-1}(x,x-k)=c^{-x}(-1)^{k+1}x^{\underline{k+1}}f_k(k-x).
\end{eqnarray*}
The $f_k(x),\ k>0,$ will be called the \textit{f}-polynomial
sequence (\textit{f}-polseq).\\
One can verify that if $f$ has weight $c$, then $f^{-1}$ has
\textit{weight} $1/c$, and that changes of \textit{scale} keep
\textit{weights} invariant, hence, they cannot modify the
characteristics of an \textit{extension}.

For $g\in\{(g(u),\ u)\}$, we have simply:
\begin{eqnarray*}
g(n,m)&=&{n\choose n}(n-m)!g_{n-m},\quad g^{-1}(n,m)=
{n\choose n}(n-m)!\widehat{g}_{n-m},\\
g(x,x-k)&=&g_kx^{\underline{k}},\quad
 g^{-1}(x,x-k)=\hat{g}_kx^{\underline{k}}.
\end{eqnarray*}
For $\rho(n,m)$, in addition to (\ref{can}) and (\ref{polext}), and
for $\rho^{-1}(n,m)$ we write relations that, when $g(u)\equiv1$,
reduce to the corresponding expressions for $f(n,m)$ and
$f^{-1}(n,m)$:
\begin{eqnarray*}
\rho(n,m)&=&(-1)^{n-m}\frac{n!}{(m-1)!}\rho_{n-m}(n),\\
\rho^{-1}(n,m)&=&{n\choose
 m}c^{-n}G_{n-m}(m)=(-1)^{n-m+1}\frac{n!}{(m-1)!}\rho^{-1}_{n-m}(-m),\\
\rho^{-1}(x,x-k)&=&(-1)^{k+1}x^{\underline{k+1}}\rho^{-1}_k(k-x),\quad
\rho^{-1}_k(x)=c^{-x}\frac{(-1)^k}{k!}\frac{G_k(-x)}{x}.
\end{eqnarray*}

When the nonzero entries in a $f$-refseq are of the same sign, we
say that $\rho=(g(u),\ f(u))$ is of the \textit{second} kind,
otherwise, of the \textit{first} kind. In a pair $\{\rho,\
\rho^{-1}\}$, at most one element can be of the 2nd kind; when such
an element exists, it will be denoted $\rho^{-1}$. Similarly,
capital letters in an inverse pair $\{\phi,\ \Phi\}$ will, in
general, indicate elements of the 2nd kind.

\textit{Duality} defines a \textit{dual} element $\widetilde{\rho}$
that extends $\rho(n,m)$ to all integers $n,\ m$, and since, as we
have seen, $\widetilde{\rho}(n,m)$, $\rho(n,m)$ and $\rho^{-1}(n,m)$
are tied together, it is natural to include these numbers in a
single extended $\rho$\textit{-array} that will represent the pair
$\{\rho,\ \rho^{-1}\}$.

We adopt the term \textit{generalized} numbers for the $\rho(n,m)$
in the sense that these numbers extend to all integer values $n,m,$.
The $\rho$-refseq: $c\sum^n_{i=0}g_ia_{n-i}$ given by
$$\frac{\rho(n+1,1)}{(n+1)!}=c[u^n]g(u)\frac{f(u)}{cu}=c[u^n]g(u)a(u)=c\sum^n_{i=0}g_ia_{n-i}.$$
\section{Riordan Arrays and ``Orthogonal'' Polynomials}
Consider a Sheffer sequence $\mathcal{S}_k(x)$ for $(g(u),\
 f(u))$ and the elements
$\textrm{U}=\left(1/g(\overline{f}(u)),\
 ue^{\overline{f}(u)}\right)$ and $C=(1/g(\overline{f}(u)),\
\overline{f}(u))$, then we can write:
\begin{eqnarray*}
&&\sum_{k\geq0}\frac{\mathcal{S}_k(x)}{k!}u^k=\frac{e^{x\overline{f}(u)}}{g(\overline{f}(u))}=
\sum_{i\geq0}x^i\frac{1}{i!}\frac{\overline{f}(u)^i}{g(\overline{f}(u))}=
\sum_{i\geq0}x^i\sum_{k\geq0}C(k,i)\frac{u^k}{k!},\\
&&\mathcal{S}_k(x)=\sum^k_{i=0}C(k,i)x^i,\quad C(n,m)={n\choose
 m}\left[\frac{u^{n-m}}{(n-m)!}\right]\frac{\left(\frac{\overline{f}(u)}{u}\right)^m}{g(\overline{f}(u))},\\
&&\textrm{U}(n,m)={n\choose
m}\left[\frac{u^{n-m}}{(n-m)!}\right]\frac{e^{m\overline{f}(u)}}{g(\overline{f}(u))}=
{n\choose m}\mathcal{S}_{n-m}(m)=\frac{(-1)^{n-m+1}n!}{(m-1)!}\textrm{U}_{n-m}(-m),\\
&&\textrm{U}(x,x-k)=(-1)^{k+1}x^{\underline{k+1}}\textrm{U}_k(k-x),
\quad\textrm{U}_k(x)=\frac{(-1)^k}{k!}\frac{\mathcal{S}_k(-x)}{x},\
 k>0,\ \textrm{U}_0(x)=\frac{1}{x}.
\end{eqnarray*}
An immediate consequence of these equations is the following
important relationship between Riordan group elements and Sheffer
sequences.

\begin{proposition}\label{prop1}
$S_k(x)=\sum^k_{i=0}C(k,i)x^i$ are polynomials forming a Sheffer
sequence for $(g(u),\ f(u))$ iff the coefficients $C(n,m)$ are
\emph{generalized} numbers for $C=(1/g(\overline{f}(u)),\
\overline{f}(u))$.
\end{proposition}
We now give examples where the $S_k(x)$ are classical ``orthogonal''
polynomials, currently found in mathematical
physics and combinatorial analysis.

\bigskip

\noindent \textit{Example 1}.  In the typical
\underline{\textit{Stirling}-array} $\{s=(1,\ \ln(1+u)),\ S=(1,\
e^u-1)\}$, we have:
\begin{eqnarray*}
&&s(n,m)=(-1)^{n-m}\frac{n!}{(m-1)!}\sigma_{n-m}(n),\quad
 S(n,m)=(-1)^{n-m+1}\frac{n!}{(m-1)!}\sigma_{n-m}(-m);\\
&&s(x,x-k)=(-1)^kx^{\underline{k+1}}\sigma_k(x),\quad
 S(x,x-k)=(-1)^{k+1}x^{\underline{k+1}}\sigma_k(k-x);\\
&&Stirling\mbox{-polseq}:\sigma_k(x);\
s\mbox{-refseq}:\frac{(-1)^n}{n+1},\
 S\mbox{-refseq}:\frac{1}{(n+1)!};
\end{eqnarray*}
$$\left(\frac{\ln(1+u)}{u}\right)^x=
x\sum_{k\geq0}(-1)^k\sigma_k(x+k)u^k,\
 \left(\frac{e^u-1}{u}\right)^{-x}=
 x\sum_{k\geq0}(-1)^k\sigma_k(x)u^k.$$
The \underline{\textit{Stirling}-array} is related to the
\textbf{Exponential Polynomials $\phi_k(x)$} \cite[p.\ 63]{Roman1984}
forming the \textit{associated} sequence for
$f(u)=\ln(1+u)$,\quad$\overline{f}(u)=e^u-1$,\quad$\gamma^{-1}=(1,\
e^u-1)$:
\begin{eqnarray*}
\sum_k\frac{\phi_k(x)}{k!}u^k&=&
e^{x\overline{f}(u)}=e^{x(e^u-1)},\quad C(n,m)=S(n,m):\ \mbox{Stirling numbers (2d kind)},\\
\Phi=\textrm{U}&=&\left(1,\ ue^{e^u-1}\right),\quad\phi_k(x)=
\sum^k_{i=0}S(k,i)x^i:\ \mbox{Touchard polynomials};\\
\Phi(n,m)&=&{n\choose
m}\phi_{n-m}(m)=\frac{(-1)^{n-m+1}n!}{(m-1)!}\Phi_{n-m}(-m),
\quad\Phi_k(x)=\frac{(-1)^k}{k!}\frac{\phi_k(-x)}{x};\\
\phi_k(1)&=&\varpi_n=\sum^k_{i=0}S(k,i):\ \mbox{Bell numbers, given
by}\ e^{e^u-1}=\sum_k\varpi_nu^k/k!.
\end{eqnarray*}
The \textbf{Iterated Exponential Polynomials} $\phi^{[q]}_k(x)$ form
the \textit{associated} sequence for $f(u)=
s^{[q]}(u)=s(s^{[q-1]}(u)),\ s^{[0]}=u$;\quad$\overline{f}(u)=
S^{[q]}(u)=S(S^{[q-1]}(u)),\ S^{[0]}=u$:
\begin{eqnarray*}
\sum_k\frac{\phi^{[q]}_k(x)}{k!}u^k&=&e^{xS^{[q]}},\quad
\phi^{[q]}_k(x)=\sum^k_{i=0}S^{[q]}(k,i)x^i;\quad
 S^{[q]}=\underbrace{S\star S\star\ldots\star S}_{q\ terms},\\
\Phi^{[q]}=\textrm{U}^{[q]}&=&\left(1,\ ue^{S^{[q]}(u)}\right),\quad
\Phi^{[1]}=\Phi,\quad C^{[q]}=(1,\
 S^{[q]}(u))\\
S^{[q]}(n,m)&=&{n\choose m}\phi^{[q]}_{n-m}(m),\quad\Phi^{[q]}_k(x)=
\frac{(-1)^k}{k!}\frac{\phi^{[q]}_k(-x)}{x}.
\end{eqnarray*}
The \underline{\textit{Stirling}-array} corresponds to $q=1$, and
for $q=2$, we have the \underline{$Stir^{[2]}$-array} $\{\beta=(1,\
\beta(u)),\ \mathcal{B}\}$,\quad $\beta(u)=\ln(1+\ln(1+u))$,\quad
$\mathcal{B}(u)=e^{e^u-1}-1$,\quad
$\mathcal{B}\mbox{-refseq}:\varpi_{n+1}/(n+1)!$,
\begin{eqnarray*}
&&\beta(n,m)=\sum^n_{i=m}s(n,i)s(i,m),\quad\mathcal{B}(n,m)=
\sum^n_{i=m}S(n,i)S(i,m),\\
&&\sum_k\frac{\phi^{[2]}_k(x)}{k!}u^k=e^{x\ln(1+\ln(1+u))}=
(1+\ln(1+u))^x,\quad\phi^{[2]}_k(x)=\sum^k_{j\geq
 i\geq0}s(k,j)s(j,i)x^i,\\
&&Stir^{[2]}\mbox{-polseq}:\sigma^{[2]}_k(x)=
\sum^k_{i=0}(x-i)\sigma_{k-i}(x-i)\sigma_i(x).
\end{eqnarray*}

\noindent \textit{Example 2}. For the \underline{\textit{Lah}-array}
$\{\lambda,\ \Lambda=(1,\
 \Lambda(u))\}\footnote{Erratum: in
Della Riccia \cite[p.\ 3,\ line before last]{Della Riccia2006}:
$\Lambda(u)=u/(1+u/2)$ should be
$\Lambda(u)=u/(1-u/2)$.}$,\quad$\Lambda(u)=-\lambda(-u)=u/(1-u/2)$:
\begin{eqnarray*}
&&\Lambda(n,m)=\frac{(-1)^{n-m+1}n!}{(m-1)!}Lah_{n-m}(-m)={n\choose
m}\frac{(n-m)!}{2^{n-m}}{n-1\choose m-1}:\mbox{scaled Lah numbers},
\end{eqnarray*}
and for the \underline{\textit{Pascal}-array} $\{p,\ P=(1/(1-u),
u/(1-u))\}$,\quad$p=P^{-1}=(-)P$:
\begin{eqnarray*}
&&P(n,m)=\frac{(-1)^{n-m+1}n!}{(m-1)!}Pas_k(-m)=
\frac{n!}{m!}[u^{n-m}]\frac{1}{(1-u)^{m+1}}= {n\choose
m}\frac{n!}{m!},\ \mbox{by \cite[(5.56)]{Graham1994}},\\
&&Pascal\mbox{-polseq}:Pascal_k(x)=
\frac{(-1)^k}{k!}(x-1)^{\underline{k-1}}.
\end{eqnarray*}
The above arrays are related to the \textbf{Laguerre Polynomials
$L^{(a)}_k(x)$ of order $a$} \cite[p.\ 31, p.\ 108]{Roman1984} forming
the Sheffer sequence for $(g(u)=(1+u)^{-a-1},\
 f(u)=u/(u-1)=\overline{f}(u))$, given by
 $\sum_kL^{(a)}_k(x)u^k/k!=(1-u)^{-a-1}e^{xu/(u-1)}$;
\quad$L^{\{0\}}_k(x)$: (simple) Laguerre polynomials.
\begin{eqnarray}
\textrm{Lag}^{(a)}=\textrm{U}&=&((1-u)^{-a-1},\
ue^{\overline{f}(u)}),\quad C^{(a)}=\left((1-u)^{-a-1},\
 \frac{u}{u-1}\right),\nonumber\\
\textrm{Lag}^{(a)}(n,m)&=&{n\choose
 m}L^{\{a\}}_{n-m}(m)=
(-1)^{n-m+1}\frac{n!}{(m-1)!}\textrm{Lag}^{(a)}_{n-m}(-m),\nonumber\\
C^{(a)}(n,m)&=&
(-1)^m\frac{n!}{m!}[u^{n-m}](1-u)^{-a-1-m}=(-1)^m\frac{n!}{m!}{a+n\choose
 n-m},\ \mbox{by \cite[(5.56)]{Graham1994}},\nonumber\\
L^{(a)}_k(x)&=&\sum^k_{i=0}C^{(a)}(k,i)x^i=
\sum^k_{i=0}\frac{k!}{i!}{a+k\choose k-i}(-x)^i.\label{laguerre}
\end{eqnarray}
The \underline{\textit{Lah}-array} corresponds to $a=-1$ and the
\underline{\textit{Pascal}-array} to $a=0$.

\bigskip

\noindent \textit{Example 3}. Let us consider the
\underline{\textit{Tanh}-array} $\{\theta=(1,\ \theta(u)),\
\Theta=(1,\ \Theta(u))\}$, with
\begin{eqnarray*}
&&\theta(u)=2\frac{e^u-1}{e^u+1}=2\tanh\frac{u}{2},\
 \Theta(u)=\ln\frac{1+u/2}{1-u/2}=
2\arg\tanh\frac{u}{2},\ \Theta=\Lambda\star s,\ S=\Lambda\star\theta:\\
&&\theta(n,m)=(-1)^{n-m}\frac{n!}{(m-1)!}\delta_{n-m}(n),\quad
\Theta(n,m)=(-1)^{n-m+1}\frac{n!}{(m-1)!}\delta_{n-m}(-m),\\
&&\delta_k(x)=\sum^k_{i=0}\frac{(-1)^{i+1}}{2^i}
\frac{(k-x-1)^{\underline{i}}}{i!}\sigma_{k-i}(k-x-i),\quad\sigma_k(x)=
-\sum^k_{i=0}\frac{(x-1)^{\underline{i}}}{2^ii!}\delta_{k-i}(k-x),\\
&&Tanh\mbox{-polseq}:\ \delta_k(x).
\end{eqnarray*}
The \underline{\textit{Tanh}-array} and the
\underline{\textit{Tangent}-array} $\{\arctan=(1,\ \arctan u),\
\tan=(1,\ \tan u)\}$ are related by the elementary trigonometric
formulas:
$$\tan u=\frac{1}{\imath}\tanh(\imath
 u)=\frac{1}{2\imath}2\tanh\left(2\imath\frac{u}{2}\right)=
\frac{\theta(2\imath u)}{2\imath},\quad\arctan
u=\frac{1}{\imath}\arg\tanh(\imath u)=\frac{\Theta(2\imath
u)}{2\imath},$$ which look like a scaling of $\theta$ and $\Theta$
with $\mu=2\imath,\ \imath^2=-1$:
$tan(n,m)=(2\imath)^{n-m}\theta(n,m)$,
$arctan(n,m)=(2\imath)^{n-m}\Theta(n,m)$. The numbers $tan(n,m)$ and
$\arctan(n,m)$ appear in Comtet \cite[p.p.259-260]{Comtet1974} as
$T(n,k)$ and $t(n,k)$. Putting
$T(n,m)=(2\imath)^{n-m}(n!/(m-1)!)\delta_{n-m}(n)$ in the recursion
for $T(n,k)$ and after common factors are divided out, we are left
with
\begin{eqnarray}
&&(x+1)\delta_k(x+1)=(x-k)\delta_k(x)-\frac{x-k+2}{4}\delta_{k-2}(x),
\quad\delta_k(x)\equiv0,\ k<0,\ x\delta_0(x)\equiv1,\nonumber\\
&&(\mbox{compare with}
(x+1)\sigma_k(x+1)=(x-k)\sigma_k(x)+x\sigma_{k-1}(x)\
\mbox{\cite[Exercise 6.18]{Graham1994}.}\label{stirec}
\end{eqnarray}
One can prove by induction that the $\delta_k(x),\ k=2j>0,$ have
degree $j-1$ and $\delta_k(x)\equiv0$ when $k$ is odd; hence factors
$(-1)^k$ may be omitted, leaving us with simplified formulas:
$$\left(\frac{2\tanh\frac{u}{2}}{u}\right)^x=
x\sum_{k\geq0}\delta_k(x+k)u^k,\quad
\left(\frac{1}{u}\ln\frac{1+\frac{u}{2}}{1-\frac{u}{2}}\right)^{-x}=
x\sum_{k\geq0}\delta_k(x)u^k;\quad\Theta=\widetilde{\theta}.$$ The
\underline{\textit{Tanh}-array} is related to the
\textbf{Mittag-Leffler Polynomials $M_k(x)$} \cite[p.75]{Roman1984}
forming the \textit{associated} sequence for
$$f(u)=\frac{e^u-1}{e^u+1}=\tanh\frac{u}{2},\quad\overline{f}(u)=
\ln \frac{1+u}{1-u}=2\arg\tanh
 u,\quad\sum_k\frac{M_k(x)}{k!}u^k=\left(\frac{1+u}{1-u}\right)^x.$$
\begin{eqnarray*}
\textrm{MiLef}=\textrm{U}&=&\left(1,\ u\frac{1+u}{1-u}\right), \quad
C=(1,\ 2\arg\tanh
 u),\quad\textrm{MiLef}_k(x)=
\frac{(-1)^k}{k!}\frac{M_k(-x)}{x},\\
\textrm{MiLef}(n,m)&=&{n\choose
 m}M_{n-m}(m)=\frac{(-1)^{n-m+1}n!}{(m-1)!}\textrm{MiLef}_{n-m}(-m),
\end{eqnarray*}
\begin{equation}
M_k(x)=2^k\sum^k_{i=0}\Theta(k,i)x^i=2^k\sum^k_{i=0}\sum_j\Lambda(k,j))s(j,i)x^i=
\sum^k_{j=0}\frac{k!}{j!}{k-1\choose
k-j}2^jx^{\underline{j}},\label{mittag}
\end{equation}
where we used the identity $\sum_is(j,i)x^i=x^{\underline{j}}$.

\bigskip

\noindent \textit{Example 4}. The  simple \underline{\textit{Binom}-array}
$\{bin=(\exp^{-u},\ u),\ Bin\}$,
$$bin(n,m)=(-1)^{n-m}{n\choose m},\
 Bin(n,m)={n\choose m},\quad Bin(x,x-k)=(-1)^kbin(x,x-k)=\frac{x^{\underline{k}}}{k!},$$
is related to the \textbf{Poisson-Charlier Polynomials
$c^{\{a\}}_k(x)=a^{-k}L^{\{x-k\}}_k(a)$} \cite[p.119]{Roman1984},
since the egf of $c^{\{a\}}_k(0)=a^{-k}L^{\{-k\}}_k(a)=(-1)^k$ is
$e^{-u}$, by using $L^{\{-k\}}_k(a)=(-1)^ka^k$ from
(\ref{laguerre}). The \underline{\textit{Tree}-array} $\{r=(1,\
ue^{-u}),\ R=(1,\ \overline{ue^{-u}})\}$,
\begin{eqnarray*}
r(n,m)&=&(-1)^{n-m}{n\choose m}m^{n-m}, \quad R(n,m)={n\choose
 m}\frac{m}{n}n^{n-m},\\
\left(\frac{r(u)}{u}\right)^x&=&e^{-xu},\
\left(\frac{R(u)}{u}\right)^{-x}=x\sum_k\frac{(-1)^k}{k!}(x-k)^{k-1}u^k,\quad
Tree_k(x)=\frac{1}{k!}(x-k)^{k-1},\\
R(n,1)&=&n^{n-1}:\ \mbox{number of rooted trees of $n$ vertices
\cite[p.152]{Comtet1974}}.
\end{eqnarray*}
can be related to the case $a=-1$ of the \textbf{Abel Polynomials
$A_k(x;a)$, $a\neq0$}, \cite[p.73]{Roman1984}, forming the
\textit{associated} sequence for $f(u)=ue^{au}$;
$\overline{f}(u)=\overline{ue^{au}}$, $A_k(x;a)=x(x-ak)^{k-1}$.
\begin{eqnarray*}
&&\textrm{Abel}^{(a)}=\textrm{U}^{(a)}=(1,\
ue^{\overline{ue^{au}}}),\quad\textrm{Abel}^{(a)}_k(x)=
\frac{(-1)^k}{k!}\frac{A_k(-x;a)}{x},\quad C^{(a)}= (1,\
\overline{ue^{au}});\\
&&\textrm{Abel}^{(a)}(n,m)=(-1)^{n-m+1}\frac{n!}{(m-1)!}\textrm{Abel}^{(a)}_{n-m}(-m)={n\choose
m}A_{n-m}(m;a).\\
&&\mbox{For}\
 a=-1,\ \textrm{Abel}^{(-1)}_k(x)=(-1)^k\sum^k_{i=0}{k\choose
i}i^{k-i}(-x)^i,\quad{k\choose i}i^{k-i}:\ \mbox{idempotent
numbers}.
\end{eqnarray*}

\noindent \textit{Example 5}. The \underline{\textit{Sinh}-array}
$\{argshin=(1,\ 2\arg\sinh(u/2)/b),\ shin=(1,\ 2\sinh(bu/2))\}$ is
related to the \textbf{Gould Polynomials $G_k(x;a,b)$, $b\neq0$},
\cite[p.67]{Roman1984} forming the \textit{associated} sequence for
$f(u)=e^{au}(e^{bu}-1)$, $b\neq0$. In fact, when $a=-b/2$,
$G_k(x;b)=G_k(x;-b/2,b)$,\quad$f(u)=
e^{bu/2}-e^{-bu/2}=2\sinh(bu/2)$,\ $c=b$,\quad
$\overline{f}(u)=2\arg\sinh(u/2)/b$,
\begin{eqnarray*}
&&\sum_kG_k(x;b)u^k/k!=e^{x(2/b)\arg\sinh(
u/2)}=\mathcal{B}_{1/2}(u)^{x/b}= (u/2+\sqrt{1+u^2/4})^{2x/b}\
 \mbox{\cite[p.71]{Knuth1992b}},\\
&&\mbox{where}\ \mathcal{B}_{1/2}(u)=(u/2+\sqrt{1+u^2/4})^2\
\mbox{is a \emph{generalized binomial series}\
\cite[p.203]{Graham1994}}.
\end{eqnarray*}
\begin{eqnarray*}
\textrm{Gould}^{(b)}=\textrm{U}^{(b)}&=&\left(1,\
 ue^{\frac{2}{b}\arg\sinh
\frac{u}{2}}\right),\quad\textrm{Gould}^{(b)}_k(x)=
\frac{(-1)^k}{k!}\frac{G_k(-x;b)}{x},\\
\textrm{Gould}^{(b)}(n,m)&=&(-1)^{n-m+1}\frac{n!}{(m-1)!}\textrm{Gould}^{(b)}_{n-m}(-m)=
{n\choose m}G_{n-m}(m;b);\\
G_k(x;1)&=&x\left(x+\frac{1}{2}k-1)\right)^{\underline{k-1}}:\
 \mbox{central factorial polynomials\ \cite[p.68]{Roman1984}}.
\end{eqnarray*}

\noindent \textit{Example 6}. The \textbf{Bernoulli Polynomials $B^{(a)}_k(x)$
of order $a$, $a\neq0$}, \cite[p.93]{Roman1984} form the
\textit{Appell} sequence for $g(u)=((e^u-1)/u)^a$, given by
$\sum_kB_k^{(a)}(x)u^k/k!= (u/(e^u-1))^ae^{xu}$.\\
$B_k^{(m)}=B_k^{(m)}(0)$ are the higher order Bernoulli numbers and
$B_k=B_k^{(1)}$ are the Bernoulli numbers given by
$\sum_kB_ku^k/k!=u/(e^u-1)=B(u)$.
\begin{eqnarray*}
&&\textrm{Bern}^{(a)}=\textrm{U}=\left(\left(\frac{u}{e^u-1}\right)^a,\
ue^u\right),\quad\textrm{Bern}^{(a)}(n,m)={n\choose
 m}B_{n-m}^{(a)}(m)\\
&&C^{(a)}=\left(\left(\frac{u}{e^u-1}\right)^a,\ u\right),\quad
 C^{(a)}(n,m)={n\choose
m}B^{(a)}_{n-m},\\
&&B^{(a)}_k(x)=\sum^k_{i=0}{k\choose i}B^{(a)}_{k-i}x^i,\quad
 B_k(x)=\sum^k_{i=0}{k\choose i}B_{k-i}x^i:\
\mbox{Bernoulli polynomials}.
\end{eqnarray*}
The \textit{{N}{\"{o}}rlund polynomials} $B_k^{(x)}=B_k^{(x)}(0)$
\cite{Adelberg1999} form the \textit{associated} sequence for
$\overline{\ln B(u)}$,
\begin{eqnarray}
&&\sum_k\frac{B_k^{(x)}}{k!}u^k=e^{x\ln
B(u)}=\left(\frac{e^u-1}{u}\right)^{-x}=
x\sum_{k\geq0}(-1)^k\sigma_k(x)u^k,\nonumber\\
&&\frac{B_k^{(x)}}{xk!}=(-1)^k\sigma_k(x)=
\frac{s(x,x-k)}{x^{\underline{k+1}}},\
 \frac{B_k^{(-x)}}{xk!}=(-1)^{k+1}\sigma_k(-x)=
\frac{S(x+k,x)}{(x+k)^{\underline{k+1}}},\label{norpol}\\
&&\frac{B_k}{k!}=\sigma_k(1)-[k=1]=-k\sigma_k(0),\ k>0,\quad\mbox{by
the recursion}\ (\ref{stirec}).\label{norind}
\end{eqnarray}
For the \underline{\textit{Bern}-array} $\{ber,\ Ber=(1,\
uB(u))\}$,\quad\textit{Ber}-refseq: $B_n/n!$, we have
\begin{eqnarray*}
&&Ber(n,m)={n\choose
 m}B_{n-m}^{(m)},\quad Bern_k(x)= \frac{(-1)^k}{k!}\frac{B^{(-x)}_k}{x}= -\sigma_k(-x),\
 k\geq0.
\end{eqnarray*}
In passing, we remark that the \textit{associated} sequence
$\mathcal{U}_k(x)$ for $f(u)=\overline{uB(u)}$ is given by
$$\sum_k\frac{\mathcal{U}_k(x)}{k!}u^k=e^{xuB(u)}=
\sum_ix^i\frac{(uB(u))^i}{i!},\quad\mathcal{U}_k(x)=\sum^k_{i=0}{k\choose
i}B_{k-i}^{(i)}x^i.$$

Now consider the Stirling polynomials convolution formula
\cite[(6.46)]{Graham1994}, written with $t=1$, $i\rightarrow
 i-m, r\rightarrow r+m$ and $s\rightarrow s+m$:
$$-(r+m)(s+n)\sum^n_{i=m}\sigma_{i-m}(r+i)\sigma_{n-i}(-s-i)=
(r-s+m-n)\sigma_{n-m}(r-s).$$ Using (\ref{norpol}), we can write
$\sigma_{n-m}(r-s)$, $\sigma_{i-m}(r+i)$ and $\sigma_{n-i}(-s-i)$ in
terms of Stirling numbers and \textit{{N}{\"{o}}rlund} polynomials;
after substitution in the convolution relation, we get
$$\sum^n_{i=m}\frac{s(r+i,r+m)}{(r+i)^{\underline{i-m}}}
\frac{S(s+n,s+i)}{(s+n)^{\underline{n-i+1}}}=\frac{r-s+m-n}{s+n}\frac{B^{(r-s)}_{n-m}}{(r-s)(n-m)!}.$$
\begin{eqnarray*}
&&\mbox{For}\ a=r-s,\ \mbox{integers}\ r,s\geq0,\
 \frac{1}{(r+i)^{\underline{i-m}}}=\frac{(r+m)!}{(r+i)!},\ \frac{1}{(s+n)^{\underline{n-i+1}}}=
\frac{(s+i-1)!}{(s+n)!}:\\
&&\sum^n_{i=m}s(r+i,r+m)S(s+n,s+i)
\frac{(s+i-1)^{\underline{i-1}}}{(r+i)!}=
\frac{a+m-n}{s+n}\frac{(s+n)!}{(r+m)!}\frac{B^{(a)}_{n-m}}{a(n-m)!}.
\end{eqnarray*}
With $r=s$, $a=0$, we get from (\ref{norpol}): and (\ref{norind}),
$$\sum^n_{i=m}s(r+i,r+m)S(r+n,r+i)
\frac{1}{i}=\frac{(r+n)^{\underline{n-m}}}{r+n}\left(\frac{B_{n-m}}{(n-m)!}+[m=n-1]\right),\
 m>0,$$ which is a generalization of the case $r=s=0$, that corresponds to the known identity:
\begin{equation}
\sum^n_{i=m}s(i,m)S(n,i)\frac{1}{i}=\frac{1}{n}{n\choose
 m}\frac{B_{n-m}}{(n-m)!}+[m=n-1],\ m>0\quad\cite[(6.100)]{Graham1994},\label{knuth}
\end{equation}
The identity (\ref{knuth}), in turn, generalizes the identity
$B_n=\sum^n_{i=0}(-1)^ii!S(n,i)/(i+1)$ appearing in Comtet
\cite[p.220]{Comtet1974} and which is the special case $m=1$ of
(\ref{knuth}) and, at the same time, of  Kaneko's identity
$B_{n-m}=(-1)^{n-m+i}S(n-m,i)i!/(i+1)^m\quad\mbox{\cite[Theorem
1]{Kaneko1997}}$.

\bigskip

\noindent\textit{Example 7}. The \textbf{Euler Polynomials $E^{(a)}_k(x)$ of
order $a$, $a\neq0$}, \cite[p.100]{Roman1984} form the
\textit{Appell} sequence for $g(u)=((e^u+1)/2)^a$,
\begin{eqnarray*}
&&\sum_k\frac{E^{(a)}_k(x)}{k!}u^k=\left(\frac{2}{e^{u}+1}\right)^ae^{xu}=\left(\frac{e^{-\frac{u}{2}}}{\cosh
 \frac{u}{2}}\right)^ae^{xu},\ E_k(x)= E^{(1)}_k(x):\mbox{Euler
 polynomials},\\
&&\mbox{for}\ u\rightarrow 2u,\
x=\frac{a}{2}:\sum_k\frac{2^kE^{(a)}_k(\frac{a}{2})}{k!}u^k=
\left(\frac{1}{\cosh u}\right)^a,\
 E_k=2^kE_k\left(\frac{1}{2}\right):\mbox{Euler numbers}.
\end{eqnarray*}
\begin{eqnarray*}
&&\textrm{Euler}^{(a)}=\textrm{U}=\left(\left(\frac{2}{e^u+1}\right)^a,\
ue^u\right),\quad\textrm{Euler}^{(a)}(n,m)={n\choose
m}E_{n-m}^{(a)}(m),\\
&&C^{(a)}=\left(\left(\frac{2}{e^u+1}\right)^a,\ u\right),\quad
 C^{(a)}(n,m)={n\choose m}E^{(a)}_{n-m}(0),\\
&&E^{(a)}_k(x)=\sum^k_{i=0}C^{(a)}(k,i)x^i=\sum^k_{i=0}{k\choose
i}E^{(a)}_{k-i}(0)x^i.
\end{eqnarray*}
For the \underline{\textit{Euler}-array} $\{eul,\ Eul=(1,\
uE(u))\}$, $E(u)=1/\cosh u$, \textit{Eul}-refseq: $E_n/n!$, we find
\begin{eqnarray*}
Eul(n,m)&=&\frac{n!}{m!}[u^{n-m}]E(u)^m=
{n\choose m}2^{n-m}E^{(m)}_{n-m}\left(\frac{m}{2}\right)\\
2^{n-m}E^{(m)}_{n-m}\left(\frac{m}{2}\right)&=&\sum^{n-m}_{i=0}{n-m\choose
i}2^im^iE^{(m)}_i(0),\quad\frac{Eul(n+1,1)}{n+1}=E_n=\sum^n_{i=0}{n\choose
i}2^iE_i(0).
\end{eqnarray*}

The egf of Euler polynomials evaluated at $x=0$ and the gf $G(u)$
for the Genocchi numbers $G_n$ are related by
$$\sum_n\frac{E_n(0)}{n!}u^n=\left(1-\tanh\frac{u}{2}\right)=\frac{2}{e^u+1}=
\sum_{n\geq0}\frac{G_{n+1}}{(n+1)!}u^n=\frac{G(u)}{u}\quad\mbox{\cite[p.49]{Comtet1974}},$$
thus, for the \underline{\textit{Geno}-array} $\{gen,\ Gen=(1,\
G(u))\}$, \textit{Gen}-refseq: $G_{n+1}/(n+1)!$, we have
$$Gen(n,m)=\frac{n!}{m!}[u^n]\left(\frac{G(u)}{u}\right)^m={n\choose
m}E^{(m)}_{n-m}(0),\quad
Gen_k(x)=\frac{(-1)^k}{k!}\frac{E^{(-x)}_k}{x}.$$
\begin{eqnarray*}
\mbox{With}\ G(u)&=&u\left(1-\tanh\frac{u}{2}\right)=
u\left(1-\frac{\theta(u)}{2}\right)\ \mbox{and}\
 \frac{\theta(u)^i}{i!}=\sum_j\theta(j,i)\frac{u^j}{j!},\ \mbox{we find}:\\
Gen(n,m)&=&\frac{n!}{m!}[u^{n-m}]\sum_i{m\choose
i}\frac{(-1)^i}{2^i}\theta(u)^i={n\choose
m}\sum^{n-m}_{i=0}{m\choose
i}\frac{(-1)^ii!}{2^i}\theta(n-m,i)\\
&=&{n\choose m}(-1)^{n-m}(n-m)!\sum^{n-m}_{i=0}{m\choose
 i}\frac{i}{2^i}\delta_{n-m-i}(n-m),\\
Gen(n+1,1)&=&G_{n+1}=(-1)^n\sum^{n-m}_{i=0}{1\choose
 i}\frac{i}{2^i}\delta_{n-i}(n)=[n=0]+\frac{(-1)^n(n+1)!}{2}\delta_{n-1}(n).
\end{eqnarray*}
\begin{eqnarray*}
\mbox{With}\ G(u)&=&\frac{2u}{e^u+1}=\frac{2u(e^u-1)}{e^{2u}-1}=
B(2u)S(u),\ \mbox{we get the binomial convolution}\\
Gen(n,m)&=&\sum^{n-m}_{i=0}{n\choose i}2^iB^{(m)}_iS(n-i,m).\\
\mbox{With}\ \frac{Ber(2u)}{2}&=&u\frac{2u}{e^{2u}-1}=
\frac{u}{e^u-1}\frac{2u}{e^u+1}=B(u)G(u),\ \mbox{we get}:\\
2^{n-m}Ber(n,m)&=&\sum^{n-m}_{i=0}{n\choose
i}B_i^{(m)}Gen(n-i,m)={n\choose m}\sum^{n-m}_{i=0}{n-m\choose
i}B_i^{(m)}E^{(m)}_{n-m-i}(0);
\end{eqnarray*}
\begin{eqnarray*}
&&Gen(n+1,1)=G_{n+1}=(n+1)E_n(0)=\sum^n_{i=0}{n+1\choose i}2^iB_i,\\
&&Ber(n+1,1)=(n+1)B_n=\frac{1}{2^n}\sum^n_{i=0}{n+1\choose
 i}B_iG_{n+1-i}.
\end{eqnarray*}
Writing $G(2u)/2=2u/(e^{2u}+1)=e^{-u}u/\cosh u=e^{-u}Eul(u)$, and
applying the binomial coefficients \textit{inversion formula}, we
obtain a pair of \textit{inverse relations}, presumably original,
\begin{eqnarray*}
\frac{2^{n-m}}{m^n}Gen(n,m)&=&\sum_i{n\choose
i}(-1)^{n-i}\frac{Eul(i,m)}{m^i}\leftrightarrow
 \frac{Eul(n,m)}{m^n}=\sum_i{n\choose
 i}\frac{2^{i-m}Gen(i,m)}{m^i},\\
2^nG_{n+1}&=&\sum^n_{i=0}{n+1\choose
 i+1}(-1)^{n-i}(i+1)E_i\ \leftrightarrow\ (n+1)E_n=\sum^n_{i=0}{n+1\choose
 i+1}2^iG_{i+1}.
\end{eqnarray*}

\noindent \textit{Example 8}. We consider the \underline{\textit{Harm1}-array}
$\{har1,\ Har1=(1,\ Har1(u))\}$ and the
\underline{\textit{Harm2}-array} $\{har2,\ Har2=(1,\ Har2(u))\}$,
where $Har1(u)$ and $Har2(u)$ are gf's related to the harmonic
numbers $H_n$:
\begin{eqnarray*}
\sum_{k>0}H_nu^n&=&\frac{-\ln(1-u)}{1-u}=Har1(u)\quad\cite[(7.43)]{Graham1994},\\
2\int\frac{-\ln(1-u)}{1-u}du&=&\frac{\ln^2(1-u)}{u}=
u\left(\frac{\ln(1-u)}{-u}\right)^2=\sum_{n>0}\frac{2H_n}{n+1}u^n=Har2(u),
\end{eqnarray*}
$$Har1(n,m)=\frac{n!}{m!}[u^{n-m}]\left(\frac{-\ln(1-u)}{u(1-u)}\right)^m,\
 Har2(n,m)=
\frac{n!}{m!}[u^{n-m}]\left(\frac{\ln(1-u)}{-u}\right)^{2m}.$$Using
the inverse pair of group representations $Har1=(-)s\star (-)r$ and
$(-)r=(-)S\star Har1$, and $Har1(n+1,1)/(n+1)!=H_{n+1}$, we get:
$$Har1(n,m)=\sum_i(-1)^{n-i}s(n,i){i\choose
m}m^{i-m}\leftrightarrow{n\choose
m}m^{n-m}=\sum_i(-1)^{n-i}S(n,i)Har1(i,m),$$
$$(n+1)!H_{n+1}=\sum^n_{i=0}(-1)^{n-i}s(n+1,i+1)(i+1)\leftrightarrow n+1=
\sum^n_{i=0}(-1)^{n-i}S(n+1,i+1)(i+1)!H_{i+1}.$$ Similarly, with
$(-)Har2=s\star Ber$ and $Ber=S\star(-)Har2$, and
$Har2(n+1,1)/(n+1)!=2H_{n+1}/(n+2)$, we find:
\begin{eqnarray*}
&&(-1)^{n-m}Har2(n,m)=\sum^n_{i=m}s(n,i){i\choose
m}B^{(m)}_{i-m}=\frac{n!}{(m-1)!}2\sigma_{n-m}(n+m)\\
&&\quad\quad\leftrightarrow\quad{n\choose
 m}B^{(m)}_{n-m}=\sum^n_{i=m}S(n,i)(-1)^{i-m}Har2(i,m);
\end{eqnarray*}
\begin{equation}\label{harm}
\frac{2H_{n+1}}{n+2}=\frac{(-1)^n}{(n+1)!}\sum^n_{i=0}s(n+1,i+1)(i+1)B_i=2\sigma_n(n+2),
\end{equation}
$$B_n=\frac{2}{n+1}\sum^n_{i=0}S(n+1,i+1)
\frac{(-1)^i(i+1)!}{i+2}H_{i+1}.$$ The identity (\ref{harm}) appears
in \cite[p.100]{Roman1984}, written in a different form. Finally:
$$Har1_k(x)=(-1)^k\sum^k_{i=0}(x-i)\sigma_{k-i}(k-x)(-x)^{i-1},\quad
 Har2_k(x)=(-1)^{k+1}2\sigma_k(k-2x).$$
The elements $Har1$ and $Har2$ are related to the \textbf{Narumi
Polynomials $Nar^{(a)}_k(x)$} \cite[p.127]{Roman1984}, forming the
Sheffer sequence for $(g(u)=(u/(e^u-1))^a,\ f(u)=e^u-1)$,
$\overline{f}(u)=\ln(1+u)$:
\begin{equation}
\sum_k\frac{Nar^{(a)}_k(x)}{k!}u^k=
\left(\frac{u}{\ln(1+u)}\right)^a(1+u)^x.\label{nar}
\end{equation}
In fact, from (\ref{nar}) and the definitions of $Har1$ and $Har2$,
we derive the \textit{canonical} forms:
$$Har1(n,m)={n\choose m}(-1)^{n-m}Nar^{(-m)}_{n-m}(-m),\quad
Har2(n,m)={n\choose m}(-1)^{n-m}Nar^{(-2m)}_{n-m}(0).$$
\begin{eqnarray*}
&&\textrm{Nar}^{(a)}=\textrm{U}^{(a)}=
\left(\left(\frac{u}{\ln(1+u)}\right)^a,\ u(1+u)\right),\quad
C^{(a)}=
\left(\left(\frac{u}{\ln(1+u)}\right)^a,\ \ln(1+u)\right);\\
&&\textrm{Nar}^{(a)}(n,m)={n\choose m}Nar^{(a)}_{n-m}(m),\quad
Nar^{(a)}_k(x)=\sum^k_{i=0}C(k,i)x^i,\\
&&C^{(a)}(n,m)=\frac{n!}{m!}[u^{n-m}]\left(\frac{\ln(1+u)}{u}\right)^{m-a}=
(-1)^{n-m}\frac{n!}{m!}(m-a)\sigma_{n-m}(n-a).
\end{eqnarray*}
In addition, with $u\rightarrow -u$, $v=-\ln(1-u)$ and $x=-a=m$,
(\ref{nar}) yields:
\begin{eqnarray}
\frac{(-1)^kNar^{(-m)}_k(m)}{k!}&=&
[u^k]\left(\frac{\ln(1-u)(1-u)}{-u}\right)^m=
[u^k]\left(\frac{v}{e^v-1}\right)^m=[u^k]\sum_iB^{(m)}_i\frac{v^i}{i!}\nonumber\\
&=&\sum^k_{i=1}\frac{(-1)^{k-i}}{k!}B^{(m)}_is(k,i);\label{narwim}
\end{eqnarray}
$$\mbox{when}\ m=1,\ \sum^k_{i=1}(-1)^{k-i}s(k,i)B_i=
\left[\frac{u^k}{k!}\right]\frac{\ln(1-u)(1-u)}{-u}=
k!\left(\frac{1}{k+1}-\frac{1}{k}\right)=\frac{-(k-1)!}{k+1},$$ that
is an identity which appears in Wilf \cite[(4.3.21)]{Wilf1990}.

Consider the \underline{$\mathcal{T}au$-array} $\{\tau,\
\mathcal{T}\}$,\quad$\tau=s\star(-)r$,\quad
$\tau(u)=-r(-s(u))=(1+u)\ln(1+u)$:
\begin{eqnarray*}
&&\tau(n,m)=\textrm{Nar}^{(-m)}(n,m)={n\choose
m}Nar^{(-m)}_{n-m}(m)=\sum^n_{i=m}s(n,i){i\choose i-m}m^{i-m},\\
&&\mathcal{T}(n,m)=(-1)^{n-m}\widetilde{\tau}(n,m)=\sum^n_{i=m}S(i,m){n-1\choose
 n-i}n^{n-i}.
\end{eqnarray*}
The numbers $\tau(n,m)$ are denoted $b(n,m)$ in Comtet
\cite[pp.\ 139--140]{Comtet1974}, wherein they are used in the
computation of the $n$-th derivative of $x^{ax},\ x>0,\ a\
\mbox{real}\neq0$. For completeness, we also mention the
group representation $\tau=(2)\lambda\star Har1$.

\section{Concluding remarks}

In this paper we obtained several known or original identities
between sequences by setting $m=1$ in identities between
generalized numbers, for example, (\ref{harm}). Conversely,
we found identities between generalized numbers, extending
identities between numerical sequences like, for instance,
(\ref{narwim}) which is an original generalization of Wilf's
identity \cite[(4.3.21)]{Wilf1990}. The Akiyama-Tanigawa algorithm
and the Euler-Seidel construction are also based on extensions of
numerical sequences, but the matrices have properties different from
those of Riordan matrices and, at any rate, the purpose there is to
develop efficient methods for calculating special sequences. The
interested reader may consult the literature on the subject, for
instance, \cite{Diletal2006,MerSprVer2005,Chen2001,Dumont1981},
in order to compare the various approaches.

\textit{\textit{Canonical}} forms implied Corollary \ref{cor1},
readily proving the existence of \textit{extensions} $\rho(x,x-k)$,
that are of \textit{polynomial} type when $\rho$ has \textit{weight}
$c=1$, and they established a connection with the family of
``orthogonal'' polynomials, but some care should be exercised
because certain ``orthogonal'' polynomials may have different names,
for instance, Mittag-Leffler polynomials and Meixner polynomials of
the second kind $M(x;0,0)$ \cite[p.\ 126]{Roman1984} coincide,
similarly, Bernoulli polynomials of the second kind $b_k(x)$
\cite[p.\ 113]{Roman1984} and Narumi polynomials $Nar^{(-1)}_k(x)$ are
the same. Recall also that, applying Proposition (\ref{prop1}), we
easily obtained the formula (\ref{laguerre}) for $L^{(\alpha)}_k(x)$
and (\ref{mittag}) for $M_k(x)$, which are found in \cite[p.\ 109 and
p.\ 76]{Roman1984} after longer algebraic manipulations (and a
printing error in the last expression of $M_k(x)$).

Finally, we remember that \textit{transformation rules} greatly
simplified algebraic manipulations, and that the \textit{duality
law} played an important role in the computation of inverse numbers,
especially when one of the two gf's: $f(u)$, $\overline{f}(u)$, was
not available in closed form.
\section{Acknowledgement}
The author wishes to thank an anonymous referee for their careful
reading of this paper and their constructive suggestions.

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\noindent 2000 {\it Mathematics Subject Classification}:
Primary 05A19; Secondary 05A15.

\noindent \emph{Keywords: } Riordan group, convolution polynomials,
polynomial extensions, Sheffer sequences, orthogonal polynomials.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received October 17 2008;
revised version received December 11 2008. 
Published in {\it Journal of Integer Sequences}, December 11 2008.

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\noindent
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