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\begin{center}
\vskip 1cm{\LARGE\bf 
On a Number Pyramid Related to the \\
\vskip .05in
Binomial,
Deleham, Eulerian, MacMahon and \\
\vskip .1in
Stirling number triangles}
\vskip 1cm
\large
Ghislain R. Franssens\\
Belgian Institute for Space Aeronomy \\
Ringlaan 3\\
B-1180 Brussels\\
Belgium \\
\href{mailto:ghislain.franssens@oma.be}{\tt ghislain.franssens@oma.be}
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\vskip .2 in

\begin{abstract}
We study a particular number pyramid $b_{n,k,l}$ that relates the
binomial,
Deleham, Eulerian, MacMahon-type and Stirling number triangles. The
numbers $b_{n,k,l}$ are generated by a function $B^{c}(x,y,t)$, $c\in \mathbb{C}$, that
appears in the calculation of derivatives of a class of functions whose
derivatives can be expressed as polynomials in the function itself or a related function. Based on
the properties of the numbers $b_{n,k,l}$, we derive several new relations related to these triangles. 
In particular, we show that the
number triangle $T_{n,k}$, recently constructed by Deleham (Sloane's
\seqnum{A088874}),
is generated by the Maclaurin series of $\func{sech}^{c}t$,
$ c\in \mathbb{C}$.
We also give explicit expressions and various partial sums for
the triangle $T_{n,k}$. Further, we find that $e_{2p}^{m}$, the
numbers appearing in the Maclaurin series of $\cosh ^{m}t$, for all $m\in 
\mathbb{N}$, equal the number of closed walks, based at a vertex, of length $
2p$ along the edges of an $m$-dimensional cube.
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
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%\begin{document}




\section{Introduction}

In this work we study a function $B^{c}(x,y,t)$, the $c$-th power of $%
B(x,y,t)$ defined in Eq. (\ref{1.1}), that plays a central role in the
calculation of derivatives, of a class of functions whose derivatives can be
expressed as polynomials in the function itself or a related function. The construction of these
polynomials, in terms of the function $B^{c}(x,y,t)$, is treated in a
separate paper \cite{GF}. Here we focus on $B^{c}(x,y,t)$ as a generating
function in its own right, and derive from it some interesting
number-theoretic results.

We show that the function $B^{c}(x,y,t)$ generates a number pyramid $b_{n,k,l}$%
, of which various partial sums are closely related to some important number
triangles, including the binomial coefficients $\tbinom{n}{k}$,
a number triangle $%
T_{n,k}$ recently constructed by Deleham \cite[\seqnum{A088874}]{S}, the Eulerian
numbers $A_{n,k}$ \cite{C},
a particular kind of MacMahon numbers $B_{n,k}$ \cite[p. 331]{M},
and Stirling numbers of the first kind $s(n,k)$ \cite[p. 824, 24.1.3]{AS}.

We derive several new expressions related to these triangles. 
For the triangles $A_{n,k}$ and $B_{n,k}$, we obtain new
generating functions. We show in particular that the so far unstudied triangle $T_{n,k}$
 is generated by the Maclaurin series of $\func{sech}%
^{c}t$, for all $c\in \mathbb{C}$. The numbers $T_{n,k}$ are thus as
fundamental for $\func{sech}^{c}t$ as the Euler numbers $E_{n}$ are for $%
\func{sech}t$ \cite[p. 804, 23.1.2]{AS}. We give explicit expressions and various
partial sums for the numbers $T_{n,k}$.

Moreover, the special cases $c=m\in \mathbb{Z}_{+}$ and $c=-m\in \mathbb{Z}%
_{-}$ give rise to a particular generalization of the Euler numbers $E_{n}$,
here denoted $E_{n}^{m}$ and called \textquotedblleft multinomial Euler
numbers\textquotedblright , and a generalization of even parity numbers $e_{n}$ (defined in Eq. (\ref{D.2.2})),
here denoted $e_{n}^{m}$ and called \textquotedblleft even multinomial
parity numbers\textquotedblright , respectively. The $E_{n}^{m}$ are generated by the
Maclaurin series of $\func{sech}^{m}t$ (so $E_{n}^{1}=E_{n}$) and the $%
e_{n}^{m}$ by the Maclaurin series of $\cosh ^{m}t$ (so $e_{n}^{1}=e_{n}$). 
Obviously, $E_{2p+1}^{m}=0$ and $e_{2p+1}^{m}=0$, 
for all $p\in \mathbb{N}$, because $\func{sech}^{m}t$ and $\cosh ^{m}t$ are
even functions of $t$. We obtain explicit formulas for the numbers $%
E_{2p}^{m}$ and $e_{2p}^{m}$, as well as relations between them. The numbers 
$e_{2p}^{m}$ turn out to have as combinatorial interpretation, the number of
closed walks, based at a vertex, of length $2p$ along the edges of an $m$%
-dimensional cube.

\section{Notation and definitions}

\begin{enumerate}
\item Define the sets of positive odd and even integers $\mathbb{Z}%
_{o,+}$ and $\mathbb{Z}_{e,+}$, the negative odd and even integers $%
\mathbb{Z}_{o,-}$ and $\mathbb{Z}_{e,-}$, the odd integers $\mathbb{Z}%
_{o}\triangleq \mathbb{Z}_{o,-}\cup \mathbb{Z}_{o,+}$ and the even integers $%
\mathbb{Z}_{e}\triangleq \mathbb{Z}_{e,-}\cup \left\{ 0\right\} \cup \mathbb{%
Z}_{e,+}$, the positive integers $\mathbb{Z}_{+}\triangleq \mathbb{Z}%
_{o,+}\cup \mathbb{Z}_{e,+}$ and negative integers $\mathbb{Z}%
_{-}\triangleq \mathbb{Z}_{o,-}\cup \mathbb{Z}_{e,-}$, the natural numbers $%
\mathbb{N}\triangleq \left\{ 0\right\} \cup \mathbb{Z}_{+}$ and the integers 
$\mathbb{Z}\triangleq \mathbb{Z}_{-}\cup \left\{ 0\right\} \cup \mathbb{Z}%
_{+}$. Let $\mathbb{Z}_{+,n}\triangleq \left\{ 1,2,...,n\right\} $, $\mathbb{%
Z}_{-,n}\triangleq \left\{ -n,-(n-1),...,-1\right\} $, $\mathbb{N}%
_{n}\triangleq \left\{ 0\right\} \cup \mathbb{Z}_{+,n}$ and denote by $%
\mathbb{C}$ the complex numbers.

\item Define%
\begin{equation}
\delta _{condition}\triangleq \left\{ 
\begin{array}{cc}
1, &amp; \text{if }condition\text{ is true;} \\ 
0, &amp; \text{if }condition\text{ is false,}%
\end{array}%
\right.  \label{D.2.1}
\end{equation}%
and for all $n\in \mathbb{Z}$ the even and odd parity numbers%
\begin{eqnarray}
e_{n} &amp;\triangleq &amp;\delta _{n\in \mathbb{Z}_{e}},  \label{D.2.2} \\
o_{n} &amp;\triangleq &amp;\delta _{n\in \mathbb{Z}_{o}}.  \label{D.2.3}
\end{eqnarray}

\item Denote the $n$-th derivative with respect to $x$ by $D_{x}^{n}$.

\item We define $0^{n}\triangleq \delta _{n=0}$, for all $n\in \mathbb{N}$,
and $z^{0}\triangleq 1$, for all $z\in \mathbb{C}$.

\item Let $n\in \mathbb{N}$ and $z\in \mathbb{C}$. Denote by%
\begin{eqnarray}
z^{(n)} &amp;\triangleq &amp;\delta _{n=0}+\delta _{n&gt;0}\,z(z+1)(z+2)...(z+(n-1)),
\label{D.5.1a} \\
&amp;=&amp;\frac{\Gamma (z+n)}{\Gamma (z)},  \label{D.5.1b} \\
&amp;=&amp;\sum_{k=0}^{n}(-1)^{n-k}s\left( n,k\right) z^{k},  \label{D.5.1c}
\end{eqnarray}%
the rising factorial polynomial (Pochhammer's symbol). In particular, $%
0^{(n)}=\delta _{n=0}$ and $%
m^{(n)}=(m-1+n)!/(m-1)!$ for $m\in \mathbb{Z}_{+}$.

Also, denote by%
\begin{eqnarray}
z_{(n)} &amp;\triangleq &amp;\delta _{n=0}+\delta _{n&gt;0}\,z(z-1)(z-2)...(z-(n-1)),
\label{D.5.2a} \\
&amp;=&amp;\frac{\Gamma (z+1)}{\Gamma (z+1-n)},  \label{D.5.2b} \\
&amp;=&amp;\sum_{k=0}^{n}s\left( n,k\right) z^{k},  \label{D.5.2c}
\end{eqnarray}%
the falling factorial polynomial. In particular, $0_{(n)}=\delta _{n=0}$ and
$m_{(n)}=\left( m!/(m-n)!\right) \delta
_{n\leq m}$ for $m\in \mathbb{Z}_{+}$. In Eqs. (\ref{D.5.1c}) and (\ref{D.5.2c}), $s\left( n,k\right) $ are
Stirling numbers of the first kind. We have $z_{(n)}=(-1)^{n}\left(
-z\right) ^{(n)}$.

\item We will need%
\begin{eqnarray}
\frac{1}{\left( 1-z\right) ^{c}} &amp;=&amp;\sum_{n=0}^{+\infty }c^{(n)}\frac{z^{n}}{%
n!},  \label{D.6.1} \\
\left( 1+z\right) ^{c} &amp;=&amp;\sum_{n=0}^{+\infty }c_{(n)}\frac{z^{n}}{n!},
\label{D.6.2}
\end{eqnarray}%
being absolutely and uniformly convergent series for all $z\in \left\{
z\in \mathbb{C}:\left\vert z\right\vert &lt;1\right\} $ and for all $c\in 
\mathbb{C}$. We have for all $n\in \mathbb{N}$ and for all $a,b\in 
\mathbb{C}$,%
\begin{eqnarray}
\left( a+b\right) ^{(n)} &amp;=&amp;\sum_{k=0}^{n}\tbinom{n}{k}a^{(n-k)}b^{(k)},
\label{D.6.4} \\
\left( a+b\right) _{(n)} &amp;=&amp;\sum_{k=0}^{n}\tbinom{n}{k}a_{(n-k)}b_{(k)}.
\label{D.6.5}
\end{eqnarray}%
In particular, for $a=c=-b$, we get the orthogonality relations, for all
$n\in \mathbb{N}$ and for all $c\in \mathbb{C}$,%
\begin{eqnarray}
\sum_{k=0}^{n}\tbinom{n}{k}c^{(n-k)}\left( -c\right) ^{(k)} &amp;=&amp;\delta _{n=0},
\label{D.6.6} \\
\sum_{k=0}^{n}\tbinom{n}{k}c_{(n-k)}\left( -c\right) _{(k)} &amp;=&amp;\delta _{n=0}.
\label{D.6.7}
\end{eqnarray}

\item With $m,n\in \mathbb{N}$ and $K\triangleq \left\{
k_{1},k_{2},...,k_{m}\in \mathbb{N}\right\} $, define $\left\vert
K\right\vert \triangleq k_{1}+k_{2}+...+k_{m}$, $\#\left( K\right)
\triangleq m$ and $\tbinom{n}{K}\triangleq n!/\left(
k_{1}!k_{2}!...k_{m}!\right) $, expressions that are used in the last
section.
\end{enumerate}

\section{The generating function $B^{c}(x,y,t)$}

For all $x,y,t\in \mathbb{C}$ define%
\begin{equation}
B(x,y,t)\triangleq \left\{ 
\begin{array}{cc}
\frac{x-y}{xe^{-\frac{x-y}{2}t}-ye^{+\frac{x-y}{2}t}}, &amp; \text{if }x\neq y%
\text{;} \\ 
\frac{1}{1-xt}, &amp; \text{if }x=y\text{.}%
\end{array}%
\right.  \label{1.1}
\end{equation}%

\begin{proposition}
\label{P.1}
The Maclaurin series of the $c$-th power of $B(x,y,t)$, for all
$c\in \mathbb{C}$, is given by%
\begin{equation}
B^{c}(x,y,t)=\sum_{n=0}^{+\infty }2^{-n}B_{n}(x,y;c)\frac{t^{n}}{n!},
\label{1.2}
\end{equation}%
and converges absolutely and uniformly for $\left\vert t\right\vert &lt;\left\vert 
\frac{\ln x-\ln y}{x-y}\right\vert $. For all $n\in \mathbb{N}$,%
\begin{equation}
B_{n}(x,y;c)=\sum_{k=0}^{n}B_{n,k}(c)x^{n-k}y^{k},  \label{1.3}
\end{equation}%
with the coefficients $B_{n,k}(c)$ satisfying, %
for all $k\in \mathbb{N}_{n}$,%
\begin{equation}
B_{n+1,k+1}(c)=\left( 2(k+1)+c\right) B_{n,k+1}(c)+\left( 2(n-k)+c\right)
B_{n,k}(c),  \label{1.4}
\end{equation}%
with $B_{0,0}(c)=1$ and we define $B_{n,k}(c)\triangleq 0$, for all $k\notin 
\mathbb{N}_{n}$.
\end{proposition}

\begin{proof}
The point $t=0$ is an ordinary point of $B^{c}(x,y,t)$, so $B^{c}(x,y,t)$
has a Maclaurin power series, converging absolutely and uniformly for $%
\left| t\right| &lt;\left| \frac{\ln x-\ln y}{x-y}\right| $.

Define the partial differential operator%
\begin{equation}
D(x,y,t;c)\triangleq \left( 1-\frac{x+y}{2}t\right) \frac{\partial }{%
\partial t}+\frac{x-y}{2}\left( x\frac{\partial }{\partial x}-y\frac{%
\partial }{\partial y}\right) -c\frac{x+y}{2}.  \label{P.1.1}
\end{equation}%
A direct calculation shows that 
\begin{equation}
D(x,y,t;c)B^{c}(x,y,t)=0.  \label{P.1.2}
\end{equation}

Substituting in Eq. (\ref{P.1.2}) for $B^{c}(x,y,t)$ the uniformly convergent
series (\ref{1.2}) gives%
\begin{equation*}
\sum_{n=0}^{+\infty }2^{-n}\left( D_{n}(x,y;c)B_{n}(x,y;c)\right) \frac{t^{n}%
}{n!}=0,
\end{equation*}%
wherein%
\begin{equation}
D_{n}(x,y;c)\triangleq \frac{1}{2}T_{1}+\frac{x-y}{2}\left( x\frac{\partial 
}{\partial x}-y\frac{\partial }{\partial y}\right) -(n+c)\frac{x+y}{2}
\label{P.1.3}
\end{equation}%
and $T_{p}$ is the difference shift operator such that $%
T_{p}B_{n}(x,y;c)=B_{n+p}(x,y;c)$. This holds for any $t$, so we have%
\begin{equation}
D_{n}(x,y;c)B_{n}(x,y;c)=0.  \label{P.1.4}
\end{equation}

Substituting in Eq. (\ref{P.1.4}) for $B_{n}(x,y;c)$ the bivariate homogeneous
polynomial (\ref{1.3}) gives%
\begin{equation*}
\sum_{k=0}^{n}\left( D_{n,k}(x,y;c)B_{n,k}(c)\right) x^{n-k}y^{k}=0,
\end{equation*}%
wherein%
\begin{equation}
D_{n,k}(x,y;c)\triangleq \frac{1}{2}T_{1,0}-\left( \left( n-k+1\right)
+c/2\right) T_{0,-1}-\left( k+c/2\right)  \label{P.1.5}
\end{equation}%
and $T_{p,q}$ is the bivariate difference shift operator such that $%
T_{p,q}B_{n,k}(c)=B_{n+p,k+q}(c)$. This holds for any $x$ and $y$, so we have%
\begin{equation}
D_{n,k}(x,y;c)B_{n,k}(c)=0,  \label{P.1.6}
\end{equation}%
which is just Eq. (\ref{1.4}). 

&gt;From the fact that $B^{c}(x,y,0)=1$, we obtain $%
B_{0}(x,y;c)=B_{0,0}(c)=1$.
\end{proof}

We have that $B(x,y,t)=B(y,x,t)$ for all $x,y,t\in \mathbb{C}$, hence $%
B_{n}(x,y;c)=B_{n}(y,x;c)$ for all $n\in \mathbb{N}$, and $%
B_{n,k}(c)=B_{n,n-k}(c)$ for all $c\in \mathbb{C}$.

\subsection{Special cases}

(i) For $x=0$ or $y=0$, we get%
\begin{equation}
B^{c}(0,z,t)=B^{c}(z,0,t)=e^{\frac{c}{2}zt}.  \label{1.8b}
\end{equation}%
This implies that%
\begin{equation}
B_{n}(0,z;c)=B_{n}(z,0;c)=\left( cz\right) ^{n},  \label{1.9b}
\end{equation}%
and this yields in turn that%
\begin{equation}
B_{n,k}(c)=c^{n}\delta _{n=k}.  \label{1.10b}
\end{equation}

(ii) For $y=\pm x$, we get%
\begin{eqnarray}
B^{c}(x,x,t) &amp;=&amp;\frac{1}{\left( 1-xt\right) ^{c}},  \label{1.5} \\
B^{c}(x,-x,t) &amp;=&amp;\func{sech}^{c}(xt).  \label{1.6}
\end{eqnarray}%
This gives%
\begin{eqnarray}
B_{n}(x,x;c) &amp;=&amp;2^{n}c^{(n)}x^{n},  \label{1.7} \\
B_{n}(x,-x;c) &amp;=&amp;2^{n}\left( \lim_{t\rightarrow 0}D_{t}^{n}\func{sech}%
^{c}t\right) x^{n},  \label{1.8}
\end{eqnarray}%
and this yields in turn%
\begin{eqnarray}
\sum_{k=0}^{n}B_{n,k}(c) &amp;=&amp;2^{n}c^{(n)},  \label{1.9} \\
\sum_{k=0}^{n}(-1)^{k}B_{n,k}(c) &amp;=&amp;2^{n}\left( \lim_{t\rightarrow
0}D_{t}^{n}\func{sech}^{c}t\right) .  \label{1.10}
\end{eqnarray}

\subsection{The numbers $B_{n,k}(1)$ and $B_{n,k}(2)$}

Putting $c=0$ in Eq. (\ref{1.2}) shows that $B_{n}(x,y;0)=\delta _{n=0}$, so $%
B_{n,k}(0)=\delta _{n=0}$.

(i) For $c=1$, Eq. (\ref{1.4}) becomes%
\begin{equation*}
B_{n+1,k+1}(1)=\left( 2(k+1)+1\right) B_{n,k+1}(1)+\left( 2(n-k)+1\right)
B_{n,k}(1),
\end{equation*}%
so%
\begin{equation}
B_{n,k}(1)=B_{n+1,k+1},  \label{1.11}
\end{equation}%
with $B_{n,k}\ $the numbers derived by MacMahon \cite[p. 331]{M},
(Sloane's \seqnum{A060187}), cf. Table \ref{B}.


\begin{table}[tbp] \centering%
\begin{tabular}{|r|r|r|r|r|r|r|}
\hline
{\tiny n\verb?\? k} &amp; {\tiny 1} &amp; {\tiny 2} &amp; {\tiny 3} &amp; 
{\tiny 4} &amp; {\tiny 5} &amp; {\tiny 6} \\ \hline
{\tiny 1} &amp; {\tiny 1} &amp;  &amp;  &amp;  &amp;  &amp;  \\ \hline
{\tiny 2} &amp; {\tiny 1} &amp; {\tiny 1} &amp;  &amp;  &amp;  &amp;  \\ \hline
{\tiny 3} &amp; {\tiny 1} &amp; {\tiny 6} &amp; {\tiny 1} &amp;  &amp;  &amp;  \\ \hline
{\tiny 4} &amp; {\tiny 1} &amp; {\tiny 23} &amp; {\tiny 23} &amp; {\tiny 1} &amp;  &amp;  \\ \hline
{\tiny 5} &amp; {\tiny 1} &amp; {\tiny 76} &amp; {\tiny 230} &amp; {\tiny 76} &amp; {\tiny 1} &amp; 
\\ \hline
{\tiny 6} &amp; {\tiny 1} &amp; {\tiny 237} &amp; {\tiny 1682} &amp; {\tiny 1682} &amp; {\tiny %
237} &amp; {\tiny 1} \\ \hline
\end{tabular}
\caption{The number triangle $B_{n,k}$\label{B}}%
\end{table}

In this case, Eqs. (\ref{1.9}) and (\ref{1.10}) become%
\begin{eqnarray}
\sum_{k=0}^{n}B_{n+1,k+1} &amp;=&amp;2^{n}n!,  \label{1.12'} \\
\sum_{k=0}^{n}(-1)^{k}B_{n+1,k+1} &amp;=&amp;2^{n}E_{n},  \label{1.13'}
\end{eqnarray}%
with $E_{n}$ the Euler (or secant) numbers \cite[p. 804, 23.1.2]{AS}, ($%
\left\vert E_{2n}\right\vert $ are Sloane's \seqnum{A000364}).
The numbers $B_{n,k}$
are thus (also) generated by (for $\left\vert t\right\vert &lt;\frac{1}{2}%
\left\vert \frac{\ln y}{1-y}\right\vert $ and $y\neq 1$)%
\begin{equation}
\frac{1-y}{e^{-\left( 1-y\right) t}-ye^{+\left( 1-y\right) t}}%
=\sum_{n=0}^{+\infty }\sum_{k=0}^{n}B_{n+1,k+1}y^{k}\frac{t^{n}}{n!}.
\label{1.19a'}
\end{equation}%
We can also obtain from Eq. (\ref{1.19a'}) the following more standard generating function for the $B_{n,k}$, (i.e., on the same
footing as Eq. (\ref{1.18}) below), (for $%
\left\vert t\right\vert &lt;\left\vert \frac{\ln y}{1-y}\right\vert $ and $%
y\neq 1$)%
\begin{equation}
\frac{1}{2}\ln \frac{\frac{e^{-\frac{1-y^{2}}{2}t}+ye^{+\frac{1-y^{2}}{2}t}}{%
1+y}}{\frac{e^{-\frac{1-y^{2}}{2}t}-ye^{+\frac{1-y^{2}}{2}t}}{1-y}}%
=\sum_{n=1}^{+\infty }\sum_{k=1}^{n}B_{n,k}y^{2k-1}\frac{t^{n}}{n!}.
\label{1.19b'}
\end{equation}%
Eqs. (\ref{1.19a'}) and (\ref{1.19b'}) appear to be new.

(ii) For $c=2$, Eq. (\ref{1.4}) becomes%
\begin{equation*}
B_{n+1,k+1}(2)=\left( k+2\right) 2B_{n,k+1}(2)+\left( n-k+1\right)
2B_{n,k}(2),
\end{equation*}%
so%
\begin{equation}
B_{n,k}(2)=2^{n}A_{n+1,k+1},  \label{1.14}
\end{equation}%
with $A_{n,k}\ $the Eulerian numbers \cite{C}, (Sloane's \seqnum{A008292}), cf.
Table \ref{A}. Another notation for the Eulerian numbers is $\QTATOPD\langle
\rangle {n}{k}=A_{n,k+1}$.

\begin{table}[tbp] \centering
{\normalsize 
\begin{tabular}{|r|r|r|r|r|r|r|}
\hline
{\tiny n\verb?\? k} &amp; {\tiny 1} &amp; {\tiny 2} &amp; {\tiny 3} &amp; 
{\tiny 4} &amp; {\tiny 5} &amp; {\tiny 6} \\ \hline
{\tiny 1} &amp; {\tiny 1} &amp;  &amp;  &amp;  &amp;  &amp;  \\ \hline
{\tiny 2} &amp; {\tiny 1} &amp; {\tiny 1} &amp;  &amp;  &amp;  &amp;  \\ \hline
{\tiny 3} &amp; {\tiny 1} &amp; {\tiny 4} &amp; {\tiny 1} &amp;  &amp;  &amp;  \\ \hline
{\tiny 4} &amp; {\tiny 1} &amp; {\tiny 11} &amp; {\tiny 11} &amp; {\tiny 1} &amp;  &amp;  \\ \hline
{\tiny 5} &amp; {\tiny 1} &amp; {\tiny 26} &amp; {\tiny 66} &amp; {\tiny 26} &amp; {\tiny 1} &amp; 
\\ \hline
{\tiny 6} &amp; {\tiny 1} &amp; {\tiny 57} &amp; {\tiny 302} &amp; {\tiny 302} &amp; {\tiny 57}
&amp; {\tiny 1} \\ \hline
\end{tabular}%
}

\caption{The number triangle $A_{n,k}$\label{A}}%

\end{table}
In this case, Eqs. (\ref{1.9}) and (\ref{1.10}) become%
\begin{eqnarray}
\sum_{k=0}^{n}A_{n+1,k+1} &amp;=&amp;(n+1)!,  \label{1.15} \\
\sum_{k=0}^{n}(-1)^{k}A_{n+1,k+1} &amp;=&amp;2^{n+2}\left( 2^{n+2}-1\right) \frac{%
B_{n+2}}{n+2},  \label{1.16}
\end{eqnarray}%
with $B_{n}$ the Bernoulli numbers \cite[p. 804, 23.1.2]{AS}, ($\left\vert
B_{n}\right\vert $ are Sloane's \seqnum{A027641} and \seqnum{A027642}).
In Eq.\ (\ref{1.16}) we used $%
D_{t}^{n}\func{sech}^{2}t=D_{t}^{n+1}\tanh t$. The Eulerian numbers $A_{n,k}$
are thus (also) generated by%
\begin{equation}
\frac{1}{\left( \frac{e^{-\frac{1-y}{2}t}-ye^{+\frac{1-y}{2}t}}{1-y}\right)
^{2}}=\sum_{n=0}^{+\infty }\sum_{k=0}^{n}A_{n+1,k+1}y^{k}\frac{t^{n}}{n!}.
\label{1.17}
\end{equation}%
The well-known standard generating function for the Eulerian numbers is%
\begin{equation}
\frac{1-y}{1-ye^{(1-y)t}}=1+\sum_{n=1}^{+\infty }\sum_{k=1}^{n}A_{n,k}y^{k}%
\frac{t^{n}}{n!}.  \label{1.18}
\end{equation}

For further convenience we define $A_{n,k}\triangleq 0$ and $%
B_{n,k}\triangleq 0$, for all $k\notin \mathbb{Z}_{+,n}$.

\subsection{Examples of some $B_{n}(x,y;c)$}

The first six $B_{n}(x,y;c)$ are:%
\begin{equation*}
B_{0}(x,y;c)=1,
\end{equation*}%
\begin{equation*}
B_{1}(x,y;c)=cx+cy,
\end{equation*}%
\begin{equation*}
B_{2}(x,y;c)=c^{2}x^{2}+2c\left( 2+c\right) xy+c^{2}y^{2},
\end{equation*}%
\begin{equation*}
B_{3}(x,y;c)=c^{3}x^{3}+c\left( 3c^{2}+12c+8\right) x^{2}y+c\left(
3c^{2}+12c+8\right) xy^{2}+c^{3}y^{3},
\end{equation*}%
\begin{equation*}
B_{4}(x,y;c)=%
\begin{array}{c}
c^{4}x^{4} \\ 
+4c\left( c^{3}+6c^{2}+8c+4\right) x^{3}y \\ 
+2c\left( 3c^{3}+24c^{2}+56c+32\right) x^{2}y^{2} \\ 
+4c\left( c^{3}+6c^{2}+8c+4\right) xy^{3} \\ 
+c^{4}y^{4}%
\end{array}%
,
\end{equation*}%
\begin{equation*}
B_{5}(x,y;c)=%
\begin{array}{c}
c^{5}x^{5} \\ 
+c\left( 5c^{4}+40c^{3}+80c^{2}+80c+32\right) x^{4}y \\ 
+c\left( 10c^{4}+120c^{3}+480c^{2}+720c+352\right) x^{3}y^{2} \\ 
+c\left( 10c^{4}+120c^{3}+480c^{2}+720c+352\right) x^{2}y^{3} \\ 
+c\left( 5c^{4}+40c^{3}+80c^{2}+80c+32\right) xy^{4} \\ 
+c^{5}y^{5}%
\end{array}%
.
\end{equation*}

\section{Properties of the $B_{n}(x,y;c)$}

\subsection{Additive property with respect to the parameter $c$}

Obviously, for all $a,b\in \mathbb{C}$,%
\begin{equation}
B^{a+b}(x,y,t)=B^{a}(x,y,t)B^{b}(x,y,t),  \label{1.20}
\end{equation}%
and from this follows, for all $n\in \mathbb{N}$,%
\begin{equation}
B_{n}(x,y;a+b)=\sum_{k=0}^{n}\tbinom{n}{k}B_{n-k}(x,y;a)B_{k}(x,y;b).
\label{1.21}
\end{equation}%
Substituting Eq. (\ref{1.3}) in Eq. (\ref{1.21}) gives%
\begin{equation}
B_{n,k}(a+b)=\sum_{p=0}^{n}\tbinom{n}{p}%
\sum_{q=0}^{k}B_{n-p,k-q}(a)B_{p,q}(b).  \label{1.22}
\end{equation}%
For instance, by letting $a=b=1$, Eq. (\ref{1.22}) yields the following quadratic
expansion of Eulerian numbers $A_{n,k}$ in the MacMahon numbers $B_{n,k}$,%
\begin{equation}
A_{n+1,k+1}=\frac{1}{2^{n}}\sum_{p=0}^{n}\tbinom{n}{p}%
\sum_{q=0}^{k}B_{n-p+1,k-q+1}B_{p+1,q+1}.  \label{1.23}
\end{equation}

\subsection{Infinite series}

\begin{proposition}
\label{P.2}For all $c\in \mathbb{C}$ and for all $n\in \mathbb{N%
}$,%
\begin{equation}
B_{n}(x,y;c)=\left\{ 
\begin{array}{ccc}
\frac{\left( x-y\right) ^{n+c}}{x^{c}}\sum_{k=0}^{+\infty }c^{(k)}\left(
2k+c\right) ^{n}\frac{\left( y/x\right) ^{k}}{k!} &amp; \text{if} &amp; \left\vert
y\right\vert &lt;\left\vert x\right\vert \text{;} \\ 
\frac{\left( y-x\right) ^{n+c}}{y^{c}}\sum_{k=0}^{+\infty }c^{(k)}\left(
2k+c\right) ^{n}\frac{\left( x/y\right) ^{k}}{k!} &amp; \text{if} &amp; \left\vert
x\right\vert &lt;\left\vert y\right\vert \text{,}%
\end{array}%
\right.  \label{2.1}
\end{equation}%
where $x,y\in \mathbb{C}$ and the series converges absolutely.
\end{proposition}

\begin{proof}
(i) Applying Eq. (\ref{D.6.1}) to $B^{c}(x,y,t)$ gives, for all $(x,y)\in
D_{\left| y\right| &lt;\left| x\right| }\triangleq \left\{ (x,y)\in \mathbb{C}%
^{2}:\left| y\right| &lt;\left| x\right| \right\} $ and for all $t\in \Lambda
_{t}\left( x,y\right) \triangleq \left\{ t\in \mathbb{C}:\func{Re}%
((1-y/x)t)&lt;\left( \ln \left| x\right| -\ln \left| y\right| \right) \right\} $%
, the absolutely convergent series%
\begin{equation*}
B^{c}(x,y,t)=\frac{\left( x-y\right) ^{c}}{x^{c}}\sum_{k=0}^{+\infty }c^{(k)}%
\frac{\left( y/x\right) ^{k}}{k!}e^{+\left( k+c/2\right) \left( x-y\right)
t}.
\end{equation*}%
Expanding herein $e^{+\left( k+c/2\right) \left( x-y\right) t}$ in Maclaurin
series gives%
\begin{equation*}
B^{c}(x,y,t)=\frac{\left( x-y\right) ^{c}}{x^{c}}\sum_{k=0}^{+\infty }c^{(k)}%
\frac{\left( y/x\right) ^{k}}{k!}\sum_{n=0}^{+\infty }\left( k+c/2\right)
^{n}\left( x-y\right) ^{n}\frac{t^{n}}{n!}.
\end{equation*}%
Both series are absolutely convergent, so we may interchange the order of summation \cite[p. 175, Theorem 8.3]{Ru}%
, yielding%
\begin{equation*}
B^{c}(x,y,t)=\sum_{n=0}^{+\infty }\left( \frac{\left( x-y\right) ^{n+c}}{%
x^{c}}\sum_{k=0}^{+\infty }c^{(k)}\left( k+c/2\right) ^{n}\frac{\left(
y/x\right) ^{k}}{k!}\right) \frac{t^{n}}{n!}.
\end{equation*}%
On the other hand holds by Proposition \ref{P.1}, for all $t\in \Omega
_{t}\left( x,y\right) \triangleq \left\{ t\in \mathbb{C}:\left| t\right|
&lt;\left| \frac{\ln x-\ln y}{x-y}\right| \right\} $, that%
\begin{equation*}
B^{c}(x,y,t)=\sum_{n=0}^{+\infty }2^{-n}B_{n}(x,y;c)\frac{t^{n}}{n!}.
\end{equation*}

For $(x,y)\in D_{\left\vert y\right\vert &lt;\left\vert x\right\vert }$, $%
\Lambda _{t}\left( x,y\right) \cap \Omega _{t}\left( x,y\right) \neq
\varnothing $. Then for all $t\in \Lambda _{t}\left( x,y\right) \cap \Omega
_{t}\left( x,y\right) $ holds%
\begin{equation*}
\sum_{n=0}^{+\infty }2^{-n}B_{n}(x,y;c)\frac{t^{n}}{n!}=\sum_{n=0}^{+\infty
}\left( \frac{\left( x-y\right) ^{n+c}}{x^{c}}\sum_{k=0}^{+\infty
}c^{(k)}\left( k+c/2\right) ^{n}\frac{\left( y/x\right) ^{k}}{k!}\right) 
\frac{t^{n}}{n!},
\end{equation*}%
and the first part of Eq. (\ref{2.1}) follows.

(ii) Similar.
\end{proof}

In particular, Eq. (\ref{2.1}) becomes, for $c=m\in \mathbb{Z}_{+}$, %
\begin{equation}
B_{n}(x,y;m)=\left( x-y\right) ^{n+m}\sum_{k=0}^{+\infty }\tbinom{m-1+k}{k}%
\left( 2k+m\right) ^{n}x^{-(k+m)}y^{k},  \label{2.3}
\end{equation}%
and for $c=-m\in \mathbb{Z}_{-}$,%
\begin{equation}
B_{n}(x,y;-m)=\left( x-y\right) ^{n-m}\sum_{k=0}^{m}(-1)^{k}\tbinom{m}{k}%
\left( 2k-m\right) ^{n}x^{m-k}y^{k}.  \label{2.4}
\end{equation}

Moreover, Eq. (\ref{2.1}) reduces to the following special form, for all $c\in \mathbb{C}$,%
\begin{eqnarray}
\sum_{k=0}^{+\infty }c^{(k)}\left( 2k+c\right) ^{n}\frac{z^{k}}{k!} &amp;=&amp;\frac{%
B_{n}(1,z;c)}{\left( 1-z\right) ^{n+c}},\left\vert z\right\vert &lt;1,
\label{2.5} \\
\sum_{k=0}^{+\infty }c^{(k)}\left( 2k+c\right) ^{n}\frac{z^{-k}}{k!} &amp;=&amp;z^{c}%
\frac{B_{n}(z,1;c)}{\left( z-1\right) ^{n+c}},\left\vert z\right\vert &gt;1.
\label{2.6}
\end{eqnarray}

In particular, Eqs. (\ref{2.5}) and (\ref{2.6}) become,

(i) for $c=1$, for all $n\in \mathbb{N}$,%
\begin{eqnarray}
\sum_{k=0}^{+\infty }\left( 2k+1\right) ^{n}z^{k} &amp;=&amp;\frac{B_{n}(1,z)}{%
\left( 1-z\right) ^{n+1}},\left| z\right| &lt;1,  \label{2.7} \\
\sum_{k=0}^{+\infty }\left( 2k+1\right) ^{n}z^{-k} &amp;=&amp;z\frac{B_{n}(z,1)}{%
\left( z-1\right) ^{n+1}},\left| z\right| &gt;1.  \label{2.8}
\end{eqnarray}%
Herein is $B_{n}(x,y)$ the MacMahon homogeneous bivariate polynomial,%
\begin{equation}
B_{n}(x,y;1)\triangleq B_{n}(x,y)=\sum_{k=0}^{n}B_{n+1,k+1}x^{n-k}y^{k}.
\label{2.11a}
\end{equation}

(ii) for $c=2$, for all $n\in \mathbb{Z}_{+}$,%
\begin{eqnarray}
\sum_{l=1}^{+\infty }l^{n}z^{l} &amp;=&amp;\frac{z}{\left( 1-z\right) ^{n+1}}%
A_{n-1}(1,z),\left| z\right| &lt;1,  \label{2.9} \\
\sum_{l=1}^{+\infty }l^{n}z^{-l} &amp;=&amp;\frac{z}{\left( z-1\right) ^{n+1}}%
A_{n-1}(z,1),\left| z\right| &gt;1.  \label{2.10}
\end{eqnarray}%
Herein is $A_{n}(x,y)$ the Eulerian homogeneous bivariate polynomial,%
\begin{equation}
2^{-n}B_{n}(x,y;2)\triangleq
A_{n}(x,y)=\sum_{k=0}^{n}A_{n+1,k+1}x^{n-k}y^{k}.  \label{2.11b}
\end{equation}%
Notice that the left hand side of Eq. (\ref{2.9}) is by definition the
polylogarithm of negative integer order, $\func{Li}_{-n}(z)$ \cite{L}%
. Further, combining Eq. (\ref{2.9}) with \cite[Eq. (14)]{W.St2}, we get the
interesting identity, for all $n\in \mathbb{N}$,%
\begin{equation}
\sum_{p=1}^{n}(-1)^{n-p}p!S(n,p)z^{p-1}=A_{n-1}(z,z-1).  \label{2.14}
\end{equation}

Taking in Eq. (\ref{2.4}) the $\lim_{y\rightarrow -x}$ and using Eq. (\ref{1.6}) we
obtain, for all $m,n\in \mathbb{N}$,%
\begin{equation}
\lim_{t\rightarrow 0}D_{t}^{n}\cosh ^{m}t=\frac{1}{2^{m}}\sum_{k=0}^{m}%
\tbinom{m}{k}\left( 2k-m\right) ^{n}.  \label{2.12}
\end{equation}%
Taking in Eq. (\ref{2.4}) the $\lim_{y\rightarrow x}$ and using Eq. (\ref{1.7})
yields, for all $m\in \mathbb{N}$,%
\begin{equation}
\frac{1}{2^{m}}\sum_{k=0}^{m}(-1)^{k}\tbinom{m}{k}\left( 2k-m\right)
^{m}=(-1)^{m}m!.  \label{2.13}
\end{equation}

\subsection{Generating expression}

\begin{proposition}
\label{P.5}For all $n\in \mathbb{N}$ and for all $c,z\in 
\mathbb{C}$,%
\begin{equation}
B_{n}(1,z;c)=\left( 1-z\right) ^{n+c}\left( c+2zD_{z}\right) ^{n}\left(
1-z\right) ^{-c}.  \label{5.1}
\end{equation}
\end{proposition}

\begin{proof}
It is easy to show from Eq. (\ref{D.6.2}) that, for all $n\in \mathbb{N}$, for all 
$b,x\in \mathbb{C}$ and for all $z\in \left\{ z\in \mathbb{C}:\left|
z\right| &lt;1\right\} $, the following identity holds%
\begin{equation*}
\left( x+2zD_{z}\right) ^{n}\left( 1+z\right) ^{b}=\sum_{k=0}^{+\infty
}\left( 2k+x\right) ^{n}b_{(k)}\frac{z^{k}}{k!}.
\end{equation*}%
Then%
\begin{equation*}
\left( 1+z\right) ^{a}\left( x+2zD_{z}\right) ^{n}\left( 1+z\right)
^{b}=\sum_{k=0}^{+\infty }\left( \sum_{l=0}^{k}\tbinom{k}{l}%
a_{(k-l)}b_{(l)}\left( 2l+x\right) ^{n}\right) \frac{z^{k}}{k!}.
\end{equation*}%
Putting $x=c$, $a=n+c$, $b=-c$ and substituting $z\rightarrow -z$, we get%
\begin{eqnarray*}
&amp;&amp;\left( 1-z\right) ^{n+c}\left( c+2zD_{z}\right) ^{n}\left( 1-z\right) ^{-c}
\\
&amp;=&amp;\sum_{k=0}^{+\infty }\left( (-1)^{k}\sum_{l=0}^{k}\tbinom{k}{l}\left(
n+c\right) _{(k-l)}\left( -c\right) _{(l)}\left( 2l+c\right) ^{n}\right) 
\frac{z^{k}}{k!}.
\end{eqnarray*}%
Due to the fact that $\left( 1-z\right) ^{n+c}\left( c+2zD_{z}\right) ^{n}\left(
1-z\right) ^{-c}$ is a polynomial of degree $n$ in $z$, we must have that%
\begin{equation*}
(-1)^{k}\sum_{l=0}^{k}\tbinom{k}{l}\left( n+c\right) _{(k-l)}\left(
-c\right) _{(l)}\left( 2l+c\right) ^{n}=0,
\end{equation*}%
for all $k\notin \mathbb{Z}_{+,n}$.\ Hence using Eq. (\ref{4.1b}) below, Eq. (\ref%
{1.3}) and the fact that $B_{n,k}(c)\triangleq 0$, for all $k\notin \mathbb{Z%
}_{+,n}$, Eq. (\ref{5.1}) follows.
\end{proof}

In particular, for $c=1$, we obtain%
\begin{equation}
\left( 1-z\right) ^{n+1}\left( 1+2zD_{z}\right) ^{n}\left( 1-z\right)
^{-1}=\sum_{k=0}^{n}B_{n+1,k+1}z^{k},  \label{5.2}
\end{equation}%
and for $c=2$, we obtain%
\begin{equation}
\left( 1-z\right) ^{n+2}\left( 1+zD_{z}\right) ^{n}\left( 1-z\right)
^{-2}=\sum_{k=0}^{n}A_{n+1,k+1}z^{k}.  \label{5.3}
\end{equation}%
These appear to be new generating expressions for the MacMahon and Eulerian
numbers.

\section{Properties of the $B_{n,k}(c)$}

\begin{proposition}
\label{P.3}For all $m,n\in \mathbb{N}$ and for all $c\in 
\mathbb{C}$,%
\begin{equation}
\sum_{k=0}^{m}\tbinom{m}{k}\left( n+c\right) ^{(m-k)}\left(
k!B_{n,k}(c)\right) =c^{(m)}\left( 2m+c\right) ^{n}.  \label{3.1}
\end{equation}
\end{proposition}

\begin{proof}
For all $c\in \mathbb{C}$ and for all $z\in \mathbb{C}$ such that $\left|
z\right| &lt;1$ we have the absolutely convergent series (\ref{2.5}). As $%
\left| z\right| &lt;1$, we can apply Eq. (\ref{D.6.1}) and get%
\begin{equation*}
\sum_{m=0}^{+\infty }c^{(m)}\left( 2m+c\right) ^{n}\frac{z^{m}}{m!}%
=\sum_{k=0}^{n}k!B_{n,k}(c)\frac{z^{k}}{k!}\sum_{l=0}^{+\infty }\left(
n+c\right) ^{(l)}\frac{z^{l}}{l!}.
\end{equation*}%
Interchanging the summation order gives%
\begin{equation*}
\sum_{m=0}^{+\infty }c^{(m)}\left( 2m+c\right) ^{n}\frac{z^{m}}{m!}%
=\sum_{l=0}^{+\infty }\sum_{k=0}^{n}\tbinom{k+l}{k}\left( n+c\right)
^{(l)}k!B_{n,k}(c)\frac{z^{k+l}}{\left( k+l\right) !}.
\end{equation*}%
With the definition $B_{n,k}(c)\triangleq 0$, for all $k\notin \mathbb{Z}%
_{+,n}$, we can write this as%
\begin{equation*}
\sum_{m=0}^{+\infty }c^{(m)}\left( 2m+c\right) ^{n}\frac{z^{m}}{m!}%
=\sum_{l=0}^{+\infty }\sum_{k=0}^{+\infty }\tbinom{k+l}{k}\left( n+c\right)
^{(l)}k!B_{n,k}(c)\frac{z^{k+l}}{\left( k+l\right) !}.
\end{equation*}%
This is equivalent to%
\begin{equation*}
\sum_{m=0}^{+\infty }c^{(m)}\left( 2m+c\right) ^{n}\frac{z^{m}}{m!}%
=\sum_{m=0}^{+\infty }\sum_{k=0}^{m}\tbinom{m}{k}\left( n+c\right)
^{(m-k)}k!B_{n,k}(c)\frac{z^{m}}{m!},
\end{equation*}%
and since $z$ is arbitrary, Eq. (\ref{3.1}) follows.
\end{proof}

In particular, for $c=1$, we obtain%
\begin{equation}
\sum_{k=0}^{m}\tbinom{n+m-k}{n}B_{n+1,k+1}=\left( 2m+1\right) ^{n},
\label{3.2}
\end{equation}%
and for $c=2$, we obtain%
\begin{equation}
\sum_{k=0}^{m}\tbinom{n+1+m-k}{n+1}A_{n+1,k+1}=\left( m+1\right) ^{n+1}.
\label{3.3}
\end{equation}%
These are well-known partial sums of the MacMahon and Eulerian number
triangles \cite[p. 328 and p. 331]{M}.

\subsection{Expressions}

\begin{proposition}
\label{P.4}For all $n\in \mathbb{N}$, for all $k\in \mathbb{N}%
_{n}$ and for all $c\in \mathbb{C}$,%
\begin{eqnarray}
k!B_{n,k}(c) &amp;=&amp;\sum_{l=0}^{k}\tbinom{k}{l}\left( -\left( n+c\right) \right)
^{(k-l)}c^{(l)}\left( 2l+c\right) ^{n},  \label{4.1a} \\
&amp;=&amp;(-1)^{k}\sum_{l=0}^{k}\tbinom{k}{l}\left( n+c\right) _{(k-l)}\left(
-c\right) _{(l)}\left( 2l+c\right) ^{n}.  \label{4.1b}
\end{eqnarray}
\end{proposition}

\begin{proof}
(i) We will show that Eq. (\ref{4.1a}) is a solution of Eq. (\ref{3.1}).\ Substitute
Eq. (\ref{4.1a}) in Eq. (\ref{3.1}) and get%
\begin{equation*}
\sum_{k=0}^{m}\tbinom{m}{k}\left( n+c\right) ^{(m-k)}\sum_{l=0}^{k}\tbinom{k%
}{l}\left( -\left( n+c\right) \right) ^{(k-l)}c^{(l)}\left( 2l+c\right)
^{n}=c^{(m)}\left( 2m+c\right) ^{n}.
\end{equation*}%
Interchanging the summation order gives%
\begin{equation*}
\sum_{l=0}^{m}\tbinom{m}{l}\left( \sum_{q=0}^{m-l}\tbinom{m-l}{q}\left(
n+c\right) ^{(m-l-q)}\left( -\left( n+c\right) \right) ^{(q)}\right)
c^{(l)}\left( 2l+c\right) ^{n}=c^{(m)}\left( 2m+c\right) ^{n}.
\end{equation*}%
Due to the orthogonality relation (\ref{D.6.6}) this simplifies to%
\begin{equation*}
\sum_{l=0}^{m}\tbinom{m}{l}\delta _{l=m}c^{(l)}\left( 2l+c\right)
^{n}=c^{(m)}\left( 2m+c\right) ^{n},
\end{equation*}%
and this is an identity.

(ii) Use $z_{(n)}=(-1)^{n}\left( -z\right) ^{(n)}$.
\end{proof}

In particular, for $c=-m\in \mathbb{Z}_{-}$,

(i) for $n\geq m$%
\begin{equation}
B_{n,k}(-m)=(-1)^{k}\sum_{l=\max (0,k+m-n)}^{\min (k,m)}\tbinom{n-m}{k-l}%
\tbinom{m}{l}\left( 2l-m\right) ^{n},  \label{4.5a}
\end{equation}

(ii) for $n&lt;m$%
\begin{equation}
B_{n,k}(-m)=\sum_{l=0}^{\min (k,m)}(-1)^{l}\tbinom{m-n-1+(k-l)}{k-l}\tbinom{m%
}{l}\left( 2l-m\right) ^{n}.  \label{4.5b}
\end{equation}

An equivalent form of Eqs. (\ref{4.1a}) and (\ref{4.1b}) is%
\begin{equation}
k!B_{n,k}(c)=\sum_{l=0}^{k}(-1)^{k-l}\tbinom{k}{l}\frac{\Gamma \left(
n+c+1\right) \Gamma \left( c+l\right) }{\Gamma \left( n+c+1-(k-l)\right)
\Gamma \left( c\right) }\left( 2l+c\right) ^{n}.  \label{4.1}
\end{equation}%
In particular, for $c=1$, we obtain%
\begin{equation}
B_{n+1,k+1}=\sum_{l=0}^{k}(-1)^{k-l}\tbinom{n+1}{k-l}\left( 2l+1\right) ^{n}.
\label{4.2}
\end{equation}%
Expression (\ref{4.2}) coincides with that given by MacMahon \cite[p. 331]{M}. For $c=2$
and using Eq. (\ref{1.14}), we get the familiar result%
\begin{equation}
A_{n+1,k+1}=\sum_{l=0}^{k}(-1)^{k-l}\tbinom{n+2}{k-l}\left( l+1\right)
^{n+1},  \label{4.3}
\end{equation}%
or equivalently, for all $n-1\in \mathbb{Z}_{+}$ and for all $k\in \mathbb{Z}%
_{+,n-1}$,%
\begin{equation}
\QTATOPD\langle \rangle {n-1}{k-1}=A_{n-1,k}=\sum_{l=0}^{k}(-1)^{l}\tbinom{n%
}{l}\left( k-l\right) ^{n-1}.  \label{4.3b}
\end{equation}

Let $S\left( j,i\right) $ denote the Stirling numbers of the second kind (Sloane's \seqnum{A008277}).
\begin{proposition}
\label{P.15}For all $n\in \mathbb{N}$, for all $k\in \mathbb{N}%
_{n}$ and for all $c\in \mathbb{C}$,%
\begin{equation}
B_{n,k}(c)=(-1)^{k}\sum_{j=0}^{n}\tbinom{n}{j}2^{j}c^{n-j}\sum_{i=0}^{\min
(k,j)}(-1)^{i}\tbinom{n-i}{k-i}S\left( j,i\right) c^{(i)}.  \label{15.1}
\end{equation}%
\end{proposition}

\begin{proof}
Using Eqs. (\ref{6.1}) and (\ref{6.2}) from Proposition \ref{P.6} below, we have%
\begin{eqnarray*}
B_{n,k}(c) &amp;=&amp;\sum_{l=0}^{n}b_{n,k,l}c^{l}, \\
&amp;=&amp;(-1)^{k}\sum_{l=0}^{n}(-1)^{l}\sum_{p=0}^{l}(-1)^{p}\tbinom{n}{p}%
2^{n-p}\sum_{i=l-p}^{\min (k,n-p)}\tbinom{n-i}{k-i}S\left( n-p,i\right)
s\left( i,l-p\right) c^{l}.
\end{eqnarray*}%
Interchanging the order of the first two summation gives%
\begin{eqnarray*}
B_{n,k}(c) &amp;=&amp;(-1)^{k}\sum_{p=0}^{n}\sum_{l=p}^{n}(-1)^{l-p}\tbinom{n}{p}%
2^{n-p}\sum_{i=l-p}^{\min (k,n-p)}\tbinom{n-i}{k-i}S\left( n-p,i\right)
s\left( i,l-p\right) c^{l}, \\
&amp;=&amp;(-1)^{k}\sum_{p=0}^{n}\sum_{l-p=0}^{n-p}(-1)^{l-p}\tbinom{n}{n-p}%
2^{n-p}\sum_{i=l-p}^{\min (k,n-p)}\tbinom{n-i}{k-i}S\left( n-p,i\right)
s\left( i,l-p\right) c^{l-p+p}, \\
&amp;=&amp;(-1)^{k}\sum_{p=0}^{n}\sum_{j=0}^{n-p}(-1)^{j}\tbinom{n}{n-p}%
2^{n-p}\sum_{i=j}^{\min (k,n-p)}\tbinom{n-i}{k-i}S\left( n-p,i\right)
s\left( i,j\right) c^{j+p}, \\
&amp;=&amp;(-1)^{k}\sum_{q=0}^{n}\tbinom{n}{q}2^{q}c^{n-q}\sum_{j=0}^{q}(-c)^{j}%
\sum_{i=j}^{\min (k,q)}\tbinom{n-i}{k-i}S\left( q,i\right) s\left(
i,j\right) .
\end{eqnarray*}%
Taking into account that $S\left( n,k\right) =s\left( n,k\right) \triangleq
0 $, for all $k\notin \mathbb{N}_{n}$, we can write this as%
\begin{equation*}
B_{n,k}(c)=(-1)^{k}\sum_{q=0}^{n}\tbinom{n}{q}2^{q}c^{n-q}%
\sum_{j=0}^{q}(-c)^{j}\sum_{i=0}^{k}\tbinom{n-i}{k-i}S\left( q,i\right)
s\left( i,j\right) .
\end{equation*}%
Interchanging the two last summations yields%
\begin{equation*}
B_{n,k}(c)=(-1)^{k}\sum_{q=0}^{n}\tbinom{n}{q}2^{q}c^{n-q}\sum_{i=0}^{k}%
\tbinom{n-i}{k-i}S\left( q,i\right) \sum_{j=0}^{i}s\left( i,j\right)
(-c)^{j}.
\end{equation*}%
Using the fundamental property of the Stirling numbers of the first kind 
\cite[p. 824, 24.1.3, I, B, 1]{AS},%
\begin{equation*}
(-c)_{(i)}=\sum_{j=0}^{i}s\left( i,j\right) (-c)^{j},
\end{equation*}%
we obtain%
\begin{equation*}
B_{n,k}(c)=(-1)^{k}\sum_{q=0}^{n}\tbinom{n}{q}2^{q}c^{n-q}\sum_{i=0}^{k}%
\tbinom{n-i}{k-i}S\left( q,i\right) (-c)_{(i)}.
\end{equation*}%
By using the identity $(-c)_{(i)}=(-1)^{i}c^{(i)}$, writing $j$ for $q$ and replacing the
upper limit in the second sum with $\min (k,j)$, Eq. (\ref{15.1}) follows.
\end{proof}

In particular, for $c=1$, we get%
\begin{equation}
B_{n+1,k+1}=(-1)^{k}\sum_{j=0}^{n}\tbinom{n}{j}2^{j}\sum_{i=0}^{\min
(k,j)}(-1)^{i}\tbinom{n-i}{k-i}i!S\left( j,i\right) ,  \label{15.2}
\end{equation}%
and for $c=2$, we get%
\begin{equation}
A_{n+1,k+1}=(-1)^{k}\sum_{j=0}^{n}\tbinom{n}{j}\sum_{i=0}^{\min
(k,j)}(-1)^{i}\tbinom{n-i}{k-i}(i+1)!S\left( j,i\right) .  \label{15.3}
\end{equation}%
These appear to be new expressions for the MacMahon and Eulerian numbers, in
terms of Stirling numbers of the second kind.

\subsection{Polynomial expression}

\begin{proposition}
\label{P.6}For all $n\in \mathbb{N}$ and for all $z\in \mathbb{C%
}$,%
\begin{equation}
B_{n,k}(c)=\sum_{l=0}^{n}b_{n,k,l}c^{l},  \label{6.1}
\end{equation}%
where, for all $k,l\in \mathbb{N}_{n}$,%
\begin{equation}
b_{n,k,l}=(-1)^{k+l}\sum_{p=\max (0,l-k)}^{l}(-1)^{p}\tbinom{n}{p}%
2^{n-p}\sum_{i=l-p}^{\min (k,n-p)}\tbinom{n-i}{k-i}s\left( i,l-p\right)
S\left( n-p,i\right) .  \label{6.2}
\end{equation}%
\end{proposition}

\begin{proof}
Using%
\begin{equation*}
\left( c+2zD_{z}\right) ^{n}=\sum_{l=0}^{n}\tbinom{n}{l}c^{n-l}2^{l}\left(
zD_{z}\right) ^{l},
\end{equation*}%
we get%
\begin{equation*}
\left( 1-z\right) ^{n+c}\left( c+2zD_{z}\right) ^{n}\left( 1-z\right)
^{-c}=\left( 1-z\right) ^{n+c}\sum_{l=0}^{n}\tbinom{n}{l}c^{n-l}2^{l}\left(
zD_{z}\right) ^{l}\left( 1-z\right) ^{-c}.
\end{equation*}%
Using herein the formula \cite[p. 144]{Ro1984},%
\begin{equation*}
\left( zD_{z}\right) ^{l}=\sum_{p=0}^{l}S\left( l,p\right) z^{p}D_{z}^{p},
\end{equation*}%
we obtain%
\begin{eqnarray*}
&amp;&amp;\left( 1-z\right) ^{n+c}\left( c+2zD_{z}\right) ^{n}\left( 1-z\right) ^{-c}
\\
&amp;=&amp;\left( 1-z\right) ^{n+c}\sum_{l=0}^{n}\tbinom{n}{l}c^{n-l}2^{l}%
\sum_{p=0}^{l}S\left( l,p\right) z^{p}D_{z}^{p}\left( 1-z\right) ^{-c}, \\
&amp;=&amp;\sum_{l=0}^{n}\tbinom{n}{l}c^{n-l}2^{l}\sum_{p=0}^{l}S\left( l,p\right)
z^{p}c^{(p)}\left( 1-z\right) ^{n-p}, \\
&amp;=&amp;\sum_{l=0}^{n}\tbinom{n}{l}c^{n-l}2^{l}\sum_{p=0}^{l}S\left( l,p\right)
\left( -c\right) _{(p)}\left( 1-z\right) ^{n-p}\left( -z\right) ^{p}, \\
&amp;=&amp;\sum_{p=0}^{n}\sum_{l=p}^{n}\tbinom{n}{l}c^{n-l}2^{l}S\left( l,p\right)
\left( -c\right) _{(p)}\left( 1-z\right) ^{n-p}\left( -z\right) ^{p}.
\end{eqnarray*}%
Substituting herein the binomial expansion for $\left( 1-z\right) ^{n-p}$
gives%
\begin{eqnarray*}
&amp;&amp;\left( 1-z\right) ^{n+c}\left( c+2zD_{z}\right) ^{n}\left( 1-z\right) ^{-c}
\\
&amp;=&amp;\sum_{p=0}^{n}\sum_{l=p}^{n}\tbinom{n}{l}c^{n-l}2^{l}S\left( l,p\right)
\left( -c\right) _{(p)}\sum_{q=0}^{n-p}\tbinom{n-p}{q}\left( -z\right)
^{q}\left( -z\right) ^{p}, \\
&amp;=&amp;\sum_{p=0}^{n}\sum_{l=p}^{n}\sum_{m=p}^{n}\tbinom{n}{l}\tbinom{n-p}{m-p}%
c^{n-l}2^{l}\left( -c\right) _{(p)}S\left( l,p\right) \left( -z\right) ^{m},
\\
&amp;=&amp;\sum_{p=0}^{n}\sum_{m=p}^{n}\sum_{l=p}^{n}\tbinom{n}{l}\tbinom{n-p}{m-p}%
c^{n-l}2^{l}\left( -c\right) _{(p)}S\left( l,p\right) \left( -z\right) ^{m},
\\
&amp;=&amp;\sum_{m=0}^{n}\sum_{p=0}^{m}\tbinom{n-p}{m-p}\left( -c\right)
_{(p)}\sum_{l=p}^{n}\tbinom{n}{l}c^{n-l}2^{l}S\left( l,p\right) \left(
-z\right) ^{m}.
\end{eqnarray*}%
Making use of the fundamental relation of the Stirling numbers of the first
kind,%
\begin{equation*}
\left( -c\right) _{(p)}=\sum_{k=0}^{p}s\left( p,k\right) \left( -c\right)
^{k},
\end{equation*}%
we get%
\begin{eqnarray*}
&amp;&amp;\left( 1-z\right) ^{n+c}\left( c+2zD_{z}\right) ^{n}\left( 1-z\right) ^{-c}
\\
&amp;=&amp;\sum_{m=0}^{n}\sum_{p=0}^{m}\tbinom{n-p}{m-p}\sum_{k=0}^{p}(-1)^{k}s%
\left( p,k\right) \sum_{q=p}^{n}\tbinom{n}{q}2^{q}S\left( q,p\right)
c^{n-q+k}\left( -z\right) ^{m}, \\
&amp;=&amp;\sum_{m=0}^{n}\sum_{p=0}^{m}\tbinom{n-p}{m-p}\sum_{k=0}^{p}(-1)^{k}s%
\left( p,k\right) \sum_{l=0}^{n-p}\tbinom{n}{n-l}2^{n-l}S\left( n-l,p\right)
c^{l+k}\left( -z\right) ^{m}.
\end{eqnarray*}%
Summing over diagonals in the two last summations and putting $S\left(
n,k\right) =s\left( n,k\right) \triangleq 0$, for all $k\notin \mathbb{N}%
_{n} $, this becomes%
\begin{eqnarray*}
&amp;&amp;\left( 1-z\right) ^{n+c}\left( c+2zD_{z}\right) ^{n}\left( 1-z\right) ^{-c}
\\
&amp;=&amp;\sum_{m=0}^{n}\sum_{p=0}^{m}\tbinom{n-p}{m-p}\sum_{q=0}^{n}%
\sum_{l=0}^{n-p}(-1)^{q-l}\tbinom{n}{l}2^{n-l}S\left( n-l,p\right) s\left(
p,q-l\right) c^{q}\left( -z\right) ^{m}, \\
&amp;=&amp;\sum_{m=0}^{n}\sum_{q=0}^{n}\sum_{p=0}^{m}\sum_{l=0}^{n-p}\tbinom{n-p}{m-p%
}(-1)^{q-l}\tbinom{n}{l}2^{n-l}S\left( n-l,p\right) s\left( p,q-l\right)
c^{q}\left( -z\right) ^{m}, \\
&amp;=&amp;\sum_{m=0}^{n}\sum_{q=0}^{n}\sum_{p=0}^{m}\sum_{l=0}^{n}\tbinom{n-p}{m-p}%
(-1)^{q-l}\tbinom{n}{l}2^{n-l}S\left( n-l,p\right) s\left( p,q-l\right)
c^{q}\left( -z\right) ^{m}, \\
&amp;=&amp;\sum_{m=0}^{n}\sum_{q=0}^{n}\sum_{l=0}^{n}\sum_{p=0}^{m}\tbinom{n-p}{m-p}%
(-1)^{q-l}\tbinom{n}{l}2^{n-l}S\left( n-l,p\right) s\left( p,q-l\right)
c^{q}\left( -z\right) ^{m}, \\
&amp;=&amp;\sum_{m=0}^{n}\sum_{q=0}^{n}\sum_{l=0}^{n}(-1)^{q-l}\tbinom{n}{l}%
2^{n-l}\sum_{p=\max (0,q-l)}^{\min (m,n-l)}\tbinom{n-p}{m-p}S\left(
n-l,p\right) s\left( p,q-l\right) c^{q}\left( -z\right) ^{m}.
\end{eqnarray*}%
Renaming indexes gives%
\begin{eqnarray*}
&amp;&amp;\left( 1-z\right) ^{n+c}\left( c+2zD_{z}\right) ^{n}\left( 1-z\right) ^{-c}
\\
&amp;=&amp;\sum_{k=0}^{n}(-1)^{k}\sum_{l=0}^{n}(-1)^{l}\sum_{j=0}^{n}\tbinom{n}{j}%
(-1)^{j} \\
&amp;&amp;\left( 2^{n-j}\sum_{i=\max (0,l-n+n-j)}^{\min (k,n-j)}\tbinom{n-i}{k-i}%
S\left( n-j,i\right) s\left( i,l-n+n-j\right) \right) c^{l}z^{k}, \\
&amp;=&amp;\sum_{k=0}^{n}(-1)^{k}\sum_{l=0}^{n}(-1)^{l}\sum_{j=0}^{n}\tbinom{n}{j}%
(-1)^{n-j} \\
&amp;&amp;\left( 2^{j}\sum_{i=\max (0,j+l-n)}^{\min (k,j)}\tbinom{n-i}{k-i}S\left(
j,i\right) s\left( i,j+l-n\right) \right) c^{l}z^{k}.
\end{eqnarray*}%
Applying Proposition \ref{P.5} and
identification with%
\begin{equation*}
B_{n}(1,z;c)=\sum_{k=0}^{n}\left( \sum_{l=0}^{n}b_{n,k,l}c^{l}\right) z^{k},
\end{equation*}%
yields, for all $k,l\in \mathbb{N}_{n}$,%
\begin{equation*}
b_{n,k,l}=(-1)^{n+k+l}\sum_{j=n-l}^{n}(-1)^{j}\tbinom{n}{j}\left(
2^{j}\sum_{i=j-(n-l)}^{\min (k,j)}\tbinom{n-i}{k-i}S\left( j,i\right)
s\left( i,j-(n-l)\right) \right) .
\end{equation*}%
This can be rewritten as%
\begin{eqnarray*}
b_{n,k,l} &amp;=&amp;(-1)^{n+l+k}\sum_{j=n-l}^{n}(-1)^{j}\tbinom{n}{j}%
2^{j}\sum_{i=j-(n-l)}^{\min (k,j)}\tbinom{n-i}{k-i}S\left( j,i\right)
s\left( i,j-(n-l)\right) , \\
&amp;=&amp;(-1)^{k}\sum_{j-(n-l)=0}^{l}(-1)^{j-(n-l)}\tbinom{n}{j-(n-l)+(n-l)} \\
&amp;&amp;2^{j-(n-l)+(n-l)}\sum_{i=j-(n-l)}^{\min (k,j-(n-l)+(n-l))}\tbinom{n-i}{k-i}%
S\left( j-(n-l)+(n-l),i\right) s\left( i,j-(n-l)\right) , \\
&amp;=&amp;(-1)^{k}\sum_{q=0}^{l}(-1)^{q}\tbinom{n}{q+(n-l)} \\
&amp;&amp;2^{q+(n-l)}\sum_{i=q}^{\min (k,q+(n-l))}\tbinom{n-i}{k-i}S\left(
q+(n-l),i\right) s\left( i,q\right) , \\
&amp;=&amp;(-1)^{k}\sum_{q=0}^{l}(-1)^{q}\tbinom{n}{l-q}2^{n-(l-q)}\sum_{i=q}^{\min
(k,n-(l-q))}\tbinom{n-i}{k-i}S\left( n-(l-q),i\right) s\left( i,q\right) , \\
&amp;=&amp;(-1)^{k}\sum_{p=\max (0,l-k)}^{l}(-1)^{l-p}\tbinom{n}{p}%
2^{n-p}\sum_{i=l-p}^{\min (k,n-p)}\tbinom{n-i}{k-i}S\left( n-p,i\right)
s\left( i,l-p\right) .
\end{eqnarray*}
\end{proof}

Using basic properties of the Stirling numbers, it is easy to derive the
following special values for the $b_{n,k,l}$,%
\begin{eqnarray}
b_{n,n,l} &amp;=&amp;b_{n,0,l}=\delta _{n=l},  \label{6.3a} \\
b_{n,k,n} &amp;=&amp;\tbinom{n}{k},  \label{6.3b} \\
b_{n,k,0} &amp;=&amp;\delta _{k=0}\delta _{n=0}.  \label{6.3c}
\end{eqnarray}

\subsection{Symmetry}

Recall Eq. (\ref{4.1b}) and a variant of it obtained by replacing $k$ with $n-k$,%
\begin{eqnarray*}
k!B_{n,k}(c) &amp;=&amp;(-1)^{k}\sum_{l=0}^{k}\tbinom{k}{l}\left( n+c\right)
_{(k-l)}\left( -c\right) _{(l)}\left( 2l+c\right) ^{n}, \\
\left( n-k\right) !B_{n,n-k}(c) &amp;=&amp;(-1)^{n-k}\sum_{l=0}^{n-k}\tbinom{n-k}{l}%
\left( n+c\right) _{(n-k-l)}\left( -c\right) _{(l)}\left( 2l+c\right) ^{n}.
\end{eqnarray*}%
Due to the symmetry $B_{n,k}(c)=B_{n,n-k}(c)$, there must hold, for all
$n\in \mathbb{N}$, for all $k\in \mathbb{N}_{n}$ and for all $c\in \mathbb{C}$%
, that%
\begin{eqnarray}
&amp;&amp;\frac{(-1)^{k}}{k!}\sum_{l=0}^{k}\tbinom{k}{l}\left( n+c\right)
_{(k-l)}\left( -c\right) _{(l)}\left( 2l+c\right) ^{n}  \notag \\
&amp;=&amp;\frac{(-1)^{n-k}}{\left( n-k\right) !}\sum_{l=0}^{n-k}\tbinom{n-k}{l}%
\left( n+c\right) _{(n-k-l)}\left( -c\right) _{(l)}\left( 2l+c\right) ^{n}.
\label{4.4}
\end{eqnarray}

In particular, for $k=0$, Eq. (\ref{4.4}) yields%
\begin{equation}
\frac{(-1)^{n}}{n!}\sum_{l=0}^{n}(-1)^{l}\tbinom{n}{l}\frac{c+n}{c+l}\left(
c+2l\right) ^{n}=\frac{c^{n}}{c^{(n)}},  \label{4.4a}
\end{equation}%
and for $k=1$,%
\begin{eqnarray}
&amp;&amp;\frac{(-1)^{n}}{n!}\sum_{l=0}^{n}(-1)^{l}\tbinom{n}{l}\frac{\left(
c+n+1\right) \left( c+n\right) }{(c+l+1)(c+l)}\left( c+2l\right) ^{n+1} 
\notag \\
&amp;=&amp;\frac{c\left( c+2\right) ^{n+1}-\left( c+n+1\right) c^{n+1}}{c^{(n)}}.
\label{4.4b}
\end{eqnarray}

\subsection{A result of Ruiz}

Summing Eq. (\ref{4.1}) over $k$ from $0$ to $n$ and using result (\ref{1.9}), we get, for all $n\in \mathbb{N}$ and for all $c\in \mathbb{C}$,%
\begin{eqnarray*}
c^{(n)} &amp;=&amp;2^{-n}\sum_{k=0}^{n}B_{n,k}(c), \\
&amp;=&amp;\sum_{k=0}^{n}\sum_{l=0}^{k}(-1)^{k-l}\tbinom{n+c}{k-l}\tbinom{c-1+l}{l}%
\left( l+c/2\right) ^{n},
\end{eqnarray*}%
or%
\begin{equation*}
\sum_{l=0}^{n}\left( \sum_{q=0}^{n-l}(-1)^{q}\tbinom{n+c}{q}\right) \tbinom{%
c-1+l}{l}\left( l+c/2\right) ^{n}=c^{(n)}.
\end{equation*}%
Using the binomial identity%
\begin{equation*}
\sum_{q=0}^{n-l}(-1)^{q}\tbinom{n+c}{q}=\left( -1\right) ^{n-l}\tbinom{n+c-1%
}{n-l},
\end{equation*}%
we get%
\begin{equation*}
\sum_{l=0}^{n}\left( -1\right) ^{n-l}\tbinom{n+c-1}{n-l}\tbinom{c-1+l}{l}%
\left( l+c/2\right) ^{n}=c^{(n)},
\end{equation*}%
or%
\begin{equation*}
\tbinom{n+c-1}{n}\sum_{l=0}^{n}\left( -1\right) ^{n-l}\tbinom{n}{l}\left(
l+c/2\right) ^{n}=c^{(n)},
\end{equation*}%
or%
\begin{equation*}
\sum_{l=0}^{n}\left( -1\right) ^{n-l}\tbinom{n}{l}\left( l+c/2\right)
^{n}=n!,
\end{equation*}%
or finally%
\begin{equation}
\frac{1}{n!}\sum_{l=0}^{n}\left( -1\right) ^{l}\tbinom{n}{l}\left( \left(
-c/2\right) -l\right) ^{n}=1.  \label{4.5}
\end{equation}%
Identity (\ref{4.5}) is a result of Ruiz \cite{R}. Written in the form%
\begin{equation}
\frac{1}{n!}\sum_{l=0}^{n}\left( -1\right) ^{n-l}\tbinom{n}{l}\left(
c+2l\right) ^{n}=2^{n},  \label{4.5c}
\end{equation}%
it can be regarded as the first identity in a series of identities of which %
Eqs. (\ref{4.4a}) and (\ref{4.4b}) are the next two successors. Ruiz's result
however is special because the sum in Eq. (\ref{4.5c}) is independent of $c$.

In addition, applying $D_{c}^{m}$ to Eq. (\ref{4.5c}) we obtain the following derived
identities, for all $n\in \mathbb{N}$, for all $m\in \mathbb{N}_{n}$ and for all $c\in \mathbb{C}$,%
\begin{equation}
\sum_{l=0}^{n}\left( -1\right) ^{l}\tbinom{n}{l}\left( c+2l\right)
^{n-m}=\left( -1\right) ^{n}2^{n}n!\delta _{m=0}.  \label{4.5f}
\end{equation}

\section{Properties of the $b_{n,k,l}$}

Due to the symmetry relation $B_{n,k}(c)=B_{n,n-k}(c)$, we have that %
$b_{n,n-k,l}=b_{n,k,l}$, for all $n\in \mathbb{N}$ and for all $k,l\in 
\mathbb{N}_{n}$.

\subsection{Recursion relation for the $b_{n,k,l}$}

\begin{proposition}
\label{P.16}For all $n\in \mathbb{N}$ and for all $k,l\in 
\mathbb{N}_{n}$,%
\begin{equation}
b_{n+1,k+1,l+1}=2(k+1)b_{n,k+1,l+1}+b_{n,k+1,l}+2(n-k)b_{n,k,l+1}+b_{n,k,l},
\label{16.1}
\end{equation}%
with $b_{0,0,0}=1$ and $b_{n,k,l}\triangleq 0$ if $k,l\notin \mathbb{Z}%
_{+,n} $.
\end{proposition}

\begin{proof}
&gt;From the fact that $B_{n,k}(0)=\delta _{n=0}$ we find that $b_{n,k,0}=\delta _{k=0}\delta _{n=0}$
and hence $b_{0,0,0}=1$.

Substituting Eq. (\ref{6.1}) in Eq. (\ref{1.4}), we get%
\begin{equation*}
\sum_{l=0}^{n+1}b_{n+1,k+1,l}c^{l}=\left( 2(k+1)+c\right)
\sum_{l=0}^{n}b_{n,k+1,l}c^{l}+\left( 2(n-k)+c\right)
\sum_{l=0}^{n}b_{n,k,l}c^{l}.
\end{equation*}%
This is equivalent to%
\begin{eqnarray*}
\sum_{l=0}^{n+1}b_{n+1,k+1,l}c^{l}
&amp;=&amp;\sum_{l=0}^{n}b_{n,k+1,l}c^{l+1}+\sum_{l=0}^{n}b_{n,k,l}c^{l+1} \\
&amp;&amp;+2(k+1)\sum_{l=0}^{n}b_{n,k+1,l}c^{l}+2(n-k)\sum_{l=0}^{n}b_{n,k,l}c^{l},
\end{eqnarray*}%
or%
\begin{eqnarray*}
\sum_{l=0}^{n+1}b_{n+1,k+1,l}c^{l}
&amp;=&amp;\sum_{q=1}^{n+1}b_{n,k+1,q-1}c^{q}+\sum_{q=1}^{n+1}b_{n,k,q-1}c^{q} \\
&amp;&amp;+2(k+1)\sum_{l=0}^{n}b_{n,k+1,l}c^{l}+2(n-k)\sum_{l=0}^{n}b_{n,k,l}c^{l},
\end{eqnarray*}%
or, because $b_{n,k,l}\triangleq 0$ if $k,l\notin \mathbb{Z}_{+,n}$, we get%
\begin{eqnarray*}
\sum_{l=0}^{n+1}b_{n+1,k+1,l}c^{l}
&amp;=&amp;\sum_{q=0}^{n+1}b_{n,k+1,q-1}c^{q}+\sum_{q=0}^{n+1}b_{n,k,q-1}c^{q} \\
&amp;&amp;+2(k+1)\sum_{l=0}^{n+1}b_{n,k+1,l}c^{l}+2(n-k)%
\sum_{l=0}^{n+1}b_{n,k,l}c^{l}.
\end{eqnarray*}%
This holds for all $c$, so we obtain%
\begin{equation*}
b_{n+1,k+1,l}=2(k+1)b_{n,k+1,l}+b_{n,k+1,l-1}+2(n-k)b_{n,k,l}+b_{n,k,l-1}.
\end{equation*}
\end{proof}

\subsection{Partial sums of the $b_{n,k,l}$}

The following result shows that various partial sums over the number
pyramid $b_{n,k,l}$ are related to several important number triangles.

\begin{proposition}
\label{P.8}For all $n\in \mathbb{N}$ and for all $k,l\in 
\mathbb{N}_{n}$,%
\begin{eqnarray}
\sum_{l=0}^{n}b_{n,k,l} &amp;=&amp;B_{n+1,k+1},  \label{8.1} \\
\sum_{l=0}^{n}(-1)^{l}b_{n,k,l} &amp;=&amp;(-1)^{n-k}\left( \tbinom{n}{k}%
-2o_{n}\delta _{k&gt;0}\tbinom{n-1}{k-1}\right) ,  \label{8.2} \\
\sum_{k=0}^{n}b_{n,k,l} &amp;=&amp;(-1)^{n-l}2^{n}s(n,l),  \label{8.3} \\
\sum_{k=0}^{n}(-1)^{k}b_{n,k,l} &amp;=&amp;e_{n}(-1)^{n/2}2^{n}\delta _{l\leq
n/2}T_{n/2,l}. \label{8.4}
\end{eqnarray}%
\end{proposition}

\begin{proof}
(i) This immediately follows from Eq. (\ref{1.11}).

(ii) We have, for all $z,t\in \mathbb{C}$,%
\begin{eqnarray*}
&amp;&amp;B^{-1}(1,z,t) \\
&amp;=&amp;\frac{1}{1-z}e^{-\frac{1-z}{2}t}-\frac{z}{1-z}e^{+\frac{1-z}{2}t}, \\
&amp;=&amp;\sum_{n=0}^{+\infty }2^{-n}(-1)^{n}\left( 1-z\right) ^{n-1}\frac{t^{n}}{n!%
}-z\sum_{n=0}^{+\infty }2^{-n}\left( 1-z\right) ^{n-1}\frac{t^{n}}{n!}, \\
&amp;=&amp;\sum_{n=0}^{+\infty }2^{-n}\left( (-1)^{n}-z\right) \left( 1-z\right)
^{n-1}\frac{t^{n}}{n!}, \\
&amp;=&amp;\sum_{n=0}^{+\infty }2^{-n}\left( \delta _{n=0}+\delta _{n&gt;0}\left(
(-1)^{n}-z\right) \sum_{k=0}^{n-1}\tbinom{n-1}{k}(-z)^{k}\right) \frac{t^{n}%
}{n!}, \\
&amp;=&amp;\sum_{n=0}^{+\infty }2^{-n}\left( \delta _{n=0}+\delta _{n&gt;0}\left(
(-1)^{n}\sum_{k=0}^{n-1}\tbinom{n-1}{k}(-z)^{k}+\sum_{k=0}^{n-1}\tbinom{n-1}{%
k}(-z)^{k+1}\right) \right) \frac{t^{n}}{n!}, \\
&amp;=&amp;\sum_{n=0}^{+\infty }2^{-n}\left( \delta _{n=0}+\delta _{n&gt;0}\left(
(-1)^{n}\sum_{k=0}^{n-1}\tbinom{n-1}{k}(-z)^{k}+\sum_{l=1}^{n}\tbinom{n-1}{%
l-1}(-z)^{l}\right) \right) \frac{t^{n}}{n!},
\end{eqnarray*}%
or%
\begin{eqnarray*}
&amp;&amp;B^{-1}(1,z,t) \\
&amp;=&amp;\sum_{n=0}^{+\infty }2^{-n}\left( \delta _{n=0}+\delta _{n&gt;0}\left( 
\begin{array}{c}
(-1)^{n} \\ 
+\sum_{k=1}^{n-1}\left( (-1)^{n}\tbinom{n-1}{k}+\tbinom{n-1}{k-1}\right)
(-z)^{k} \\ 
+(-z)^{n}%
\end{array}%
\right) \right) \frac{t^{n}}{n!}.
\end{eqnarray*}%
With the identity%
\begin{equation*}
(-1)^{n}\tbinom{n-1}{k}+\tbinom{n-1}{k-1}=\left( \left( -1\right)
^{n}+\left( 1-\left( -1\right) ^{n}\right) \frac{k}{n}\right) \tbinom{n}{k},
\end{equation*}%
the expression inside the first pair of parentheses becomes%
\begin{eqnarray*}
&amp;&amp;\delta _{n=0}+\delta _{n&gt;0}\left( (-1)^{n}+\sum_{k=1}^{n-1}\left( \left(
-1\right) ^{n}+\left( 1-\left( -1\right) ^{n}\right) \frac{k}{n}\right) 
\tbinom{n}{k}(-z)^{k}+(-z)^{n}\right) \\
&amp;=&amp;\sum_{k=0}^{n}\left( \delta _{n=0}+\delta _{n&gt;0}\left( \left( -1\right)
^{n}+\left( 1-\left( -1\right) ^{n}\right) \frac{k}{n}\right) \right) 
\tbinom{n}{k}(-z)^{k}, \\
&amp;=&amp;\sum_{k=0}^{n}(-1)^{n-k}\tbinom{n}{k}\left( 1-\delta _{n&gt;0}\left(
1-\left( -1\right) ^{n}\right) \frac{k}{n}\right) z^{k},
\end{eqnarray*}%
and we get for the full expression%
\begin{equation*}
B^{-1}(1,z,t)=\sum_{n=0}^{+\infty }2^{-n}\sum_{k=0}^{n}(-1)^{n-k}\tbinom{n}{k%
}\left( 1-\delta _{n&gt;0}\left( 1-\left( -1\right) ^{n}\right) \frac{k}{n}%
\right) z^{k}\frac{t^{n}}{n!},
\end{equation*}%
or%
\begin{equation*}
B^{-1}(1,z,t)=\sum_{n=0}^{+\infty }2^{-n}\sum_{k=0}^{n}(-1)^{n-k}\left( 
\tbinom{n}{k}-2o_{n}\delta _{k&gt;0}\tbinom{n-1}{k-1}\right) z^{k}\frac{t^{n}}{%
n!}.
\end{equation*}%
Identifying this expression with%
\begin{equation*}
B^{-1}(1,z,t)=\sum_{n=0}^{+\infty }\sum_{k=0}^{n}\left(
2^{-n}\sum_{l=0}^{n}b_{n,k,l}(-1)^{l}\right) z^{k}\frac{t^{n}}{n!}
\end{equation*}%
yields%
\begin{equation*}
\sum_{l=0}^{n}(-1)^{l}b_{n,k,l}=(-1)^{n-k}\left( \tbinom{n}{k}-2o_{n}\delta
_{k&gt;0}\tbinom{n-1}{k-1}\right) .
\end{equation*}

(iii) Applying Eq. (\ref{1.9}) to the following double sum we obtain, for all $c\in \mathbb{C}$,%
\begin{equation*}
\sum_{k=0}^{n}\sum_{l=0}^{n}b_{n,k,l}c^{l}=%
\sum_{k=0}^{n}B_{n,k}(c)=2^{n}c^{(n)},
\end{equation*}%
or%
\begin{equation*}
\sum_{l=0}^{n}\left( (-1)^{n-l}2^{-n}\sum_{k=0}^{n}b_{n,k,l}\right)
c^{l}=c_{(n)}.
\end{equation*}%
Identifying this with the definition of the Stirling numbers of the first
kind gives%
\begin{equation*}
(-1)^{n-l}2^{-n}\sum_{k=0}^{n}b_{n,k,l}=s(n,l).
\end{equation*}

(iv) This result is related to the Maclaurin series of $\func{sech}^{c}t$,
considered in the next section. There it is shown that (i)%
\begin{equation*}
\sum_{k=0}^{2m+1}(-1)^{k}b_{2m+1,k,l}=0,
\end{equation*}%
and (ii) (see Proposition \ref{P.9} below)%
\begin{equation*}
\sum_{k=0}^{n}(-1)^{k}b_{n,k,l}=(-1)^{n/2}2^{n}e_{n}\delta _{l\leq
n/2}T_{n/2,l},
\end{equation*}%
where the number triangle $T_{m,l}$ is Sloane's sequence \seqnum{A088874}.
\end{proof}

We can add to the sums given by Proposition \ref{P.8}, Eq. (\ref{1.14}), which expressed in terms of the $b_{n,k,l}$ reads%
\begin{equation}
\sum_{l=0}^{n}b_{n,k,l}2^{l}=2^{n}A_{n+1,k+1}.  \label{8.5}
\end{equation}

\subsection{The numbers $T_{n,k}$}

We now give an explicit expression for the numbers \seqnum{A088874}.

\begin{proposition}
\label{P.7}For all $n\in \mathbb{N}$ and for all $l\in \mathbb{N%
}_{n}$,%
\begin{equation}
e_{n}\delta _{l\leq n/2}(-1)^{n/2-l}T_{n/2,l}=\sum_{p=0}^{l}(-1)^{p}\tbinom{n%
}{p}w_{n-p,l-p},  \label{7.1}
\end{equation}%
wherein%
\begin{equation}
w_{n,m}\triangleq 2^{n}\sum_{k=m}^{n}S\left( n,k\right) s\left( k,m\right)
\left( 1/2\right) ^{k}.  \label{7.2}
\end{equation}
\end{proposition}

\begin{proof}
Summing over $k$ from $0$ to $n$ in the expression for $b_{n,k,l}$ given by Proposition \ref{P.6}, we get%
\begin{equation*}
\sum_{k=0}^{n}(-1)^{k}b_{n,k,l}=\sum_{p=0}^{l}(-1)^{l-p}\tbinom{n}{p}%
2^{n-p}\sum_{k=0}^{n}\sum_{i=l-p}^{\min (k,n-p)}\tbinom{n-i}{k-i}S\left(
n-p,i\right) s\left( i,l-p\right) .
\end{equation*}%
Using $S\left( n,k\right) =s\left( n,k\right) \triangleq 0$, $\forall
k\notin \mathbb{N}_{n}$, we can write this as%
\begin{equation*}
\sum_{k=0}^{n}(-1)^{k}b_{n,k,l}=\sum_{p=0}^{l}(-1)^{l-p}\tbinom{n}{p}%
2^{n-p}\sum_{k=0}^{n}\sum_{i=0}^{k}\tbinom{n-i}{k-i}S\left( n-p,i\right)
s\left( i,l-p\right) .
\end{equation*}%
Interchanging the last two summations gives%
\begin{eqnarray*}
\sum_{k=0}^{n}(-1)^{k}b_{n,k,l} &amp;=&amp;\sum_{p=0}^{l}(-1)^{l-p}\tbinom{n}{p}%
2^{n-p}\sum_{i=0}^{n}\left( \sum_{k=i}^{n}\tbinom{n-i}{k-i}\right) S\left(
n-p,i\right) s\left( i,l-p\right) , \\
&amp;=&amp;\sum_{p=0}^{l}(-1)^{l-p}\tbinom{n}{p}2^{n-p}\sum_{i=0}^{n}2^{n-i}S\left(
n-p,i\right) s\left( i,l-p\right) ,
\end{eqnarray*}%
or%
\begin{equation*}
2^{-n}\sum_{k=0}^{n}(-1)^{k}b_{n,k,l}=\sum_{p=0}^{l}(-1)^{l-p}\tbinom{n}{p}%
\sum_{i=l-p}^{n-p}2^{n-p-i}S\left( n-p,i\right) s\left( i,l-p\right) .
\end{equation*}%
Using herein definition (\ref{7.2}) and the result (\ref%
{9.10}) from Proposition \ref{P.9} below, Eq. (\ref{7.1}) follows.
\end{proof}

We give the onset of the number triangle $T_{n,k}$ in Table \ref{T}. In
particular, $T_{n,0}=\delta _{n=0}$, $T_{n,1}$ are the tangent numbers
(Sloane's \seqnum{A000182}), and $T_{n,n}=1.3.5...(2n-1)$ are the double factorial
numbers (Sloane's \seqnum{A001147}).


\begin{table}[tbp] \centering

{\normalsize $%
\begin{tabular}{|r|r|r|r|r|r|r|r|r|r|}
\hline
{\tiny n\verb?\? k} &amp; {\tiny 0} &amp; {\tiny 1} &amp; {\tiny 2} &amp; 
{\tiny 3} &amp; {\tiny 4} &amp; {\tiny 5} &amp; {\tiny 6} &amp; {\tiny 7} &amp; {\tiny 8} \\ 
\hline
{\tiny 0} &amp; {\tiny 1} &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  \\ \hline
{\tiny 1} &amp; {\tiny 0} &amp; {\tiny 1} &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  \\ \hline
{\tiny 2} &amp; {\tiny 0} &amp; {\tiny 2} &amp; {\tiny 3} &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  \\ \hline
{\tiny 3} &amp; {\tiny 0} &amp; {\tiny 16} &amp; {\tiny 30} &amp; {\tiny 15} &amp;  &amp;  &amp;  &amp;  &amp; 
\\ \hline
{\tiny 4} &amp; {\tiny 0} &amp; {\tiny 272} &amp; {\tiny 588} &amp; {\tiny 420} &amp; {\tiny 105}
&amp;  &amp;  &amp;  &amp;  \\ \hline
{\tiny 5} &amp; {\tiny 0} &amp; {\tiny 7\thinspace 936} &amp; {\tiny 18\thinspace 960} &amp; 
{\tiny 16\thinspace 380} &amp; {\tiny 6\thinspace 300} &amp; {\tiny 945} &amp;  &amp;  &amp;  \\ 
\hline
{\tiny 6} &amp; {\tiny 0} &amp; {\tiny 353\thinspace 792} &amp; {\tiny 911\thinspace 328}
&amp; {\tiny 893\thinspace 640} &amp; {\tiny 429\thinspace 660} &amp; {\tiny %
103\thinspace 950} &amp; {\tiny 10\thinspace 395} &amp;  &amp;  \\ \hline
{\tiny 7} &amp; {\tiny 0} &amp; {\tiny 22\thinspace 368\thinspace 256} &amp; {\tiny %
61\thinspace 152\thinspace 000} &amp; {\tiny 65\thinspace 825\thinspace 760} &amp; 
{\tiny 36\thinspace 636\thinspace 600} &amp; {\tiny 11\thinspace 351\thinspace
340} &amp; {\tiny 1\thinspace 891\thinspace 890} &amp; {\tiny 135\thinspace 135} &amp; 
\\ \hline
{\tiny 8} &amp; {\tiny 0} &amp; {\tiny 1\thinspace 903\thinspace 757\thinspace 312}
&amp; {\tiny 5\thinspace 464\thinspace 904\thinspace 448} &amp; {\tiny 6\thinspace
327\thinspace 135\thinspace 360} &amp; {\tiny 3\thinspace 918\thinspace
554\thinspace 640} &amp; {\tiny 1\thinspace 427\thinspace 025\thinspace 600} &amp; 
{\tiny 310\thinspace 269\thinspace 960} &amp; {\tiny 37\thinspace 837\thinspace
800} &amp; {\tiny 2\thinspace 027\thinspace 025} \\ \hline
\end{tabular}%
$}

\caption{The number triangle $T_{n,k}$\label{T}}%

\end{table}

\section{The Maclaurin series of $\func{sech}^{c}t$}

Substituting $x=-y=1$ in Eqs. (\ref{1.2}) and (\ref{1.3}), and using Eq. (\ref{1.6}), shows that the
Maclaurin series of the $c$-th power of $\func{sech}t$ is given by, for all $c\in 
\mathbb{C}$,%
\begin{equation}
\func{sech}^{c}t=\sum_{n=0}^{+\infty }p_{n}(c)\frac{t^{n}}{n!},  \label{9.1}
\end{equation}%
where%
\begin{eqnarray}
p_{n}(c) &amp;=&amp;2^{-n}\sum_{k=0}^{n}(-1)^{k}B_{n,k}(c),  \label{9.2} \\
&amp;=&amp;\sum_{l=0}^{n}\left( 2^{-n}\sum_{k=0}^{n}(-1)^{k}b_{n,k,l}\right) c^{l}.
\label{9.3}
\end{eqnarray}

The polynomials $p_{n}(c)$ have the following properties.

(i) $p_{2k+1}(c)=0$, for all $k\in \mathbb{N}$ and for all $c\in \mathbb{C}$%
, because $\func{sech}^{c}t$ is even in $t$.

(ii) $p_{n}(c)$ has degree $n/2$. This can be seen as follows. We have, for all 
$m\in \mathbb{N}$, $\lim_{c\rightarrow 0}D_{c}^{m}\func{sech}%
^{c}t=\left( \ln \func{sech}t\right) ^{m}$ and $\left( \ln \func{sech}%
t\right) ^{m}=O\left( t^{2m}\right) $. Hence $p_{n}(c)$ must have degree $%
n/2$.

(iii) $p_{n}(0)=\delta _{n=0}$.

Then due to (i) we have, for all $m\in \mathbb{N}$,%
\begin{equation}
\sum_{k=0}^{2m+1}(-1)^{k}b_{2m+1,k,l}=0,  \label{9.4}
\end{equation}%
and we define, in accordance with (ii),%
\begin{equation}
\sum_{k=0}^{2m}(-1)^{k}b_{2m,k,l}\triangleq (-1)^{m}2^{2m}\delta _{l\leq
m}T_{m,l}.  \label{9.5}
\end{equation}%
Hence%
\begin{equation}
p_{2m}(c)=\lim_{t\rightarrow 0}D_{t}^{2m}\func{sech}^{c}t=(-1)^{m}%
\sum_{l=0}^{m}T_{m,l}c^{l},  \label{9.6b}
\end{equation}%
so%
\begin{equation}
\func{sech}^{c}t=\sum_{m=0}^{+\infty }\left(
(-1)^{m}\sum_{l=0}^{m}T_{m,l}c^{l}\right) \frac{t^{2m}}{(2m)!}.  \label{9.6}
\end{equation}%
Due to (iii), $T_{n,0}=\delta _{n=0}$, for all $n\in \mathbb{N}$.

We now clarify the nature of the numbers $T_{m,l}$ defined in Eq. (\ref{9.5}).

\begin{proposition}
\label{P.9}The number triangle $T_{n,l}$, for all $n\in \mathbb{N}$ and for all 
$l\in \mathbb{N}_{n}$, satisfies the recursion relation%
\begin{equation}
T_{n+1,l+1}=\sum_{p=0}^{n-l}2^{p}\left( \tbinom{p+l+1}{p+1}+\delta _{l&gt;0}%
\tbinom{p+l}{p+1}\right) T_{n,l+p},  \label{9.10}
\end{equation}%
with $T_{n,0}=\delta _{n=0}$.
\end{proposition}

\begin{proof}
Starting from the identity%
\begin{equation*}
D_{t}^{2}\func{sech}^{c}t=c^{2}\func{sech}^{c}t-c(c+1)\func{sech}^{c+2}t,
\end{equation*}%
and substituting herein the series expansion for $\func{sech}^{c}t$, Eq. (\ref%
{9.6}), we get%
\begin{eqnarray*}
&amp;&amp;\sum_{n=1}^{+\infty }(-1)^{n}\sum_{l=0}^{n}T_{n,l}c^{l}\frac{t^{2(n-1)}}{%
(2(n-1))!} \\
&amp;=&amp;c^{2}\sum_{n=0}^{+\infty }(-1)^{n}\sum_{l=0}^{n}T_{n,l}c^{l}\frac{t^{2n}}{%
(2n)!}-c(c+1)\sum_{n=0}^{+\infty }(-1)^{n}\sum_{l=0}^{n}T_{n,l}(c+2)^{l}%
\frac{t^{2n}}{(2n)!}.
\end{eqnarray*}%
As this holds for all $t$, we must have that%
\begin{equation*}
-\sum_{l=0}^{n+1}T_{n+1,l}c^{l}=c^{2}\sum_{l=0}^{n}T_{n,l}c^{l}-c(c+1)%
\sum_{l=0}^{n}T_{n,l}(c+2)^{l}.
\end{equation*}%
This can be rearranged in the form%
\begin{equation*}
\sum_{l=0}^{n}T_{n+1,l+1}c^{l}=(c+1)\sum_{l=0}^{n}T_{n,l}(c+2)^{l}-%
\sum_{l=0}^{n}T_{n,l}c^{l+1}.
\end{equation*}%
By the binomial theorem,%
\begin{eqnarray*}
\sum_{l=0}^{n}T_{n+1,l+1}c^{l} &amp;=&amp;\sum_{l=0}^{n}T_{n,l}\sum_{k=0}^{l}\tbinom{%
l}{k}2^{l-k}c^{k+1} \\
&amp;&amp;+\sum_{l=0}^{n}T_{n,l}\sum_{k=0}^{l}\tbinom{l}{k}2^{l-k}c^{k}-%
\sum_{l=0}^{n}T_{n,l}c^{l+1}.
\end{eqnarray*}%
Interchanging the order of summation in the double sum terms gives%
\begin{eqnarray*}
\sum_{l=0}^{n}T_{n+1,l+1}c^{l} &amp;=&amp;\sum_{k=0}^{n}\sum_{q=0}^{n-k}2^{q}\tbinom{%
q+k}{k}T_{n,q+k}c^{k+1} \\
&amp;&amp;+\sum_{k=0}^{n}\sum_{q=0}^{n-k}2^{q}\tbinom{q+k}{k}T_{n,q+k}c^{k}-%
\sum_{k=1}^{n+1}T_{n,k-1}c^{k}.
\end{eqnarray*}%
We can rearrange this further into%
\begin{eqnarray*}
\sum_{l=0}^{n}T_{n+1,l+1}c^{l} &amp;=&amp;\sum_{l=1}^{n+1}\sum_{q=0}^{n-(l-1)}2^{q}%
\tbinom{q+l-1}{l-1}T_{n,q+l-1}c^{l} \\
&amp;&amp;+\sum_{l=0}^{n}\sum_{q=0}^{n-l}2^{q}\tbinom{q+l}{l}T_{n,q+l}c^{l}-%
\sum_{l=1}^{n+1}T_{n,l-1}c^{l}.
\end{eqnarray*}%
As this holds for all $c$, we must have that%
\begin{eqnarray*}
T_{n+1,1} &amp;=&amp;\sum_{q=0}^{n}2^{q}T_{n,q}; \\
\sum_{l=1}^{n}T_{n+1,l+1}c^{l} &amp;=&amp;\sum_{l=1}^{n}\left( 
\begin{array}{c}
\sum_{q=0}^{n-(l-1)}2^{q}\tbinom{q+l-1}{l-1}T_{n,q+l-1} \\ 
+\sum_{q=0}^{n-l}2^{q}\tbinom{q+l}{l}T_{n,q+l}-T_{n,l-1}%
\end{array}%
\right) c^{l} \\
&amp;&amp;+T_{n,n}c^{n+1}-T_{n,n}c^{n+1},
\end{eqnarray*}%
or%
\begin{eqnarray*}
T_{n+1,1} &amp;=&amp;\sum_{q=0}^{n}2^{q}T_{n,q}; \\
T_{n+1,l+1} &amp;=&amp;\sum_{q=0}^{n-(l-1)}2^{q}\tbinom{q+l-1}{l-1}%
T_{n,q+l-1}+\sum_{q=0}^{n-l}2^{q}\tbinom{q+l}{l}T_{n,q+l}-T_{n,l-1},l\in 
\mathbb{Z}_{+,n},
\end{eqnarray*}%
or%
\begin{eqnarray*}
T_{n+1,1} &amp;=&amp;\sum_{q=0}^{n}2^{q}T_{n,q}; \\
T_{n+1,l+1} &amp;=&amp;\sum_{q=1}^{n-l+1}2^{q}\tbinom{q+l-1}{l-1}T_{n,q+l-1}+%
\sum_{q=0}^{n-l}2^{q}\tbinom{q+l}{l}T_{n,q+l},l\in \mathbb{Z}_{+,n},
\end{eqnarray*}%
or%
\begin{eqnarray*}
T_{n+1,1} &amp;=&amp;\sum_{q=0}^{n}2^{q}T_{n,q}; \\
T_{n+1,l+1} &amp;=&amp;\sum_{p=0}^{n-l}2^{p+1}\tbinom{p+l}{l-1}T_{n,p+l}+%
\sum_{p=0}^{n-l}2^{p}\tbinom{p+l}{l}T_{n,p+l},l\in \mathbb{Z}_{+,n},
\end{eqnarray*}%
or%
\begin{eqnarray*}
T_{n+1,1} &amp;=&amp;\sum_{p=0}^{n}2^{p}T_{n,p},l=0; \\
T_{n+1,l+1} &amp;=&amp;\sum_{p=0}^{n-l}2^{p}\left( 2\tbinom{p+l}{l-1}+\tbinom{p+l}{l}%
\right) T_{n,p+l},l\in \mathbb{Z}_{+,n},
\end{eqnarray*}%
or%
\begin{eqnarray*}
T_{n+1,1} &amp;=&amp;\sum_{p=0}^{n}2^{p}T_{n,p},l=0; \\
T_{n+1,l+1} &amp;=&amp;\sum_{p=0}^{n-l}2^{p}\left( 2\tbinom{p+l}{p+1}+\tbinom{p+l}{p}%
\right) T_{n,p+l},l\in \mathbb{Z}_{+,n}.
\end{eqnarray*}%
With the basic additive (Pascal's first) binomial identity%
\begin{equation*}
\tbinom{p+l}{p+1}+\tbinom{p+l}{p}=\tbinom{p+l+1}{p+1},
\end{equation*}%
we can write this also as%
\begin{eqnarray*}
T_{n+1,1} &amp;=&amp;\sum_{p=0}^{n}2^{p}T_{n,p}; \\
T_{n+1,l+1} &amp;=&amp;\sum_{p=0}^{n-l}2^{p}\left( \tbinom{p+l+1}{p+1}+\tbinom{p+l}{%
p+1}\right) T_{n,l+p},l\in \mathbb{Z}_{+,n}.
\end{eqnarray*}%
Both these equations can be combined into the following single equation, for all $l\in 
\mathbb{N}_{n}$,%
\begin{equation*}
T_{n+1,l+1}=\sum_{p=0}^{n-l}2^{p}\left( \tbinom{p+l+1}{p+1}+\delta _{l&gt;0}%
\tbinom{p+l}{p+1}\right) T_{n,l+p}.
\end{equation*}%
Finally, from the fact that $\func{sech}^{0}t=1$ we 
conclude that $T_{n,0}=\delta _{n=0}$.
\end{proof}

Identifying our results with those given in Sloane \cite[\seqnum{A085734}
and \seqnum{A088874}]{S}\ we see that the $T_{n,k}$, satisfying
recursion relation (\ref{9.10}) and boundary condition $T_{n,0}=\delta
_{n=0} $, are indeed Sloane's sequence \seqnum{A088874} (there constructed by Deleham using a
delta operator). Thus Eq. (\ref{9.6}) shows that the number triangle $T_{n,k}$
is as fundamental for the Maclaurin series of $\func{sech}^{c}t$ as the
Euler numbers $E_{n}$ are for the Maclaurin series of $\func{sech}t$.

Eq. (\ref{9.6}) also immediately leads to the orthogonality relation, for all
$n,k\in \mathbb{N}$,%
\begin{equation}
\sum_{p=0}^{n}\tbinom{2n}{2p}\sum_{l=\max \left( 0,k+p-n\right) }^{\min
\left( k,p\right) }\left( -1\right) ^{l}T_{n-p,k-l}T_{p,l}=\delta
_{n=0}\delta _{k=0}.  \label{9.7}
\end{equation}

The series (\ref{9.6}) shows that the numbers $E_{2m}^{(c)}\triangleq
(-1)^{m}\sum_{l=0}^{m}T_{m,l}c^{l}$, for all $m\in \mathbb{N}$ and for all
$c\in \mathbb{C}$, are a set of generalized Euler numbers (although they are
polynomials in $c$) in the sense of Luo, et al. \cite{Lu}. It seems more natural
however, to regard the integer numbers $T_{m,l}$ as a more fundamental set,
because they are independent of $c$.

\subsection{Multinomial Euler numbers $E_{n}^{m}$}

For the important special case that $c=m\in \mathbb{N}$, it might be
convenient to introduce a generalization of the Euler numbers that are
also integers.

By identifying Eq. (\ref{9.6}) with the Maclaurin series of $\func{sech}^{m}t$, written in the form%
\begin{equation}
\func{sech}^{m}t=\sum_{n=0}^{+\infty }E_{n}^{m}\frac{t^{n}}{n!},
\label{10.1}
\end{equation}%
we get, for all $p\in \mathbb{N}$, $E_{2p+1}^{m}=0$ and%
\begin{equation}
E_{2p}^{m}=(-1)^{p}\sum_{l=0}^{p}T_{p,l}m^{l}.  \label{10.2}
\end{equation}

On the other hand, we also have that $\func{sech}^{m}t=\left( \sum_{n=0}^{+\infty }E_{n}\frac{t^{n}}{n!}%
\right) ^{m}$, so we obtain,%
\begin{equation}
E_{n}^{m}=\delta _{m=0}\delta _{n=0}+\delta _{m&gt;0}\sum_{K:\left\vert
K\right\vert =n}\tbinom{n}{K}\prod_{i=1}^{\#\left( K\right) }E_{k_{i}}.
\label{10.3}
\end{equation}%
Expression (\ref{10.3}) suggests that the $E_{n}^{m}$ be called \emph{%
multinomial Euler numbers}. Equating the right hand sides of Eq. (\ref{10.2})
and Eq. (\ref{10.3}) gives%
\begin{equation}
(-1)^{n}\sum_{l=0}^{n}T_{n,l}m^{l}=\delta _{m=0}\delta _{n=0}+\delta
_{m&gt;0}\sum_{K:\left\vert K\right\vert =n}\tbinom{n}{K}\prod_{i=1}^{\#\left(
K\right) }E_{k_{i}}.  \label{10.5}
\end{equation}%
This reduces, for $m=1$, to%
\begin{equation}
(-1)^{n}\sum_{l=0}^{n}T_{n,l}=E_{2n}.  \label{10.4}
\end{equation}

Some particular multinomial Euler numbers are mentioned in Sloane \cite{S}. Eq. (%
\ref{10.2}) reduces to the following particular cases: $E_{0}^{m}=1$, $E_{2}^{m}=-m$, $%
E_{4}^{m}=m\left( 3m+2\right) $ (rhombic matchstick numbers, Sloane's
\seqnum{A045944}) and $E_{6}^{m}=-m\left( 15m^{2}+30m+16\right) $ (not in Sloane).
Also, $E_{n}^{0}=\delta _{n=0}$, $E_{n}^{1}=E_{n}$ (Euler or $\func{sech}$
numbers, $\left\vert E_{n}\right\vert $ is Sloane's \seqnum{A000364}), $E_{n-1}^{2}$
are the $\tanh $ numbers (due to $D_{t}^{n}\func{sech}^{2}t=D_{t}^{n+1}%
\tanh t$, with $\left\vert E_{n-1}^{2}\right\vert $ being Sloane's \seqnum{A000182}).

We give the onset of the number square $E_{n}^{m}$, for even $n$, in Table %
\ref{E}.


\begin{table}[tbp] \centering
\begin{tabular}{|r|r|r|r|r|r|r|}
\hline
{\tiny n\verb?\? m} &amp; {\tiny 1} &amp; {\tiny 2} &amp; {\tiny 3} &amp; 
{\tiny 4} &amp; {\tiny 5} &amp; {\tiny 6} \\ \hline
{\tiny 0} &amp; {\tiny 1} &amp; {\tiny 1} &amp; {\tiny 1} &amp; {\tiny 1} &amp; {\tiny 1} &amp; 
{\tiny 1} \\ \hline
{\tiny 2} &amp; {\tiny -1} &amp; {\tiny -2} &amp; {\tiny -3} &amp; {\tiny -4} &amp; {\tiny -5} &amp; 
{\tiny -6} \\ \hline
{\tiny 4} &amp; {\tiny 5} &amp; {\tiny 16} &amp; {\tiny 33} &amp; {\tiny 56} &amp; {\tiny 85} &amp; 
{\tiny 120} \\ \hline
{\tiny 6} &amp; {\tiny -61} &amp; {\tiny -272} &amp; {\tiny -723} &amp; {\tiny -1\thinspace
504} &amp; {\tiny -2\thinspace 705} &amp; {\tiny -4\thinspace 416} \\ \hline
{\tiny 8} &amp; {\tiny 1\thinspace 385} &amp; {\tiny 7\thinspace 936} &amp; {\tiny %
25\thinspace 953} &amp; {\tiny 64\thinspace 256} &amp; {\tiny 134\thinspace 185} &amp; 
{\tiny 249\thinspace 600} \\ \hline
{\tiny 10} &amp; {\tiny -50\thinspace 521} &amp; {\tiny -353\thinspace 792} &amp; {\tiny %
-1\thinspace 376\thinspace 643} &amp; {\tiny -3\thinspace 963\thinspace 904} &amp; 
{\tiny -9\thinspace 451\thinspace 805} &amp; {\tiny -19\thinspace 781\thinspace
376} \\ \hline
\end{tabular}

\caption{The number square $E_{n}^{m}$\label{E}}%

\end{table}

\subsection{Even multinomial parity numbers $e_{n}^{m}$}

By identifying Eq. (\ref{9.6}) with the Maclaurin series of $\cosh ^{m}t$, written in the form%
\begin{equation}
\cosh ^{m}t=\sum_{n=0}^{+\infty }e_{n}^{m}\frac{t^{n}}{n!},  \label{11.1}
\end{equation}%
we get, for all $p\in \mathbb{N}$, $e_{2p+1}^{m}=0$ and%
\begin{equation}
e_{2p}^{m}=(-1)^{p}\sum_{l=0}^{p}T_{p,l}(-m)^{l}.  \label{11.2}
\end{equation}

On the other hand, we also have that $\cosh ^{m}t=\left( \sum_{n=0}^{+\infty }e_{n}\frac{t^{n}}{n!}%
\right) ^{m}$, so we obtain,%
\begin{equation}
e_{n}^{m}=\delta _{m=0}\delta _{n=0}+\delta _{m&gt;0}\sum_{K:\left\vert
K\right\vert =n}\tbinom{n}{K}\prod_{i=1}^{\#\left( K\right) }e_{k_{i}}.
\label{11.3}
\end{equation}%
Expression (\ref{11.3}) suggests that the $e_{n}^{m}$ be called \emph{even
multinomial parity numbers}. Eq. (\ref{2.12}) gives us the following explicit
expression for the $e_{n}^{m}$ numbers,%
\begin{equation}
e_{n}^{m}=\frac{1}{2^{m}}\sum_{j=0}^{m}\tbinom{m}{j}(m-2j)^{n}.  \label{13}
\end{equation}

Equating the right hand sides of Eq. (\ref{11.2}) and Eq. (\ref{13}) gives%
\begin{equation}
(-1)^{n}\sum_{l=0}^{n}T_{n,l}(-m)^{l}=\frac{1}{2^{m}}\sum_{j=0}^{m}\tbinom{m%
}{j}(m-2j)^{2n}.  \label{14}
\end{equation}%
In particular, for $m=1$, Eq. (\ref{14}) reduces to%
\begin{equation}
(-1)^{n}\sum_{l=0}^{n}(-1)^{l}T_{n,l}=1,  \label{15a}
\end{equation}%
and, for $m=2$, to%
\begin{equation}
(-1)^{n}\sum_{l=0}^{n}T_{n,l}\,(-2)^{l}=2^{2n-1}.  \label{15b}
\end{equation}

Various particular even multinomial parity numbers are mentioned in Sloane \cite{S}%
. For instance, $e_{0}^{m}=1$, $e_{2}^{m}=m$, $e_{4}^{m}=m\left( 3m-2\right) 
$ (octagonal numbers, Sloane's \seqnum{A000567}) and \newline
$e_{6}^{m}=m\left( 15m^{2}-30m+16\right) $ (not in Sloane). Also, $%
e_{n}^{0}=\delta _{n=0}$, $e_{n}^{1}=e_{n}$, $e_{n}^{2}=\delta _{n=0}+\delta
_{n&gt;0}e_{n}2^{n-1}$ ($e_{2p}^{2}$ is Sloane's \seqnum{A009117}), $%
e_{n}^{3}=e_{n}(3^{n}+3)/4$ ($e_{2p}^{3}$ is Sloane's \seqnum{A054879}), $%
e_{n}^{4}=e_{n}\left( 4^{n}+4.2^{n}\right) /8$ ($e_{2p}^{4}$ is Sloane's
\seqnum{A092812}) and $e_{n}^{5}=e_{n}\left( 10+5.3^{n}+5^{n}\right) /16$ (not in
Sloane). In general, expression (\ref{13}) shows that $e_{2p}^{m}$ equals
the number of closed walks, based at a vertex, of length $2p$ along the
edges of an $m$-dimensional cube \cite{St}.

We give the onset of the number square $e_{n}^{m}$, for even $n$, in Table %
\ref{P}.


\begin{table}[tbp] \centering
\begin{tabular}{|r|r|r|r|r|r|r|}
\hline
{\tiny n\verb?\? m} &amp; {\tiny 1} &amp; {\tiny 2} &amp; {\tiny 3} &amp; 
{\tiny 4} &amp; {\tiny 5} &amp; {\tiny 6} \\ \hline
{\tiny 0} &amp; {\tiny 1} &amp; {\tiny 1} &amp; {\tiny 1} &amp; {\tiny 1} &amp; {\tiny 1} &amp; 
{\tiny 1} \\ \hline
{\tiny 2} &amp; {\tiny 1} &amp; {\tiny 2} &amp; {\tiny 3} &amp; {\tiny 4} &amp; {\tiny 5} &amp; 
{\tiny 6} \\ \hline
{\tiny 4} &amp; {\tiny 1} &amp; {\tiny 8} &amp; {\tiny 21} &amp; {\tiny 40} &amp; {\tiny 65} &amp; 
{\tiny 96} \\ \hline
{\tiny 6} &amp; {\tiny 1} &amp; {\tiny 32} &amp; {\tiny 183} &amp; {\tiny 544} &amp; {\tiny %
1\thinspace 205} &amp; {\tiny 2\thinspace 256} \\ \hline
{\tiny 8} &amp; {\tiny 1} &amp; {\tiny 128} &amp; {\tiny 1\thinspace 641} &amp; {\tiny %
8\thinspace 320} &amp; {\tiny 26\thinspace 465} &amp; {\tiny 64\thinspace 896} \\ 
\hline
{\tiny 10} &amp; {\tiny 1} &amp; {\tiny 512} &amp; {\tiny 14\thinspace 763} &amp; {\tiny %
131\thinspace 584} &amp; {\tiny 628\thinspace 805} &amp; {\tiny 2\thinspace
086\thinspace 656} \\ \hline
\end{tabular}

\caption{The number square $e_{n}^{m}$\label{P}}%

\end{table}

\subsection{Relations between the $E_{n}^{m}$ and the $e_{n}^{m}$ numbers}

(i) Evidently, due to the fact that $\cosh ^{m}t\,\func{sech}^{m}t=1$, for all $m\in 
\mathbb{N}$, holds the following orthogonality relation, for all $n,m\in \mathbb{N}$,%
\begin{equation}
\sum_{i=0}^{n}\tbinom{n}{i}e_{n-i}^{m}E_{i}^{m}=\delta _{n=0}.  \label{12}
\end{equation}

Combining Eqs. (\ref{12}) and (\ref{13}), and using the fact that $e_{0}^{m}=1$, for all $m\in \mathbb{N}
$, we obtain the following recursion relation for the $E_{n}^{m}$,%
\begin{equation}
E_{n}^{m}=\delta _{n=0}-\delta _{n&gt;0}\sum_{i=0}^{n-1}\tbinom{n}{i}\frac{1}{%
2^{m}}\sum_{j=0}^{m}\tbinom{m}{j}(m-2j)^{n-i}E_{i}^{m}.  \label{16a}
\end{equation}%
In particular, for $m=1$, Eq. (\ref{16a}) yields%
\begin{equation}
E_{2p}=\delta _{p=0}-\delta _{p&gt;0}\sum_{j=0}^{p-1}\tbinom{2p}{2j}E_{2j}.
\end{equation}%
Due to the symmetry of the binomial expression in Eq. (\ref{12}) and because $%
E_{0}^{m}=1 $, for all $m\in \mathbb{N}$, we get equivalently a recursion
relation for the $e_{n}^{m}$ in terms of the $E_{n}^{m}$ numbers,%
\begin{equation}
e_{n}^{m}=\delta _{n=0}-\delta _{n&gt;0}\sum_{i=0}^{n-1}\tbinom{n}{i}%
E_{n-i}^{m}e_{i}^{m}.  \label{16b}
\end{equation}%
In particular, for $m=1$, by using $e_{n}^{1}=e_{n}$ and $e_{0}=1=E_{0}$,
we get 
\begin{equation}
E_{0}=\delta _{p=0}-\delta _{p&gt;0}\sum_{j=0}^{p-1}\tbinom{2p}{2j}E_{2(p-j)}.
\end{equation}

(ii) Comparing Eq. (\ref{10.2}) with Eq. (\ref{11.2}) we see that the numbers $%
E_{2n}^{m}$ are the extension to negative integers $m$ of the numbers $%
e_{2n}^{m}$, as expected from their generating functions given in Eqs. (\ref{10.1}) and (%
\ref{11.1}).

(iii) Recall Faa di Bruno's formula for the $n$-th derivative of a
composition of two functions \cite[p. 823, 24.1.2, II, C.]{AS}, for all $n\in \mathbb{Z}_{+}$,%
\begin{equation*}
D_{t}^{n}f\left( g\left( t\right) \right) =\sum_{k=1}^{n}\left(
D_{g}^{k}f\left( g\right) \right) \left( t\right) \sum_{P(n):\left\vert
K\right\vert =k}n!\prod_{i=1}^{n}\frac{\left( D_{t}^{i}g\left( t\right)
\right) ^{k_{i}}}{\left( i!\right) ^{k_{i}}k_{i}!},
\end{equation*}%
where $P(n)\triangleq \left\{ K\triangleq \left\{ k_{1},k_{2},...,k_{n}\in 
\mathbb{N}\right\} :1k_{1}+2k_{2}+...+nk_{n}=n\right\} $. An element $K\in
P(n)$ represents a partition of a set of cardinality $n$ into $k_{1}$
classes of cardinality $1$, $k_{2}$ classes of cardinality $2$, up to $k_{n}$
classes of cardinality $n$.

Applied to $f\circ g$, with $g\left( t\right) =\cosh ^{m}t$ and $f(g)=1/g$,
we get%
\begin{eqnarray*}
E_{n}^{m} &amp;=&amp;\lim_{t\rightarrow 0}D_{t}^{n}f\left( g\left( t\right) \right) ,
\\
&amp;=&amp;\sum_{k=1}^{n}(-1)^{k}k!\sum_{P(n):\left\vert K\right\vert
=k}n!\prod_{i=1}^{n}\frac{\left( e_{i}^{m}\right) ^{k_{i}}}{\left( i!\right)
^{k_{i}}k_{i}!}.
\end{eqnarray*}%
Define, for all $n\in \mathbb{N}$, $S_{e}^{m}(n,0)\triangleq \delta _{n=0}$
and if $n&gt;0$, for all $k\in \mathbb{Z}_{+,n}$,%
\begin{equation}
S_{e}^{m}(n,k)\triangleq \sum_{P(n):\left\vert K\right\vert
=k}n!\prod_{i=1}^{n}\frac{\left( e_{i}^{m}\right) ^{k_{i}}}{\left( i!\right)
^{k_{i}}k_{i}!}.  \label{17.1}
\end{equation}%
Then%
\begin{equation}
E_{n}^{m}=\sum_{k=1}^{n}(-1)^{k}k!S_{e}^{m}(n,k).  \label{17.2}
\end{equation}%
Eq. (\ref{17.2}) expresses the multinomial Euler numbers in terms of
the even multinomial parity numbers through the intermediate numbers $%
S_{e}^{m}(n,k)$.

The even parity symbol $e_{i}$ in the product in Eq. (\ref{17.1}) makes that all 
$k_{i}$ with odd index $i$ must be taken zero, so $S_{e}^{m}(n,k)$ and hence 
$E_{n}^{m}$ are both zero for odd $n$. With $n=2p$, we get, for all $p\in 
\mathbb{Z}_{+}$,%
\begin{equation*}
S_{e}^{m}(2p,k)=\sum_{P_{e}(2p):\left\vert K_{e}\right\vert
=k}(2p)!\prod_{j=1}^{p}\frac{\left( e_{2j}^{m}\right) ^{k_{2j}}}{\left(
(2j)!\right) ^{k_{2j}}k_{2j}!},
\end{equation*}%
where $P_{e}(2p)\triangleq \left\{ K_{e}\triangleq \left\{
k_{2},k_{4},...,k_{n}\in \mathbb{N}\right\}
:k_{2}+2k_{4}+...+pk_{2p}=p\right\} $. An element $K_{e}\in P_{e}(2p)$
represents a partition of a set of $2p$ elements into $0$ classes of
cardinality $1$, $k_{2}$ classes of cardinality $2$, $0$ classes of
cardinality $3$, $k_{4}$ classes of cardinality $4$, up to $k_{2p}$ classes
of cardinality $2p$.

In particular, for $m=1$, we\ get from Eq. (\ref{17.2}), for all $p\in \mathbb{Z}%
_{+}$,%
\begin{equation*}
S_{e}(2p,k)\triangleq S_{e}^{1}(2p,k)=\sum_{P_{e}(2p):\left\vert
K_{e}\right\vert =k}(2p)!\prod_{j=1}^{p}\frac{1}{\left( (2j)!\right)
^{k_{2j}}k_{2j}!},
\end{equation*}%
i.e., the number of ways of partitioning a set of $2p$ elements into $k$
non-empty subsets, each of even cardinality, and%
\begin{equation}
E_{2p}=\sum_{k=1}^{2p}(-1)^{k}k!S_{e}(2p,k).  \label{18}
\end{equation}%
This seems to be a new expression for the (even) Euler numbers. Here the sum
involves partitions into subsets of even cardinality. A similar sum,
involving partitions into subsets of any cardinality, is the well-known
result for the Stirling numbers of the second kind,%
\begin{equation*}
1=\sum_{k=0}^{2p}(-1)^{k}k!S(2p,k).
\end{equation*}

It thus turns out that the numbers $S_{e}^{m}(n,k)$, (which by comparing %
Eq. (\ref{17.1}) with Eq. (\ref{19}) might be called \textquotedblleft even
multinomial Stirling numbers of the second kind\textquotedblright ), are
more natural to the $E_{n}^{m}$ than the $S(n,k)$. This can be seen by
applying Faa di Bruno's formula to $f\circ g$, with $g\left( t\right) =e^{t}$
and $f(g)=\left( \frac{1}{2}\left( g+1/g\right) \right) ^{-m}$, and using 
\cite[p. 823, 24.1.2, II B]{AS},%
\begin{equation}
S(n,k)=\sum_{P(n):\left\vert K\right\vert =k}\frac{n!}{\prod_{i=1}^{n}\left(
i!\right) ^{k_{i}}k_{i}!}.  \label{19}
\end{equation}%
We get%
\begin{equation}
E_{n}^{m}=\sum_{k=1}^{n}\left( \lim_{t\rightarrow 0}D_{t}^{k}\func{sech}%
^{m}\left( \ln \left( 1+t\right) \right) \right) S(n,k),  \label{19a}
\end{equation}%
an expression more complicated than $E_{n}^{m}=\lim_{t\rightarrow 0}D_{t}^{n}%
\func{sech}^{m}\left( t\right) $. For $m=1$, the numbers defined by the expression inside the parentheses in (%
\ref{19a}) are Sloane's \seqnum{A009014}.

We can derive another expression for the $E_{n}^{m}$ in terms of the $S(n,k)$,
directly from the generating function $\func{sech}^{m}\left( x\right) $, as
was done in Luo, et al. \cite{Lu}, but it turns out to involve a double sum. In the particular case $m=1$ however,
we can obtain this other expression from our results by combining Eqs. (\ref%
{1.13'}) and (\ref{15.2}), and then it reads%
\begin{equation}
E_{n}=\sum_{k=0}^{n}\tbinom{n}{k}\sum_{l=0}^{k}(-1)^{l}l!2^{k-l}S\left(
k,l\right) .  \label{20}
\end{equation}


\begin{thebibliography}{99}
\bibitem{AS} M. Abramowitz and I. A. Stegun, \textit{Handbook of
Mathematical Functions}, 9th ed., Dover, 1970.

\bibitem{C} L. Comtet, \textit{Advanced Combinatorics: The Art of Finite and
Infinite Expansions}, rev. enl. ed., Reidel, 1974.

\bibitem{GF} G. R. Franssens, Functions with derivatives given by polynomials in the function
itself or a related function, 2005, to appear.

\bibitem{GR} I. S. Gradshteyn and I. M. Ryzhik, \textit{Table of Integrals,
Series and Products}, Academic Press, 1980.

\bibitem{M} P. A. MacMahon, The divisors of numbers, \textit{Proc.\ London
Math. Soc.} \textbf{19} (1920),\ 305--340.

\bibitem{L} L. Lewin, \textit{Polylogarithms and Associated Functions},
North-Holland, 1981.

\bibitem{Lu} Q. M. Luo, F. Qi and L. Debnath, Generalizations of Euler
numbers and polynomials, \textit{Int.\ J. Math. Sci.} (2003),\ 3893--3901.

\bibitem{R} S. M. Ruiz, An algebraic identity leading to Wilson's
Theorem, \textit{Math. Gaz.} \textbf{80} (1996),\ 579--582.

\bibitem{Ro1984} S. Roman, \textit{The Umbral Calculus}, Academic Press,
1984.

\bibitem{Ru} W. Rudin, \textit{Principles of Mathematical Analysis},
McGraw-Hill, 1976.

\bibitem{S} N. J. A. Sloane, On-Line Encyclopedia of Integer Sequences, 
available at \href{http://www.research.att.com/~njas/sequence}{\tt http://www.research.att.com/\symbol{126}njas/sequences}.

\bibitem{St} R. P. Stanley, \textit{Enumerative Combinatorics}, Wadsworth
and Brooks/Cole, 1986.

\bibitem{W.St2} E. W. Weisstein, Stirling Number of the Second Kind,
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\end{thebibliography}



\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11B83, 05A15; Secondary  11Y55, 33B10, 30B10.

\bigskip

\noindent \emph{Keywords: } 
Generating function, binomial coefficients, Deleham numbers,
Eulerian numbers, MacMahon numbers, Stirling numbers, Euler numbers.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000182},
\seqnum{A000364},
\seqnum{A000567},
\seqnum{A001147},
\seqnum{A008277},
\seqnum{A008292},
\seqnum{A009014},
\seqnum{A009117},
\seqnum{A027641},
\seqnum{A027642},
\seqnum{A045944},
\seqnum{A054879},
\seqnum{A060187},
\seqnum{A085734},
\seqnum{A088874}, and
\seqnum{A092812}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received June 17 2005;
revised version received  August 8 2006.
Published in {\it Journal of Integer Sequences}, August 21 2006.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}

                                                                                
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