\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://www.research.att.com/cgi-bin/access.cgi/as/~njas/sequences/eisA.cgi?Anum=#1}{\underline{#1}}} \def\func#1{\mathop{\rm #1}\nolimits} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \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} \end{center} \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] \newtheorem{proposition}{Proposition}[section] \newtheorem{corollary}{Corollary}[section] \newtheorem{lemma}{Lemma}[section] %\usepackage{amssymb} %\usepackage[reqno]{amsmath} %\usepackage[a4paper]{geometry} \setcounter{MaxMatrixCols}{10} \theoremstyle{plain} %\newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}{Acknowledgement} %\newtheorem{corollary}{Corollary} \newtheorem{definition}{Definition} \newtheorem{example}{Example} %\newtheorem{lemma}{Lemma} \newtheorem{problem}{Problem} %\newtheorem{proposition}{Proposition} \newtheorem{remark}{Remark} \numberwithin{equation}{section} % macros for user - defined functions \def\QTATOPD#1#2#3#4{{\textstyle {#3 \atopwithdelims#1#2 #4}}} %\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, & \text{if }condition\text{ is true;} \\ 0, & \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} &\triangleq &\delta _{n\in \mathbb{Z}_{e}}, \label{D.2.2} \\ o_{n} &\triangleq &\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)} &\triangleq &\delta _{n=0}+\delta _{n>0}\,z(z+1)(z+2)...(z+(n-1)), \label{D.5.1a} \\ &=&\frac{\Gamma (z+n)}{\Gamma (z)}, \label{D.5.1b} \\ &=&\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)} &\triangleq &\delta _{n=0}+\delta _{n>0}\,z(z-1)(z-2)...(z-(n-1)), \label{D.5.2a} \\ &=&\frac{\Gamma (z+1)}{\Gamma (z+1-n)}, \label{D.5.2b} \\ &=&\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}} &=&\sum_{n=0}^{+\infty }c^{(n)}\frac{z^{n}}{% n!}, \label{D.6.1} \\ \left( 1+z\right) ^{c} &=&\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 <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)} &=&\sum_{k=0}^{n}\tbinom{n}{k}a^{(n-k)}b^{(k)}, \label{D.6.4} \\ \left( a+b\right) _{(n)} &=&\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)} &=&\delta _{n=0}, \label{D.6.6} \\ \sum_{k=0}^{n}\tbinom{n}{k}c_{(n-k)}\left( -c\right) _{(k)} &=&\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}}, & \text{if }x\neq y% \text{;} \\ \frac{1}{1-xt}, & \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 <\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| <\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}). >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) &=&\frac{1}{\left( 1-xt\right) ^{c}}, \label{1.5} \\ B^{c}(x,-x,t) &=&\func{sech}^{c}(xt). \label{1.6} \end{eqnarray}% This gives% \begin{eqnarray} B_{n}(x,x;c) &=&2^{n}c^{(n)}x^{n}, \label{1.7} \\ B_{n}(x,-x;c) &=&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) &=&2^{n}c^{(n)}, \label{1.9} \\ \sum_{k=0}^{n}(-1)^{k}B_{n,k}(c) &=&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} & {\tiny 1} & {\tiny 2} & {\tiny 3} & {\tiny 4} & {\tiny 5} & {\tiny 6} \\ \hline {\tiny 1} & {\tiny 1} & & & & & \\ \hline {\tiny 2} & {\tiny 1} & {\tiny 1} & & & & \\ \hline {\tiny 3} & {\tiny 1} & {\tiny 6} & {\tiny 1} & & & \\ \hline {\tiny 4} & {\tiny 1} & {\tiny 23} & {\tiny 23} & {\tiny 1} & & \\ \hline {\tiny 5} & {\tiny 1} & {\tiny 76} & {\tiny 230} & {\tiny 76} & {\tiny 1} & \\ \hline {\tiny 6} & {\tiny 1} & {\tiny 237} & {\tiny 1682} & {\tiny 1682} & {\tiny % 237} & {\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} &=&2^{n}n!, \label{1.12'} \\ \sum_{k=0}^{n}(-1)^{k}B_{n+1,k+1} &=&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 <\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 <\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} & {\tiny 1} & {\tiny 2} & {\tiny 3} & {\tiny 4} & {\tiny 5} & {\tiny 6} \\ \hline {\tiny 1} & {\tiny 1} & & & & & \\ \hline {\tiny 2} & {\tiny 1} & {\tiny 1} & & & & \\ \hline {\tiny 3} & {\tiny 1} & {\tiny 4} & {\tiny 1} & & & \\ \hline {\tiny 4} & {\tiny 1} & {\tiny 11} & {\tiny 11} & {\tiny 1} & & \\ \hline {\tiny 5} & {\tiny 1} & {\tiny 26} & {\tiny 66} & {\tiny 26} & {\tiny 1} & \\ \hline {\tiny 6} & {\tiny 1} & {\tiny 57} & {\tiny 302} & {\tiny 302} & {\tiny 57} & {\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} &=&(n+1)!, \label{1.15} \\ \sum_{k=0}^{n}(-1)^{k}A_{n+1,k+1} &=&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!} & \text{if} & \left\vert y\right\vert <\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!} & \text{if} & \left\vert x\right\vert <\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| <\left| x\right| }\triangleq \left\{ (x,y)\in \mathbb{C}% ^{2}:\left| y\right| <\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)<\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| <\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 $(x,y)\in D_{\left\vert y\right\vert <\left\vert x\right\vert }$, $% \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!} &=&\frac{% B_{n}(1,z;c)}{\left( 1-z\right) ^{n+c}},\left\vert z\right\vert <1, \label{2.5} \\ \sum_{k=0}^{+\infty }c^{(k)}\left( 2k+c\right) ^{n}\frac{z^{-k}}{k!} &=&z^{c}% \frac{B_{n}(z,1;c)}{\left( z-1\right) ^{n+c}},\left\vert z\right\vert >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} &=&\frac{B_{n}(1,z)}{% \left( 1-z\right) ^{n+1}},\left| z\right| <1, \label{2.7} \\ \sum_{k=0}^{+\infty }\left( 2k+1\right) ^{n}z^{-k} &=&z\frac{B_{n}(z,1)}{% \left( z-1\right) ^{n+1}},\left| z\right| >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} &=&\frac{z}{\left( 1-z\right) ^{n+1}}% A_{n-1}(1,z),\left| z\right| <1, \label{2.9} \\ \sum_{l=1}^{+\infty }l^{n}z^{-l} &=&\frac{z}{\left( z-1\right) ^{n+1}}% A_{n-1}(z,1),\left| z\right| >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| <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*} &&\left( 1-z\right) ^{n+c}\left( c+2zD_{z}\right) ^{n}\left( 1-z\right) ^{-c} \\ &=&\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 $\left| z\right| <1$ we have the absolutely convergent series (\ref{2.5}). As $% \left| z\right| <1$, 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) &=&\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} \\ &=&(-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 $nFrom 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} &=&\sum_{l=0}^{n}b_{n,k+1,l}c^{l+1}+\sum_{l=0}^{n}b_{n,k,l}c^{l+1} \\ &&+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} &=&\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} \\ &&+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} &=&\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} \\ &&+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} &=&B_{n+1,k+1}, \label{8.1} \\ \sum_{l=0}^{n}(-1)^{l}b_{n,k,l} &=&(-1)^{n-k}\left( \tbinom{n}{k}% -2o_{n}\delta _{k>0}\tbinom{n-1}{k-1}\right) , \label{8.2} \\ \sum_{k=0}^{n}b_{n,k,l} &=&(-1)^{n-l}2^{n}s(n,l), \label{8.3} \\ \sum_{k=0}^{n}(-1)^{k}b_{n,k,l} &=&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*} &&B^{-1}(1,z,t) \\ &=&\frac{1}{1-z}e^{-\frac{1-z}{2}t}-\frac{z}{1-z}e^{+\frac{1-z}{2}t}, \\ &=&\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!}, \\ &=&\sum_{n=0}^{+\infty }2^{-n}\left( (-1)^{n}-z\right) \left( 1-z\right) ^{n-1}\frac{t^{n}}{n!}, \\ &=&\sum_{n=0}^{+\infty }2^{-n}\left( \delta _{n=0}+\delta _{n>0}\left( (-1)^{n}-z\right) \sum_{k=0}^{n-1}\tbinom{n-1}{k}(-z)^{k}\right) \frac{t^{n}% }{n!}, \\ &=&\sum_{n=0}^{+\infty }2^{-n}\left( \delta _{n=0}+\delta _{n>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!}, \\ &=&\sum_{n=0}^{+\infty }2^{-n}\left( \delta _{n=0}+\delta _{n>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*} &&B^{-1}(1,z,t) \\ &=&\sum_{n=0}^{+\infty }2^{-n}\left( \delta _{n=0}+\delta _{n>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*} &&\delta _{n=0}+\delta _{n>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) \\ &=&\sum_{k=0}^{n}\left( \delta _{n=0}+\delta _{n>0}\left( \left( -1\right) ^{n}+\left( 1-\left( -1\right) ^{n}\right) \frac{k}{n}\right) \right) \tbinom{n}{k}(-z)^{k}, \\ &=&\sum_{k=0}^{n}(-1)^{n-k}\tbinom{n}{k}\left( 1-\delta _{n>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>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>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>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} &=&\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) , \\ &=&\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} & {\tiny 0} & {\tiny 1} & {\tiny 2} & {\tiny 3} & {\tiny 4} & {\tiny 5} & {\tiny 6} & {\tiny 7} & {\tiny 8} \\ \hline {\tiny 0} & {\tiny 1} & & & & & & & & \\ \hline {\tiny 1} & {\tiny 0} & {\tiny 1} & & & & & & & \\ \hline {\tiny 2} & {\tiny 0} & {\tiny 2} & {\tiny 3} & & & & & & \\ \hline {\tiny 3} & {\tiny 0} & {\tiny 16} & {\tiny 30} & {\tiny 15} & & & & & \\ \hline {\tiny 4} & {\tiny 0} & {\tiny 272} & {\tiny 588} & {\tiny 420} & {\tiny 105} & & & & \\ \hline {\tiny 5} & {\tiny 0} & {\tiny 7\thinspace 936} & {\tiny 18\thinspace 960} & {\tiny 16\thinspace 380} & {\tiny 6\thinspace 300} & {\tiny 945} & & & \\ \hline {\tiny 6} & {\tiny 0} & {\tiny 353\thinspace 792} & {\tiny 911\thinspace 328} & {\tiny 893\thinspace 640} & {\tiny 429\thinspace 660} & {\tiny % 103\thinspace 950} & {\tiny 10\thinspace 395} & & \\ \hline {\tiny 7} & {\tiny 0} & {\tiny 22\thinspace 368\thinspace 256} & {\tiny % 61\thinspace 152\thinspace 000} & {\tiny 65\thinspace 825\thinspace 760} & {\tiny 36\thinspace 636\thinspace 600} & {\tiny 11\thinspace 351\thinspace 340} & {\tiny 1\thinspace 891\thinspace 890} & {\tiny 135\thinspace 135} & \\ \hline {\tiny 8} & {\tiny 0} & {\tiny 1\thinspace 903\thinspace 757\thinspace 312} & {\tiny 5\thinspace 464\thinspace 904\thinspace 448} & {\tiny 6\thinspace 327\thinspace 135\thinspace 360} & {\tiny 3\thinspace 918\thinspace 554\thinspace 640} & {\tiny 1\thinspace 427\thinspace 025\thinspace 600} & {\tiny 310\thinspace 269\thinspace 960} & {\tiny 37\thinspace 837\thinspace 800} & {\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) &=&2^{-n}\sum_{k=0}^{n}(-1)^{k}B_{n,k}(c), \label{9.2} \\ &=&\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>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*} &&\sum_{n=1}^{+\infty }(-1)^{n}\sum_{l=0}^{n}T_{n,l}c^{l}\frac{t^{2(n-1)}}{% (2(n-1))!} \\ &=&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} &=&\sum_{l=0}^{n}T_{n,l}\sum_{k=0}^{l}\tbinom{% l}{k}2^{l-k}c^{k+1} \\ &&+\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} &=&\sum_{k=0}^{n}\sum_{q=0}^{n-k}2^{q}\tbinom{% q+k}{k}T_{n,q+k}c^{k+1} \\ &&+\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} &=&\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} \\ &&+\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} &=&\sum_{q=0}^{n}2^{q}T_{n,q}; \\ \sum_{l=1}^{n}T_{n+1,l+1}c^{l} &=&\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} \\ &&+T_{n,n}c^{n+1}-T_{n,n}c^{n+1}, \end{eqnarray*}% or% \begin{eqnarray*} T_{n+1,1} &=&\sum_{q=0}^{n}2^{q}T_{n,q}; \\ T_{n+1,l+1} &=&\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} &=&\sum_{q=0}^{n}2^{q}T_{n,q}; \\ T_{n+1,l+1} &=&\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} &=&\sum_{q=0}^{n}2^{q}T_{n,q}; \\ T_{n+1,l+1} &=&\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} &=&\sum_{p=0}^{n}2^{p}T_{n,p},l=0; \\ T_{n+1,l+1} &=&\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} &=&\sum_{p=0}^{n}2^{p}T_{n,p},l=0; \\ T_{n+1,l+1} &=&\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} &=&\sum_{p=0}^{n}2^{p}T_{n,p}; \\ T_{n+1,l+1} &=&\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>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>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>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} & {\tiny 1} & {\tiny 2} & {\tiny 3} & {\tiny 4} & {\tiny 5} & {\tiny 6} \\ \hline {\tiny 0} & {\tiny 1} & {\tiny 1} & {\tiny 1} & {\tiny 1} & {\tiny 1} & {\tiny 1} \\ \hline {\tiny 2} & {\tiny -1} & {\tiny -2} & {\tiny -3} & {\tiny -4} & {\tiny -5} & {\tiny -6} \\ \hline {\tiny 4} & {\tiny 5} & {\tiny 16} & {\tiny 33} & {\tiny 56} & {\tiny 85} & {\tiny 120} \\ \hline {\tiny 6} & {\tiny -61} & {\tiny -272} & {\tiny -723} & {\tiny -1\thinspace 504} & {\tiny -2\thinspace 705} & {\tiny -4\thinspace 416} \\ \hline {\tiny 8} & {\tiny 1\thinspace 385} & {\tiny 7\thinspace 936} & {\tiny % 25\thinspace 953} & {\tiny 64\thinspace 256} & {\tiny 134\thinspace 185} & {\tiny 249\thinspace 600} \\ \hline {\tiny 10} & {\tiny -50\thinspace 521} & {\tiny -353\thinspace 792} & {\tiny % -1\thinspace 376\thinspace 643} & {\tiny -3\thinspace 963\thinspace 904} & {\tiny -9\thinspace 451\thinspace 805} & {\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>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, $% e_{n}^{0}=\delta _{n=0}$, $e_{n}^{1}=e_{n}$, $e_{n}^{2}=\delta _{n=0}+\delta _{n>0}e_{n}2^{n-1}$ ($e_{2p}^{2}$ 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} & {\tiny 1} & {\tiny 2} & {\tiny 3} & {\tiny 4} & {\tiny 5} & {\tiny 6} \\ \hline {\tiny 0} & {\tiny 1} & {\tiny 1} & {\tiny 1} & {\tiny 1} & {\tiny 1} & {\tiny 1} \\ \hline {\tiny 2} & {\tiny 1} & {\tiny 2} & {\tiny 3} & {\tiny 4} & {\tiny 5} & {\tiny 6} \\ \hline {\tiny 4} & {\tiny 1} & {\tiny 8} & {\tiny 21} & {\tiny 40} & {\tiny 65} & {\tiny 96} \\ \hline {\tiny 6} & {\tiny 1} & {\tiny 32} & {\tiny 183} & {\tiny 544} & {\tiny % 1\thinspace 205} & {\tiny 2\thinspace 256} \\ \hline {\tiny 8} & {\tiny 1} & {\tiny 128} & {\tiny 1\thinspace 641} & {\tiny % 8\thinspace 320} & {\tiny 26\thinspace 465} & {\tiny 64\thinspace 896} \\ \hline {\tiny 10} & {\tiny 1} & {\tiny 512} & {\tiny 14\thinspace 763} & {\tiny % 131\thinspace 584} & {\tiny 628\thinspace 805} & {\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>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>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>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>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} &=&\lim_{t\rightarrow 0}D_{t}^{n}f\left( g\left( t\right) \right) , \\ &=&\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 $n>0$, 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, available from \href{http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html}{\tt http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html} \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}