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\begin{center}
\vskip 1cm{\LARGE\bf Two Formulas for Successive Derivatives and \\
\vskip .1in
Their Applications
}
\vskip 1cm
\large
Grzegorz Rz\c{a}dkowski\\
Faculty of Mathematics and Natural Sciences \\
Cardinal Stefan Wyszy\'nski University in Warsaw \\
Dewajtis 5 \\
01 - 815 Warsaw \\
Poland \\
\href{mailto:g.rzadkowski@uksw.edu.pl}{\tt g.rzadkowski@uksw.edu.pl} \\
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\vskip .2 in

\begin{abstract}
We recall two formulas, due to C. Jordan, for the
successive derivatives of functions with an exponential or logarithmic
inner function. We apply them to get addition formulas for the Stirling
numbers of the second kind and for the Stirling numbers of the first
kind. Then we show how one can obtain, in a simple way, explicit
formulas for the generalized Euler polynomials, generalized Euler
numbers, generalized Bernoulli polynomials and the Bell polynomials.
\end{abstract}

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\section{Introduction}

By  $ { n \brace k}$ we mean the Stirling number of the second kind
(the number of ways of partitioning a set of $n$ elements into $k$
nonempty subsets; see Graham et al.  \cite{GKP} and sequence \seqnum{A008277} of
Sloane's {\it On-line Encyclopedia} \cite{Sl}). As usual, we set ${ n
\brace 0} =0 $ if $n&gt;0$, ${ 0 \brace 0} =1 $, and  ${ n \brace k} =0 $ for
$k&gt;n$ or $k&lt;0$.  Let us recall that the Stirling numbers satisfy the identities
{\setlength\arraycolsep{2pt} \begin{eqnarray} &amp;&amp;{ n \brace k}
	=\frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}{k \choose j}j^n=
	\frac{1}{k!}\sum_{j=0}^{k}(-1)^{j}{k \choose
	j}(k-j)^{n},\label{bin}\\ &amp;&amp;      { n+1 \brace k} =k{ n \brace
k}+ { n \brace k-1},\label{rec} \end{eqnarray}} and appear in the
Taylor expansion \begin{equation}\label{Ta}
\frac{(e^{w}-1)^{k}}{k!}=       \sum_{n=k}^{\infty }{ n \brace k}
\frac{w^{n}}{n!}.  \end{equation} 

Peregrino \cite{RSP} proved the
following addition formula for the Stirling numbers of the second kind
\begin{equation}\label{ad}
	{ u+v \brace k} =\sum_{j=0}^{v}\sum_{i=0}^{v} {v \choose i}
	k^{i}(-1)^{v+i+j}{ v-i \brace j} { u \brace k-j}.
\end{equation} 
By ${n \brack k}$ we denote the Stirling number of the
first kind (number of ways of partitioning a set of $n$ elements into
$k$ nonempty cycles, see \cite{GKP}, sequence \seqnum{A008275} in \cite{Sl}).
Similarly ${ n \brack 0} =0 $ if $n&gt;0$, ${ 0 \brack 0} =1 $,  ${ n
\brack k}=0 $ for $k&gt;n$ or $k&lt;0$. The Stirling numbers of the first
kind fulfil the recurrence formula \begin{equation}\label{rec2}
	{ n+1 \brack k} =n{ n \brack k} + { n \brack k-1}.
\end{equation} 

We use common notation for the falling factorial
	\[(x)_{k}=x(x-1)\cdots (x-k+1) \]
 and for the rising factorial (Pochhammer's symbol)
	\[x^{(k)}=x(x+1)\cdots (x+k-1).  \] 
	
The paper is organized as
follows.  We recall and prove two formulas, due to C. Jordan \cite{J},
in section 2. The formulas involve Stirling numbers of both kinds.
Since the original Peregrino's proof of (\ref{ad}), by induction on
$v$, is long we give shorter and more direct proof of his formula and
similar formulas in section 3. We prove addition formulas for Stirling
numbers of the first kind in section 4. Sections 5,6,7 are devoted to
show explicit formulas respectively for  generalized Euler polynomials,
generalized Bernoulli polynomials and the Bell polynomials. All proofs,
in the last three sections, are more simple and direct than the proofs
which exist in the literature. In the case of generalized Bernoulli
polynomials (and generalized Bernoulli numbers) our results seem to be
new.

\section{Formulas for successive derivatives}

We have the
following formulas for successive derivatives of composite functions
with the exponential, or the logarithmic, inner function.

\begin{lemma} If $f \in C^{\infty }(R)$ then the following formulas for
the $n$th order ($n=1,2,3,\ldots $) derivatives hold
{\setlength\arraycolsep{2pt} \begin{eqnarray}
\frac{d^{n}}{dt^{n}}(f(e^{t}))  &amp;=&amp;\sum_{k=1}^{n}{ n \brace
k}f^{(k)}(e^{t})e^{kt},\label{deriv2}\\
	\frac{d^{n}}{dt^{n}}(f(\log t))
	&amp;=&amp;\frac{1}{t^{n}}\sum_{k=1}^{n}(-1)^{n-k}{ n \brack
	k}f^{(k)}(\log t).\label{deriv3} \end{eqnarray}} \end{lemma}

\begin{proof} 
To prove formula (\ref{deriv2}), we proceed by induction
with respect to $n$. Denote $g(t)=f(e^{t})$. For $n=1$, (\ref{deriv2})
is obviously true.  Let us suppose that for an integer $n$ formula
(\ref{deriv2}) holds. Then using (\ref{rec}) we have \begin{eqnarray*}
\lefteqn{g^{(n+1)}(t)=\frac{d}{dt}\left( \sum_{k=1}^{n} { n \brace
k}f^{(k)}(e^{t})e^{kt}\right) } \\ &amp;&amp; =\sum_{k=1}^{n}{ n \brace k}
(f^{(k+1)}(e^{t})e^{(k+1)t}+kf^{(k)}(e^{t})e^{kt}) \\ &amp;&amp; =
f'(e^{t})e^{t} +f^{(n+1)}(e^{t})e^{(n+1)t} \\ &amp;&amp; + \sum_{k=2}^{n}
\left( k{ n \brace k} +{ n \brace k-1}\right) f^{(k)}(e^{t})e^{kt}
\\ &amp;&amp; = \sum_{k=1}^{n+1}{ n+1 \brace k}f^{(k)}(e^{t})e^{kt},
\end{eqnarray*} which ends the proof of (\ref{deriv2}). Analogously,
using (\ref{rec2}), formula (\ref{deriv3}) can be shown.  \end{proof}
Formulas (\ref{deriv2}), (\ref{deriv3}) are known and can be found,
with proofs based on finite differences, in the C. Jordan's book
\cite[pp.\ 205--206]{J}. The formulas are given, as exercises without
proofs and without direct referencing to \cite{J}, also in the L.
Comtet's book \cite[Ex.\ 6, p.\ 157]{C}.  It is easy to see that formula
(\ref{deriv2}) holds, with the same proof, for complex variable $t$ and
a holomorphic function $f$. Formula (\ref{deriv3}) holds for complex
$t$, a branch of logarithm and a  holomorphic function $f$.
\vspace{2mm}\\ For example if $f(x)=x^{m},\; g(t)=e^{mt} $ then using
(\ref{deriv2}) we get \begin{equation}\label{ex1}
	g^{(n)}(t)=\sum_{k=1}^{n}{ n \brace k} m(m-1)\cdots
(m-k+1)e^{(m-k)t}e^{kt}.  \end{equation} From the other side
\begin{equation}\label{ex12}
	g^{(n)}(t)= m^{n}e^{mt}, \end{equation} and comparing
(\ref{ex1}) with (\ref{ex12}) we obtain the well-known generating
function for the Stirling numbers of the second kind
		\[\sum_{k=1}^{n}{ n \brace k}(m)_{k}=m^{n}.
\] \section{Addition formulas for Stirling numbers of the second kind}
Let us substitute $f(x)=\exp(x)$ in (\ref{deriv2}). We have
\begin{equation}\label{ex31}
	(e^{e^{t}})^{(u+v)}=\sum_{k=1}^{u+v}{ u+v \brace k}
	e^{e^{t}}\cdot e^{kt}.  \end{equation} From the other side
	\[(e^{e^{t}})^{(u)}=\sum_{n=1}^{u}{ u \brace n}
		e^{e^{t}}\cdot e^{nt}
\] and \begin{eqnarray}
\lefteqn{(e^{e^{t}})^{(u+v)}=[(e^{e^{t}})^{(u)}]^{(v)} =\sum_{n=1}^{u}
{u \brace n}(e^{e^{t}}\cdot e^{nt})^{(v)}}\nonumber \\
&amp;&amp;=\sum_{n=1}^{u} {u \brace n} \sum_{m=0}^{v}{v\choose
m}(\sum_{i=0}^{m}
 {m \brace i}
		e^{e^{t}}\cdot e^{it})\cdot n^{v-m}e^{nt}\nonumber \\
&amp;&amp;=\sum_{n=1}^{u} {u \brace n} \sum_{m=0}^{v}{v\choose
m}(\sum_{i=0}^{m}
 {m \brace i}
		e^{e^{t}}\cdot e^{(i+n)t} n^{v-m}).\label{ex32}
\end{eqnarray} \begin{theorem} 
The following addition formula
for the Stirling numbers of the second kind holds.
\begin{equation}\label{ex33}
	 {u\!+\!v \brace k}=\sum_{n=1}^{k} {u \brace n}
	\sum_{m=k-n}^{v}{v \choose m} {m \brace k-n}n^{v-m}.
\end{equation} \end{theorem} \begin{proof} Formula (\ref{ex33}) follows
by comparing the coefficients of $e^{e^{t}}e^{kt}$ in (\ref{ex31}) and
(\ref{ex32}) for $i+n=k$.  \end{proof} Denoting in (\ref{ex33})
$m=v-i$, ($i=0,1,2,\ldots,v-k+n$) we have
	\[ {u\!+\!v \brace k}=\sum_{n=1}^{k} {u \brace n}
	\sum_{i=0}^{v-k+n}{v \choose i} {v-i \brace k-n}n^{i}, \] and
then letting $n=k-j$, ($j=0,1,2,\ldots,k-1$) we obtain
\begin{equation}\label{ex34}
	 {u\!+\!v \brace k}=\sum_{j=0}^{k-1}\sum_{i=0}^{v-j}{v \choose
	 i}(k-j)^{i} {v-i \brace j} {u \brace k-j}.
\end{equation} Formula (\ref{ad}) follows easily from (\ref{ex34}). To
see this let us observe that the range of the variables $i,j$ in
(\ref{ex34}) can be changed to the same range as in (\ref{ad}) i.e.,
$i,j=0,1,2,\ldots,v$. This is allowed because the number $  {u \brace
k-j}$ is zero if $j\ge k$ and $ {v-i \brace j} $ equals zero for
$j&gt;v-i$ i.e., if $i+j&gt;v$. We rearrange (\ref{ex34}) into (\ref{ad})
using formulas (\ref{bin}), for the symbol $ {v-i \brace j} $,
respectively in the start and end of the following calculation:
{\setlength\arraycolsep{2pt} \begin{eqnarray*}
 {u\!+\!v \brace k}&amp;=&amp;\sum_{j=0}^{v}\sum_{i=0}^{v}{v \choose
 i}(k-j)^{i} {v-i \brace j}{u \brace k-j}\\ &amp;=&amp;\sum_{j=0}^{v}{u \brace
k-j}\sum_{i=0}^{v}{v \choose
i}(k-j)^{i}\frac{1}{j!}\sum_{l=0}^{j}(-1)^{j-l}{j \choose l}l^{v-i}\\
&amp;=&amp; \sum_{j=0}^{v}{u \brace k-j}\frac{1}{j!}\sum_{l=0}^{j}(-1)^{j-l}{j
\choose l} \sum_{i=0}^{v}{v \choose i}(k-j)^{i}l^{v-i}\\
&amp;=&amp;\sum_{j=0}^{v}{u \brace k-j}\frac{1}{j!}\sum_{l=0}^{j}(-1)^{j-l}{j
\choose l}(k-j+l)^{v}\\ &amp;=&amp;\sum_{j=0}^{v} {u \brace
k-j}\frac{1}{j!}\sum_{l=0}^{j}(-1)^{j-l}{j \choose l}
\sum_{i=0}^{v}(-1)^{v-i}{v \choose i}k^{i}(j-l)^{v-i}\\
&amp;=&amp;\sum_{j=0}^{v} {u \brace k-j} \sum_{i=0}^{v}{v \choose
i}k^{i}(-1)^{v+i+j}\frac{1}{j!} \sum_{l=0}^{j}(-1)^{l}{j \choose
l}(j-l)^{v-i}\\ &amp;=&amp;\sum_{j=0}^{v}\sum_{i=0}^{v}
	{v \choose i} k^{i}(-1)^{v+i+j} {v-i \brace j} {u \brace k-j}.
\end{eqnarray*}} \section{Addition formulas for Stirling numbers of the
first kind} Let us substitute $f(x)=\log x$ in (\ref{deriv3}). We have
($t&gt;1$) \begin{equation}\label{ex41}
(\log{\log{t}})^{(u+v)}=\frac{(-1)^{u+v+1}}{t^{u+v}}\sum_{k=1}^{u+v}
{u+v \brack k}
	\frac{(k-1)!}{\log^{k}t}.  \end{equation} From the other side
	\[(\log{\log{t}})^{(u)}=\frac{(-1)^{u+1}}{t^{u}}\sum_{n=1}^{u}
	{u \brack n} \frac{(n-1)!}{\log^{n}t} \] and by using the
Leibniz formula and formula (\ref{deriv3}) for $g(t)=(\log t)^{-n}$ we
get \begin{eqnarray}
\lefteqn{(\log{\log{t}})^{(u+v)}=[(\log{\log{t}})^{(u)}]^{(v)}}\nonumber
\\ &amp;&amp;=(-1)^{u+1}\sum_{n=1}^{u} {u \brack
n}(n-1)!\left(\frac{1}{\log^{n}t}\cdot\frac{1}{t^{u}}\right)^{(v)}\nonumber
\\ &amp;&amp;=(-1)^{u+1}\sum_{n=1}^{u}(n-1)! {u \brack n}\sum_{m=1}^{v} {v
\choose m}
\left(\frac{1}{\log^{n}t}\right)^{(m)}\left(\frac{1}{t^{u}}\right)^{(v-m)}\nonumber
\\ &amp;&amp;=(-1)^{u+1}\sum_{n=1}^{u}(n-1)! {u \brack n} \sum_{m=1}^{v} {v
\choose m}\frac{(-u)_{v-m}}{t^{u+v-m}}\nonumber \\ &amp;&amp;\hspace{4mm}\times
\frac{1}{t^{m}} \sum_{i=0}^{m}(-1)^{m-i} {m \brack
i}\frac{(-n)_{i}}{(\log t)^{n+i}}.\label{ex42} \end{eqnarray}
\begin{theorem} The following addition formula for the
Stirling numbers of the first kind holds.
\begin{equation}\label{ex44}
	 {u+v \brack k}=\sum_{j=0}^{v}\sum_{i=0}^{v} {v \choose
	i}u^{(i)} {v-i \brack j}
	 {u \brack k-j}.
\end{equation} \end{theorem} \begin{proof}   By comparing the
coefficients of $(t^{u+v}\log^{k}t)^{-1}$ in (\ref{ex41}) and
(\ref{ex42}) for $i+n=k,\; i=k-n$ we get {\setlength\arraycolsep{2pt}
\begin{eqnarray*}
	 {u+v \brack k}(k-1)!&amp;=&amp;(-1)^{v}\sum_{n=1}^{k}(n-1)! {u \brack
	 n} (-n)_{k-n}\nonumber \\ &amp;&amp;      \times\sum_{m=k-n}^{v}{v
\choose m}(-u)_{v-m}(-1)^{m-k+n} {m \brack k-n}\nonumber
\end{eqnarray*}} and then using the identities
	\[\frac{(n-1)!(-n)_{k-n}}{(k-1)!}=(-1)^{k-n},\quad
	(-u)_{v-m}(-1)^{v-m}=u^{(v-m)} \] we obtain the formula
\begin{equation}\label{ex43}
	 {u+v \brack k}=\sum_{n=1}^{k} {u \brack n}\sum_{m=k-n}^{v}{v
	 \choose m}u^{(v-m)} {m \brack k-n}.
\end{equation} By the same manner  as for the Stirling numbers of the
second kind, formula (\ref{ex43}) can be rearranged to the form
(\ref{ex44}).  \end{proof} \section{Generalized Euler polynomials} The
generalized Euler polynomials $E_{n}^{\mu}(z)$ of degree
$n=0,1,2,\ldots $, complex order $\mu$ and complex argument $z$ (see
N\"orlund \cite{N}) can be defined by the generating function
	\[\sum_{n=0}^{\infty }\frac{E_{n}^{\mu}(z)}{n!}w^{n}=
	\frac{2^{\mu}e^{wz}}{(e^{w}+1)^{\mu }}, \qquad |w|&lt;\pi.  \] The
generalized Euler polynomials play an important role in the calculus of
finite differences. We will show, in a very easy way, an explicit
formula for the Euler polynomial $E_{n}^{\mu}(z)$. By (\ref{deriv2}) we
get
	\[\frac{d^{m}}{dw^{m}}\frac{1}{(e^{w}+1)^{\mu }}=
	\sum_{k=1}^{m} {m \brace k}(-\mu
	)_{k}\frac{e^{kw}}{(e^{w}+1)^{\mu +k}} \] and then by the
Leibniz formula
	\[\frac{d^{n}}{dw^{n}}\frac{2^{\mu}e^{wz}}{(e^{w}+1)^{\mu }}=
	2^{\mu}\left(\frac{z^{n}e^{wz}}{(e^{w}+1)^{\mu }}+
	\sum_{m=1}^{n}{n \choose m}\sum_{k=1}^{m} {m \brace k}(-\mu
	)_{k}\frac{e^{kw}e^{wz}z^{n-m}}{(e^{w}+1)^{\mu +k}}\right).  \]
Thus \begin{equation}\label{Ep} E_{n}^{\mu}(z)=\left.
\frac{d^{n}}{dw^{n}}\frac{2^{\mu}e^{wz}}{(e^{w}+1)^{\mu }}\right|_{w=0}
=z^{n}+\sum_{m=1}^{n}{n \choose m}z^{n-m}\sum_{k=1}^{m} {m \brace
k}\frac{(-\mu )_{k}}{2^{k}}.  \end{equation} Another approach to
formula (\ref{Ep}) is presented by Howard \cite{H}. Putting in
(\ref{Ep}) $z=0$ we obtain the following explicit formula (see Todorov
\cite{T}) for the generalized Euler number  $E_{n}^{\mu}$
\begin{equation}\label{En}
	E_{n}^{\mu}=E_{n}^{\mu}(0)=\sum_{k=1}^{n} {n \brace
	k}\frac{(-\mu )_{k}}{2^{k}}.  \end{equation} 

Luo \cite{Q} deals with the Apostol--Euler polynomials, which are a further
generalization of the polynomials $\{E_{n}^{\mu}(z)\}$. Formula
(\ref{En}) coincides, as a particular case, with formula (32) of this
paper.

\section{Generalized Bernoulli polynomials} The generalized
Bernoulli polynomials $B_{n}^{\mu}(z)$ of degree $n=0,1,2,\ldots $,
complex order $\mu$ and complex argument $z$ (see N\"orlund \cite{N} )
can be defined by the generating function
	\[\sum_{n=0}^{\infty }\frac{B_{n}^{\mu}(z)}{n!}w^{n}=
	\frac{w^{\mu}e^{wz}}{(e^{w}-1)^{\mu }},  \qquad |w|&lt;2\pi.  \]
We will show that by applying formula (\ref{deriv2}) one can obtain
explicit formulas for the polynomials and then for the generalized
Bernoulli numbers $B_{n}^{\mu}= B_{n}^{\mu}(0)$. Our approach can be
seen as consistent with the spirit of the paper of Gould
\cite{G}.\\ Using (\ref{deriv2}) and the Leibniz formula we get
successively \[\frac{d^{j}}{dw^{j}}\frac{1}{(e^{w}-1)^{\mu }}=
	\sum_{k=0}^{j} {j \brace k}(-\mu
	)_{k}\frac{e^{kw}}{(e^{w}-1)^{\mu +k}}, \]
\begin{equation}\label{md}
	\frac{d^{m}}{dw^{m}}\frac{w^{\mu }}{(e^{w}-1)^{\mu }}=
	\sum_{j=0}^{m}{m \choose j}\sum_{k=0}^{j} {j \brace
	k}\frac{(-\mu )_{k}e^{kw}}{(e^{w}-1)^{\mu +k}} (\mu
	)_{m-j}w^{\mu-m+j} \end{equation} and
\begin{equation}\label{nd} \frac{d^{n}}{dw^{n}}\frac{w^{\mu
}e^{wz}}{(e^{w}-1)^{\mu }}=
	\frac{w^{\mu }z^{n}e^{wz}}{(e^{w}-1)^{\mu }}+\sum_{m=1}^{n}{n
	\choose m} \left(\frac{d^{m}}{dw^{m}}\frac{w^{\mu
	}}{(e^{w}-1)^{\mu }}\right)z^{n-m}e^{wz}.  \end{equation}
Rewriting the right hand side of (\ref{md}) into the form
	\[\frac{w^{\mu+m }}{(e^{w}-1)^{\mu +m}}\left(
\frac{1}{w^{2m}}\sum_{j=0}^{m}{m \choose j}(\mu )_{m-j}\sum_{k=0}^{j}
{j \brace k}(-\mu )_{k}e^{kw}(e^{w}-1)^{m-k}
	w^{j}       \right), \] using (\ref{Ta}) in the expression
\begin{eqnarray*}
\lefteqn{\hspace{-9mm}w^{j}e^{kw}(e^{w}-1)^{m-k}=w^{j}(e^{w}-1)^{m}\frac{e^{kw}}{(e^{w}-1)^{k}}=w^{j}(e^{w}-1)^{m}\left(1\!+\!\frac{1}{e^{w}\!-\!1}\right)^{k}}\\
&amp;&amp; = w^{j}(e^{w}-1)^{m}\sum_{l=0}^{k} {k \choose
l}\frac{1}{(e^{w}-1)^{l}} =w^{j}\sum_{l=0}^{k} {k \choose
l}(e^{w}-1)^{m-l}\\ &amp;&amp; =w^{j}\sum_{l=0}^{k} {k \choose
l}\sum_{i=m-l}^{\infty } (m-l)! {i \brace m-l}\frac{w^{i}}{i!},
\end{eqnarray*} and grouping terms of power $w^{2m}$ we get
{\setlength\arraycolsep{2pt} \begin{eqnarray*}
\lim\limits_{w\rightarrow 0}\frac{d^{m}}{dw^{m}}\frac{w^{\mu
}}{(e^{w}-1)^{\mu }}&amp;=&amp;
	\sum_{j=0}^{m}{m \choose j}(\mu )_{m-j}\sum_{k=0}^{j} {j \brace
	k}(-\mu )_{k}\\ &amp;&amp;\times\sum_{l=0}^{k} {k \choose l}
	 {2m-j \brace m-l}\frac{(m-l)!}{(2m-j)!}.
\end{eqnarray*}} Thus by (\ref{nd}) we obtain the following explicit
formula for the generalized Bernoulli polynomials
{\setlength\arraycolsep{2pt} \begin{eqnarray} \hspace{-5mm}
B_{n}^{\mu}(z)&amp;=&amp;\frac{d^{n}}{dw^{n}}\frac{w^{\mu
}e^{wz}}{(e^{w}-1)^{\mu }}\left|_{w=0}\right.=z^{n}+\sum_{m=1}^{n}{n
\choose m}z^{n-m} \sum_{j=0}^{m}{m \choose j}(\mu )_{m-j}\nonumber \\
&amp;&amp;\quad \times\sum_{k=0}^{j} {j \brace k}(-\mu )_{k}\sum_{l=0}^{k} {k
\choose l}
	 {2m-j \brace m-l}\frac{(m-l)!}{(2m-j)!}.\label{gB}
\end{eqnarray}} Comparing it with the formula given by Srivastava and
Todorov  \cite[Eq.~(3), p.\ 510]{ST}, we see that formula (\ref{gB})
does not involve any hypergeometric function. 

In particular, for $\mu
=1$ we obtain the common Bernoulli polynomials
{\setlength\arraycolsep{2pt} \begin{eqnarray*} \hspace{-5mm}
B_{n}(z)&amp;=&amp;z^{n}-\frac{1}{2}z^{n-1}+\sum_{m=2}^{n}{n \choose
m}z^{n-m}\\ &amp;&amp;\times\left(m\sum_{k=1}^{m-1} {m-1 \brace
k}(-1)^{k}k!\sum_{l=0}^{k} {k \choose l}
	 {m+1 \brace m-l}\frac{(m-l)!}{(m+1)!}\right.\\
&amp;&amp;\left.\quad +\sum_{k=1}^{m} {m \brace k}(-1)^{k}k!\sum_{l=0}^{k} {k
\choose l}
	 {m \brace m-l}\frac{(m-l)!}{m!}\right)
\end{eqnarray*}} and putting $z=0$, the common Bernoulli numbers
{\setlength\arraycolsep{2pt} \begin{eqnarray} \hspace{-5mm}   B_{n}&amp;=&amp;
n\sum_{k=1}^{n-1} {n-1 \brace k}(-1)^{k}k!\sum_{l=0}^{k} {k \choose l}
	 {n+1 \brace n-l}\frac{(n-l)!}{(n+1)!} \nonumber \\
&amp;&amp;+\sum_{k=1}^{n} {n \brace k}(-1)^{k}k!\sum_{l=0}^{k} {k \choose l}
	 {n \brace n-l}\frac{(n-l)!}{n!}.\label{bn}
\end{eqnarray}} Applying, to the expression (\ref{bn}), the two
identities \[\sum_{l=0}^{k}{k \choose l} {n+1 \brace
n-l}(n-l)!=\sum_{j=0}^{n-k}{n-k \choose j}(-1)^{j}(n-j)^{n+1}, \]
\[\sum_{l=0}^{k} {k \choose l} {n \brace
n-l}(n-l)!=\sum_{j=0}^{n-k}{n-k \choose j}(-1)^{j}(n-j)^{n}, \] and
then the next two \[\sum_{k=1}^{n-1}\! {n\!-\!1 \brace
k}(-1)^{k}k!{n\!-\!k \choose j}=(-1)^{j+n+1}\sum_{k=0}^{j}\!{n\!+\!1
\choose k}(-1)^{k}(j\!+\!1\!-\!k)^{n-1}, \]
{\setlength\arraycolsep{2pt} \begin{eqnarray*}
	\sum_{k=1}^{n} {n \brace k}(-1)^{k}k!{n-k \choose
	j}&amp;=&amp;(-1)^{j+n}\sum_{k=0}^{j}{n+1 \choose
	k}(-1)^{k}(j+1-k)^{n}\nonumber \\ &amp;=&amp;(-1)^{j+n}\left&lt; {n \atop
	j}\right&gt;, \end{eqnarray*}}
where $\left&lt; {n \atop j}\right&gt;$
are Eulerian numbers (see \cite{GKP}, and 
\seqnum{A008292} in \cite{Sl}),
we obtain the following formula for the $n$th Bernoulli number
\begin{eqnarray*} &amp;&amp;
B_{n}=\frac{(-1)^{n}}{(n+1)!}\left[n\sum_{j=0}^{n-1}(n-j)^{n+1}\sum_{k=0}^{j}(-1)^{k+1}{n+1
\choose k}(j+1-k)^{n-1}\right.\\ &amp;&amp; \hspace{70pt}\left.
+(n+1)\sum_{j=0}^{n-1}(n-j)^{n}\left&lt; {n \atop j}\right&gt;\right].
\end{eqnarray*} \section{Bell polynomials} The Bell polynomials
$B_{n}(z)$ can be defined by the generating function (see Bell
\cite{B}) \begin{equation}\label{Bg}
	\sum_{n=0}^{\infty }\frac{B_{n}(z)}{n!}w^{n}=e^{(e^{w}-1)z}.
\end{equation} Using (\ref{deriv2}) we compute the $n$th derivative of
the right hand side of (\ref{Bg}) \begin{equation}\label{Bg2}
	\frac{d^{n}}{dw^{n}}e^{(e^{w}-1)z}=\sum_{k=1}^{n} {n \brace
	k}e^{(e^{w}-1)z}e^{kw}z^{k}.  \end{equation} The well known
explicit formula for $B_{n}(z)$
	\[B_{n}(z)=\sum_{k=1}^{n} {n \brace k}z^{k}, \] follows
immediately from (\ref{Bg2}) by putting $w=0$.

\section{Acknowledgements} I would like to thank M. Skwarczy\'nski, J.
A. Rempa{\l}a and K. Jezuita for valuable discussions during
preparation of the paper and the anonymous referee for his/her helpful
comments.

\begin{thebibliography}{13}

\bibitem{B} E. T. Bell,
Exponential polynomials, \emph{Ann. Math.} \textbf{35} (1934),
258--277.  

\bibitem{C} L. Comtet, \emph{Advanced Combinatorics}, D.
Reidel Publishing Co., 1974.  

\bibitem{G} H. W. Gould, Explicit formula
for Bernoulli numbers, \emph{Amer. Math. Monthly} \textbf{78} (1972),
44--51.  

\bibitem{GKP} R. L. Graham, D. E. Knuth  and O. Patashnik,
\emph{ Concrete Mathematics. A Foundation for Computer Science},
Addison-Wesley Company, Inc., 1994.  

\bibitem{H} F. T. Howard, A
theorem relating Potential and Bell polynomials, \emph{Discrete
Math.} \textbf{39} (1982), 129--143.  

\bibitem{J} C. Jordan,
\emph{Calculus of Finite Differences}, Chelsea P. Company, 1950.

\bibitem{Q} Qiu-Ming Luo, Apostol-Euler polynomials of higher order and
Gaussian hypergeometric functions, \emph{Taiwanese J. 
Math.} {\bf 10} (2006), 917--925.

\bibitem{N} N. E.
N\"orlund, \emph{Vorlesungen \"uber Differenzrechnung}, Springer,
1924.  

\bibitem{RSP} R. S. Peregrino, The Lucas congruence for Stirling
numbers of the second kind, \emph{Acta Arith.} \textbf{94} (2000),
41--52.  

\bibitem{Sl} N. J. A. Sloane, The On-Line Encyclopedia of
Integer Sequences, 2009.  {\tt
http://www.research.att.com/\char'176njas/sequences/}.  

\bibitem{ST} H.
M. Srivastava, P. G. Todorov, An explicit formula for the generalized
Bernoulli polynomials, \emph{J. Math. Anal. Appl.}, \textbf{130}
(1988), 509--513.  

\bibitem{T} P. G. Todorov, Explicit and recurrence
formulas for generalized Euler numbers, \emph{Funct. Approx. Comment.
Math.} \textbf{22} (1993), 13--17 (1994).

\end{thebibliography}
\bigskip \hrule \bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B73; Secondary 11B68.

\noindent \emph{Keywords: } Stirling 
numbers, addition formulas, generalized Euler polynomials, generalized Bernoulli polynomials.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences 
\seqnum{A008275}, \seqnum{A008277} and
\seqnum{A008292}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received May 8 2009;
revised version received  November 12 2009.
Published in {\it Journal of Integer Sequences}, November 16 2009.

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