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\begin{center}
\vskip 1cm{\LARGE\bf Algorithms for Bernoulli and \\
\vskip .1in Related Polynomials}

\vskip 1cm \large
Ayhan Dil, Veli Kurt and Mehmet Cenkci\\
Department of Mathematics \\
Akdeniz University\\
Antalya, 07058 \\
Turkey\\
\href{mailto:adil@akdeniz.edu.tr}{\tt adil@akdeniz.edu.tr}\\
\href{mailto:vkurt@akdeniz.edu.tr}{\tt vkurt@akdeniz.edu.tr}\\
\href{mailto:cenkci@akdeniz.edu.tr}{\tt cenkci@akdeniz.edu.tr}
\end{center}

\vskip .3in

\begin{abstract}
We investigate some algorithms that produce
Bernoulli, Euler and Genocchi polynomials. We also give closed
formulas for Bernoulli, Euler and Genocchi polynomials in terms of
weighted Stirling numbers of the second kind, which are extensions
of known formulas for Bernoulli, Euler and Genocchi numbers
involving Stirling numbers of the second kind.
\end{abstract}


\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}{Proposition}[section]
\newtheorem{corollary}{Corollary}[section]
\newtheorem{lemma}{Lemma}[section]
\newtheorem{definition}{Definition}[section]

\section{Introduction and Primary Concepts}


Bernoulli numbers $B_{n}$, $n\in \mathbb{Z}$,
$n\geqslant 0$, originally arise in the study of finite sums of a
given power of consecutive
integers. They are given by $B_{0}=1$, $B_{1}=-1/2$, $B_{2}=1/6$, $%
B_{4}=-1/30,\ldots $, with $B_{n}=0$ for odd $n&gt;1$, and%
\begin{equation*}
B_{n}=-\frac{1}{n+1}\sum_{m=0}^{n-1}\binom{n+1}{m}B_{m}
\end{equation*}%
\noindent for all even $n\geqslant 2$. In the symbolic notation, Bernoulli
numbers are given recursively by%
\begin{equation*}
\left( B+1\right) ^{n}-B_{n}=\left\{
\begin{array}{ll}
1, &amp; \text{if }n=1; \\
0, &amp; \text{if }n&gt;1.%
\end{array}%
\right.
\end{equation*}%
\noindent with the usual convention about replacing $B^{j}$ by $B_{j}$ after
expansion. The Bernoulli polynomials $B_{n}\left( x\right) $, $n\in \mathbb{Z%
}$, $n\geqslant 0$, can be expressed in the form%
\begin{equation}
B_{n}\left( x\right) =\left( B+x\right) ^{n}=\sum_{m=0}^{n}\binom{n}{m}%
B_{m}x^{n-m}.  \label{EK2}
\end{equation}

The generating functions of Bernoulli numbers and polynomials are
respectively given by%
\begin{equation*}
\frac{t}{e^{t}-1}=\sum_{n=0}^{\infty }B_{n}\frac{t^{n}}{n!},
\end{equation*}%
\noindent and%
\begin{equation*}
\frac{te^{xt}}{e^{t}-1}=\sum_{n=0}^{\infty }B_{n}\left( x\right) \frac{t^{n}%
}{n!},
\end{equation*}%
\noindent for $|t|&lt;2\pi $ (\cite{Apostol}).  The Euler polynomials $%
E_{n}\left( x\right) $, $n\in \mathbb{Z}$, $n\geqslant 0$, may be defined by
the generating function \cite{N1,N}
\begin{equation}
\frac{2e^{xt}}{e^{t}+1}=\sum_{n=0}^{\infty }E_{n}\left( x\right) \frac{t^{n}%
}{n!},  \label{2.b}
\end{equation}%
\noindent for $|t|&lt;\pi $. The Euler numbers $E_{n}$ are defined by%
\begin{equation*}
E_{n}=2^{n}E_{n}\left( \frac{1}{2}\right) \text{, }n\geqslant 0.
\end{equation*}%
\noindent The Genocchi numbers and polynomials, $G_{n}$ and $G_{n}\left(
x\right) $, $n\in \mathbb{Z}$, $n\geqslant 0$, are defined respectively 
as follows \cite[p.49]{Com}:
\begin{equation}
\frac{2t}{e^{t}+1}=\sum_{n=0}^{\infty }G_{n}\frac{t^{n}}{n!},  \label{3.a}
\end{equation}%
\noindent and%
\begin{equation*}
\frac{2te^{xt}}{e^{t}+1}=\sum_{n=0}^{\infty }G_{n}\left( x\right) \frac{t^{n}%
}{n!},
\end{equation*}%
\noindent for $|t|&lt;\pi $, which have several combinatorial interpretations
in terms of certain surjective maps of finite sets \cite{D1,DV,DuR}. The 
well-known identity
\begin{equation*}
G_{n}=2\left( 1-2^{n}\right) B_{n}\text{, }n\geqslant 0,
\end{equation*}%
\noindent shows the relation between Genocchi and Bernoulli numbers. From (%
\ref{2.b}) and (\ref{3.a}), it is easy to see that%
\begin{equation*}
G_{n}=nE_{n-1}\left( 0\right) \text{, }n\geqslant 0,
\end{equation*}

\noindent with $E_{-1}:=0$.

Several arithmetical properties of Bernoulli, Euler and Genocchi polynomials
can be obtained from the generating functions. We list here three of them,
which will be used in the next section.%
\begin{eqnarray}
B_{n}\left( 1+x\right)  &amp;=&amp;B_{n}\left( x\right) +nx^{n-1}\text{, }n\geqslant
0,  \label{4} \\
E_{n}\left( 1+x\right)  &amp;=&amp;-E_{n}\left( x\right) +2x^{n}\text{, }n\geqslant
0,  \label{5} \\
G_{n}\left( 1+x\right)  &amp;=&amp;-G_{n}\left( x\right) +2nx^{n-1}\text{, }%
n\geqslant 0.  \label{6}
\end{eqnarray}

Bernoulli, Euler and Genocchi numbers can be directly computed from
generating functions by dividing numerator to denominator (after expanding $%
e^{t}$) and comparing the coefficients on either sides. Besides this method,
many recurrence formulas for Bernoulli, Euler and Genocchi numbers are
obtained \cite{CCK,CH,Com,DR,H1,J,K1,M}.
Alternative methods for computing these numbers are
developed as well. In this paper, we concern two of them, the Euler-Seidel
matrices and the Akiyama-Tanigawa algorithm.

\begin{definition}
\label{def1}(\cite{D2}) Let $a_{0}$, $a_{1}$, $a_{2}$,$\ldots $,$a_{n}$,$%
\ldots $ be an initial sequence. The Euler-Seidel matrix corresponding to
the sequence $\left\{ a_{n}\right\} $ is determined by the sequence $\left\{
a_{n}^{k}\right\} $, whose elements are recursively given by%
\begin{eqnarray}
a_{n}^{0} &amp;=&amp;a_{n}\text{ \ \ \ \ }\left( n\geqslant 0\right) ,  \notag \\
a_{n}^{k} &amp;=&amp;a_{n}^{k-1}+a_{n+1}^{k-1}\text{ \ \ \ \ }\left( n\geqslant
0,k\geqslant 1\right) .  \label{7}
\end{eqnarray}
\end{definition}

\noindent The sequence $\left\{ a_{n}^{0}\right\} $ is the first row and the
sequence $\left\{ a_{0}^{k}\right\} $ is the first column of the matrix.
From the recurrence relation (\ref{7}), it can be seen that%
\begin{equation}
a_{n}^{k}=\sum_{i=0}^{k}\binom{k}{i}a_{n+i}^{0}\text{, }\left( n\geqslant
0,k\geqslant 1\right) .  \label{8}
\end{equation}%
\noindent The first row and column can be determined from (\ref{8}) as
follows:%
\begin{eqnarray}
a_{0}^{n} &amp;=&amp;\sum_{i=0}^{n}\binom{n}{i}a_{i}^{0}\text{, }n\geqslant 0,
\notag \\
a_{n}^{0} &amp;=&amp;\sum_{i=0}^{n}\binom{n}{i}\left( -1\right) ^{n-i}a_{0}^{i}\text{%
, }n\geqslant 0.  \label{9}
\end{eqnarray}%
\noindent Thus, as the first row is given, the first column can be found and
vice versa. Next two propositions show the relationship between the
generating functions of $a_{n}^{0}$ and $a_{0}^{n}$.

\begin{proposition}
\label{prop1}Let%
\begin{equation*}
a\left( t\right) =\sum_{n=0}^{\infty }a_{n}^{0}t^{n}
\end{equation*}%
\noindent be the generating function of the initial sequence $\left\{
a_{n}^{0}\right\} $. Then the generating function of the sequence $\left\{
a_{0}^{n}\right\} $ is given by%
\begin{equation*}
\overline{a}\left( t\right) =\sum_{n=0}^{\infty }a_{0}^{n}t^{n}=\frac{1}{1-t}%
a\left( \frac{t}{1-t}\right) .
\end{equation*}
\end{proposition}

\noindent The proof of this proposition is given by Euler \cite{E}.

\begin{proposition}
\label{prop2}Let%
\begin{equation*}
A\left( t\right) =\sum_{n=0}^{\infty }a_{n}^{0}\frac{t^{n}}{n!}
\end{equation*}%
\noindent be the exponential generating function of the initial sequence $%
\left\{ a_{n}^{0}\right\} $. Then the exponential generating function of the
sequence $\left\{ a_{0}^{n}\right\} $ is given by%
\begin{equation*}
\overline{A}\left( t\right) =\sum_{n=0}^{\infty }a_{0}^{n}\frac{t^{n}}{n!}%
=e^{t}A\left( t\right) .
\end{equation*}
\end{proposition}

\noindent The proof of this proposition is given by Seidel \cite{S}.

Of notable interest of Euler-Seidel matrices is the examples for Bernoulli
and related numbers. Let%
\begin{equation*}
A\left( t\right) =\frac{t}{e^{t}-1}.
\end{equation*}%
\noindent Then%
\begin{equation*}
\overline{A}\left( t\right) =\frac{te^{t}}{e^{t}-1}=t+\frac{t}{e^{t}-1},
\end{equation*}%
\noindent and thus%
\begin{equation*}
\overline{A}\left( t\right) =t+A\left( t\right) .
\end{equation*}%
\noindent Therefore,%
\begin{equation*}
a_{0}^{0}=B_{0}=1\text{, }a_{0}^{1}=\frac{1}{2}\text{, and }%
a_{n}^{0}=a_{0}^{n}\text{ for }n\geqslant 2,
\end{equation*}%
\noindent and the corresponding Euler-Seidel matrix is%
\begin{equation*}
\left(
\begin{array}{cccccc}
1 &amp; -\frac{1}{2} &amp; \frac{1}{6} &amp; 0 &amp; -\frac{1}{30} &amp; \cdots  \\
\frac{1}{2} &amp; -\frac{1}{3} &amp; \frac{1}{6} &amp; -\frac{1}{30} &amp; \cdots  &amp;  \\
\frac{1}{6} &amp; -\frac{1}{6} &amp; \frac{2}{15} &amp; \cdots  &amp;  &amp;  \\
0 &amp; -\frac{1}{30} &amp; \cdots  &amp;  &amp;  &amp;  \\
-\frac{1}{30} &amp; \cdots  &amp;  &amp;  &amp;  &amp;  \\
\cdots  &amp;  &amp;  &amp;  &amp;  &amp;
\end{array}%
\right)
\end{equation*}

For%
\begin{equation*}
A\left( t\right) =\frac{2t}{e^{t}+1},
\end{equation*}%
\noindent the initial sequence is%
\begin{equation*}
a_{0}^{0}=0\text{, }a_{1}^{0}=1\text{, }a_{2n+1}^{0}=0\text{ and }%
a_{2n}^{0}=G_{2n}\text{ for }n\geqslant 1.
\end{equation*}%
\noindent Since%
\begin{equation*}
\overline{A}\left( t\right) =\frac{2te^{t}}{e^{t}+1}=2t-\frac{2t}{e^{t}+1}%
=2t-A\left( t\right) ,
\end{equation*}%
\noindent it is seen that%
\begin{equation*}
a_{0}^{0}=0\text{, }a_{0}^{1}=1\text{, }a_{0}^{2n+1}=0\text{ and }%
a_{0}^{2n}=-G_{2n}\text{ for }n\geqslant 1.
\end{equation*}%
\noindent Thus, the corresponding Euler-Seidel matrix is%
\begin{equation*}
\left(
\begin{array}{cccccccccc}
0 &amp; 1 &amp; -1 &amp; 0 &amp; 1 &amp; 0 &amp; -3 &amp; 0 &amp; 17 &amp; \cdots  \\
1 &amp; 0 &amp; -1 &amp; 1 &amp; 1 &amp; -3 &amp; -3 &amp; 17 &amp; \cdots  &amp;  \\
1 &amp; -1 &amp; 0 &amp; 2 &amp; -2 &amp; -6 &amp; 14 &amp; \cdots  &amp;  &amp;  \\
0 &amp; -1 &amp; 2 &amp; 0 &amp; -8 &amp; 8 &amp; \cdots  &amp;  &amp;  &amp;  \\
-1 &amp; 1 &amp; 2 &amp; -8 &amp; 0 &amp; \cdots  &amp;  &amp;  &amp;  &amp;  \\
0 &amp; 3 &amp; -6 &amp; -8 &amp; \cdots  &amp;  &amp;  &amp;  &amp;  &amp;  \\
3 &amp; -3 &amp; -14 &amp; \cdots  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  \\
0 &amp; -17 &amp; \cdots  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  \\
-17 &amp; \cdots  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  \\
\cdots  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;
\end{array}%
\right)
\end{equation*}

In their study of values at nonpositive integer arguments of multiple zeta
functions, Akiyama and Tanigawa \cite{AT} found an algorithm for computing
Bernoulli numbers in a manner similar to Pascal's triangle for binomial
coefficients. The algorithm is as follows: Starting with the sequence $%
\left\{ 1/n\right\} $, $n\geqslant 1$, as the $0$-th row, the $m$-th number
in the $\left( n+1\right) $-st row $a_{m}^{n+1}$ is determined recursively by%
\begin{equation*}
a_{m}^{n+1}=\left( m+1\right) \left( a_{m}^{n}-a_{m+1}^{n}\right) ,
\end{equation*}%
\noindent for $m=0,1,2,\ldots $ and $n=0,1,2,\ldots $. Then, $a_{0}^{n}$ of
each row is the $n$-th Bernoulli number $B_{n}$ with $B_{1}=1/2$. Kaneko 
\cite{K2} reformulated the Akiyama-Tanigawa algorithm as follows:

\begin{proposition}
\label{prop3}Given an initial sequence $a_{m}^{0}$ with $m=0,1,2,\ldots $
define sequence $a_{m}^{n}$ for $n\geqslant 1$ recursively by%
\begin{equation*}
a_{m}^{n}=\left( m+1\right) \left( a_{m}^{n-1}-a_{m+1}^{n-1}\right) .
\end{equation*}%
\noindent Then the leading elements are given by%
\begin{equation*}
a_{0}^{n}=\sum_{m=0}^{n}\left( -1\right) ^{m}m!S\left( n+1,m+1\right)
a_{m}^{0}.
\end{equation*}
\end{proposition}

\noindent Here, $S\left( n,m\right) $ denotes the Stirling numbers of the
second kind which are defined as follows  \cite[Chap.\ 5]{Com}:
\begin{equation}
\frac{\left( e^{t}-1\right) ^{m}}{m!}=\sum_{n=m}^{\infty }S\left( n,m\right)
\frac{t^{n}}{n!}.  \label{Ek3}
\end{equation}%
\noindent For $1\leqslant m\leqslant n$, $S\left( n,m\right) &gt;0$ and for $%
1\leqslant n&lt;m$, $S\left( n,m\right) =0$. The Stirling numbers of the second
kind $S\left( n,m\right) $ satisfy the recurrence relation%
\begin{equation}
S\left( n,m\right) =S\left( n-1,m-1\right) +mS\left( n-1,m\right)
\label{Ek1.1}
\end{equation}

\noindent for $n,m\geqslant 1$ with $S\left( n,0\right) =S\left( 0,m\right)
=0$, except $S\left( 0,0\right) =1$.

If the initial sequence is $a_{m}^{0}=1/\left( m+1\right) $, $m\geqslant 0$,
in Proposition \ref{prop3}, then the leading elements are Bernoulli numbers
with $B_{1}=1/2$. By replacing the initial sequence $a_{m}^{0}=1/\left(
m+1\right) $ by $a_{m}^{0}=1/\left( m+1\right) ^{k}$, $m\geqslant 0$, and
applying same algorithm, Kaneko also derived the resulting sequence as
poly-Bernoulli numbers \cite{K2}.

Chen \cite{KWC} changed the recursive step in Proposition \ref{prop3} to%
\begin{equation*}
a_{m}^{n}=ma_{m}^{n-1}-\left( m+1\right) a_{m+1}^{n-1}\text{ \ \ \ \ }\left(
n\geqslant 1,m\geqslant 0\right) ,
\end{equation*}%
\noindent and proved the following:

\begin{proposition}
\label{prop4}Given an initial sequence $a_{m}^{0}$ with $m=0,1,2,\ldots $
define sequence $a_{m}^{n}$ for $n\geqslant 1$ recursively by%
\begin{equation*}
a_{m}^{n}=ma_{m}^{n-1}-\left( m+1\right) a_{m+1}^{n-1}.
\end{equation*}%
\noindent Then%
\begin{equation*}
a_{0}^{n}=\sum_{m=0}^{n}\left( -1\right) ^{m}m!S\left( n,m\right) a_{m}^{0}.
\end{equation*}
\end{proposition}

\noindent Given initial sequences $1/\left( m+1\right) $, $m\geqslant 0$, $%
1/2^{m}$, $m\geqslant 0$, and $\left( -1\right) ^{\left[ m/4\right] }2^{-%
\left[ m/2\right] }\left( 1-\delta _{4,m+1}\right) $, $m\geqslant 0$, Chen
obtained the leading elements respectively as Bernoulli, Euler and tangent
numbers. Here, $\left[ x\right] $ is the greatest integer $\leqslant x$ and $%
\delta _{4,i}=1$ if $4\mid i$ and $\delta _{4,i}=0$ otherwise. Tangent
numbers are defined as follows (cf.\ \cite[p.\ 35]{N}):
\begin{equation*}
\text{tg}t=\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n+1}T_{2n+1}}{\left(
2n+1\right) !}t^{2n+1},
\end{equation*}
with $T_{0}=1$. At this point, we note that by choosing the
initial sequence $a_{m}^{0}=1/2^{m}$, $m\geqslant 0$, we obtain the leading
elements $a_{0}^{n-1}$ as $G_{n}/n$ for $n\geqslant 1$, which is not
obtained in \cite{KWC}. For other studies on the Akiyama-Tanigawa algorithm,
see \cite{I,MSV}.

In this paper, we give the Euler-Seidel matrices and the Akiyama-Tanigawa
algorithms for Bernoulli, Euler and Genocchi polynomials. These matrices and
algorithms are polynomial extensions of the corresponding matrices and
algorithms for Bernoulli, Euler and Genocchi numbers. In particular, the
Akiyama-Tanigawa algorithms for these polynomials lead new closed formulas
for Bernoulli, Euler and Genocchi polynomials in terms of weighted Stirling
numbers of the second kind.

\section{Euler-Seidel Matrices for Bernoulli, Euler and Genocchi Polynomials}


We start with the polynomial extension of
Definition \ref{def1}.

\begin{definition}
\label{def2}Let $a_{0}\left( x\right) $, $a_{1}\left( x\right) $, $%
a_{2}\left( x\right) $,$\ldots $,$a_{n}\left( x\right) $,$\ldots $ be
initial sequence of polynomials in $x$. The Euler-Seidel matrix
corresponding to the sequence $\left\{ a_{n}\left( x\right) \right\} $ is
determined by the sequence $\left\{ a_{n}^{k}\left( x\right) \right\} $ for $%
n\geqslant 0$ and $k\geqslant 1$, whose elements are recursively given by%
\begin{eqnarray*}
a_{n}^{0}\left( x\right) &amp;=&amp;a_{n}\left( x\right) , \\
a_{n}^{k}\left( x\right) &amp;=&amp;a_{n}^{k-1}\left( x\right) +a_{n+1}^{k-1}\left(
x\right) .
\end{eqnarray*}
\end{definition}

\noindent From the recurrence relation above, we have%
\begin{eqnarray}
a_{n}^{k}\left( x\right)  &amp;=&amp;\sum_{i=0}^{k}\binom{k}{i}a_{n+i}^{0}\left(
x\right) \text{, }n\geqslant 0,  \notag \\
a_{0}^{n}\left( x\right)  &amp;=&amp;\sum_{i=0}^{n}\binom{n}{i}a_{i}^{0}\left(
x\right) \text{, }n\geqslant 0,  \label{EK5} \\
a_{n}^{0}\left( x\right)  &amp;=&amp;\sum_{i=0}^{n}\binom{n}{i}\left( -1\right)
^{n-i}a_{0}^{i}\left( x\right) \text{, }n\geqslant 0.  \notag
\end{eqnarray}%
\noindent Let $A\left( x,t\right) $ and $\overline{A}\left( x,t\right) $ be
the exponential generating functions of $a_{n}^{0}\left( x\right) $ and $%
a_{0}^{n}\left( x\right) $, respectively. Then, we have%
\begin{equation*}
\overline{A}\left( x,t\right) =e^{t}A\left( x,t\right) .
\end{equation*}%
\noindent Note that for $x=0$, the above equalities reduce to Dumont's.

Now, we construct the Euler-Seidel matrices corresponding to Bernoulli,
Euler and Genocchi polynomials.

Let $a_{n}^{0}\left( x\right) =B_{n}\left( x\right) $, $n\geqslant 0$. Then%
\begin{equation*}
A\left( x,t\right) =\sum_{n=0}^{\infty }B_{n}\left( x\right) \frac{t^{n}}{n!}%
=\frac{te^{xt}}{e^{t}-1},
\end{equation*}%
\noindent and%
\begin{equation}
\overline{A}\left( x,t\right) =e^{t}A\left( x,t\right) =\frac{te^{\left(
x+1\right) t}}{e^{t}-1}=\sum_{n=0}^{\infty }B_{n}\left( x+1\right) \frac{%
t^{n}}{n!}.  \label{EK1}
\end{equation}%
\noindent Using (\ref{4}), we obtain%
\begin{equation*}
\overline{A}\left( x,t\right) =A\left( x,t\right) +te^{xt},
\end{equation*}%
\noindent and%
\begin{equation*}
a_{0}^{n}\left( x\right) =B_{n}\left( x\right) +nx^{n-1}\text{, }n\geqslant
0.
\end{equation*}%
\noindent Therefore, given Bernoulli polynomials as the first row, then the
first column of the corresponding Euler-Seidel matrix is again Bernoulli
polynomials except the opposite sign for the coefficient of the second
highest power. This is because%
\begin{eqnarray*}
a_{0}^{n}\left( x\right)  &amp;=&amp;B_{n}\left( x\right) +nx^{n-1} \\
&amp;=&amp;\sum_{m=0}^{n}\binom{n}{m}B_{m}x^{n-m}+nx^{n-1} \\
&amp;=&amp;B_{0}x^{n}+B_{1}x^{n-1}+nx^{n-1}+\sum_{m=2}^{n}\binom{n}{m}B_{m}x^{n-m} \\
&amp;=&amp;B_{0}x^{n}-\frac{n}{2}x^{n-1}+nx^{n-1}+\sum_{m=2}^{n}\binom{n}{m}%
B_{m}x^{n-m} \\
&amp;=&amp;B_{0}x^{n}+\frac{n}{2}x^{n-1}+\sum_{m=2}^{n}\binom{n}{m}B_{m}x^{n-m},
\end{eqnarray*}%
\noindent by using (\ref{EK2}) and $B_{1}=-1/2$. The Euler-Seidel matrix
corresponding to Bernoulli polynomials is as follows:%
\begin{equation*}
\left(
\begin{array}{lllll}
1 &amp; x-\frac{1}{2} &amp; x^{2}-x+\frac{1}{6} &amp; x^{3}-\frac{3}{2}x^{2}+\frac{1}{2}x
&amp; \cdots  \\
x+\frac{1}{2} &amp; x^{2}-\frac{1}{3} &amp; x^{3}-\frac{1}{2}x^{2}-\frac{1}{2}x+%
\frac{1}{6} &amp; \cdots  &amp;  \\
x^{2}+x+\frac{1}{6} &amp; x^{3}+\frac{1}{2}x^{2}-\frac{1}{2}x-\frac{1}{6} &amp;
\cdots  &amp;  &amp;  \\
x^{3}+\frac{3}{2}x^{2}+\frac{1}{2}x &amp; \cdots  &amp;  &amp;  &amp;  \\
\cdots  &amp;  &amp;  &amp;  &amp;
\end{array}%
\right)
\end{equation*}

Now let $a_{n}^{0}\left( x\right) =E_{n}\left( x\right) $, $n\geqslant 0$.
Then%
\begin{equation*}
A\left( x,t\right) =\sum_{n=0}^{\infty }E_{n}\left( x\right) \frac{t^{n}}{n!}%
=\frac{2e^{xt}}{e^{t}+1},
\end{equation*}%
\noindent and%
\begin{equation}
\overline{A}\left( x,t\right) =e^{t}A\left( x,t\right) =\frac{2e^{\left(
x+1\right) t}}{e^{t}+1}=\sum_{n=0}^{\infty }E_{n}\left( x+1\right) \frac{%
t^{n}}{n!}.  \label{EK3}
\end{equation}%
\noindent Using (\ref{5}), we get%
\begin{equation*}
\overline{A}\left( x,t\right) =-A\left( x,t\right) +2e^{xt},
\end{equation*}%
\noindent and%
\begin{equation*}
a_{0}^{n}\left( x\right) =-E_{n}\left( x\right) +2x^{n}\text{, }n\geqslant 0.
\end{equation*}%
\noindent Thus, if Euler polynomials are taken as the first row, then the
first column of the corresponding Euler-Seidel matrix is negative Euler
polynomials except the coefficient of the highest power. The matrix is%
\begin{equation*}
\left(
\begin{array}{lllll}
1 &amp; x-\frac{1}{2} &amp; x^{2}-x &amp; x^{3}-\frac{3}{2}x^{2}+\frac{1}{4} &amp; \cdots
\\
x+\frac{1}{2} &amp; x^{2}-\frac{1}{2} &amp; x^{3}-\frac{1}{2}x^{2}-x+\frac{1}{4} &amp;
\cdots  &amp;  \\
x^{2}+x &amp; x^{3}+\frac{1}{2}x^{2}-x-\frac{1}{4} &amp; \cdots  &amp;  &amp;  \\
x^{3}+\frac{3}{2}x^{2}-\frac{1}{4} &amp; \cdots  &amp;  &amp;  &amp;  \\
\cdots  &amp;  &amp;  &amp;  &amp;
\end{array}%
\right)
\end{equation*}

Finally, let $a_{n}^{0}\left( x\right) =G_{n}\left( x\right) $, $n\geqslant 0
$. Then%
\begin{equation*}
A\left( x,t\right) =\sum_{n=0}^{\infty }G_{n}\left( x\right) \frac{t^{n}}{n!}%
=\frac{2te^{xt}}{e^{t}+1},
\end{equation*}%
\noindent and%
\begin{equation}
\overline{A}\left( x,t\right) =e^{t}A\left( x,t\right) =\frac{2te^{\left(
x+1\right) t}}{e^{t}+1}=\sum_{n=0}^{\infty }G_{n}\left( x+1\right) \frac{%
t^{n}}{n!}.  \label{EK4}
\end{equation}

\noindent Using (\ref{6}) yields%
\begin{equation*}
\overline{A}\left( x,t\right) =-A\left( x,t\right) +2te^{xt},
\end{equation*}%
\noindent and%
\begin{equation*}
a_{0}^{n}\left( x\right) =-G_{n}\left( x\right) +2nx^{n-1}\text{, }%
n\geqslant 0.
\end{equation*}%
\noindent The corresponding matrix is%
\begin{equation*}
\left(
\begin{array}{lllll}
0 &amp; 1 &amp; 2x-1 &amp; 3x^{2}-3x &amp; \cdots  \\
1 &amp; 2x &amp; 3x^{2}-x-1 &amp; \cdots  &amp;  \\
2x+1 &amp; 3x^{2}+x-1 &amp; \cdots  &amp;  &amp;  \\
3x^{2}+3x &amp; \cdots  &amp;  &amp;  &amp;  \\
\cdots  &amp;  &amp;  &amp;  &amp;
\end{array}%
\right)
\end{equation*}

We conclude this section by giving alternative proofs of the equations%
\begin{eqnarray*}
B_{n}\left( x+1\right)  &amp;=&amp;\sum_{m=0}^{n}\binom{n}{m}B_{m}\left( x\right)
\text{, }n\geqslant 0, \\
E_{n}\left( x+1\right)  &amp;=&amp;\sum_{m=0}^{n}\binom{n}{m}E_{m}\left( x\right)
\text{, }n\geqslant 0, \\
G_{n}\left( x+1\right)  &amp;=&amp;\sum_{m=0}^{n}\binom{n}{m}G_{m}\left( x\right)
\text{, }n\geqslant 0.
\end{eqnarray*}

\noindent From (\ref{EK1}), we see that%
\begin{equation*}
\sum_{n=0}^{\infty }a_{0}^{n}\left( x\right) \frac{t^{n}}{n!}=\overline{A}%
\left( x,t\right) =\sum_{n=0}^{\infty }B_{n}\left( x+1\right) \frac{t^{n}}{n!%
}.
\end{equation*}

\noindent Thus, by comparing the coefficients of power series on both sides,
we get $a_{0}^{n}\left( x\right) =B_{n}\left( x+1\right) $. By (\ref{EK5}),
we have%
\begin{equation*}
B_{n}\left( x+1\right) =a_{0}^{n}\left( x\right) =\sum_{m=0}^{n}\binom{n}{m}%
a_{m}^{0}\left( x\right) =\sum_{m=0}^{n}\binom{n}{m}B_{m}\left( x\right) .
\end{equation*}

\noindent Other two equalities can be proved in the same way by using
appropriate relations (\ref{EK3}) or (\ref{EK4}).

\section{The Akiyama-Tanigawa Algorithms for Bernoulli, Euler and Genocchi
Polynomials}


In this section, we derive the Akiyama-Tanigawa
algorithms for Bernoulli, Euler and Genocchi polynomials and give
closed formulas for these polynomials in terms of weighted
Stirling numbers of the second kind.

The weighted Stirling numbers of the second kind, $S\left( n,m,x\right) $,
are defined  as follows \cite{C1,C2}:
\begin{equation}
\frac{e^{xt}\left( e^{t}-1\right) ^{m}}{m!}=\sum_{n=m}^{\infty }S\left(
n,m,x\right) \frac{t^{n}}{n!}.  \label{Ek2}
\end{equation}
When $x=0$, $S\left( n,m,0\right) =S\left( n,m\right) $ are the
Stirling numbers of the second kind. For $n=m=0$, $S\left( 0,0,x\right) =1$,
for $1\leqslant m&lt;n$, $S\left( n,m,x\right) =0$ and for other values of $m$
and $n$, $S\left( n,m,x\right) $ are determined by the following recurrence
formula:%
\begin{equation}
S\left( n+1,m,x\right) =\left( m+x\right) S\left( n,m,x\right) +S\left(
n,m-1,x\right) .  \label{10}
\end{equation}
The relationship between $S\left( n,m,x\right) $ and $S\left(
n,m\right) $ can be given by the following equation:

\begin{lemma}
\label{le1}We have%
\begin{equation*}
S\left( n,m,x\right) =\sum_{k=0}^{n-m}\binom{n}{k}x^{k}S\left( n-k,m\right) .
\end{equation*}
\end{lemma}

\begin{proof}
By (\ref{Ek2}) and (\ref{Ek3}), we have%
\begin{eqnarray*}
\sum_{n=m}^{\infty }S\left( n,m,x\right) \frac{t^{n}}{n!} &amp;=&amp;\frac{%
e^{xt}\left( e^{t}-1\right) ^{m}}{m!}=e^{xt}\sum_{n=m}^{\infty }S\left(
n,m\right) \frac{t^{n}}{n!} \\
&amp;=&amp;\sum_{n=0}^{\infty }x^{n}\frac{t^{n}}{n!}\sum_{n=0}^{\infty }S\left(
n+m,m\right) \frac{t^{n+m}}{\left( n+m\right) !} \\
&amp;=&amp;\sum_{n=0}^{\infty }\left( \sum_{k=0}^{n}\frac{x^{k}}{k!}\frac{S\left(
n-k+m,m\right) }{\left( n-k+m\right) !}\right) t^{n+m}.
\end{eqnarray*}%
Comparing the coefficients of power series yields the result.
\end{proof}

Given an initial sequence $a_{m}^{0}$ with $m=0,1,2,\ldots $, let the
sequence $a_{m}^{n}$ for $n\geqslant 1$ be given recursively by%
\begin{equation}
a_{m}^{n}=ma_{m}^{n-1}-\left( m+1\right) a_{m+1}^{n-1}  \label{11}
\end{equation}
as in Proposition \ref{prop4}. Let $g_{n}\left( t\right) $ be the
generating function of $a_{m}^{n}$,%
\begin{equation}
g_{n}\left( t\right) =\sum_{m=0}^{\infty }a_{m}^{n}t^{m}.  \label{12}
\end{equation}%
We define $a_{m}^{n}\left( x\right) $ and $g_{n}\left( x,t\right) $
by means of%
\begin{eqnarray}
a_{m}^{n}\left( x\right)  &amp;=&amp;\left( x+a_{m}\right) ^{n}=\sum_{k=0}^{n}\binom{%
n}{k}x^{k}a_{m}^{n-k}\text{, }n\geqslant 0,  \label{13} \\
g_{n}\left( x,t\right)  &amp;=&amp;\left( x+g\left( t\right) \right)
^{n}=\sum_{k=0}^{n}\binom{n}{k}x^{k}g_{n-k}\left( t\right) \text{, }%
n\geqslant 0,  \label{14}
\end{eqnarray}
where we use the usual convention about replacing $\left(
a_{m}\right) ^{j}$ by $a_{m}^{j}$ and $\left( g\left( t\right) \right) ^{j}$
by $g_{j}\left( t\right) $ in the binomial expansions. Note that for $x=0$, (%
\ref{13}) reduces to (\ref{11}) and (\ref{14}) reduces to (\ref{12}).

\begin{proposition}
\label{prop5}Given an initial sequence $a_{0,m}$ with $m=0,1,2,\ldots $, let
the sequence $a_{m}^{n}$ for $n\geqslant 1$ be given as (\ref{11}) and $%
a_{m}^{n}\left( x\right) $ be given as (\ref{13}). Then%
\begin{equation}
a_{0}^{n}\left( x\right) =\sum_{m=0}^{n}\left( -1\right) ^{m}m!S\left(
n,m,x\right) a_{m}^{0},  \label{15}
\end{equation}
where $S\left( n,m,x\right) $ are the weighted Stirling numbers of
the second kind.
\end{proposition}

\begin{proof}
From (\ref{14}) and (\ref{12}), we have%
\begin{equation*}
g_{n}\left( x,t\right) =\sum_{k=0}^{n}\binom{n}{k}x^{k}g_{n-k}\left(
t\right) =\sum_{k=0}^{n}\binom{n}{k}x^{k}\sum_{m=0}^{\infty
}a_{m}^{n-k}t^{m}.
\end{equation*}
By the recurrence formula (\ref{11}), we get%
\begin{eqnarray*}
g_{n}\left( x,t\right) &amp;=&amp;\sum_{k=0}^{n}\binom{n}{k}x^{k}\sum_{m=0}^{\infty
}\left\{ ma_{m}^{n-k-1}-\left( m+1\right) a_{m+1}^{n-k-1}\right\} t^{m} \\
&amp;=&amp;\sum_{k=0}^{n}\binom{n}{k}x^{k}\sum_{m=0}^{\infty }\left( m+1\right)
a_{m+1}^{n-k-1}t^{m+1} \\
&amp;&amp;-\sum_{k=0}^{n}\binom{n}{k}x^{k}\left( m+1\right) a_{m+1}^{n-k-1}t^{m} \\
&amp;=&amp;\sum_{k=0}^{n}\binom{n}{k}x^{k}\left( t-1\right) \sum_{m=0}^{\infty
}\left( m+1\right) a_{m+1}^{n-k-1}t^{m} \\
&amp;=&amp;\sum_{k=0}^{n}\binom{n}{k}x^{k}\left( t-1\right) \frac{d}{dt}\left(
g_{n-k-1}\left( t\right) \right) \\
&amp;=&amp;\sum_{k=0}^{n}\binom{n}{k}x^{k}\left( \left( t-1\right) \frac{d}{dt}%
\right) ^{n-k}g_{0}\left( t\right) .
\end{eqnarray*}
Applying the formula (cf.\ \cite[p.\ 310]{GKP})
\begin{equation*}
\left( \left( t-1\right) \frac{d}{dt}\right) ^{n-k}=\sum_{m=0}^{n-k}S\left(
n-k,m\right) \left( t-1\right) ^{m}\left( \frac{d}{dt}\right) ^{m},
\end{equation*}
we have
\begin{equation*}
g_{n}\left( x,t\right) =\sum_{k=0}^{n}\binom{n}{k}x^{k}\sum_{m=0}^{n-k}S%
\left( n-k,m\right) \left( t-1\right) ^{m}\left( \frac{d}{dt}\right)
^{m}g_{0}\left( t\right) .
\end{equation*}
Setting $t=0$ and interchanging summations, we obtain
\begin{equation*}
a_{0}^{n}\left( x\right) =\sum_{m=0}^{n}\left( -1\right)
^{m}m!a_{m}^{0}\sum_{k=0}^{n-m}\binom{n}{k}x^{k}S\left( n-k,m\right) .
\end{equation*}%
Using Lemma \ref{le1}, the result follows.
\end{proof}

\begin{theorem}
\label{th1}Let $a_{m}^{0}=1/\left( m+1\right) $, $m\geqslant 0$, in (\ref{15}%
). Then the leading polynomials $a_{0}^{n}\left( x\right) $, $n\geqslant 0$,
are given by%
\begin{equation*}
a_{0}^{n}\left( x\right) =\sum_{m=0}^{n}\left( -1\right) ^{m}m!\frac{S\left(
n,m,x\right) }{m+1}=B_{n}\left( x\right) .
\end{equation*}
\end{theorem}

\begin{proof}
We have%
\begin{eqnarray*}
&amp;&amp;\hspace{-1.0in}\sum_{n=0}^{\infty }\left( \sum_{m=0}^{n}\left( -1\right)
^{m}m!\frac{S\left( n,m,x\right) }{m+1}\right) \frac{t^{n}}{n!} \\
&amp;=&amp;\sum_{m=0}^{\infty }\frac{\left( -1\right) ^{m}m!}{m+1}\sum_{n=m}^{\infty
}S\left( n,m,x\right) \frac{t^{n}}{n!} \\
&amp;=&amp;\sum_{m=0}^{\infty }\frac{\left( -1\right) ^{m}m!}{m+1}\frac{e^{xt}\left(
e^{t}-1\right) ^{m}}{m!} \\
&amp;=&amp;\frac{e^{xt}}{1-e^{t}}\sum_{m=1}^{\infty }\frac{\left( 1-e^{t}\right) ^{m}%
}{m} \\
&amp;=&amp;\frac{e^{xt}}{1-e^{t}}\left( -\text{log}\left( 1-\left( 1-e^{t}\right)
\right) \right) =\frac{te^{xt}}{e^{t}-1}=\sum_{n=0}^{\infty }B_{n}\left(
x\right) \frac{t^{n}}{n!}.
\end{eqnarray*}%
This proves the theorem.
\end{proof}

\begin{theorem}
\label{th2}Let $a_{m}^{0}=1/2^{m}$, $m\geqslant 0$, in (\ref{15}). Then the
leading polynomials $a_{0}^{n}\left( x\right) $, $n\geqslant 0$, are given by%
\begin{equation*}
a_{0}^{n}\left( x\right) =\sum_{m=0}^{n}\left( -1\right) ^{m}m!\frac{S\left(
n,m,x\right) }{2^{m}}=E_{n}\left( x\right) .
\end{equation*}
\end{theorem}

\begin{proof}
We have%
\begin{eqnarray*}
&amp;&amp;\hspace{-1.0in}\sum_{n=0}^{\infty }\left( \sum_{m=0}^{n}\left( -1\right)
^{m}m!\frac{S\left( n,m,x\right) }{2^{m}}\right) \frac{t^{n}}{n!} \\
&amp;=&amp;\sum_{m=0}^{\infty }\frac{\left( -1\right) ^{m}m!}{2^{m}}%
\sum_{n=m}^{\infty }S\left( n,m,x\right) \frac{t^{n}}{n!} \\
&amp;=&amp;\sum_{m=0}^{\infty }\frac{\left( -1\right) ^{m}m!}{2^{m}}\frac{%
e^{xt}\left( e^{t}-1\right) ^{m}}{m!} \\
&amp;=&amp;e^{xt}\sum_{m=0}^{\infty }\left( \frac{1-e^{t}}{2}\right) ^{m}=\frac{%
2e^{xt}}{e^{t}+1}=\sum_{n=0}^{\infty }E_{n}\left( x\right) \frac{t^{n}}{n!}.
\end{eqnarray*}%
Comparing the coefficients of power series yields the result.
\end{proof}

\begin{theorem}
\label{th3}Let $a_{m}^{0}=1/2^{m}$, $m\geqslant 0$, as in Theorem \ref{th2}.
Then the polynomials $a_{0}^{n}\left( x\right) $, $n\geqslant 0$, are given
by%
\begin{equation*}
\left( n+1\right) a_{0}^{n}\left( x\right) =\sum_{m=0}^{n}\left( -1\right)
^{m+1}\left( m+1\right) !\frac{S\left( n+1,m+1,x\right) }{2^{m+1}}%
=G_{n}\left( x\right) .
\end{equation*}
\end{theorem}

\begin{proof}
The proof follows from Theorem \ref{th2} and the relations $G_{n}\left(
x\right) =nE_{n-1}\left( x\right) $, $n\geqslant 0$, and $S\left(
n,0,x\right) =0$.
\end{proof}

\section{Acknowledgment}

The authors would like to thank the anonymous
referee for valuable comments.

This work was supported by Akdeniz University Scientific Research
Project Unit.

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\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}: Primary
11B68.\\

\noindent {\it Keywords}: Bernoulli polynomials, Euler
polynomials, Genocchi polynomials, weighted Stirling numbers of
the second kind, Euler-Seidel matrices, Akiyama-Tanigawa
algorithm.

\bigskip
\hrule
\bigskip


\noindent (Concerned with sequences 
\seqnum{A000110},
\seqnum{A000182},
\seqnum{A000364},
\seqnum{A001898},
\seqnum{A027641}, 
\seqnum{A027642}, and
\seqnum{A100616}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received September 18 2006;
revised version received  May 10 2007.
Published in {\it Journal of Integer Sequences}, May 10 2007.

\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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