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\begin{center}
\vskip 1cm{\LARGE\bf A Note on a Theorem of Guo, Mez\H{o}, and Qi}
\vskip 1cm \large Ant\^onio Francisco Neto\footnote{This work was
supported by Conselho Nacional de Desenvolvimento Cient\'ifico e
Tecnol\'ogico
(CNPq-Brazil) under grant 307617/2012-2.}\\
DEPRO, Escola de Minas\\
Campus Morro do Cruzeiro, UFOP\\
35400-000 Ouro Preto MG \\
Brazil \\
\href{mailto:antfrannet@gmail.com}{\tt antfrannet@gmail.com}\\
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\begin{abstract}
In a recent paper, Guo, Mez\H{o}, and Qi proved an identity
representing the Bernoulli polynomials at non-negative integer points
$m$ in terms of the $m$-Stirling numbers of the second kind. In this
note, using a new representation of the Bernoulli polynomials in the
context of the Zeon algebra, we give an alternative proof of the
aforementioned identity.
\end{abstract}

\section{Introduction}

In an interesting recent paper \cite{GMQ},
Guo, Mez\H{o}, and Qi found the following identity
\begin{equation}\label{Bm-S2nd}
B_n(m)=\sum_{l=0}^n(-1)^l\frac{l!}{l+1}S_m(n+m,l+m)
\end{equation}
relating the Bernoulli polynomials $B_n(m)$ at non-negative integer points $m$ with $m$-Stirling numbers of the second kind $S_m(n+m,l+m)$.
Eq.~(\ref{Bm-S2nd}) is a generalization of the identity \cite[p.\ 560]{GKP}
$$
B_n=\sum_{m=0}^n(-1)^m\frac{m!}{m+1}S(n,m).
$$ Note that $B_n\equiv B_n(0)$ and $S\equiv S_0$ are the usual Bernoulli numbers \cite[Chap.\ 2]{Wilf} and Stirling numbers of the second kind \cite[Chap.\ 1]{Wilf}, respectively.

In this work, we will give another proof of (\ref{Bm-S2nd}) by showing
that (\ref{Bm-S2nd}) is a straightforward consequence of a new Zeon
representation \cite{Fein,NetodAnjos}, \cite[Chap.\ 5]{MansourSchork}
of the Bernoulli polynomials.

We believe the approach here is of interest because it gives a straightforward demonstration of Guo, Mez\H{o}, and Qi result, and, as a consequence, it provides another instance where computations involving Zeons and/or Grassmann variables provide direct and interesting results \cite{Abde, Bedi,Cara,NetodAnjos,Neto1,Neto2,Scho}.
For more on Grassmann variables we refer the reader to the books of Berezin \cite[Chap.\ 1]{BerIS}, DeWitt \cite[Chap.\ 1]{DeWitt}, and Rogers \cite[Chap.\ 3]{Rogers}.

For completeness, we recall some basic definitions and results
already stated in previous work \cite{NetodAnjos}. Throughout this
work we let $\mathbb{R}$ denote the real numbers, $\mathbb{N}$ the
positive integers, and $\mathbb{N}_0=\{0\}\cup \mathbb{N}$ the
non-negative integers.

\section{Zeon Algebra and Grassmann-Berezin Integral}

\begin{definition}\label{Def1} The
\textit{Zeon algebra} $\mathcal{Z}_n \supset \mathbb{R}$ is
defined as the associative algebra generated by the collection
$\{\varepsilon_i\}_{i=1}^n$ ($n&lt;\infty$) and the scalar $1 \in
\mathbb{R}$, such that
$1\varepsilon_i=\varepsilon_i=\varepsilon_i1$, $\varepsilon_i
\varepsilon_j = \varepsilon_j \varepsilon_i$ $\forall$ $i$, $j$
and $\varepsilon_i^2=0$ $\forall$ $i$.
\end{definition}

For $\{i,j,\ldots,k\} \subset \{1,2,\ldots,n\}$ and
$\varepsilon_{ij\cdots k}\equiv \varepsilon_i\varepsilon_j\cdots
\varepsilon_k$ the most general element with $n$ generators
$\varepsilon_i$ can be written as (with the convention of sum over
repeated indices implicit)
\begin{equation}\label{phin}\phi_n=
a+a_i\varepsilon_i+a_{ij}\varepsilon_{ij}+\cdots+ a_{12\cdots
n}\varepsilon_{12\cdots n}=\sum_{\mathbf{i} \in
2^{[n]}}a_{\mathbf{i}}\varepsilon_\mathbf{i},\end{equation} with
$a$, $a_i$, $a_{ij}$, $\ldots$, $a_{12\cdots n}$ $\in$
$\mathbb{R}$, $2^{[n]}$ being the power set of
$[n]:=\{1,2,\ldots,n\}$, and $1\leq i&lt;j&lt; \cdots \leq n$. We define
the soul of $\phi_n$ by $s\left(\phi_n\right):=\phi_n-a$
\cite[Chap.\ 1]{DeWitt}.

\begin{definition}\label{Def2} The \textit{Grassmann-Berezin integral} on $\mathcal{Z}_n$, denoted
by $\int$, is the linear functional $\int: \mathcal{Z}_n
\rightarrow \mathbb{R}$ such that (we use throughout this work the
compact notation $d\nu_n:=d\varepsilon_n \cdots d\varepsilon_1$)
$$
d\varepsilon_id\varepsilon_j=d\varepsilon_jd\varepsilon_i,\,\,
\int \phi_n\bigl(\hat{\varepsilon}_i\bigr)d\varepsilon_i=0\,\,{\rm and}
\int \phi_n\bigl(\hat{\varepsilon}_i\bigr)\varepsilon_id\varepsilon_i=\phi_n\bigl(\hat{\varepsilon}_i\bigr),
$$ where $\phi_n\bigl(\hat{\varepsilon}_i\bigr)$ means any element of $\mathcal{Z}_n$ with no dependence on
$\varepsilon_i$. Multiple integrals are iterated integrals, i.e.,
$$
\int f(\phi_n) d\nu_n = \int \cdots \biggl(\int \biggl(\int f(\phi_n) d\varepsilon_n\biggr)
d \varepsilon_{n-1}\biggr)\cdots d\varepsilon_1.
$$\end{definition}

Many functions of ordinary calculus admit extensions to the realm of Zeon algebra \cite{NetodAnjos}. For instance, if $\phi_n=a+s\left(\phi_n\right)$ in Eq.~(\ref{phin}), we have
\begin{equation}\label{defe}
e^{\phi_n}:= e^{a}\sum_{m=0}^n\frac{s^m\left(\phi_n\right)}{m!}
\end{equation} and only a finite number of terms is present in the sum on the right-hand side of Eq.~(\ref{defe}), since $s^m\left(\phi_n\right)=0$ for $m&gt; n$.

Likewise, we have
\begin{equation}\label{definv}
\frac{1}{\phi_n}=\frac{1}{a+s\left(\phi_n\right)}:=\frac{1}{a}\sum_{m=0}^n\left(-\frac{1}{a}\right)^ms^m\left(\phi_n\right).
\end{equation}

\section{Proof of Eq.~(\ref{Bm-S2nd})}

We are now ready to prove Eq.~(\ref{Bm-S2nd}). We take $\varphi_n:= \varepsilon_1+\cdots + \varepsilon_n \in \mathcal{Z}_n$ from now on. We start with
\begin{equation}\label{ZBm-S2nd}
B_n(x)=\sum_{m=0}^n
\frac{(-1)^m}{m+1}\int e^{x\varphi_n}\left(e^{\varphi_n}-1\right)^m d\nu_n.
\end{equation}
We will proceed by showing that
\begin{equation}\label{Bn}
\sum_{m=0}^n{n+1 \choose  m}B_m=0
\end{equation} with $n \in \mathbb{N}$, $B_0\equiv 1$, ${n \choose m}:=n!/\bigl(m!(n-m)!\bigr)$,
 and
\begin{equation}\label{Bp}
B_n(x)=\sum_{m=0}^n{n \choose m}B_mx^{n-m}.
\end{equation} Eqs.~(\ref{Bn}) and (\ref{Bp}) can be regarded as the definitions of the Bernoulli numbers \cite[Eq.\ (6.79)]{GKP} and the Bernoulli polynomials \cite[Eq.\ (7.80)]{GKP}, respectively.

We will first show by induction on $n$ that
\begin{equation}\label{varphiind}
\varphi_n=\displaystyle\sum_{m=0}^{n-1}\frac{\left(-1\right)^m}{m+1}\left(e^{\varphi_n}-1\right)^{m+1}.
\end{equation} Indeed, it is easy to see that both sides of Eq.~(\ref{varphiind}) give $\varphi_1\equiv \varepsilon_1$ for $n=1$. Next, using Eqs.~(\ref{defe}) and (\ref{definv}), we have
$$\begin{array}{lcl}
\varphi_{n+1}&amp;=&amp;\varphi_n+\varepsilon_{n+1}\\
&amp;=&amp;\varphi_n+\varepsilon_{n+1}\displaystyle\frac{e^{\varphi_n}}{1+\left(e^{\varphi_n}-1\right)}\\
&amp;=&amp;\varphi_n+\varepsilon_{n+1}\displaystyle\sum_{m=0}^n(-1)^me^{\varphi_n}\left(e^{\varphi_n}-1\right)^m\\
&amp;=&amp;\displaystyle\sum_{m=0}^{n-1}\frac{\left(-1\right)^m}{m+1}\left(e^{\varphi_n}-1\right)^{m+1}+\varepsilon_{n+1}\sum_{m=0}^n(-1)^me^{\varphi_n}\left(e^{\varphi_n}-1\right)^m\\
&amp;=&amp;\displaystyle\sum_{m=0}^{n-1}\frac{\left(-1\right)^m}{m+1}{m+1 \choose 0}\left(e^{\varphi_n}-1\right)^{m+1}+\varepsilon_{n+1}\sum_{m=0}^n\frac{(-1)^m}{m+1}{m+1 \choose 1}
e^{\varphi_n}\left(e^{\varphi_n}-1\right)^m\\
&amp;=&amp;\displaystyle\sum_{m=0}^n\frac{(-1)^m}{m+1}
\left(e^{\varphi_n}+\varepsilon_{n+1}e^{\varphi_n}-1\right)^{m+1}\\
&amp;=&amp;\displaystyle\sum_{m=0}^n\frac{(-1)^m}{m+1}\left(e^{\varphi_{n+1}}-1\right)^{m+1},
\end{array}$$ and the result follows, i.e., Eq.~(\ref{varphiind}) is true for all $n\geq1$.

Now we can prove Eq.~(\ref{Bn}). Starting with Eq.~(\ref{varphiind}) and using Eq.~(\ref{defe}) we have
$$
\varphi_n=\displaystyle\sum_{m=0}^{n-1}\frac{\left(-1\right)^m}{m+1}\sum_{l=1}^n\frac{\varphi_n^l}{l!}\left(e^{\varphi_n}-1\right)^m.
$$ Integrating, we get ($n \geq 2$)
$$\begin{array}{lcl}
0=\displaystyle\int\varphi_nd\nu_n
&amp;=&amp;\displaystyle\sum_{l=1}^n\sum_{1\leq k_1,k_2,\ldots,k_l\leq n}\displaystyle\sum_{m=0}^{n-1}\frac{\left(-1\right)^m}{m+1}\int\varepsilon_{k_1k_2\cdots k_l}\left(e^{\varphi_n}-1\right)^md\nu_n\\
&amp;=&amp;\displaystyle\sum_{l=1}^n{n \choose l}\displaystyle\sum_{m=0}^{n-l}\frac{\left(-1\right)^m}{m+1}\int\left(e^{\varphi_{n-l}}-1\right)^md\nu_{n-l}\\
&amp;=&amp;\displaystyle\sum_{l=1}^n{n \choose l}B_{n-l}
\end{array}$$ and making the change of variables $n-l \mapsto l$ we obtain Eq.~(\ref{Bn}).

We will now show Eq.~(\ref{Bp}). Indeed, from Eq.~(\ref{ZBm-S2nd}) we have
$$\begin{array}{lcl}
B_n(x)&amp;=&amp;\displaystyle\sum_{l=0}^n\displaystyle\sum_{m=0}^n
\frac{(-1)^l}{l+1}\frac{x^m}{m!}\int \varphi_n^m\left(e^{\varphi_n}-1\right)^l d\nu_n\\
&amp;=&amp;\displaystyle\sum_{l=0}^n
\frac{(-1)^l}{l+1}\int \left(e^{\varphi_n}-1\right)^l d\nu_n\\
&amp;&amp;\hspace{2cm}+\displaystyle\sum_{l=0}^n\sum_{m=1}^n\displaystyle\sum_{1\leq l_1,l_2,\ldots,l_m\leq n}
\frac{(-1)^l}{l+1}x^m\int \varepsilon_{l_1l_2\cdots l_m}\left(e^{\varphi_n}-1\right)^l d\nu_n\\
&amp;=&amp;\displaystyle\sum_{m=0}^n{n \choose m}x^m\displaystyle\sum_{l=0}^{n-m}\frac{(-1)^l}{l+1}\int \left(e^{\varphi_{n-m}}-1\right)^l d\nu_{n-m}\\
&amp;=&amp;\displaystyle\sum_{m=0}^n{n \choose m}x^mB_{n-m}
\end{array}$$ and making the change of variables $n-m \mapsto m$ we obtain Eq.~(\ref{Bp}).

We recall the generating function for the $m$-Stirling numbers \cite[Thm.\ 16]{Broder}
\begin{equation}\label{GFm-S2nd}
\sum_{n=l}^{\infty}S_m(n+m,l+m)\frac{x^n}{n!}=\frac{1}{l!}e^{mx}\left(e^x-1\right)^l
\end{equation} with $m \in \mathbb{N}_0$. The Zeon representation
of $S_m(n+m,l+m)$ comes from
the generating function in Eq.~(\ref{GFm-S2nd})
taking, as in  previous work \cite{NetodAnjos}, $x \rightarrow \varphi_n \in \mathcal{Z}_n$ and doing a Grassmann-Berezin integration over the Zeon algebra to get the representation
\begin{equation}\label{Zm-S2nd}
S_m(n+m,l+m)=\sum_{k=l}^{n}S_m(k+m,l+m)\underbrace{\int \frac{\varphi_n^k}{k!}d\nu_n}_{\delta_{k,n}} =\frac{1}{l!}\int e^{m\varphi_n}\left(e^{\varphi_n}-1\right)^ld\nu_n
\end{equation} with $\delta_{k,n}$ meaning the Kronecker delta. We note that the representation in Eq.~(\ref{Zm-S2nd}) is a generalization of the representation of the usual Stirling numbers of the second kind \cite[Prop.\ 2.1]{Scho} obtained by setting $m=0$ in Eq.~(\ref{Zm-S2nd}). Therefore, by setting $x\equiv m$ in Eq.~(\ref{ZBm-S2nd}), we conclude that
$$
B_n(m)=\sum_{l=0}^n
\frac{(-1)^l}{l+1}\int e^{m\varphi_n}\left(e^{\varphi_n}-1\right)^l d\nu_n,
$$ which is equivalent to Eq.~(\ref{Bm-S2nd})
using the Zeon representation of the numbers $S_m(n+m,l+m)$ in Eq.~(\ref{Zm-S2nd}).

\section{Acknowledgments}
The author thanks the anonymous referee for suggestions that
improved the paper.

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\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}: Primary
11B68; Secondary 11B73; 33B10; 05A15; 05A19.

\noindent \emph{Keywords: } Zeon algebra, Berezin integration,
Bernoulli number, $m$-Stirling number, generating function.

\bigskip
\hrule
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\noindent (Concerned with sequences 
\seqnum{A027641},
\seqnum{A027642},
\seqnum{A143494},
\seqnum{A143495}, and
\seqnum{A143496}.)

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\vspace*{+.1in}
\noindent
Received March 13 2016;
revised version received  April 13 2016.
Published in {\it Journal of Integer Sequences}, May 11 2016.

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