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\begin{center}
\vskip 1cm{\LARGE\bf A New Identity for Complete Bell \\
\vskip 0in
Polynomials Based on a Formula of\\
\vskip 0.05in
Ramanujan} \vskip 1cm
\large
Sadek Bouroubi\\
University of Science and Technology Houari Boumediene\\
Faculty of Mathematics\\
Laboratory LAID3\\
P. O. Box 32 \\
16111 El-Alia, Bab-Ezzouar, Algiers \\
Algeria\\
\href{mailto:sbouroubi@usthb.dz}{\tt sbouroubi@usthb.dz} \\
\href{mailto:bouroubis@yahoo.fr}{\tt bouroubis@yahoo.fr} \\
\ \\
Nesrine Benyahia Tani \\
Faculty of Economics and Management Sciences\\
Laboratory LAID3\\
Ahmed Waked Street \\
Dely Brahim, Algiers \\
Algeria\\
\href{mailto:benyahiatani@yahoo.fr}{\tt benyahiatani@yahoo.fr}
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\vskip .2in

\begin{abstract}
Let $p(n)$ be the number of partitions 
of $n$. In this
paper, we give a new identity for complete Bell polynomials based on a
sequence related to the generating function of
$p(5n+4)$ established by Srinivasa Ramanujan.
\end{abstract}

\vskip .2in

\newtheorem{theorem}{Theorem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{Remark}[theorem]{Remark}

\section{Introduction}
Let us first present some necessary definitions related to the Bell
polynomials, which are quite general and have numerous applications
in combinatorics. For a more complete \mbox{exposition}, the reader
is referred to the excellent books
of Comtet \cite{co}, Riordan \cite{ri1} and Stanley \cite{st}.

Let $(a_{1}, a_{2},\ldots)$ be a sequence of
real or complex numbers. Its partial (exponential) Bell polynomial
$B_{n,k}(a_{1}, a_{2},\ldots)$, is defined as follows:
\begin{equation*}
\underset{n=k}{\overset{\infty }{\dsum }}B_{n,k}\left(
a_{1},a_{2},\ldots\right) \dfrac{t^{n}}{n!}=\dfrac{1}{k!}\left(
\underset{m=1}{\overset{\infty }{\dsum
}}a_{m}\dfrac{t^{m}}{m!}\right) ^{k}\cdot
\end{equation*}
Their exact expression is
\begin{equation*}
B_{n,k}\left( a_{1},a_{2},\ldots\right) =\underset{\pi \left( n,k\right) }{%
\dsum }\dfrac{n!}{k_{1}!k_{2}!\cdots}\left( \dfrac{a_{1}}{1!}\right)
^{k_{1}}\left( \dfrac{a_{2}}{2!}\right) ^{k_{2}}\cdots,  \label{B5}
\end{equation*}%
where $\pi \left( n,k\right)$ denotes the set of all integer
solutions $\left( k_{1},k_{2},\ldots\right)$ of the system
\begin{equation*}
\left\{
\begin{array}{l}
k_{1}+\cdots+k_{j}+\cdots=k;\vspace{0.2cm}\\
k_{1}+\cdots+jk_{j}+\cdots=n.%
\end{array}%
\right.
\end{equation*}
The (exponential) complete Bell polynomials are given by
\begin{equation*}
\exp \left( \overset{\infty }{\underset{m=1}{\dsum }}a_{m}\dfrac{t^{m}}{m!}%
\right) =\overset{\infty }{\underset{n=0}{\dsum
}}A_{n}(a_{1},a_{2},\ldots) \dfrac{t^{n}}{n!}\cdot
\end{equation*}%
In other words,
\begin{equation*}
A_{0}(a_{1},a_{2},\ldots)=1 \text{ \ \ and \ \ }
A_{n}(a_{1},a_{2},\ldots)=\dsum_{k=1}^{n}B_{n,k}(a_{1},a_{2},\ldots),
\forall n\geq1.
\end{equation*}
Hence
\begin{equation*}
A_{n}\left( a_{1},a_{2},\ldots\right) =\underset{k_{1}+\ldots+jk_{j}+\ldots=n}{\dsum }%
\dfrac{n!}{k_{1}!k_{2}!\cdots}\left( \dfrac{a_{1}}{1!}\right)
^{k_{1}}\left( \dfrac{a_{2}}{2!}\right) ^{k_{2}}\cdots\ \cdot
\end{equation*}
The main tool used to prove our main result in the next section is
the following formula of Ramanujan, about which G. H. Hardy
\cite{ha} said: ``... but here Ramanujan must take second place to
Prof.\ Rogers; and if I had to select one formula from all of
\mbox{Ramanujan's} work, I would agree with Major MacMahon in
selecting ...
\begin{equation}\label{eq1}
\dsum_{n=0}^{\infty}p(5n+4)\
x^{n}=\dfrac{5\{(1-x^{5})(1-x^{10})(1-x^{15})\cdots\}^{5}}{\{(1-x)(1-x^{2})(1-x^{3})\cdots\}^{6}},
\end{equation}
where $p(n)$ is the number of partitions of $n$.''

\section{Some basic properties of the divisor function}
Let $\sigma(n)$ be the sum of the positive divisors of $n$.
It is clear that $\sigma(p)=1+p$ for any
prime number $p$, since the only positive divisors of $p$ are 1 and
$p$. Also the only divisors of $p^{2}$ are 1, $p$ and $p^{2}$. Thus
\begin{equation*}
\sigma(p^{2})=1+p+p^{2}=\dfrac{p^{3}-1}{p-1}\cdot
\end{equation*}
It is now easy to prove \cite{ro}
\begin{equation}\label{eq2}
\sigma(p^{k})=\dfrac{p^{k+1}-1}{p-1}\cdot
\end{equation}
It is well known in number theory \cite{ro} that $\sigma(n)$ is a
multiplicative function, that is, if $n$ and $m$ are relatively
prime, then
\begin{equation}\label{eq3}
\sigma(nm)=\sigma(n)\ \sigma(m).
\end{equation}
An immediate consequence of these facts is the following Lemma:
\begin{lemma}
If $5\mid n$, then it exists $\alpha \geq1$, so that
\begin{equation}\label{eq4}
\sigma(n)=\dfrac{5^{\alpha+1}-1}{5^{\alpha}-1}\
\sigma\left(\dfrac{n}{5}\right).
\end{equation}
where $\alpha$ is the power to which 5 occur in the decomposition of
$n$ into prime factors.
\end{lemma}
\begin{proof} From (\ref{eq2}) and (\ref{eq3}), we have
\begin{equation*}
\begin{array}{rll}
  \sigma(n) &amp;= &amp; \dfrac{5^{\alpha+1}-1}{4}\ \sigma\left(\dfrac{n}{5^{\alpha}}\right),\  \textrm{and}
  \\\\
 \sigma\left(\dfrac{n}{5}\right) &amp;=&amp;\dfrac{5^{\alpha}-1}{4}\
\sigma\left(\dfrac{n}{5^{\alpha}}\right).
\end{array}
\end{equation*}
Hence the result follows.
\end{proof}
\section{Main result}
Henceforth, let us express $n$ by $n=5^{\alpha}\cdot
p_{1}^{\alpha1}\cdot p_{2}^{\alpha2}\cdots p_{r}^{\alpha_{r}}$,
where the $p's$ are distinct primes different from 5, and the
$\alpha's$ are the powers to which they occur.

The present theorem is the main result of this work.
\begin{theorem}\label{th1}
Let $a(n)$ be the real number defined as follows:
\begin{equation*}
a_{n}=\left(1+\dfrac{20}{5^{\alpha+1}-1}\right)\dfrac{\sigma(n)}{n}\cdot
\end{equation*}
Then we have
\begin{equation*}
A_{n}(1!a_{1},2!a_{2},\ldots,n!a_{n})=\dfrac{n!}{5}\ p(5n+4)\cdot
\end{equation*}
\end{theorem}
\begin{proof}
Put
\begin{equation*}
g(x)=\dfrac{5\{(1-x^{5})(1-x^{10})(1-x^{15})\cdots\}^{5}}{\{(1-x)(1-x^{2})(1-x^{3})\cdots\}^{6}}\cdot
\end{equation*}
Then
\begin{equation}\label{eq5}
\begin{array}{lll}
  \ln(g(x)) &amp;=&amp;\ln5+5\ln\ \dprod_{i=1}^{\infty}(1-x^{5i})-6\ln\dprod_{i=1}^{\infty}(1-x^{i})
  \\\\
   &amp;=&amp;\ln5+5\dsum_{i=1}^{\infty}\ln(1-x^{5i})-6\dsum_{i=1}^{\infty}\ln(1-x^{i})
   \\\\
&amp;=&amp;\ln5-5\dsum_{i,j=1}^{\infty}\dfrac{x^{5ij}}{j}+6\dsum_{i,j=1}^{\infty}\dfrac{x^{ij}}{j}
\\\\ &amp;=&amp;\ln5+\dsum_{n=1}^{\infty}a_{n}x^{n},
\end{array}
\end{equation}
where
\begin{equation*}
a_{n}=\left\{
\begin{array}{ll}
  \dfrac{6\ \sigma(n)-25\ \sigma\left(\dfrac{n}{5}\right)}{n}, &amp; if\ 5\mid n;\vspace{0.1cm}\\
  \dfrac{6}{n}\ \sigma(n), &amp; otherwise.
\end{array}
\right.
\end{equation*}
If $\alpha\geq1$, i.e., $5\mid n$, then we get by (\ref{eq4})
\begin{equation*}
\sigma\left(\dfrac{n}{5}\right)=\dfrac{5^{\alpha}-1}{5^{\alpha+1}-1}\
\sigma(n).
\end{equation*}
Thus
\begin{equation*}
a_{n}=\left(1+\dfrac{20}{5^{\alpha+1}-1}\right)\dfrac{\sigma(n)}{n},
\forall \alpha\geq0.
\end{equation*}
Hence, we obtain from (\ref{eq5})
\begin{equation*}
\begin{array}{lll}
g(x)&amp;=&amp;5.\exp\Big(\dsum_{n=1}^{\infty}a_{n}x^{n}\Big)\\\\
&amp;=&amp;5\left(1+\dsum_{k=1}^{\infty}\dfrac{\left(\dsum_{n=1}^{\infty}n!a_{n}\dfrac{x^{n}}{n!}\right)^{k}}{k!}\right)\\\\
&amp;=&amp;5+5\dsum_{k=1}^{\infty}\left(\dsum_{n=k}^{\infty}B_{n,k}(1!a_{1},2!a_{2},\ldots)\dfrac{x^{n}}{n!}\right)\\\\
&amp;=&amp;5+5\dsum_{n=1}^{\infty}A_{n}(1!a_{1},2!a_{2},\ldots,n!a_{n})\dfrac{x^{n}}{n!}\\\\
&amp;=&amp;5\dsum_{n=0}^{\infty}A_{n}(1!a_{1},\ldots,n!a_{n})\dfrac{x^{n}}{n!}\cdot
\end{array}
\end{equation*}
Therefore, by comparing coefficients of the two power
series in (\ref{eq1}), we finally get
\begin{equation*}
A_{n}(1!a_{1},2!a_{2},\ldots,n!a_{n})=\dfrac{n!}{5}\ p(5n+4),\text{
\ for \ }n\geq0.
\end{equation*}
\end{proof}
\noindent Theorem \ref{th1} has the following Corollary.\\
\begin{corollary}
For $n\geq1$, we have
\begin{equation*}
\sigma(n)=\dfrac{5^{\alpha+1}-1}{5^{\alpha+1}+19}\cdot\dfrac{1}{(n-1)!}\
\dsum_{j=1}^{n}(-1)^{j-1}(j-1)!\ B_{n,j}\left(\dfrac{1!}{5}p(9),
\dfrac{2!}{5}p(14),\ldots\right).
\end{equation*}
\end{corollary}
\begin{proof}
This follows from the following inversion relation of Chaou and al
\cite{ch}:
\begin{equation*}
y_{n}=\dsum_{k=1}^{n}B_{n,k}(x_{1},x_{2},\ldots) \Leftrightarrow
x_{n}=\dsum_{k=1}^{n}(-1)^{k-1}(k-1)!\ B_{n,k}(y_{1},y_{2},\ldots).
\end{equation*}
\end{proof}

\section{Acknowledgements} The
authors would like to thank the referee for his valuable comments
which have improved the quality of the paper.

\begin{thebibliography}{12}
\bibitem{bo} M. Abbas and S. Bouroubi, On new identities for Bell's
polynomials, \emph{Discrete Mathematics} \textbf{293} (2005) 5--10.

\bibitem{be} E. T. Bell, Exponential polynomials,
\emph{Annals of Mathematics} \textbf{35} (1934), 258--277.

\bibitem{ch} W.-S. Chaou, Leetsch C. Hsu, and Peter J.-S. Shiue, Application
of Fa\`{a} di Bruno's formula in characterization of inverse
relations. \emph{Journal of Computational and Applied Mathematics}
\textbf{190} (2006), 151--169.

\bibitem{co} L. Comtet, \textit{Advanced Combinatorics.} 
D. Reidel Publishing Company, Dordrecht-Holland, Boston, 1974,
pp.\ 133--175.

\bibitem{ha} G. H. Hardy, \textit{Ramanujan}, Amer. Math. Soc., Providence,
1999.

\bibitem{ri1} J. Riordan, \textit{Combinatorial Identities}, Huntington, New
York, 1979.

\bibitem{ri2} J. Riordan, \textit{An Introduction to Combinatorial Analysis,}
John Wiley \&amp; Sons, New York, 1958; Princeton University Press,
Princeton, NJ, 1980.

\bibitem{ro} K. Rosen, \textit{Elementary Number Theory and Its 
Applications}, 4th ed., Addison-Wesley, 2000.

\bibitem{st} R. P. Stanley, \textit{Enumerative Combinatorics,} Volume 1,
Cambridge Studies in Advanced Mathematics \textbf{49}, Cambridge
University Press, Cambridge, 1997.
\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}: 05A16,
05A17, 11P81

\noindent \emph{Keywords: } Bell polynomials, integer partition,
Ramanujan's formula.

\bigskip
\hrule
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\noindent (Concerned with sequences
\seqnum{A000041} and \seqnum{A071734}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in} \noindent Received January 3 2009; revised version
received March 22 2009.
Published in {\it Journal of Integer Sequences} March 27 2009. 

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\noindent
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