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\begin{center}
\vskip 1cm{\LARGE\bf
On Sums of Powers of the \\
\vskip .1in
$p$-adic Valuation of $n!$}
\vskip 1cm
\large
Rafael Jakimczuk\\
Divisi\'on Matem\'atica \\
Universidad Nacional de Luj\'an\\
Buenos Aires \\
Argentina\\ 
\href{mailto:jakimczu@mail.unlu.edu.ar}{\tt jakimczu@mail.unlu.edu.ar}\\
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\vskip .2in

\begin{abstract} 
We study sums where the multiplicities of the
primes in the prime factorization of $n!$ appear, and obtain a strong
connection between these sums and the Riemann zeta function.
\end{abstract}


\section{Introduction}


Consider the prime factorization of a positive integer $a$,
\[a=p_1^{s_1}	p_2^{s_2}\cdots p_k^{s_k},\]
where $p_1, p_2, \ldots, p_k$ are the distinct prime factors of $a$ and $s_1, s_2, \ldots, s_k$ are their multiplicities. We let $\Omega(a)$ denote the total number of prime factors \cite{Hardy}, that is,
\[\Omega(a)=s_1+s_2+\cdots+s_k.	\]
Consider the prime factorization of $n!$. We let $E(p)$ denote the
multiplicity of a prime $p$ appearing in this factorization.
Consequently, we can write the prime
factorization of $n!$  as follows:
\[n!=\prod_{2\leq p \leq n}p^{E(p)}.    \]
The following asymptotic formula is well-known:
(e.g., \cite{Hardy,Tenenbaum})
\begin{eqnarray}\label{E1} \Omega(n!)= \sum_{2\leq p \leq n}
E(p)= n \log\log n +A n +o\left(\frac{n}{\log n}\right), \end{eqnarray}
where the constant $A$ is \[A=M+\sum_{p}\frac{1}{p(p-1)}\approx
1.034653, \] and $M$  is  Mertens's constant.

Let $k\geq 1$ an arbitrary but fixed positive integer. We study the
sequences \[\sum_{2\leq p\leq
n}\frac{1}{E(p)^{k}},       \] and  prove a strong connection between
these sequences and the Riemann zeta function $\zeta(s)$. We also study
the sequences  \[\sum_{2\leq p\leq n}E(p)^{k}  \qquad
(k\geq 2).\]

\section{Main Results}

\begin{theorem} \label{T1}
Let $k\geq 1$ be an arbitrary but fixed positive integer. For any sufficiently
large integer $n$ we have
\begin{eqnarray}\label{E2}
\sum_{2\leq p\leq n}\frac{1}{E(p)^{k}}=C_k \frac{n}{\log n}	+O_k\left(\frac{n}{\log^2 n}\right),
\end{eqnarray}
where 
\begin{eqnarray}\label{E3}
C_k=(-1)^k+\sum^{k+1}_{j=2}	(-1)^{k+j-1}\zeta(j).
\end{eqnarray}
\end{theorem} 
\begin{proof} By the prime number theorem, we have 
\begin{eqnarray}\label{E4}
\pi(x)=\sum_{2\leq p\leq x} 1 =\frac{x}{\log x}+O\left(\frac{x}{\log^2 x}\right). 
\end{eqnarray}
Consider the prime factorization of $n!$. The multiplicity $E(p)$ of the prime $p$  is, by Legendre's theorem, equal to
\[E(p)=\sum^{\infty}_{j=1}\left\lfloor \frac{n}{p^j}\right\rfloor.	\]
If $p$ satisfies the inequality 
\begin{eqnarray}\nonumber
\frac{n}{j+1}&lt;p\leq \frac{n}{j},
\end{eqnarray}
where $j$ is a fixed positive integer, and the inequality
\begin{eqnarray}\nonumber
p&gt;\sqrt{n},
\end{eqnarray}
then we obtain
\[E(p)=\sum^{\infty}_{j=1}\left\lfloor \frac{n}{p^j}\right\rfloor	=\left\lfloor \frac{n}{p}\right\rfloor=j.\]
Now we have
\begin{eqnarray}\label{E5}
\sum_{p\leq n}\frac{1}{E(p)^k}= \sum_{p\leq \sqrt{n}}\frac{1}{E(p)^k}	+\sum_{\sqrt{n}&lt;p\leq n}\frac{1}{E(p)^k},
\end{eqnarray}
where
\begin{eqnarray}\label{E6}
\sum_{p\leq \sqrt{n}}\frac{1}{E(p)^k}=O\left(\pi\left(\sqrt{n}\right)\right)=O\left(\frac{\sqrt{n}}{\log n}\right)= O\left(\frac{n}{\log^2 n}\right).
\end{eqnarray}
Let $J=\left\lfloor \sqrt{n}\right\rfloor$. If $p&gt;\sqrt{n}$ then $E(p)=\left\lfloor \frac{n}{p}\right\rfloor$. Hence,  we have 
\begin{eqnarray}\label{E7}
\sum_{\sqrt{n}&lt;p\leq n}\frac{1}{E(p)^k}=\sum^{J-1}_{j=1}\frac{1}{j^k}	\sum_{\frac{n}{j+1}&lt;p\leq\frac{n}{j}}1 = \sum^{J-1}_{j=1}\frac{1}{j^k}\left(\pi\left(\frac{n}{j}\right)-\pi\left(\frac{n}{j+1}\right)\right).
\end{eqnarray}
Note that, by Eq.~(\ref{E4}), we have
\begin{eqnarray}\label{E8}
\pi\left(\frac{n}{j}\right)= \frac{n}{j \log \frac{n}{j}}	+O\left(\frac{n}{j \log^2\frac{n}{j}}\right)= \frac{n}{j \log n}+O\left(\frac{\log j}{j}\frac{n}{\log^2 n}\right)+O\left(\frac{n}{j \log^2\frac{n}{j}}\right),
\end{eqnarray}
where we use the formula 
\begin{eqnarray}\nonumber
\frac{1}{1-f(n)}=1+\left(\frac{1}{1-f(n)}\right)f(n).
\end{eqnarray}
In the same way, we find that
\begin{eqnarray}\label{E9}
\pi\left(\frac{n}{j+1}\right)=  \frac{n}{(j+1) \log n}+O\left(\frac{\log (j+1)}{(j+1)}\frac{n}{\log^2 n}\right)+O\left(\frac{n}{(j+1) \log^2\frac{n}{(j+1)}}\right)
\end{eqnarray}
Eqs.~(\ref{E8}) and (\ref{E9}) give
\begin{eqnarray}\label{E10}
&amp;&amp;\frac{1}{j^k}\left(\pi\left(\frac{n}{j}\right)-\left(\pi\left(\frac{n}{j+1}\right)\right)\right)= \frac{1}{j^k}\left(\frac{1}{j}-\frac{1}{j+1}\right)	\frac{n}{\log n}+O\left(\frac{\log j}{j^{k+1}}\frac{n}{\log^2 n}\right)\nonumber\\&amp;-&amp;O\left(\frac{\log (j+1)}{j^{k}(j+1)}\frac{n}{\log^2 n}\right)+O\left(\frac{n}{j^{k+1} \log^2\frac{n}{j}}\right)
-O\left(\frac{n}{j^k (j+1) \log^2\frac{n}{j+1}}\right).
\end{eqnarray}
Note that
\begin{eqnarray}\label{E11}
\sum^{J-1}_{j=1}\frac{1}{j^k}\left(\frac{1}{j}-\frac{1}{j+1}\right)\frac{n}{\log n}	=C_k \frac{n}{\log n}+O\left(\frac{n}{\log^2 n}\right),
\end{eqnarray}
where
\begin{eqnarray}\label{E12}
C_k=\sum^{\infty}_{j=1}\frac{1}{j^k}\left(\frac{1}{j}-\frac{1}{j+1}\right).
\end{eqnarray}
On the other hand, we have
\begin{eqnarray}\label{E13}
\sum^{J-1}_{j=1}O\left(\frac{\log j}{j^{k+1}}\frac{n}{\log^2 n}\right)=O\left(\frac{n}{\log^2 n}\sum^{J-1}_{j=1}\frac{\log j}{j^{k+1}}\right)=O\left(\frac{n}{\log^2 n}\right),
\end{eqnarray}
Analogously, we find that
\begin{eqnarray}\label{E14}
\sum^{J-1}_{j=1}O\left(\frac{\log (j+1)}{j^{k}(j+1)}\frac{n}{\log^2 n}\right)=O\left(\frac{n}{\log^2 n}\right).
\end{eqnarray}
Besides, we have
\begin{eqnarray}\label{E15}
\sum^{J-1}_{j=1}O\left(\frac{n}{j^{k+1} \log^2\frac{n}{j}}\right)=O\left(\sum^{J-1}_{j=1}\frac{n}{j^{k+1} \log^2\frac{n}{j}}\right)	= O\left(\frac{n}{\log^2 n}\right),
\end{eqnarray}
since
\begin{eqnarray}\nonumber
\sum^{J-1}_{j=1}\frac{n}{j^{k+1} \log^2\frac{n}{j}}=O\left(\sum^{J-1}_{j=1}\frac{n}{j^{k+1} \log^2\frac{n}{J}}\right)= O\left(\frac{n}{\log^2 n}\right).
\end{eqnarray}
In the same manner, we obtain
\begin{eqnarray}\label{E16}
\sum^{J-1}_{j=1}O\left(\frac{n}{j^{k}(j+1) \log^2\frac{n}{j+1}}\right)= O\left(\frac{n}{\log^2 n}\right).
\end{eqnarray}
Substituting equations (\ref{E11}), (\ref{E13}), (\ref{E14}), (\ref{E15}) and (\ref{E16}) into (\ref{E10}) (see (\ref{E7})) we find that
\begin{eqnarray}\label{E17}
\sum_{\sqrt{n}&lt; p \leq n}	\frac{1}{E(p)^k}= C_k \frac{n}{\log n}+O\left(\frac{n}{\log^2 n}\right).
\end{eqnarray}
Equations (\ref{E5}), (\ref{E6}), and (\ref{E17}) give (\ref{E2}).

We have the equation 
\[\frac{1}{j^{k}}\left(\frac{1}{j}	-\frac{1}{(j+1)}\right)=\frac{1}{j^{k+1}}-\left(\frac{1}{j^{k-1}}\left(	\frac{1}{j}-\frac{1}{j+1} \right)\right).\]
Therefore, by (\ref{E12})
\begin{eqnarray}\label{E18}
C_k = \zeta(k+1)-C_{k-1}.
\end{eqnarray}
On the other hand, we have
\begin{eqnarray}\label{E19}
C_1= \sum^{\infty}_{j=1}\left(\frac{1}{j^{2}}	-\frac{1}{j (j+1)}\right)=\zeta(2)-1,
\end{eqnarray}
since
\[\sum^{\infty}_{j=1}\frac{1}{j (j+1)}=	\sum^{\infty}_{j=1}\left(\frac{1}{j}-\frac{1}{j+1}\right)=1.\]
Equations~(\ref{E18}) and (\ref{E19}) give (\ref{E3}). 
\end{proof}

 
\begin{example} If $k=1$ then Theorem~\ref{T1} is
\[\sum_{2\leq p\leq n}\frac{1}{E(p)}=\left(\frac{\pi^2}{6}-1\right)\frac{n}{\log n}+O\left(\frac{n}{\log^2 n}\right).\]
\end{example}

\begin{theorem}\label{T3}  
The sequence of positive numbers $C_k$ is strictly decreasing and 
\begin{eqnarray}\nonumber
C_k=\frac{1}{2}+O\left(2^{-k}\right).
\end{eqnarray}
Besides, we have the following limit
\[\lim_{k\rightarrow \infty}C_k=\frac{1}{2}.\]
\end{theorem}
\begin{proof}
Clearly the sequence of positive numbers $C_k$ is strictly decreasing
(see Eq.~(\ref{E12})). Therefore the limit of this sequence exists and
is either
positive or zero. Eq.~(\ref{E18}) then implies that the limit is
$1/2$, since $\lim_{k\rightarrow \infty}\zeta(k)=1$.

To obtain a more precise result we need the following formula \cite{Srivastava}:
\begin{eqnarray}\nonumber
\sum^{\infty}_{j=2}(-1)^j	\left(\zeta(j)-1\right)=\frac{1}{2}.
\end{eqnarray}
Therefore, we have, for $k\geq 1$, that
\begin{eqnarray}
&amp;&amp;C_k = (-1)^k\left(1-\sum^{k+1}_{j=2}(-1)^j \zeta(j)\right)=(-1)^k \left(1-\sum^{k+1}_{j=2}(-1)^j-\sum^{k+1}_{j=2}(-1)^j \left(\zeta(j)-1\right)\right)\nonumber\\&amp;=&amp; (-1)^k\left(1-\frac{1}{2}\left(1-(-1)^k\right)-\sum^{\infty}_{j=2}(-1)^j\left(\zeta(j)-1\right)+\sum_{j&gt;k+1}(-1)^j\left(\zeta(j)-1\right)\right)\nonumber\\&amp;=&amp;
(-1)^k\left(\frac{1}{2}-\frac{1}{2}\left(1-(-1)^k\right)+\sum_{j&gt;k+1}(-1)^j\left(\zeta(j)-1\right)\right)=\frac{1}{2}+O\left(2^{-k}\right)\nonumber
\end{eqnarray}
\end{proof}
 
\begin{theorem}\label{T4}
Let $k\geq 2$ a fixed but arbitrary positive integer.
For every sufficiently large integer we have
\begin{eqnarray}\nonumber
\sum_{p\leq n}E(p)^k = n^k \sum_{p}\frac{1}{(p-1)^k}+O_k\left(n^{k-1}\log n\right).
\end{eqnarray}
\end{theorem}
\begin{proof} The following equation is well-known \cite{Bordelles}. If $p\leq n$ then
\begin{eqnarray}\nonumber
E(p)=\frac{n}{p-1}+O\left(\frac{\log n}{\log p}\right).
\end{eqnarray}
Therefore, we have
\begin{eqnarray}
&amp;&amp;\sum_{p\leq n}E(p)^k =\sum_{p\leq n}\left(\frac{n}{p-1}+O\left(\frac{\log n}{\log p}\right)\right)^k = n^k \sum_{p\leq n}\frac{1}{(p-1)^k}\left(1+O\left(\frac{p \log n}{n \log p}\right)\right)^k \nonumber\\&amp;=&amp; n^k \sum_{p}\frac{1}{(p-1)^k}+O\left(n^k\sum_{p&gt;n}\frac{1}{p^k}\right)+O\left(n^{k-1}\log n \sum_{p\leq n}\frac{1}{p^{k-1}\log p}\right)\nonumber
\end{eqnarray}
To complete the proof we need the bounds
\begin{eqnarray}\nonumber
\sum_{p&gt;n}\frac{1}{p^k}	\ll_k \frac{1}{n^{k-1}\log n}
\end{eqnarray}  
and
\begin{eqnarray}\nonumber
\sum_{p\leq n}\frac{1}{p^{k-1}\log p}	\ll_k 1,
\end{eqnarray}
since $k\geq 2$.
\end{proof}
   
\begin{remark}\label{R5}
\leavevmode
\begin{enumerate}
\item Theorem~\ref{T4}  is a natural generalization of Eq.~(\ref{E1}), since we have  the well-known formula 
\begin{eqnarray}\nonumber
\sum_{p\leq n}\frac{1}{p-1}	= \log\log n+A+O\left(\frac{1}{\log n}\right) \qquad (n\geq 2).
\end{eqnarray}
\item In the case $k=2$, the constant in Theorem~\ref{T4}  is closely related to the constant $A$ in Eq.~(\ref{E1}), since
\begin{eqnarray}
&amp;&amp;\sum_{p}\frac{1}{(p-1)^2}	= \sum^{\infty}_{n=2}	\frac{J_2(n)-\varphi(n)}{n}\log \zeta(n)\nonumber\\&amp;=&amp; \sum^{\infty}_{n=2}\frac{J_2(n)}{n}\log \zeta(n)+\gamma-A\approx 1,375064994748635\ldots \nonumber
\end{eqnarray}
where $J_2(n)=n^2 \prod_{p \mid n}(1-p^{-2})$ is the second Jordan arithmetic function \cite{Glaisher}.
\end{enumerate}
\end{remark}
     

\section{Acknowledgments} The author would like to thank the anonymous
referee for his/her valuable comments and suggestions for improving the
original version of this article. The better error terms in
Theorem~\ref{T1}  and Theorem~\ref{T3} are due to the referee.
Theorem~\ref{T4}, together with Remark~\ref{R5}, are also results of the
referee. The author is  also very grateful to Universidad Nacional de
Luj\'an.

\begin{thebibliography}{99}
 
\bibitem{Bordelles}
O. Bordell\`es, \textit{Arithmetic Tales}, Springer, 2012.

\bibitem{Glaisher}
J. W. L. Glaisher, On the sums of the inverse powers of the prime
numbers, \textit{Quart. J. Pure Appl. Math. } \textbf{25} (1891),
347--362.

\bibitem{Hardy}
G. H. Hardy and E. M. Wright, \textit{An Introduction to the Theory of
Numbers}, Oxford, 1960.

\bibitem{Srivastava}
H. M. Srivastava, Sums of certain series of the Riemann zeta function,
\textit{J. Math. Anal. Appl.} \textbf{134} (1988),  129--140.

\bibitem{Tenenbaum}
G. Tenenbaum, \textit{Introduction \`a la Th\'eorie Analytique
et Probabiliste des Nombres}, Belin, 2008.



\end{thebibliography}

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\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11A99; Secondary 11B99.

\noindent \emph{Keywords: } 
factorial, prime factorization, Riemann zeta function.

\bigskip
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\noindent (Concerned with sequence
\seqnum{A0xxxxx}.)

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\vspace*{+.1in}
\noindent
Received February 10 2017;
revised version received  April 9 2017.
Published in {\it Journal of Integer Sequences}, May 1 2017.

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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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