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\begin{center}
\vskip 1cm{\LARGE\bf On an Open Problem of T\'oth } \vskip 1cm
\large Deyu Zhang\\
School of Mathematical Sciences\\
Shandong Normal University\\
Jinan 250014\\
Shandong\\ 
P. R. China\\
\href{mailto:zdy_78@hotmail.com}{\tt zdy\_78@yahoo.com.cn}\\

\ \\

Wenguang Zhai\\
Department of Mathematics\\
China University of Mining and Technology \\
Beijing, 100083\\
P. R. China\\
\href{mailto:zhaiwg@hotmail.com}{\tt zhaiwg@hotmail.com} \\
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\begin{abstract}
In a recent paper, T\'oth mentioned that it is an open problem to
give the asymptotic formula for  $\sum_{n\leq x}P^{k}(n)$, where
$P(n)$ is the well-known gcd-sum function and $k\geq 2$ is a fixed integer. In this paper, we use the analytic properties of the Dirichlet divisor function to
obtain the asymptotic formula for it.
\end{abstract}

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\section{Introduction}

In 1933, Pillai \cite{s8} introduced the gcd-sum function
\begin{equation}P(n)=\sum_{k=1}^{n}\gcd(k,n).\end{equation}
By grouping the terms according to the values of $\gcd(k,n)$ we have
\begin{equation}\label{eq:E2}P(n)=\sum_{d\mid n}d\phi(n/d)=n\sum_{d\mid n}\frac{\phi(d)}{d}, \end{equation} where $\phi$ is
Euler's function.  Many authors have studied the properties of
$P(n)$, see \cite{s1, s2, s3, s4, s5, s8, s9, s10}; it is Sloane's
sequence \seqnum{A018804}. Chidambaraswamy and Sitaramachandrarao
\cite{s5} showed that, given an arbitrary $\epsilon
&gt;0$,
$$\sum_{n\leq x}P(n)=e_1x^2\log x+e_2x^2+O(x^{1+\theta+\epsilon}),$$
where $\theta$ is the constant appearing in the error term of the
Dirichlet divisor problem, $e_1, e_2$ are certain constants.

It follows from \eqref{eq:E2} that the arithmetic mean of
$\gcd(1,n),\ldots, \gcd(n,n)$ is given by
\begin{equation}A(n)= \frac{P(n)}{n}=\sum_{d\mid n}\frac{\phi(d)}{d}.\end{equation}

T\'oth \cite{s9} showed that
\begin{equation}\sum_{n\leq x}A^{2}(n)= x(C_1\log^{3}x+C_2\log^{2}x+C_3\log x+C_4)+O(x^{1/2+\epsilon}),\end{equation}
where $C_1, C_2, C_3, C_4 $ are computable constants. Furthermore,
he listed some open problems concerning the gcd-sum function, one of
which is  to derive  the asymptotic formula for $ \sum_{n\leq
x}P^{k}(n)$, where $k\geq 2$ is a fixed integer.

In this paper, we use the analytic method to get  the asymptotic
formula for $ \sum_{n\leq x}A^{k}(n)$.

\begin{theorem}
\label{thm:1}   Let $k\geq 2$ be a fixed integer. Then
\begin{equation}\sum_{n\leq x}A^{k}(n)=xQ_{2^{k}-1}(\log x)+O(x^{\beta_k+\epsilon}),
\end{equation}
where $Q_{2^{k}-1}(t)$ is a polynomial of degree $2^{k}-1$ in $t$
and
$$\beta_2= \frac{1}{2}, \ \ \   \beta_3=\frac{5}{8}, \ \ \  \beta_4=
\frac{7}{9},\ \ \  \beta_5= \frac{31}{36}, \ \ \ \beta_6=
\frac{207}{224},\ \ \ \ \beta_k= 1-2^{-\frac{2}{3}k}/50,\ \ \ \
k\geq 7.\ \ \ \ \ \ \  $$
\end{theorem}




 \begin{corollary}
\label{cor:2}Let $k\geq 2$ be a fixed integer. Then
\begin{equation}\sum_{n\leq x}P^{k}(n)=x^{2}Q'_{2^{k}-1}(\log
x)+O(x^{1+\beta_k+\epsilon}),
\end{equation}
where $Q'_{2^{k}-1}(t)$ is a polynomial of degree $2^{k}-1$ in $t$.
\end{corollary}


\begin{theorem}
\label{thm:3}  Let $k\geq 2$ be a fixed integer and
$$E_k(x)=\sum_{n\leq x}A^{k}(n)-xQ_{2^{k}-1}(\log x).$$  Then for
$k=3,4,5,$ we have
$$\int_1^U E_k(x)dx\ll U^{1+\delta_k+\epsilon},$$
where $$\delta_3=1/2,\ \  \delta_4=0.6030739,\ \ \delta_5=0.773114.$$
\end{theorem}



\section{\bf Preliminary Lemmas}

 \begin{lemma}\label{lem:4}


Let $s$ be a complex number with $\text{Re}(s)&gt;1$. Then

$$ \sum_{n=1}^{\infty}\frac{A^{k}(n)}{n^{s}}=\zeta^{2^{k}}(s)G_k(s),$$
where $G_k(s)=\sum\limits_{n=1}^{\infty}\frac{g(n)}{n^{s}}$ is a
Dirichlet series which is absolutely convergent for
$\text{Re}(s)&gt;1/2$.

\end{lemma}
\begin{proof}
Recall that $A(n)$ is multiplicative function. Then  it follows from
the  Euler product representation that for $\text{Re}(s)&gt;1,$
\begin{equation}\label{eq:E7}F(s):=\sum_{n=1}^{\infty}\frac{A^{k}(n)}{n^{s}}
=\prod_{p}\left(1+\sum_{\alpha=1}^{\infty}\frac{A^{k}(p^{\alpha})}{p^{\alpha
s}}\right).\end{equation} where
$$A(p^{\alpha})=1+\frac{\phi(p)}{p}+\cdots+\frac{\phi(p^{\alpha})}{p^{\alpha}}=1+\alpha-\frac{\alpha}{p}, \ \ \ \alpha\geq
1.
$$Thus we have
 $$\begin{array}{llll}
 &amp;\displaystyle 1+\sum_{\alpha=1}^{\infty}\frac{A^{k}(p^{\alpha})}{p^{\alpha s}}
 &amp;\displaystyle = &amp;\displaystyle 1+\sum_{\alpha=1}^{\infty}\frac{(1+\alpha-\frac{\alpha}{p})^{k}}{p^{\alpha
 s}}\\[12pt]
 &amp;\displaystyle  &amp;\displaystyle= &amp;\displaystyle 1+\frac{2^{k}}{p^{s}}-\frac{k\cdot 2^{k-1}}{p^{s+1}}+\ldots+(-1)^k\frac{1}{p^{s+k}}\\[12pt]
&amp;\displaystyle  &amp;\displaystyle &amp;\displaystyle +\frac{3^{k}}{p^{2s}}-\frac{2k\cdot 3^{k-1}}{p^{2s+1}}+\ldots+(-1)^k\frac{2^k}{p^{2s+k}}\\[12pt]
&amp;\displaystyle  &amp;\displaystyle &amp;\displaystyle +\frac{4^{k}}{p^{3s}}-\frac{3k\cdot 4^{k-1}}{p^{3s+1}}+\ldots+(-1)^k\frac{3^k}{p^{3s+k}}\\[12pt]
&amp;\displaystyle  &amp;\displaystyle &amp;\displaystyle +\ldots=:x_k(s).
\end{array}$$
We substitute the above formula  to the formula \eqref{eq:E7} to get
$$ \sum_{n=1}^{\infty}\frac{A^{k}(n)}{n^{s}}=\prod_{p}x_k(s)=\zeta^{2^{k}}(s)\cdot\prod_{p}(1-\frac{1}{p^{s}})^{2^{k}}\cdot x_k(s)=:\zeta^{2^{k}}(s)G_k(s),$$
where $$\begin{array}{lll}
 &amp;\displaystyle G_k(s) &amp;\displaystyle=\prod_{p}(1-\frac{1}{p^{s}})^{2^{k}}\cdot x_k(s)\\[12pt]
&amp;\displaystyle &amp;\displaystyle=\prod_{p}\Bigg(1-\frac{k\cdot
2^{k-1}}{p^{s+1}}+\ldots+(-1)^k\frac{1}{p^{s+k}}\\[12pt]
&amp;\displaystyle
&amp;\displaystyle+\frac{3^{k}-4^{k}+\binom{2^{k}}{2}}{p^{2s}}+\frac{k\cdot
2^{2k-1}-2k\cdot 3^{k-1}}{p^{2s+1}}+\ldots\\[12pt]
&amp;\displaystyle &amp;\displaystyle+
\frac{4^{k}-6^{k}+\binom{2^{k}}{2}2^k-\binom{2^{k}}{3}}{p^{3s}}+\frac{2k\cdot
2^{k}3^{k-1}-3k\cdot 4^{k-1}-k\cdot 2^{k-1}\cdot
\binom{2^{k}}{2}}{p^{3s+1}}+\ldots\Bigg).
\end{array}$$ From the above formula, it is easy to see that
$G_k(s)$ can be expanded to a Dirichlet series
$\sum\limits_{n=1}^{\infty}\frac{g(n)}{n^{s}}$, which is absolutely
convergent for $\text{Re}(s)&gt;1/2$.

\end{proof}





 \begin{lemma}\label{lem:5}
Suppose $ 1/2\leq \sigma \leq 1$, then
\begin{equation}\label{eq:E8}\zeta(\sigma+it)\ll (|t|+2)^{\frac{1-\sigma}{3}}\log(|t|+2). \end{equation}
\end{lemma}
\begin{proof}
We define the function $\mu(\sigma)$ for each $\sigma$ as the
infimum of number $c \geq 0$ such that $\zeta(\sigma+it)\ll t^{c},$
or alternatively as $$\mu(\sigma)=\limsup_{t\rightarrow \infty}
\frac{\log|\zeta(\sigma+it)|}{\log t}.$$ Then $\mu(\sigma)$ is
continuous, nonincreasing and for $\sigma_1\leq \sigma \leq
\sigma_2$,
$$ \mu(\sigma)\leq \mu(\sigma_1)\frac{\sigma_2-\sigma}{\sigma_2-\sigma_1}+\mu(\sigma_2)\frac{\sigma-\sigma_1}{\sigma_2-\sigma_1}.$$
By the well-known estimates $$\zeta(1/2+it)\ll t^{1/6 }, \ \
\zeta(1+it)\ll \log t,$$  we can easily get the formula
\eqref{eq:E8}.
\end{proof}

\begin{lemma}\label{lem:6}
If $ \zeta(s)=\chi(s)\zeta(1-s)$, then the estimate
$$\chi(s)\ll (|t|+2)^{1/2-\sigma} $$
holds uniformly for $0\leq \sigma\leq 1.$
\end{lemma}
\begin{proof}
Using standard properties of the gamma-function one may write the
functional equation of $\zeta(s)$ as
$$\zeta(s)=\chi(s)\zeta(1-s), \ \ \ \
\chi(s)=(2\pi)^{s}/(2\Gamma(s)\cos(\pi s/2)).
$$
From Stirling's formula
$$|\Gamma(s)|=(2\pi)^{\frac{1}{2}}|t|^{\sigma-\frac{1}{2}}e^{-\frac{\pi
|t|}{2}}\left(1+O(|t|^{-1})\right) \ \ \ \ (|t|\geq t_0),$$ it
follows that
$$\begin{array}{lll}
 &amp;\displaystyle
\chi(s)&amp;\displaystyle=\left(\frac{2\pi}{|t|}\right)^{\sigma+i|t|-\frac{1}{2}}e^{i(|t|+\frac{\pi}{4})}\left(1+O(|t|^{-1})\right)\\[12pt]
&amp;\displaystyle      &amp;\displaystyle\ll
(|t|+2)^{1/2-\sigma}.\end{array}$$




\end{proof}











\section{\bf Proofs of Theorem \ref{thm:1} and Corollary \ref{cor:2}}

Recall that the generalized divisor function
$$d_k(n)=\sum_{n=n_1\cdots n_k}1,$$
and its Dirichlet series is
$$\zeta^k(s)=\sum_{n=1}^{\infty}\frac{d_k(n)}{n^s}.$$
 From \cite [Theorem 13.2 and 13.3]{s7} it follows that
\begin{eqnarray}\label{eq:E9}
\sum_{n\leq x}d_{2^{k}}(n)=xP_{2^{k}-1}(\log
x)+O(x^{\beta_k+\epsilon}),
\end{eqnarray}
where $P_{2^k-1}(t)$ is a polynomial of degree $2^k-1$ in $t$, and
$$\beta_2= \frac{1}{2}, \ \ \   \beta_3=\frac{5}{8}, \ \ \
\beta_4= \frac{7}{9},\ \ \  \beta_5= \frac{31}{36},  \ \ \ \beta_6=
\frac{207}{224},\ \ \ \beta_k= 1-2^{-\frac{2}{3}k}/50,\ \ \ \  k\geq
7.\ \ \ \ \ \ \  $$ Then by Lemma \ref{lem:4}, we have that

$$\sum_{n\leq x}A^{k}(n)=\sum_{m\ell\leq x}d_{2^{k}}(m)g(\ell)=\sum_{\ell\leq x}g(\ell)\sum_{m\leq
x/\ell}d_{2^{k}}(m),$$ and formula \eqref{eq:E9} applied to the
inner sum gives
$$
\sum_{n\leq x}A^{k}(n)=\sum_{\ell\leq
x}g(\ell)\left\{\frac{x}{\ell}P_{2^{k}-1}\left(\log(\frac{x}{\ell}
)\right)+O\left((\frac{x}{\ell})^{\beta_k+\epsilon}\right)\right\}
$$
$$=xQ_{2^{k}-1}(\log x)+O(x^{\beta_k+\epsilon}),\ \ \ \ \ \ \ \ \ \ $$
if we notice from Lemma \ref{lem:4} that the infinite series
$\sum\limits_{\ell=1}^{\infty}\frac{g(\ell)}{\ell}$ and
$\sum\limits_{\ell=1}^{\infty}\frac{g(\ell)\log^k\ell}{\ell}$ are
absolutely convergent, and $$\sum_{\ell\leq x}|g(\ell)|\ll
x^{1/2+\epsilon}.$$

From the definitions of $P(n)$ and  Abel's summation formula, we can
easily get $$\sum_{n\leq x}P^{k}(n)=x^{2}Q'_{2^{k}-1}(\log
x)+O(x^{1+\beta_k+\epsilon}),$$ where $Q'_{2^{k}-1}(t)$ is a
polynomial of degree $2^{k}-1$ in $t$.





\section{\bf Proof of Theorem \ref{thm:3}}


It suffices to prove that
\begin{equation}
\int_U^{2U}E_k(x)dx\ll U^{1+\delta_k +\epsilon},
\end{equation}
where$$\delta_3=1/2,\ \  \delta_4=0.6030739,\ \
\delta_5=0.773114.$$By Perron's formula (see for example, \cite
[Chapter 5]{s6}), we have for $T\leq x\leq 2T$ that
$$ \sum_{n\leq x}A^{k}(n)=\frac{1}{2\pi i}\int_{1+\varepsilon-iT}^{1+\varepsilon+iT}\zeta^{2^{k}}(s)G_k(s)\frac{x^{s}}{s}ds+O(T^{\varepsilon}).   $$
Then we move the integration to the parallel segment with
$\text{Re}(s)=1-\varepsilon$ to get
$$ E_k(x)=\frac{1}{2\pi i}\int_{1-\varepsilon-iT}^{1-\varepsilon+iT}\zeta^{2^{k}}(s)G_k(s)\frac{x^{s}}{s}ds+O(T^{\varepsilon}).   $$
So
\begin{eqnarray}\label{eq:E11}
&amp;\displaystyle\int_U^{2U} E_k(x)dx&amp;\displaystyle=\frac{1}{2\pi
i}\int_{1-\varepsilon-iT}^{1-\varepsilon+iT}
\frac{\zeta^{2^{k}}(s)G_k(s)}{s}\left(\int_U^{2U}x^{s}dx\right)ds+O(U^{1+\varepsilon})\nonumber\\[8pt]
&amp; &amp;\displaystyle=\frac{1}{2\pi
i}\int_{1-\varepsilon-iT}^{1-\varepsilon+iT}\frac{\zeta^{2^{k}}(s)G_k(s)(2^{s+1}-1)U^{s+1}}{s(s+1)}ds+O(U^{1+\varepsilon}).\end{eqnarray}
Moving the integral line in the last integral of \eqref{eq:E11} to
$\sigma=c, $ where $\frac{1}{2}&lt;c&lt;1,$ we have
\begin{equation}\label{eq:E12}
\int_U^{2U}E_k(x)dx\ll
U^{1+c}\int_{c-iT}^{c+iT}\frac{|\zeta(s)|^{2^{k}}}{|s(s+1)|}ds\ll
U^{1+c}\int_{1}^{T}\frac{|\zeta(c+it)|^{2^{k}}}{T^{2}}ds,
\end{equation}
if we notice that $G_k(s)$ is absolutely convergent in
$\text{Re}(s)&gt;\frac{1}{2}.$

For the case $k=3$, it follows from \cite [Theorem 8.3]{s7} that
$$\int_1^{T}\left|\zeta(\frac{1}{2}+it)\right|^{8}dt\ll T^{\frac{3}{2}}.$$
On taking  $ c=1/2+\varepsilon$ in \eqref{eq:E12}, we have
\begin{equation}
\int_U^{2U}E_3(x)dx\ll U^{3/2}.
\end{equation}



For the case $k=4$, from  \cite [Theorem 8.3]{s7} and \cite [Theorem
8.4]{s7}, it follows that
$$\int_1^{T}\left|\zeta(\frac{1}{2}+it)\right|^{16}dt\ll T^{1+\frac{350}{216}}$$
and
$$\int_1^{T}\left|\zeta(\sigma+it)\right|^{16}dt\ll T^{1+\varepsilon},$$
where $\sigma$ satisfies $$ \frac{12408}{4537-4890\sigma}=16,$$
which gives $\sigma=0.7692229.$ By \cite [Lemma 8.3]{s7}, we have
$$\int_1^{T}\left|\zeta(0.6030739+it)\right|^{16}dt\ll T^{2}.$$
In the formula \eqref{eq:E12}, we take $c=0.6030739$ to get
\begin{equation} \int_U^{2U}E_4(x)dx\ll U^{1+0.6030739}.
\end{equation}
Similarly, we can get $\delta_k, k\geq 5.$ For example,
$\delta_5=0.773114.$

\section{Acknowledgments}

The authors express their gratitude to the referee for a careful
reading of the manuscript and many valuable suggestions,  which
highly improve the quality of this paper.

Deyu Zhang is supported by National Natural Science Foundation of
China (Grant No.\ 11001154) and Shandong Province Natural Science
Foundation (ZR2010AQ009). Wenguang Zhai  is supported by National
Natural Science Foundation of China (Grant No.\ 11171344) and Natural
Science Foundation of Beijing (No.\ 1112010).





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{Article 07.3.3}.

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{Article 07.9.2}.

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{Article 01.2.2}.

\bibitem{s4}K. Broughan, The average order of the Dirichlet series
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07.4.2}.

\bibitem{s5} J. Chidambaraswamy and R. Sitaramachandrarao, Asymptotic
results for a class of arithmetical functions, \textit{Monatsh.
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\bibitem{s6} H. Iwaniec and E. Kowalski, {\it Analytic Number Theory},
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\bibitem{s7}A. Ivi\'c,  {\it The Riemann Zeta-Function},
John Wiley and Sons, New York, 1985; 2nd ed., Dover, 2003.


\bibitem{s8} S. S. Pillai, On an arithmetic function, \textit{J. Annamalai
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\bibitem{s9} L. T\'oth, A survey of gcd-sum functions,  \textit{J. Integer Sequences} \textbf{13} (2010),
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10.8.1}.

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08.2.3}.



\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}: Primary
11N37.

\noindent \emph{Keywords: } gcd-sum function,  Dirichlet divisor
problem, asymptotic formula.

\bigskip
\hrule
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\noindent (Concerned with sequence \seqnum{A018804}.)

\bigskip
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\vspace*{+.1in}
\noindent
Received January 10 2013;
revised versions received March 6 2013; June 14 2013. 
Published in {\it Journal of Integer Sequences}, July 25 2013.

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