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\begin{center}
\vskip 1cm{\LARGE\bf Mean Values of a Class of \\
\vskip .1in Arithmetical Functions} 
\vskip 1cm 
\large 
Deyu Zhang\footnote{This work is supported by National Natural Science
Foundation of China(Grant Nos.\ 10771127, 11001154)
and Shandong Province Natural Science Foundation (Nos.\ BS2009SF018, ZR2010AQ009).}\\
School of Mathematical Sciences\\
Shandong Normal University\\
Jinan 250014\\
Shandong\\ 
P. R. China\\
\href{mailto:zdy_78@yahoo.com.cn}{\tt zdy\_78@yahoo.com.cn}\\
\ \\
Wenguang Zhai$^1$\\
Department of Mathematics\\
China University of Mining and Technology \\
Beijing, 100083\\
P. R. China\\
\href{mailto:zhaiwg@hotmail.com}{\tt zhaiwg@hotmail.com} \\
\end{center}

\vskip .2in

\begin{abstract}
In this paper we consider a class of functions $\mathcal {U}$ of
arithmetical functions which include $\tilde{P}(n)/n$, where
$\tilde{P}(n):=n \prod_{p|n}(2-\frac{1}{p})$.  For any given
$U\in\mathcal {U}$, we obtain the asymptotic formula for
$\sum_{n\leq x}U(n)$, which improves a result of De Koninck and
K\'atai.
\end{abstract}

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\section{Introduction}

In 1933, Pillai \cite{s8} introduced the function
$$P(n)=\sum_{k=1}^{n}\gcd(k,n),$$
and proved that
$$P(n)=\sum_{d\mid n}d\varphi(n/d), \ \ \ \ \text{and} \ \ \ \ \sum_{d\mid
n}P(d)=n d(n)=\sum_{d\mid n}\sigma(d)\varphi(n/d),$$ where $\varphi$
is Euler's function, $d(n)$ and $\sigma(n)$ denote the number of
divisors of $n$ and the sum of the divisors of $n$ respectively.
Many authors investigated the properties of $P(n)$, see $\cite{s1,
s2, s3, s4, s5, s8, s11};$ 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 $e_1, e_2$ are computable constants and $0&lt;\theta&lt;1/2$ is some
exponent contained in
\begin{eqnarray}\label{eq:E1}
\sum_{n\leq x}d(n)=x\log
x+(2\gamma-1)x+O(x^{\theta+\epsilon}).\end{eqnarray} The asymptotic
formula \eqref{eq:E1} is the well-known Dirichlet divisor problem.
The latest value of $\theta$ is $\theta=131/416$ proved by
Huxley \cite{H}.


 T\'oth \cite{s10} first defined the gcd-sum function
over regular integers modulo $n$ by the relation
\begin{eqnarray}\label{eq:E2}
\tilde{P}(n)=\sum_{k\in {\rm Reg}_n } \gcd(k,n),
\end{eqnarray}
where ${\rm Reg}_n=\{k:1\leq k\leq n \ \text{and }k \ \ \text{is
regular (mod $n$})\}$, and proved that $\tilde{P}(n)$ is multiplicative
and for every $n\geq 1,$
\begin{eqnarray}\label{eq:E3}
\tilde{P}(n)=n \prod_{p|n}(2-\frac{1}{p}).
\end{eqnarray}
It is sequence \seqnum{A176345} in Sloane's Encyclopedia. He also
obtained the following asymptotic formula
\begin{eqnarray}\label{eq:E4}
\sum_{n\leq x}\tilde{P}(n)=\frac{x^{2}}{2\zeta(2)}(K_1 \log
x+K_2)+O(x^{3/2}\delta(x)),
\end{eqnarray}
where $K_1$ and $K_2$ are certain constants and $\delta(x)$  is
given by
$$\delta(x)=\exp(-A(\log x)^{3/5}(\log\log x)^{-1/5}).$$
Zhang and Zhai \cite{s13} showed that the estimate of $\sum_{n\leq
x}\tilde{P}(n)$ is closely related
 to the square-free divisor problem and improved the error term of
 \eqref{eq:E4} under RH.

 De Koninck and K\'atai \cite{s6} introduced two wide
 classes of arithmetical functions $\mathcal {R}$ and $\mathcal
 {U}$, the first of which includes the function $P(n)/n$, and the
 second of which includes $\tilde{P}(n)/n$. More precisely,
 the class $\mathcal {R}$ is made of the following functions
 $R$. Firstly let $\gamma(n)$ denote the kernel of $n\geq 2,$
 that is $\gamma(n)=\prod_{p\mid n}p$ (with $\gamma(1)=1$). Then,
 given an arbitrary positive constant $c$, an arbitrary real number
 $\alpha &gt;0$ and a multiplicative function $\kappa (n)$ satisfying
 $\mid \kappa(n)\mid\leq\frac{c}{\gamma(n)^{\alpha}}$ for all $n\geq
 2,$ let $R\in \mathcal {R}$ be defined by
\begin{eqnarray}\label{eq:E5}
R(n)=R_{\kappa,c,\alpha}(n):=d(n)\sum_{d\parallel n}\kappa
(d)=d(n)\prod_{p^{a}\parallel n}(1+\kappa (p^{a})).
\end{eqnarray}
It is easily seen that if we let $\kappa(p^{a})=-\frac{a/(a+1)}{p},$
then the corresponding function $R(n)$ is precisely $P(n)/n$.

De Koninck and K\'atai \cite{s6} showed  that
\begin{eqnarray}\label{eq:E6}
T(x):=\sum_{n\leq x}R(n)=A_0 x\log x+B_0 x+O(x^{\beta+\epsilon}),
\end{eqnarray}
with $$\beta=\begin{cases} \theta,
&amp;\text{if}\ \ \alpha\geq 1-\theta; \\
1-\alpha, &amp;\text{if}\ \ \alpha&lt; 1-\theta;
\end{cases}$$
where $\theta$ is the exponent in \eqref{eq:E1},  $A_0, B_0$ are
certain constants.




As for the class of functions $\mathcal
 {U}$, it is made of the functions
$$U(n)=U_{h,c,\alpha}(n):=2^{\omega(n)}\sum_{d\mid n}h(d),$$
where $\omega(n)$ stands for the number of distinct prime factors of
$n$, and $h$ is a multiplicative function satisfying
 $|h(n)|\leq\frac{c}{\gamma(n)^{\alpha}}$ for all $n\geq
 2.$ It is easily seen that by taking $h(p)=-\frac{1}{2p}$ and
$h(p^{a})=0,$ for $a\geq 2,$ we obtain the particular case
$U(n)=\tilde{P}(n)/n$. De Koninck and K\'atai \cite{s6} proved that
\begin{eqnarray}\label{eq:E7}
S(x):=\sum_{n\leq x}U(n)=t_1 x\log x+t_2 x+O(\frac{x}{\log x}),
\end{eqnarray}
where $t_1, t_2$ are certain constants.




 In this paper, we shall prove the following

 \begin{theorem}
\label{thm:1}   Suppose $0\leq \alpha&lt;1.$ Then we have
\begin{eqnarray}\label{eq:E8}
S(x)=t_1 x\log x+t_2 x+O(x^{1-\alpha+\epsilon}+x^{1/2+\epsilon}).
\end{eqnarray}

\end{theorem}
\begin{remark}
 (i) From our proof we see that the evaluation of $S(x)$ is
closely related to the distribution of the zeros of the Riemann zeta
function. The exponent $1/2$ can be reduced to  $4/11$ if RH is
true.


(ii) The exponent $1-\alpha $ in the error term of Theorem
\ref{thm:1} is best possible when $\alpha $ is small. For example,
if we take $h(n)=n^{-\alpha}$ with $0&lt;\alpha&lt;1/2$, then our proof
with slight modifications yields
$$\sum_{n\leq x}U(n)=t_1 x\log x+t_2
x+t_3 x^{1-\alpha}\log x+t_4
x^{1-\alpha}+O(x^{1/2+\epsilon}).$$\end{remark}

We are also interested in the short interval case. In this case, the
restrictions on $\alpha $ and RH can be removed. Actually, we have
the following Theorem \ref{thm:3}.

 \begin{theorem}
\label{thm:3} Suppose \eqref{eq:E1} holds for $1/4&lt;\theta &lt;1/3$.
Then for $x^{\theta+2\epsilon}\leq y\leq x,$ we have
\begin{eqnarray}\label{eq:E9}
\sum_{x&lt;n\leq
x+y}U(n)=H(x+y)-H(x)+O(yx^{-\frac{\epsilon}{2}}+x^{\theta+\epsilon}),
\end{eqnarray}
where $H(x)=t_1 x\log x+t_2 x$.
\end{theorem}

\section{\bf Preliminary Lemmas}

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


Let $s$ be a complex number with $\Re s&gt;1$. Then

$$ \sum_{n=1}^{\infty}\frac{U(n)}{n^{s}}=\frac{\zeta^{2}(s)}{\zeta(2s)}G(s),$$
where $G(s)$ can be written as a Dirichlet series $G(s)=
\sum\limits_{n=1}^{\infty}\frac{g(n)}{n^{s}}$, which is absolutely
convergent for $\Re s&gt;1-\alpha$. Moreover $g(n)$ satisfies
$|g(n)|\ll n^{-\alpha+\epsilon}$.

\end{lemma}
\begin{proof} For $\Re s&gt;1,$  by Euler product representation we
have
$$F(s):=\sum_{n=1}^{\infty}\frac{U(n)}{n^{s}}=\prod_{p}\left(1+\sum_{\beta=1}^{\infty}\frac{U(p^{\beta})}{p^{\beta s}}\right),$$
where $U(p^{\beta})=2(1+h(p)+\cdots+h(p^{\beta})), \beta \geq 1.$
Thus
 $$\begin{array}{lll}
 &amp;\displaystyle 1+\sum_{\beta=1}^{\infty}\frac{U(p^{\beta})}{p^{\beta s}}&amp;\displaystyle =1+\sum_{\beta=1}^{\infty}\frac{2}{p^{\beta
 s}}+2\sum_{\beta=1}^{\infty}p^{-\beta
 s}\sum_{j=1}^{\beta}h(p^{j})\\[12pt]
 &amp;\displaystyle  &amp;\displaystyle=\frac{1-p^{-2s}}{(1-p^{-s})^{2}}+2\sum_{\beta=1}^{\infty}p^{-\beta
 s}\sum_{j=1}^{\beta}h(p^{j})\\[12pt]
&amp;\displaystyle
&amp;\displaystyle=\frac{1-p^{-2s}}{(1-p^{-s})^{2}}\times\left(1+\frac{2(1-p^{-s})^{2}}{1-p^{-2s}}\sum_{\beta=1}^{\infty}p^{-\beta
 s}\sum_{j=1}^{\beta}h(p^{j})\right),
 \end{array}$$
hence we get
$$ \sum_{n=1}^{\infty}\frac{U(n)}{n^{s}}=\frac{\zeta^{2}(s)}{\zeta(2s)}G(s),$$
where
$$G(s)=\prod_{p}\left(1+\frac{2(1-p^{-s})^{2}}{1-p^{-2s}}\sum_{\beta=1}^{\infty}p^{-\beta
 s}\sum_{j=1}^{\beta}h(p^{j})\right).$$

From the above formula, it is easy to see that $G(s)$ can be
 expanded to a Dirichlet series $G(s)=\sum\limits_{n=1}^{\infty}\frac{g(n)}{n^{s}}$,
 which is absolutely
convergent for $\Re s&gt;1-\alpha$, if we notice that $|h(p)|\leq
\frac{c}{p^{\alpha}}$. Therefore $|g(n)|\ll n^{-\alpha+\epsilon}$.
\end{proof}

 \begin{lemma}\label{lem:5}
  Let
$$\sum_{n=1}^{\infty}\frac{d^{(2)}(n)}{n^{s}}=\frac{\zeta^{2}(s)}{\zeta(2s)},
\ \ \ \Re s&gt;1,
$$ where $d^{(2)}(n)$ denote the number of square-free divisors of $n$. Then for any real numbers $x\geq 1,$  we have
$$ D^{(2)}(x):=\sum_{n\leq x}d^{(2)}(n)=c_1x\log
x+c_2x+\Delta^{(2)}(x)$$ with $\Delta^{(2)}(x)=O(x^{1/2}\log x)$,
where $$c_1=\frac{1}{\zeta(2)},\ \ \
c_2=\frac{2\gamma-1}{\zeta(2)}-\frac{2\zeta'(2)}{\zeta^{2}(2)}.$$

\end{lemma}
Moreover, if RH is true, then $\Delta^{(2)}(x)=O(x^{4/11+\epsilon}
).$


\begin{proof} The first  result is due to
Mertens \cite{s7} and the second one is due to Baker \cite{s0}.
\end{proof}

 \begin{lemma}\label{lem:6}  $$ \sum_{n\leq x}|g(n)|\ll
 x^{1-\alpha+\epsilon}.$$
\end{lemma}
\begin{proof} It follows from $|g(n)|\ll n^{-\alpha+\epsilon}$.
\end{proof}

\begin{lemma}\label{lem:7}  Let $k\geq 2$ be a fixed integer , $1&lt;y\leq x$ be large
real numbers and
$${\cal A}(x,y;k,\epsilon):=\sum_{\stackrel{x&lt;nm^k\leq x+y}{m&gt;x^\epsilon}}1.$$
Then we have
$$
{\cal A}(x,y;k,\epsilon)\ll yx^{-\epsilon}+x^{1/4}.
$$
\end{lemma}
\begin{proof}  This is Lemma 3 of Zhai \cite{s12}.
\end{proof}


\section{\bf Proof of Theorem \ref{thm:1} }

 Notice that
\begin{eqnarray}\label{eq:E10}\frac{\zeta^{2}(s)}{\zeta(2s)}=
\sum\limits_{\ell=1}^{\infty}\frac{d^{(2)}(\ell)}{\ell^{s}}, \ \ \ G(s)=\sum\limits_{m=1}^{\infty}\frac{g(m)}{m^{s}}.
\end{eqnarray}
 By the Dirichlet
convolution, we have
$$\sum_{n\leq x}U(n)=\sum_{m\ell\leq x}g(m)d^{(2)}(\ell)=\sum_{m\leq x}g(m)\sum_{\ell\leq
x/m}d^{(2)}(l),$$ and  Lemma \ref{lem:5} applied to the inner sum
gives
$$
\sum_{n\leq x}U(n)=\sum_{m\leq
x}g(m)\left\{\frac{c_1x}{m}\log(\frac{x}{m})+\frac{c_2x}{m}+O\left((\frac{x}{m})^{1/2+\epsilon}\right)\right\}
$$
$$ =c_1 x\left\{\left(\log x+\frac{c_2}{c_1}\right)\sum_{m\leq
x}\frac{g(m)}{m}-\sum_{m\leq x}\frac{g(m)\log
m}{m}\right\}+O\left(x^{1/2+\epsilon}\sum_{m\leq
x}\frac{|g(m)|}{m^{1/2+\epsilon}}\right)$$
$$ =c_1 x\left\{\left(\log x+\frac{c_2}{c_1}\right)\sum\limits_{m=1}^{\infty}\frac{g(m)}{m}-\sum\limits_{m=1}^{\infty}\frac{g(m)\log
m}{m}+O(x^{-\alpha+\epsilon})\right\}+O\left(x^{1/2+\epsilon}\sum_{m\leq
x}\frac{|g(m)|}{m^{1/2+\epsilon}}\right),$$ if we notice by Lemma
\ref{lem:6} that both of the infinite series
$\sum_{m=1}^{\infty}\frac{g(m)}{m}, \ \ \
\sum_{m=1}^{\infty}\frac{g(m)\log m}{m} $ are absolutely convergent,
and
\begin{eqnarray}\label{eq:E11}\sum_{m&gt;
x}\frac{g(m)}{m}\ll x^{-\alpha+\epsilon}, \ \ \ \  \sum_{m&gt;
x}\frac{g(m)\log m}{m}\ll x^{-\alpha+\epsilon}.\end{eqnarray} Then
we have
\begin{eqnarray}\label{eq:E12}
\sum_{n\leq x}U(n)=t_1 x\log x+t_2
x+O(x^{1-\alpha+\epsilon})+O\left(x^{1/2+\epsilon}\sum_{m\leq
x}\frac{|g(m)|}{m^{1/2+\epsilon}}\right),\end{eqnarray} where
$$t_1
=\frac{1}{\zeta(2)}\sum\limits_{m=1}^{\infty}\frac{g(m)}{m}=\frac{G(1)}{\zeta(2)},\
\ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \  $$
$$\
\ \ \  \ \ \ \ \
t_2=\frac{1}{\zeta(2)}\left\{(2\gamma-1-\frac{2\zeta'(2)}{\zeta(2)})\sum\limits_{m=1}^{\infty}\frac{g(m)}{m}
-\sum\limits_{m=1}^{\infty}\frac{g(m)\log m}{m}\right\}$$

$$
=\frac{1}{\zeta(2)}\left\{(2\gamma-1-\frac{2\zeta'(2)}{\zeta(2)})G(1)
-G'(1)\right\}. \ \ \ \ \ \ \ \ $$ By Lemma \ref{lem:6}, we have
$$\sum_{m\leq x}\frac{|g(m)|}{m^{1/2+\epsilon}}\leq \sum_{m\leq
x}\frac{1}{m^{1/2+\alpha+\epsilon}}\leq \begin{cases} x^{\epsilon},
\ \ \ &amp;\alpha\geq 1/2; \\
x^{1/2-\alpha+\epsilon}, \ \ \ &amp;\alpha&lt;1/2,
\end{cases}$$
Theorem \ref{thm:1} follows from the above estimates and Eq.~\eqref{eq:E12}.






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

By Lemma \ref{lem:4}, we have
$$ U(n)=\sum_{n=n_1n_2n_3^{2}}d(n_1)g(n_2)\mu(n_3),$$
where $d(n)$ is the divisor function. Then
\begin{eqnarray}\label{eq:E13}
\sum_{x&lt;n\leq x+y}U(n)=\sum_{x&lt;n_1n_2n_3^{2}\leq
x+y}d(n_1)g(n_2)\mu(n_3) =\Sigma_1+O\left(\Sigma_2+\Sigma_3\right),
\end{eqnarray}
where
$$\ \ \ \ \Sigma_1=\sum_{{n_2\leq x^{\epsilon}}\atop{n_3\leq x^{\epsilon}}}g(n_2)\mu(n_3)\sum_{\frac{x}{n_2n_3^{2}}
&lt;n_1\leq \frac{x+y}{n_2n_3^{2}}}d(n_1),$$

$$\Sigma_2=\sum_{{x&lt;n_1n_2n_3^{2}\leq
x+y}\atop{n_2&gt; x^{\epsilon}}}d(n_1)|g(n_2)|,\ \ \ \ \ \ \ \ $$

$$\Sigma_3=\sum_{{x&lt;n_1n_2n_3^{2}\leq
x+y}\atop{n_3&gt; x^{\epsilon}}}d(n_1)|g(n_2)|.\ \ \ \ \ \ \ \  $$
Recalling \eqref{eq:E1}, the inner sum in $\Sigma_1$ is
$$
\frac{(x+y)}{n_2n_3^{2}}\log\frac{(x+y)}{n_2n_3^{2}}-
\frac{x}{n_2n_3^{2}}\log\frac{x}{n_2n_3^{2}}+(2\gamma-1)\frac{y}{n_2n_3^{2}}
+O\left(\frac{x^{\theta}}{n_2^{\theta}n_3^{2\theta}}\right)\ \ \ \ \
\ \ \ \ \
$$
$$
\ \ =\frac{(x+y)\log(x+y)-x\log x}{n_2n_3^{2}}- y\frac{\log
(n_2n_3^{2})}{n_2n_3^{2}}+(2\gamma-1)\frac{y}{n_2n_3^{2}}
+O\left(\frac{x^{\theta}}{n_2^{\theta}n_3^{2\theta}}\right).$$
Inserting the above expression into $\Sigma_1$ and after some easy
calculations, we get
\begin{eqnarray}\label{eq:E14}
\Sigma_1=H(x+y)-H(x)+O\left(yx^{-\epsilon}+y^{-\alpha\epsilon+\epsilon^{2}}+x^{\theta+\epsilon}\right).
\end{eqnarray}
For $\Sigma_2$, we have
$$|g(n_2)|\ll n_2^{-\alpha+\epsilon}\ll
x^{-\alpha\epsilon+\epsilon^{2}},$$ if we notice that
$n_2&gt;x^{\epsilon}$, and hence
$$\Sigma_2\ll x^{-\alpha\epsilon+\epsilon^{2}}\sum_{x&lt;n_1n_2n_3^{2}\leq
x+y}d(n_1)=x^{-\alpha\epsilon+\epsilon^{2}}\sum_{x&lt;n\leq
x+y}d_{\ast}(n),$$ where
$$d_{\ast}(n)=\sum_{n=n_1n_2n_3^{2}}
d(n_1)\ll n^{\epsilon^{2}}.$$ Therefore we have
\begin{eqnarray}\label{eq:E15}
\Sigma_2\ll x^{-\alpha\epsilon+\epsilon^{2}}\sum_{x&lt;n\leq
x+y}n^{\epsilon^{2}}\ll yx^{-\alpha\epsilon+\epsilon^{2}}.
\end{eqnarray}
Since $d(n)\ll n^{\epsilon^{2}}, \ \ g(n_2)\ll 1,$ by Lemma
\ref{lem:7} we have
$$\Sigma_3\ll x^{\epsilon^{2}}\sum_{{x&lt;n_1n_2n_3^{2}\leq
x+y}\atop{n_3&gt; x^{\epsilon}}}1 \ll
x^{\epsilon^{2}}\sum_{{x&lt;nn_3^{2}\leq x+y}\atop{n_3&gt;
x^{\epsilon}}}d(n)$$
$$ \ll x^{2\epsilon^{2}}\sum_{{x&lt;nn_3^{2}\leq
x+y}\atop{n_3&gt; x^{\epsilon}}}1 =x^{2\epsilon^{2}}{\cal
A}(x,y;2,\epsilon)$$
 \begin{eqnarray}\label{eq:E16}\ll
yx^{-\epsilon+2\epsilon^{2}}+x^{1/4+\epsilon^{2}}. \ \ \ \ \ \ \ \ \
\ \ \ \
\end{eqnarray}
Then Theorem \ref{thm:3} follows from Eqs.~\eqref{eq:E13}--\eqref{eq:E16}.

\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.


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

\noindent \emph{Keywords: } gcd-sum function, regular integers
modulo $n$, Riemann hypothesis, short interval result.

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\noindent (Concerned with sequences
\seqnum{A018804} and
\seqnum{A176345}.)

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\noindent
Received January 25 2011;
revised version received May 24 2011. 
Published in {\it Journal of Integer Sequences}, June 10 2011.

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