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\begin{center}
\vskip 1cm{\LARGE\bf Mean Values of a Gcd-Sum Function \\
\vskip .1in Over Regular Integers Modulo $n$} \vskip 1cm \large Deyu
Zhang and Wenguang Zhai\footnote{This work is supported by National
Natural Science Foundation of China(Grant Nos.\ 10771127, 10826028)
and Research Award Foundation for Young Scientists of Shandong
Province (No.\ BS2009SF018).}\\
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}\\
\href{mailto:zhaiwg@hotmail.com}{\tt zhaiwg@hotmail.com}\\
\end{center}

\vskip .2in

\begin{abstract}
In this paper we study the mean value of a gcd-sum function over
regular integers modulo $n$. In particular, we improve the previous
result under the Riemann hypothesis (RH). We also study the short
interval problem for it without assuming RH.
\end{abstract}

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\section{Introduction }


In general, an element $k$ of a ring $R$ is said to be (von Neumann)
regular if there is an $x\in R$ such that $k=kxk.$  Let $n&gt;1$ be an
integer with prime factorization $n=p_1^{\nu_1}\cdots p_r^{\nu_r}$.
An integer $k$ is called {\it regular}\ (mod $n$) if there
exists an integer $x$ such that $k^{2}x\equiv k \ (\text{mod $n$})$, i.e.,
the residue class of $k$ is a regular element (in the sense of J.
von Neumann) of the ring $ {\mathbb{Z}}_{n}$ of residue classes
(mod $n$).

 Let
$\text{Reg}_n=\{k:1\leq k\leq n \ \text{and }k \ \text{is
regular (mod $n$})\}$. T\'oth \cite{s10} first defined the
gcd-sum function over regular integers modulo $n$ by the relation
\begin{eqnarray}\label{eq:E1}
\tilde{P}(n)=\sum_{k\in \text{Reg}_{n}}\gcd(k,n),
\end{eqnarray}
where $\gcd(a,b)$ denotes the greatest common divisor of $a$ and
$b$.  It is sequence
\seqnum{A176345} in Sloane's Encyclopedia.
This is analogous to the gcd-function, called also Pillai's
arithmetical function,
$$P(n)=\sum_{k=1}^{n}\gcd(k,n),$$
which has been studied recently by several authors, see \cite{s1,
s2, s3, s4, s5, s8, s11}; it is Sloane's sequence
\seqnum{A018804}.  T\'oth \cite{s10} proved that
$\tilde{P}(n)$ is multiplicative and for every $n\geq 1,$
\begin{eqnarray}\label{eq:E2}
\tilde{P}(n)=n \prod_{p|n}(2-\frac{1}{p}).
\end{eqnarray}
He also obtained the following asymptotic formula
\begin{eqnarray}\label{eq:E3}
\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 the function $\delta(x)$ and constants $K_1$ and $K_2$ are
given by
$$\delta(x)=\exp(-A(\log x)^{3/5}(\log\log x)^{-1/5}),$$
\begin{eqnarray}\label{eq:E4} K_1=\sum_{n=1}^{\infty}\frac{\mu(n)}{n\psi(n)}=\prod_{p}\Bigg(1-\frac{1}{p(p+1)}\Bigg), \end{eqnarray}
\begin{eqnarray}\label{eq:E5} K_2=K_1\bigg(2\gamma-\frac{1}{2}-\frac{2\zeta'(2)}{\zeta(2)}\Bigg)-\sum_{n=1}^{\infty}\frac{\mu(n)(\log
n-\alpha(n)+2\beta(n))}{n\psi(n)},\end{eqnarray} where
$\psi(n)=n\prod_{p|n}(1+\frac{1}{p})$ denotes the Dedekind function,
and
$$\alpha(n)=\sum_{p|n}\frac{\log p}{p-1}, \ \ \beta(n)=\sum_{p|n}\frac{\log
p}{p^{2}-1}.$$

 It is very difficult to improve the exponent
$\frac{3}{2}$ in the error term of \eqref{eq:E3} unless we have
substantial progress in the study of the zero free region of
$\zeta(s).$ Therefore it is reasonable to get better improvements by
assuming the truth of the Riemann hypothesis (RH). Let $d(n)$ denote
the Dirichlet divisor function and
\begin{eqnarray}\label{eq:E6}
\Delta (x):=\sum_{n\leq x}d(n)-x(\log x+2\gamma -1).
\end{eqnarray}
Dirichlet first proved that $\Delta(x) = O(x^{1/2})$.  The exponent
$1/2$ was improved by many authors. The latest result reads
\begin{eqnarray}\label{eq:E7}
\Delta (x)\ll x^{\theta+\epsilon}, \ \ \ \ \ \theta=131/416,
\end{eqnarray}
due to Huxley \cite{s6}.  T\'oth \cite{s10} proved that if
RH is true,  then the error term of \eqref{eq:E3} can be  replaced
by $O(x^{(7-5\theta)/(5-4\theta)}\exp(B\log x(\log\log x)^{-1})).$
 For $\theta=131/416$ one has
$(7-5\theta)/(5-4\theta)\approx1.4505.$

  In this paper, we will use the Dirichlet convolution method to study the mean value of
$\tilde{P}(n)$, and we find that the estimate of $\sum_{n\leq
x}\tilde{P}(n)$ is closely related
 to the square-free divisor problem.
 Let $d^{(2)}(n)$ denote the number of square-free divisors of $n$.
 Note that $d^{(2)}(n)=2^{\omega(n)}$, where $\omega(n)$ is the number of distinct prime factors of $n$.
 Let

$$ D^{(2)}(x)=\sum_{n\leq x}d^{(2)}(n).$$
It was shown by Mertens \cite{s7} that
\begin{eqnarray}\label{eq:E8}
D^{(2)}(x)=\frac{1}{\zeta(2)}x\log
x+\Bigg(\frac{2\gamma-1}{\zeta(2)}-\frac{2\zeta'(2)}{\zeta^{2}(2)}\Bigg)x+\Delta^{(2)}(x),
\end{eqnarray}
where $\Delta^{(2)}(x)=O( x^{1/2}\log x)$.   The exponent $
\frac{1}{2}$ is also difficult to be improved, because it is related
to  the zero distribution of $\zeta(s).$ One way of making progress
is to assume the Riemann hypothesis (RH). Many authors investigated
this problem, and the best result under the Riemann hypothesis  is
\begin{eqnarray}\label{eq:E9} \Delta^{(2)}(x)\ll x^{\lambda+\epsilon},\end{eqnarray}
where $\lambda=4/11,$ due to Baker \cite{s0}.

 In this paper, we shall prove the following  results.

 \begin{theorem}
\label{thm:1}  {\it For any real numbers $x\geq 1$ and $\epsilon
&gt;0$, if $$\Delta^{(2)}(x)\ll
 x^{\lambda+\epsilon},$$ then we have
\begin{eqnarray}\label{eq:E10}
\sum_{n\leq x}\tilde{P}(n)=\frac{x^{2}}{2\zeta(2)}(K_1 \log x+K_2)
+O(x^{1+\lambda+\epsilon}),
\end{eqnarray}
where $K_1,\ K_2$ are defined by \eqref{eq:E4} and \eqref{eq:E5}. }
\end{theorem}

\begin{corollary}\label{cor:1}
 {\it If RH is true, then
\begin{eqnarray}\label{eq:E11}
\sum_{n\leq x}\tilde{P}(n)=\frac{x^{2}}{2\zeta(2)}(K_1 \log x+K_2)
+O(x^{15/11+\epsilon}).
\end{eqnarray}
}
\end{corollary}

\noindent {\bf Remark.} Note that $15/11\approx1.3636, $ which improves the
previous result.

In order to avoid assuming the truth of the Riemann hypothesis, we study
the short interval problem for it.

 \begin{theorem}
\label{thm:2}  {\it For

$$x^{\theta+3\epsilon}\leq y\leq x,$$
we have
 \begin{eqnarray} \label{eq:E12} \ \ \ \ \ \
\sum_{x&lt;n\leq
x+y}\tilde{P}(n)=\frac{1}{2\zeta(2)}\int_{x}^{x+y}u\left(2K_1\log
u+K_1+2K_2\right)du+O(yx^{1-\epsilon}+x^{1+\theta+2\epsilon}).\end{eqnarray}
where $\theta$ is defined by \eqref{eq:E7}.}
\end{theorem}
\begin{corollary}\label{cor:2}

{\it For
$$x^{131/416+3\epsilon}\leq y\leq x,$$
we have
 \begin{eqnarray}\label{eq:E13}  \ \ \ \ \ \
\sum_{x&lt;n\leq
x+y}\tilde{P}(n)=\frac{1}{2\zeta(2)}\int_{x}^{x+y}u\left(2K_1\log
u+K_1+2K_2\right)du+O(yx^{1-\epsilon}+x^{\frac{547}{416}+2\epsilon}).\end{eqnarray}

}
 \end{corollary}
{\bf Notation.} Throughout the paper $\epsilon$ always denotes
 a fixed but sufficiently small positive constant. We
write $f(x)\ll g(x)$, or $f(x)=O(g(x))$, to mean that $|f(x)|\leq C
g(x)$. For any fixed integers $1\leq a\leq b$ ,we consider the
divisor function
$$d(a,b;n)=\sum\limits_{n=m^{a}k^{b}}1.$$


\section{\bf Proof of Theorem  \ref{thm:1}}
Let $s$ be complex numbers with $\Re s&gt;1$. We consider the mean
value of the  arithmetic function $
\tilde{P}^{\ast}(n)=\frac{\tilde{P}(n)}{n}. $
 Define
\begin{eqnarray}\label{eq:E14}F(s):=\sum_{n=1}^{\infty}\frac{\tilde{P}^{\ast}(n)}{n^{s}}.\end{eqnarray}
By Euler product representation we have
\begin{eqnarray*}
F(s)&amp;&amp; =\prod _{p}\left(1+\frac{2p-1}{p^{s+1}}+\frac{2p^{2}-p}{p^{2s+2}}+\frac{2p^{3}-p^{2}}{p^{3s+3}}+\cdots\right)\\
&amp;&amp; =\zeta(s)\prod
_{p}\left(1-\frac{1}{p^{s}}\right)\left(1+\frac{2}{p^{s}}-\frac{1}{p^{s+1}}+\frac{2}{p^{2s}}-\frac{1}{p^{2s+1}}+\cdots\right)\\
&amp;&amp;=\zeta(s)\prod_{p}\left(1+\frac{1}{p^{s}}-\frac{1}{p^{s+1}}\right)\\
&amp;&amp; =\frac{\zeta^{2}(s)}{\zeta(2s)}\prod
_{p}\left(1-\frac{1}{p^{s}}\right)\prod
_{p}\left(1-\frac{1}{p^{2s}}\right)^{-1}\left(1+\frac{1}{p^{s}}-\frac{1}{p^{s+1}}\right)\\
&amp;&amp; =\frac{\zeta^{2}(s)}{\zeta(2s)}G(s),
\end{eqnarray*}
where \begin{eqnarray}\label{eq:E15}G(s)=\prod
_{p}\left(1-\frac{1}{p^{s+1}+p}\right).\end{eqnarray} From the above
formula, it is easy to see that $G(s)$ can be expanded to a
Dirichlet series, which is absolutely convergent for $\Re s&gt;0$.
Write
\begin{eqnarray}\label{eq:E16}G(s)=\sum\limits_{n=1}^{\infty}\frac{g(n)}{n^{s}},\end{eqnarray}
then we can easily get
\begin{eqnarray}\label{eq:E17} g(n)\ll n^{\epsilon}, \ \ \ \ \ \ \sum_{n\leq x}|g(n)|=O(x^{\epsilon}).\end{eqnarray}
Notice that
\begin{eqnarray}\label{eq:E18}\frac{\zeta^{2}(s)}{\zeta(2s)}=\sum\limits_{m=1}^{\infty}\frac{d^{(2)}(m)}{m^{s}}.\end{eqnarray}
 By the Dirichlet
convolution, we have
$$\sum_{n\leq x}\tilde{P}^{\ast}(n)=\sum_{m\ell\leq x}d^{(2)}(m)g(\ell)=\sum_{\ell\leq x}g(\ell)\sum_{m\leq
x/\ell}d^{(2)}(m),$$ and formula \eqref{eq:E8} applied to the inner
sum gives
$$
\sum_{n\leq x}\tilde{P}^{\ast}(n)=\sum_{\ell\leq
x}g(\ell)\left\{\frac{x}{\zeta(2)\ell}\left(\log(\frac{x}{\ell})+2\gamma-1-\frac{2\zeta'(2)}{\zeta(2)}\right)+O\left((\frac{x}{\ell})^{\lambda+\epsilon}\right)\right\}
$$
$$ =\frac{x}{\zeta(2)}\left\{\left(\log x+2\gamma-1-\frac{2\zeta'(2)}{\zeta(2)}\right)\sum_{\ell\leq
x}\frac{g(\ell)}{\ell}-\sum_{\ell\leq x}\frac{g(\ell)\log
\ell}{\ell}\right\}+O\left(x^{\lambda+\epsilon}\sum_{\ell\leq
x}\frac{|g(\ell)|}{\ell^{\lambda+\epsilon}}\right).$$
$$ =\frac{x}{\zeta(2)}\left\{\left(\log x+2\gamma-1-\frac{2\zeta'(2)}{\zeta(2)}\right)\sum\limits_{\ell=1}^{\infty}\frac{g(\ell)}{\ell}
-\sum\limits_{\ell=1}^{\infty}\frac{g(\ell)\log
\ell}{\ell}+O(x^{-1+\epsilon})\right\}+O\left(x^{\lambda+\epsilon}\right),$$
if we notice by \eqref{eq:E17} that both of the infinite series
$\sum\limits_{\ell=1}^{\infty}\frac{g(\ell)}{\ell}$ and
$\sum\limits_{\ell=1}^{\infty}\frac{g(\ell)\log \ell}{\ell}$ are
absolutely convergent.

From \eqref{eq:E15}, \eqref{eq:E16} and the definitions of $K_1,
K_2$, we have
\begin{eqnarray}\label{eq:E19}\sum\limits_{\ell=1}^{\infty}\frac{g(\ell)}{\ell}=G(1)=\prod
_{p}\left(1-\frac{1}{p^{2}+p}\right)=K_1,\end{eqnarray}

\begin{eqnarray}\label{eq:E20}\sum\limits_{\ell=1}^{\infty}\frac{g(\ell)\log \ell}{\ell}=\sum_{n=1}^{\infty}\frac{\mu(n)(\log
n-\alpha(n)+2\beta(n))}{n\psi(n)}\end{eqnarray}
$$\ \ \ \ \ \ \ \ \  =K_1\left(2\gamma-\frac{1}{2}-\frac{2\zeta'(2)}{\zeta(2)}\right)-K_2.     $$
Then
\begin{eqnarray}\label{eq:E21}
\sum_{n\leq x}\tilde{P}^{\ast}(n)=\frac{x}{\zeta(2)}\left((\log
x-\frac{1}{2})K_1+K_2\right)+O(x^{\lambda+\epsilon}).
\end{eqnarray}
From the definitions of $\tilde{P}^{\ast}(n)$and  Abel's summation
formula, we can easily get

\begin{eqnarray*}\sum_{n\leq x}\tilde{P}(n)&amp;&amp;=\sum_{n\leq
x}\tilde{P}^{\ast}(n)n=\int_{1}^{x}t d\left(\sum_{n\leq
t}\tilde{P}^{\ast}(n)\right)\\
&amp;&amp;=\frac{x^2}{2\zeta(2)}\left(K_1\log
x+K_2\right)+O(x^{1+\lambda+\epsilon}).
\end{eqnarray*}

Corollary \ref{cor:1} follows by taking $\lambda=4/11.$


\section{\bf Proof of Theorem  \ref{thm:2}}
From the proof of Theorem \ref{thm:1}, we have
\begin{eqnarray}\label{eq:E22}F(s)=\sum_{n=1}^{\infty}\frac{\tilde{P}^{\ast}(n)}{n^{s}}=\frac{\zeta^{2}(s)}{\zeta(2s)}G(s).\end{eqnarray}
Let
\begin{eqnarray}\label{eq:E23}\zeta^{2}(s)G(s)=\sum\limits_{n=1}^{\infty}\frac{h(n)}{n^{s}}, \ \ \ \Re
s&gt;1.\end{eqnarray} Then we have

\begin{lemma} \label{lem:1}{\it For any real numbers $x\geq 1$ and $\epsilon
&gt;0$, we have
\begin{eqnarray}\label{eq:E24}
\sum_{n\leq x}h(n)=x\left(\left(\log
x-\frac{1}{2}+\frac{2\zeta'(2)}{\zeta(2)}\right)K_1+K_2\right)+O(x^{\theta+\epsilon}),
\end{eqnarray}
where $\theta$ is defined in \eqref{eq:E7}.}
\end{lemma}
\begin{proof}
Recall  that$$\sum\limits_{n=1}^{\infty}\frac{g(n)}{n^{s}}=G(s), \ \
\ g(n)\ll n^{\epsilon}.$$ Then we have
\begin{eqnarray}\label{eq:E25}h(n)=\sum_{n=m\ell}d(m)g(\ell), \ \ \ h(n)\ll n^{\epsilon}.\end{eqnarray}
Thus from \eqref{eq:E6},\eqref{eq:E7} we get
\begin{eqnarray*}
\sum_{n\leq x}h(n)&amp;&amp;=\sum_{m\ell\leq x}d(m)g(\ell)=\sum_{\ell\leq
x}g(\ell)\sum_{m\leq \frac{x}{\ell}}d(m) \\
&amp;&amp;=\sum_{\ell\leq x}g(\ell)\left\{\frac{x}{\ell}\left(\log
(\frac{x}{\ell})+2\gamma-1\right)+O\left((\frac{x}{\ell})^{\theta+\epsilon}\right)\right\} \\
&amp;&amp;=x\left\{\left(\log x+2\gamma-1\right)\sum_{\ell\leq
x}\frac{g(\ell)}{\ell}-\sum_{\ell\leq x}\frac{g(\ell)\log
\ell}{\ell}\right\}+O\left(x^{\theta+\epsilon}\sum_{\ell\leq
x}\frac{|g(\ell)|}{\ell^{\theta+\epsilon}}\right)\\
&amp;&amp;=x\left\{\left(\log
x+2\gamma-1\right)\sum\limits_{\ell=1}^{\infty}\frac{g(\ell)}{\ell}
-\sum\limits_{\ell=1}^{\infty}\frac{g(\ell)\log
\ell}{\ell}+O(x^{-1+\epsilon})\right\}+O\left(x^{\theta+\epsilon}\right)\end{eqnarray*}
Then Lemma~\ref{lem:1} follows from the above formula and
\eqref{eq:E19}, \eqref{eq:E20}.

\end{proof}


\begin{lemma}\label{lem:2} { \it For any real numbers $x\geq 1$ and $x&lt;u\leq 2x$, we
have
\begin{eqnarray}\label{eq:E26}
\sum_{x&lt;n\leq u}\tilde{P}^{\ast}(n)=M(u)-M(x)+E(u,x),
\end{eqnarray}
where $$ M(x)=\frac{x}{\zeta(2)}\left((\log
x-\frac{1}{2})K_1+K_2\right)$$ is the main term of $\sum_{n\leq
x}\tilde{P}^{\ast}(n)$, and $$E(u,x)\ll
(u-x)x^{-\epsilon}+x^{\theta+2\epsilon}.$$}
\end{lemma}
\begin{proof}
From \eqref{eq:E22} and \eqref{eq:E23}, we have
$$\tilde{P}^{\ast}(n)=\sum_{n=\ell m^2}h(\ell)\mu(m).$$
Then
\begin{eqnarray}\label{eq:E27}
\sum_{x&lt;n\leq u}\tilde{P}^{\ast}(n)=\sum_{x&lt;\ell m^2\leq
u}h(\ell)\mu(m) =\sum_1+\sum_2,
\end{eqnarray}
where
$$\ \ \ \ \sum_1=\sum_{m\leq x^{2\epsilon}}\mu(m)\sum_{\frac{x}{m^2}
&lt;\ell\leq \frac{u}{m^2}}h(\ell),$$

$$\sum_2=\sum_{{x&lt;\ell m^2\leq u}\atop{m&gt; x^{2\epsilon}}}h(\ell)\mu(m).\ \ \ \ \ \ $$
By Lemma~\ref{lem:1} we have
\begin{eqnarray}\label{eq:E28}
\sum_1=\sum_{m\leq
x^{2\epsilon}}\mu(m)\left(H(\frac{u}{m^2})-H(\frac{x}{m^2})+O(\frac{x}{m^2})^{\theta+\epsilon}\right)
\end{eqnarray}
$$=\sum_{m\leq
x^{2\epsilon}}\mu(m)\left(H(\frac{u}{m^2})-H(\frac{x}{m^2})\right)+O(x^{\theta+2\epsilon}),
$$
where $$H(x):=ax\log x+bx$$ is the main term of $\sum_{n\leq
x}h(n)$, and $a=K_1, \ \
b=\left(\frac{2\zeta'(2)}{\zeta(2)}-\frac{1}{2}\right)K_1+K_2$. Then
$$\sum_{m\leq
x^{2\epsilon}}\mu(m)\left(H(\frac{u}{m^2})-H(\frac{x}{m^2})\right)\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$

$$=\sum_{m\leq
x^{2\epsilon}}\mu(m)\left(\frac{H(u)-H(x)}{m^2}+\frac{2(ax-au)}{m^2}\log
m \right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \
$$
 $$=\left(H(u)-H(x)\right)\sum_{m\leq
x^{2\epsilon}}\frac{\mu(m)}{m^2}+2(ax-au)\sum_{m\leq
x^{2\epsilon}}\frac{\mu(m)\log m}{m^2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ $$

$$=\left(H(u)-H(x)\right)\sum_{m=1}^{\infty}\frac{\mu(m)}{m^2}+2(ax-au)\sum_{m=1}^{\infty}\frac{\mu(m)\log
m}{m^2}+O\left((u-x)x^{-2\epsilon}\right).
$$
It is well known that
$$ \frac{1}{\zeta(s)}=\sum_{m=1}^{\infty}\frac{\mu(m)}{m^s}, \ \ \ \
\Re s&gt;1,$$ which gives by differentiation
$$\frac{\zeta'(s)}{\zeta^2(s)}=\sum_{m=1}^{\infty}\frac{\mu(m)\log
m}{m^s},$$ and hence
$$\sum_1=\frac{H(u)-H(x)}{\zeta
(2)}+2(ax-au)\frac{\zeta'}{\zeta^{2}}(2)+O(x^{\theta+2\epsilon}+(u-x)x^{-2\epsilon})
$$
\begin{eqnarray}\label{eq:E29}
=M(u)-M(x)+O\left(x^{\theta+2\epsilon}+(u-x)x^{-2\epsilon}\right), \
\ \ \ \ \ \ \ \ \ \ \
\end{eqnarray}
where $$ M(x)=\frac{x}{\zeta(2)}\left((\log
x-\frac{1}{2})K_1+K_2\right).$$

For $\sum_2$, if we notice that $h(n)\ll n^{\epsilon},$ then
\begin{eqnarray}\label{eq:E30}
\sum_2\ll x^{\epsilon}\sum_{{x&lt;\ell m^2\leq u}\atop{m&gt;
x^{2\epsilon}}}1:=x^{\epsilon}\sum_3,
\end{eqnarray}
where \begin{eqnarray}\label{eq:E31} \sum_3=\sum_{{x&lt;\ell m^2\leq
u}\atop{m&gt; x^{2\epsilon}}}1=\sum_{x&lt;\ell m^2\leq u}1-\sum_{{x&lt;\ell
m^2\leq u}\atop{m\leq x^{2\epsilon}}}1\end{eqnarray}
$$\ \ \ \ \ \ \ \ \ \ \ \ =\sum_{x&lt;n\leq u}d(1,2;n)-\sum_{{x&lt;\ell m^2\leq
u}\atop{m\leq x^{2\epsilon}}}1=\sum_{31}-\sum_{32},$$

\noindent say. From Richert \cite{s9} we have
$$\sum_{n\leq x}d(1,2;n)=\zeta (2)x+\zeta(1/2)x^{1/2}+O(x^{2/9}\log
x).$$ Then
\begin{eqnarray}\label{eq:E32}
\sum_{31}=\zeta(2)(u-x)+O\left((u-x)x^{-1/2}+x^{2/9}\log x\right).
\end{eqnarray}
For $\sum_{32}$ we have
\begin{eqnarray}\label{eq:E33}
\sum_{32}=\sum_{m\leq x^{2\epsilon}}\sum_{\frac{x}{m^2}&lt;\ell\leq
\frac{u}{m^2}}1=\sum_{m\leq
x^{2\epsilon}}\left(\frac{u-x}{m^2}+O(1)\right)
\end{eqnarray}
$$=\zeta(2)(u-x)+O\left((u-x)x^{-2\epsilon}+x^{2\epsilon}\right).$$
Then from \eqref{eq:E31}--\eqref{eq:E33} we have
\begin{eqnarray}\label{eq:E34}
\sum_3 \ll (u-x)x^{-2\epsilon}+x^{2/9}\log x,
\end{eqnarray}
and hence

\begin{eqnarray}\label{eq:E35}
\sum_2 \ll (u-x)x^{-\epsilon}+x^{2/9+\epsilon}.
\end{eqnarray}
Lemma~\ref{lem:2} follows from \eqref{eq:E27}, \eqref{eq:E29} and
\eqref{eq:E35}.
\end{proof}

Now we prove Theorem \ref{thm:2}.  From the definitions of
$\tilde{P}^{\ast}(n)$ and  Abel's summation formula, we have
$$\sum_{x&lt;n\leq x+y}\tilde{P}(n)=\sum_{x&lt;n\leq x+y}\tilde{P}^{\ast}(n)n=\int_{x}^{x+y}u d\left(\sum_{x&lt;n\leq
u}\tilde{P}^{\ast}(n)\right),$$ and Lemma~\ref{lem:2} applied to the
sum in the right side gives
\begin{eqnarray}\label{eq:E36}\sum_{x&lt;n\leq
x+y}\tilde{P}(n)=\int_1+\int_2,\end{eqnarray} where

\begin{eqnarray*}\int_1&amp;&amp;=\int_{x}^{x+y}u
d\left(M(u)-M(x)\right),\\
\int_2&amp;&amp;=\int_{x}^{x+y}u d\left(E(u,x)\right).
\end{eqnarray*}
In view of the definition of $M(x)$ in Lemma~\ref{lem:2}, we obtain
\begin{eqnarray}\label{eq:E37}
\int_1=\int_{x}^{x+y}u
M'(u)du=\frac{1}{2\zeta(2)}\int_{x}^{x+y}u\left(2K_1\log
u+K_1+2K_2\right)du.
\end{eqnarray}
For $\int_2$,  we integrate it  by parts,  to get
\begin{eqnarray*}\int_2&amp;&amp;=\int_{x}^{x+y}u d\left(E(u,x)\right)\\
&amp;&amp;=(x+y)E(x+y;x)-\int_{x}^{x+y}E(u,x)du.
\end{eqnarray*}
By Lemma~\ref{lem:2} we get
$$E(u,x)\ll
(u-x)x^{-\epsilon}+x^{\theta+2\epsilon}.$$ Therefore
\begin{eqnarray}\label{eq:E38}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \int_2\ll
x(yx^{-\epsilon}+x^{\theta+2\epsilon})+\int_{x}^{x+y}\left((u-x)x^{-\epsilon}+x^{\theta+2\epsilon}\right)du\end{eqnarray}
\begin{eqnarray*}
&amp;&amp;\ll
yx^{1-\epsilon}+x^{1+\theta+2\epsilon}+y^{2}x^{-\epsilon}+yx^{\theta+2\epsilon}\ \ \ \ \ \ \ \ \ \ \\
&amp;&amp; \ll yx^{1-\epsilon}+x^{1+\theta+2\epsilon},\ \ \ \ \ \ \ \ \ \
\end{eqnarray*}
if we notice that $y\leq x.$

Now Theorem \ref{thm:2} follows from \eqref{eq:E36}--\eqref{eq:E38}.
If we take $\theta=131/416$, then  we can get Corollary \ref{cor:2}.

\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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\end{thebibliography}

\bigskip
\hrule
\bigskip

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

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

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

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received January 7 2010;
revised version received April 15 2010.  
Published in {\it Journal of Integer Sequences}, April 15 2010.

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\noindent
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