\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://www.research.att.com/cgi-bin/access.cgi/as/~njas/sequences/eisA.cgi?Anum=#1}{\underline{#1}}} \DeclareMathOperator*{\Res}{Res} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{defin}[theorem]{Definition} \newenvironment{definition}{\begin{defin}\normalfont\quad}{\end{defin}} \newtheorem{examp}[theorem]{Example} \newenvironment{example}{\begin{examp}\normalfont\quad}{\end{examp}} \newtheorem{rema}[theorem]{Remark} \newenvironment{remark}{\begin{rema}\normalfont\quad}{\end{rema}} \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} \newcommand{\bvec}[1]{\mbox{\boldmath$#1$}} \renewcommand{\d}{\displaystyle} \renewcommand\thefootnote{\fnsymbol{footnote}} \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 $n>1$ 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 $\epsilon >0$, 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_{x1$. 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)&& =\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)\\ && =\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)\\ &&=\zeta(s)\prod_{p}\left(1+\frac{1}{p^{s}}-\frac{1}{p^{s+1}}\right)\\ && =\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)\\ && =\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 $\Re s>0$. 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)&&=\sum_{n\leq x}\tilde{P}^{\ast}(n)n=\int_{1}^{x}t d\left(\sum_{n\leq t}\tilde{P}^{\ast}(n)\right)\\ &&=\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>1.\end{eqnarray} Then we have \begin{lemma} \label{lem:1}{\it For any real numbers $x\geq 1$ and $\epsilon >0$, 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)&&=\sum_{m\ell\leq x}d(m)g(\ell)=\sum_{\ell\leq x}g(\ell)\sum_{m\leq \frac{x}{\ell}}d(m) \\ &&=\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\} \\ &&=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)\\ &&=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 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>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).$$ For $\sum_2$, if we notice that $h(n)\ll n^{\epsilon},$ then \begin{eqnarray}\label{eq:E30} \sum_2\ll x^{\epsilon}\sum_{{x<\ell m^2\leq u}\atop{m> x^{2\epsilon}}}1:=x^{\epsilon}\sum_3, \end{eqnarray} where \begin{eqnarray}\label{eq:E31} \sum_3=\sum_{{x<\ell m^2\leq u}\atop{m> x^{2\epsilon}}}1=\sum_{x<\ell m^2\leq u}1-\sum_{{x<\ell m^2\leq u}\atop{m\leq x^{2\epsilon}}}1\end{eqnarray} $$\ \ \ \ \ \ \ \ \ \ \ \ =\sum_{x