<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\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}

\begin{center}
\vskip 1cm{\LARGE\bf  Some Arithmetic Functions Involving\\
\vskip .1in
Exponential Divisors} \vskip 1cm \large
Xiaodong Cao\\
Department of Mathematics and Physics\\
Beijing Institute of Petro-Chemical Technology\\
Beijing, 102617\\
P. R. China \\
\href{mailto:tanigawa@math.nagoya-u.ac.jp}{\tt caoxiaodong@bipt.edu.cn}\\

\ \\

Wenguang Zhai\\
School of Mathematical Sciences \\
Shandong Normal University \\
Jinan, 250014\\
Shandong \\
 P. R. China\\
\href{mailto:zhaiwg@hotmail.com}{\tt zhaiwg@hotmail.com} \\
\end{center}

\vskip .2 in

\begin{abstract}
In this paper we study   several arithmetic functions connected with
the exponential divisors of integers. We establish some asymptotic
formulas under the Riemann hypothesis, which improve previous
results. We also prove some asymptotic lower bounds.
\end{abstract}

\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{conjecture}[theorem]{Conjecture}

\newtheorem{thm}{Theorem}
\newtheorem{cor}{Corollary}
\newtheorem{lem}{Lemma}[section]
\newcommand{\bvec}[1]{\mbox{\boldmath$#1$}}
\renewcommand{\d}{\displaystyle}
\renewcommand\thefootnote{\fnsymbol{footnote}}



%************************* section 1 *******************************************

\section{Introduction and results}


Suppose $n&gt;1$ is an integer with prime factorization
$n=p^{a_1}_{1}\cdots p^{a_r}_{r}$. An integer $d$ is called an
exponential divisor (e-divisor) of $n$ if $d=p^{b_1}_{1}\cdots
p^{b_r}_{r}$ with $b_j|a_{j}(1\leq j\leq r),$ which is denoted by
the notation $d|_e n$. For convenience $1|_e1$. The properties of
the exponential divisors attract the interests of many authors(see,
for example, \cite{ha, ks1, ks2, lu, ns, pw, sa, sw, ss, to1, to2,
to3, wu}).

An integer $n=p^{a_1}_{1}\cdots p^{a_r}_{r}$ is called exponentially
squarefree (e-squarefree) if all the exponents $a_1,\cdots, a_r$ are
squarefree. The integer $1$ is also considered to be an e-squarefree
number.

Suppose $f$ and $g$ are two arithmetic functions and
$n=p^{a_1}_{1}\cdots p^{a_r}_{r}$. Subbarao \cite{su} first
introduced the exponential convolution (e-convolution) by
$$
(f\odot
g)(n)=\sum_{b_1c_1=a_1}\cdots\sum_{b_rc_r=a_r}f(p^{b_1}_{1}\cdots
p^{b_r}_{r})g(p^{c_1}_{1}\cdots p^{c_r}_{r}),
$$
which is an analogue of the classical Dirichlet convolution.  The
e-convolution $\odot$ is commutative, associative and has the
identity element $\mu^2$, where $\mu$ is the M\"obius function.
Furthermore, a function $f$ has an inverse with respect to $\odot$
iff $f(1)\neq 0$ and $f(p_1\cdots p_r)\neq 0$ for any distinct
primes $p_1,\cdots ,p_r$.

The inverse of the constant  function $f(n)\equiv 1$ with respect to
$\odot$ is called the exponential analogue of the M\"obius function
and it is denoted by $\mu^{(e)}$. Hence
$\sum_{d|_en}\mu^{(e)}(d)=\mu^2(n)$ for $n\geq 1$, $\mu^{(e)}(1)=1$,
and $\mu^{(e)}(n)=\mu (a_1)\cdots \mu(a_r)$ for $n=p^{a_1}_{1}\cdots
p^{a_r}_{r}&gt;1$. Note that
 $|\mu^{(e)}(n)|=1$ or $0$,  according as  $n$ is  e-squarefree or not.

An integer $d$ is called an exponential squarefree exponential
divisor (e-squarefree e-divisor) of $n$ if $d=p^{b_1}_{1}\cdots
p^{b_r}_{r}$ with $b_j|a_{j}(1\leq j\leq r)$ and $b_1,\cdots,b_r$
are squarefree. Observe that the integer $1$ is an e-squarefree and
it is not an e-divisor of $n&gt;1$. Let $t^{(e)}$ denote the number of
e-sqaurefree e-divisors of $n$.


Now we introduce some notation for later use. As usual, let $\mu(n)$
and $\omega(n)$ denote the M\"obius function , and the number of
distinct prime factors of $n$ respectively.  If $t$ is real, then
$\{t\}$ denotes the fractional part of $t,\psi(t)= \{t\}-1/2$.
Throughout this paper, $ \varepsilon$ is a small fixed positive
constant and $m\sim M$ means that $cM\leq m\leq CM$ for some
constants $0&lt;c&lt;C$.       For any fixed integers $a$ and $b,$ define
the function $d(a,b;n):=\sum_{m_1^am_2^b=n}1$ and let
$\Delta(a,b;t)$ denote the error term in the asymptotic formula of
the sum $\sum_{n\leq x}d(a,b;n)$.

Many authors have studied the properties of the above three
functions; see, for example, \cite{ha, ks2, sw, su, to1, to2, wu}.
T\'{o}th \cite{to2} proved that
\begin{equation}
\label{eq:E1} \sum_{n\le x}\mu^{(e)}(n)=A_1x+(x^{1/2}exp(-c_1(\log
x)^{\Delta})),
\end{equation}
where $0&lt;\Delta&lt;9/25$ and $c_1&gt;0$ are constants and
\begin{eqnarray*}
A_1:=m(\mu^{(e)})=\prod_{p}\left(1+\sum_{k=2}^{\infty}\frac
{\mu(k)-\mu(k-1)}{p^k}\right).\end{eqnarray*}
T\'{o}th \cite{to2} proved that
if the Riemann hypothesis (RH) is true, then
\begin{equation}
\label{eq:E2}
 \sum_{n\le x}\mu^{(e)}(n)=A_1x+O(x^{91/202+\varepsilon}).
\end{equation}

Subbarao \cite{su} and Wu \cite{wu} studied the asymptotic properties
of the  sum $\sum_{n\leq x}|\mu^{e}(n)|$.
T\'{o}th \cite{to2} proved
that if RH is true, then
\begin{equation}\label{eq:E3}
\sum_{n\leq x}|\mu^{(e)}(n)|=B_1x+O(x^{1/5+\varepsilon}),
\end{equation}
 where
$$B_1:=\prod_{p}(1+\sum_{\alpha=4}^\infty\frac{\mu^2(a)-\mu^2(a-1)}{p^a}).$$


For the function $t^{(e)}(n),$ T\'{o}th \cite{to2} proved the
asymptotic formula
\begin{equation}\label{eq:E4}
\sum_{n\leq x}t^{(e)}(n)=C_1x+C_2x^{1/2}+O(x^{1/4+\varepsilon}),
\end{equation}
where
\begin{eqnarray*}
&amp;&amp;C_1:=\prod_{p}\left(1+\sum_{k=2}^{\infty}\frac
{2^{\omega(k)}-2^{\omega(k-1)}}{p^k}\right), \\&amp;&amp;C_2:=\zeta(\frac
12)\prod_{p}\left(1+\sum_{k=4}^{\infty}\frac
{2^{\omega(k)}-2^{\omega(k-1)}-2^{\omega(k-2)}+2^{\omega(k-3)}}{p^k}\right).
\end{eqnarray*}
P\'{e}termann \cite{pe} proved the formula \eqref{eq:E4} with a
better error term $O(x^{1/4})$, which is the   best unconditional
result up to date. In order to further reduce the exponent $1/4$, we
have to know more information about the distribution of the
non-trivial zeros of the Riemann zeta-function.



In this short paper we shall prove the following theorem.

\begin{theorem}
\label{thm:1}  {\it If RH is true,   then
\begin{eqnarray}
  \sum_{n\le x}\mu^{(e)}(n)&amp;&amp;=A_1x+O\left(
x^{\frac{37}{94}+\varepsilon} \right),\label{eq:E5}\\
 \sum_{n\le x}|\mu^{(e)}(n)|&amp;&amp;=B_1x+B_2x^{\frac 15}+O(x^{\frac {38}{193}+\varepsilon})
 ,\label{eq:E6}\\
\sum_{n\leq x}t^{(e)}(n)&amp;&amp;=C_1x+C_2x^{1/2}+O\left(x^{\frac
{3728}{15469}+\varepsilon}\right),\label{eq:E7}
\end{eqnarray}
where $B_2$ is a computable constant.}
\end{theorem}



{\bf Remark 1.} Numerically, we have
\begin{eqnarray*}
&amp;&amp;\frac {37}{94}=0.39361\cdots, \frac {91}{202}=0.45049\cdots,\\
&amp;&amp;\frac{38}{193}=0.1968\cdots&lt;1/5,
\frac{3728}{15469}=0.240\cdots&lt;1/4.
\end{eqnarray*}

 Let $\Delta_{\mu^{(e)}}(x),
\Delta_{|\mu^{(e)}|}(x), \Delta_{t^{(e)}}(x)$ denote the error terms
in \eqref{eq:E5}, \eqref{eq:E6} and \eqref{eq:E7} respectively. We
have the following result.

\begin{theorem}\label{thm:2} {\it We have
\begin{eqnarray}
\Delta_{\mu^{(e)}}(x)&amp;=\Omega(x^{1/4}),\label{eq:E8}\\
\Delta_{|\mu^{(e)}|}(x)&amp;=\Omega(x^{1/8}), \label{eq:E9}\\
\Delta_{t^{(e)}}(x)&amp;=\Omega(x^{1/6}).\label{eq:E10}
\end{eqnarray}}
\end{theorem}



%********************  section 2  *****************************************




\section{\bf Some generating functions}

 In this section we shall study the generating Dirichlet series of the
 functions
$\mu^{(e)}(n), |\mu^{(e)}(n)|$ and $t^{(e)}(n)$, respectively. We
consider only   $|\mu^{(e)}(n)|$ and $t^{(e)}(n)$, since T\'{o}th
\cite{to2} already proved the following formula, which is
  enough for our purpose,
\begin{equation}\label{eq:E11}
\sum_{n=1}^\infty\frac{\mu^{(e)}(n)}{n^s}=\frac{\zeta(s)}{\zeta^2(2s)}U(s)\
(\Re s&gt;1),
\end{equation}
where $U(s):=\sum_{n=1}^\infty\frac{u(n)}{n^s}$ is absolutely
convergent for $\Re s&gt;1/5.$


 We first
consider   the function $|\mu^{(e)}(n)|$.  The function $\mu^{(e)}$
is multiplicative  and $\mu^{(e)}(p^a)=\mu (a)$ for every prime
power $p^a$, namely  for every prime $p$, $\mu^{(e)}(p)=1,
\mu^{(e)}(p^2)=-1, \mu^{(e)}(p^3)=-1, \mu^{(e)}(p^4)=0,
\mu^{(e)}(p^5)=-1, \mu^{(e)}(p^6)=1,
 \mu^{(e)}(p^7)=1, \mu^{(e)}(p^8)=0, \mu^{(e)}(p^9)=0, \mu^{(e)}(p^{10})=1, \mu^{(e)}(p^{11})=-1,
\cdots.$ Hence by the Euler product  we have for $\Re s&gt;1$ that
\begin{eqnarray}\label{eq:E12}
\sum_{n=1}^{\infty}\frac{|\mu^{(e)}(n)|}{n^s}=\prod_{p}\left(1+\sum_{m=1}^{\infty}\frac
{|\mu (m)|}{p^{ms}}\right).
\end{eqnarray}
Applying the product representation of Riemann zeta-function
\begin{eqnarray}\label{eq:E13}
\zeta
(s)=\prod_{p}(1+p^{-s}+p^{-2s}+\cdots)=\prod_{p}(1-p^{-s})^{-1}\ (
\Re s&gt;1),
\end{eqnarray}
we have for $\Re s&gt;1$
\begin{eqnarray}\label{eq:E14}\zeta (s)\zeta
(5s)=\prod_{p}\left((1-p^{-s})(1-p^{-5s})\right)^{-1}.
\end{eqnarray}
Let
\begin{eqnarray}
f_{|\mu^{(e)}|}(z):&amp;&amp;=1+\sum_{m=1}^{\infty}|\mu (m)|z^m\label{eq:E15}\\
&amp;&amp;=1+z+z^2+z^3+z^5+z^6+z^7+z^{10}+z^{11}+ \sum_{m=12}^{\infty} |\mu
(m)|z^m.\nonumber
\end{eqnarray}
For $|z|&lt;1$, it is easy to verify that
\begin{eqnarray}
 &amp;&amp;f_{|\mu^{(e)}|}(z)(1-z)(1-z^5)\label{eq:E16}\\
&amp;&amp;=\left(1+z+z^2+z^3+z^5+z^6+z^7+z^{10}+z^{11}+ \sum_{m=12}^{\infty}
|\mu
(m)|z^m\right)(1-z-z^5+z^6)\nonumber\\
&amp;&amp;=1-z^4-z^{8}+z^9+\sum_{m=12}^{\infty}c_m^{(1)}z^m,\nonumber
\end{eqnarray}
where
\begin{eqnarray*}
c_m^{(1)}:=&amp;&amp;|\mu(m)|-|\mu(m-1)|-|\mu(m-5)|+|\mu(m-6)|.\ (m\ge 12)
\end{eqnarray*}
Furthermore, we have
\begin{eqnarray}\label{eq:E17}
 &amp;&amp;f_{|\mu^{(e)}|}(z)(1-z)(1-z^5)(1-z^4)^{-1}(1-z^8)^{-1}\\
&amp;&amp;=\left(1-z^4-z^{8}+z^9+\sum_{m=12}^{\infty}c_m^{(1)}z^m\right)(1+z^4+z^8+\cdots)(1+z^8+z^{16}+\cdots)\nonumber\\
&amp;&amp;=1+z^9+\sum_{m=12}^{\infty}C_m^{(1)}z^m.\nonumber
\end{eqnarray}

  We get from \eqref{eq:E12}, \eqref{eq:E13}, \eqref{eq:E15} and
\eqref{eq:E17} by taking $z=p^{-s}$ that
\begin{eqnarray}\label{eq:E18}
\sum_{n=1}^{\infty}\frac{|\mu^{(e)}(n)|}{n^s}=\frac{\zeta (s)\zeta
(5s)}{\zeta(4s)\zeta(8s)}V(s)\ ( \Re s &gt;1),
\end{eqnarray}
where $V(s):=\sum_{n=1}^{\infty}\frac{v(n)}{n^s}$ is absolutely
convergent for $\Re s&gt;\frac {1}{9}$.

We now consider the function $t^{(e)}(n)$, which is also
multiplicative  and $t^{(e)}(p^a)=2^{\omega (a)}$ for every prime
power $p^a$. Therefore for every prime $p$,
$t^{(e)}(p)=1,t^{(e)}(p^2)=t^{(e)}(p^3)=t^{(e)}(p^4)=t^{(e)}(p^5)=2,
 t^{(e)}(p^6)=4,t^{(e)}(p^7)=t^{(e)}(p^8)=t^{(e)}(p^9)=2,t^{(e)}(p^{10})=4,\cdots.$

By the Euler product  we have for $\Re s&gt;1$ that
\begin{eqnarray}\label{eq:E19}
\sum_{n=1}^{\infty}\frac{t^{(e)}(n)}{n^s}=\prod_{p}\left(1+\sum_{m=1}^{\infty}\frac
{ 2^{\omega (m)}}{p^{ms}}\right).
\end{eqnarray}

 Let
\begin{eqnarray}\label{eq:E20}
&amp;&amp;\ \ \ \ \ f_{t^{(e)}}(z):=1+\sum_{m=1}^{\infty}2^{\omega (m)}z^m \\
&amp;&amp;=1+z+2z^2+2z^3+2z^4+2z^5+4z^6+2z^7+2z^8+2z^9+4z^{10}+2z^{11}+\sum_{m=12}^{\infty}
2^{\omega (m)}z^m.\nonumber
\end{eqnarray}
For $|z|&lt;1$, it is easy to check that
\begin{eqnarray*}
 &amp;&amp;f_{t^{(e)}}(z)(1-z)(1-z^2)(1-z^6)^2\\
&amp;&amp;=1 - z^4 - 2z^7 - 2z^8+2z^9+\sum_{m=10}^{\infty}c_m^{(2)}z^m
\end{eqnarray*}
and
\begin{eqnarray}\label{eq:E21}
 &amp;&amp;f_{t^{(e)}}(z)(1-z)(1-z^2)(1-z^6)^2(1-z^4)^{-1}\\
&amp;&amp;=\left(1 - z^4 - 2z^7 -
2z^8+2z^9+\sum_{m=10}^{\infty}c_m^{(2)}z^m\right)(1+z^4+z^8+\cdots)\nonumber\\
&amp;&amp;=1 - 2z^7+\sum_{m=8}^{\infty}C_m^{(2)}z^m.\nonumber
\end{eqnarray}
Combining \eqref{eq:E13} and \eqref{eq:E19}--\eqref{eq:E21} we get
\begin{eqnarray}\label{eq:E22}
\sum_{n=1}^{\infty}\frac{t^{(e)}(n)}{n^s}=\frac{\zeta (s)\zeta
(2s)\zeta^2(6s)}{\zeta(4s)}W(s)\ ( \Re s &gt;1),
\end{eqnarray}
where $W(s):=\sum_{n=1}^{\infty}\frac{w(n)}{n^s}$ is absolutely
convergent for $\Re s&gt;\frac {1}{7}$.


%**************************\section 3***********************************************

\section{\bf   Proof of Theorem \ref{thm:2}}

 We see from \eqref{eq:E11}  that the generating Dirichlet series of the function $\mu^{(e)}(n)$ is  $\frac{\zeta
(s)}{\zeta^2(2s)}U(s)$, which has infinitely many  poles on the
 line $\Re s =\frac 14,$ whence the  estimate   \eqref{eq:E8} follows. From the expression \eqref{eq:E18}
  we see that the generating Dirichlet
 series of the function $|\mu^{(e)}(n)|$ is  $\frac{\zeta
(s)\zeta(5s)}{\zeta(4s)}V(s)$, which has infinitely many poles on
the line $\Re s =\frac 18,$ whence   the  estimate  \eqref{eq:E9}
follows. Via \eqref{eq:E22}, we get
  the  estimate  \eqref{eq:E10}  with the help of 
   Theorem 2 of K\"{u}leitner and
 Nowak \cite{kn} (or by Balasubramanian, Ramachandra and Subbarao's method
 in \cite{brs}).


%******************************\section 4*********************************************************

\section{\bf The Proofs of \eqref{eq:E5}  and \eqref{eq:E7}}

We follow  the approach of Theorem 1 in Montgomery and
Vaughan \cite{mv}. Throughout this section, we assume RH.

We first prove \eqref{eq:E5}. It is well-known that the
characteristic function of the set of squarefree integers is
\begin{eqnarray}\label{eq:E23}
\mu^2(n)=|\mu(n)|=\sum_{d^2|n}\mu(d),
\end{eqnarray}
We write
\begin{eqnarray}\label{eq:E24}
\Delta(x):=\sum_{n\le x}\mu^2
(d)-\frac{x}{\zeta(2)}:=D(x)-\frac{x}{\zeta(2)}.
\end{eqnarray}


Define the function $a_1(n)$ by
\begin{eqnarray}\label{eq:E25}
a_1(n)=\sum_{md^2=n}\mu^2(m)\mu (d),
\end{eqnarray}
it is easy to see that
\begin{eqnarray}\label{eq:E26}
\frac{\zeta(s)}{\zeta^2(2s)}=\frac{\sum_{n=1}^{\infty}\frac{\mu^2(n)}{n^s}}{\zeta(2s)}=\sum_{n=1}^{\infty}\frac{a_1(n)}{n^s},
\Re s &gt;1,
\end{eqnarray}

Thus
\begin{eqnarray}\label{eq:E27}
T_1(x):=\sum_{n\le x}a_1(n)=\sum_{md^2\le x}\mu^2(m)\mu
(d)=\sum_{d\le x^{\frac 12}}\mu (d)D(\frac {x}{d^2}).
\end{eqnarray}

Suppose $1&lt;y&lt;x^{1/2}$ is a parameter to be determined. We now write
$T_1(x)$ in the form
\begin{eqnarray}\label{eq:E28}
T_1(x)=S_1(x)+S_2(x),
\end{eqnarray}
where
\begin{eqnarray}\label{eq:E29}
S_1(x)=\sum_{d\le y}\mu (d)D\left(\frac {x}{d^2}\right),
\end{eqnarray}
and
\begin{eqnarray}\label{eq:E30}
S_2(x)=\sum_{y&lt;d\le x^{\frac 12}}\mu (d)D\left(\frac
{x}{d^2}\right)=\sum_{ \stackrel { md^2\le x}{ d&gt;y}
 } \mu^2(m)\mu(d).
\end{eqnarray}

We first evaluate $S_1(x)$. From \eqref{eq:E24}  we get

\begin{eqnarray}\label{eq:E31}
S_1(x)&amp;&amp;=\sum_{d\le y}\mu (d)\left(\frac
{x}{d^2\zeta(2)}+\Delta\left(\frac {x}{d^2}\right)\right)\\
&amp;&amp;=\frac {x}{\zeta (2)}\sum_{d\le y}\frac {\mu(d)}{d^2}+\sum_{d\le
y}\mu (d)\Delta\left(\frac {x}{d^2}\right).\nonumber
\end{eqnarray}

To treat $S_2(x)$, we let
\begin{eqnarray}\label{eq:E32}
g_y(s)=\zeta^{-1}(s)-\sum_{d\le y}\frac
{\mu(d)}{d^{s}},(s=\sigma+it),
\end{eqnarray}
so that for $\Re s=\sigma&gt;1$
\begin{eqnarray}\label{eq:E33}
g_y(s)=\sum_{d&gt; y}\frac {\mu(d)}{d^{s}}.\end{eqnarray}
 Hence
\begin{eqnarray}\label{eq:E34}
\frac {g_y(2s)\zeta(s)}{\zeta
(2s)}=\sum_{n=1}^{\infty}\frac{b_1(n)}{n^s},
\end{eqnarray}
for $\sigma &gt;1$, where
\begin{eqnarray}\label{eq:E35}
b_1(n)=\sum_{ \stackrel { md^2=n}{ d&gt;y}
 } \mu^2(m)\mu(d)
\end{eqnarray}
 Let $\frac 12&lt;\sigma &lt;2$, $\delta=\frac{\varepsilon}{10}$.
Assume RH, from 14.25 in Titchmarsh \cite{ti} we have
\begin{eqnarray}\label{eq:E36}
\sum_{d\le y}\frac {\mu(d)}{d^{s}}=\zeta^{-1}(s)+O\left(y^{\frac 12
-\sigma +\delta}(|t|^\delta +1)\right).
\end{eqnarray}
In addition, RH implies that $\zeta (s)\ll |t|^\delta+1$ and
$\zeta^{-1}(s)\ll (|t|^\delta+1)$ uniformly for $\sigma &gt;\frac 12
+\delta$ and $ |s-1|&gt;\varepsilon$, and $\sum_{n\le y}\mu(n)\ll
y^{\frac 12+\delta}$. From \eqref{eq:E32}, \eqref{eq:E33} and
\eqref{eq:E36}, we have
$$g_y(2s)\ll y^{-\frac 12}(|t|^\delta +1),( \sigma \ge \frac 12 +\delta).$$
Thus
\begin{eqnarray}\label{eq:E37}
g_y(2s)\frac {\zeta (s)}{\zeta (2s)}\ll y^{-\frac 12}(|t|^{3\delta}
+1),( \sigma \ge \frac 12 +\delta,\ |s-1|&gt;\varepsilon).
\end{eqnarray}
 From \eqref{eq:E30}, \eqref{eq:E34}, \eqref{eq:E35} and Perron's
formula(\cite{ti}, Lemma 3.19), we obtain
\begin{eqnarray}\label{eq:E38}
S_2(x)=\sum_{n\le x}b_1(n)=\frac {1}{2\pi
i}\int_{1+\varepsilon-ix^2}^{1+\varepsilon+ix^2}g_y(2s)\frac {\zeta
(s)}{\zeta (2s)}x^ss^{-1}\mathrm{d}s +O(x^\delta),
\end{eqnarray}
since $b(n)\ll n^\delta$ by a divisor argument. If we move the line
of integration to $\sigma =\frac 12+\delta$, then by the residue
theorem
\begin{eqnarray}\label{eq:E39}
&amp;&amp;\frac {1}{2\pi
i}\int_{1+\varepsilon-ix^2}^{1+\varepsilon+ix^2}g_y(2s)\frac {\zeta
(s)}{\zeta
(2s)}x^ss^{-1}\mathrm{d}s\\
&amp;&amp;=I_1+I_2-I_3+\Res_{s=1}\ g_y(2s)\frac {\zeta (s)}{\zeta
(2s)}x^ss^{-1},\nonumber
\end{eqnarray}
where
\begin{eqnarray*}
&amp;&amp;I_1=\frac {1}{2\pi i}\int_{\frac
12+\delta+ix^2}^{1+\varepsilon+ix^2}g_y(2s)\frac {\zeta (s)}{\zeta
(2s)}x^ss^{-1}\mathrm{d}s, I_2=\frac {1}{2\pi i}\int_{\frac
12+\delta-ix^2}^{\frac 12+\delta +ix^2}g_y(2s)\frac {\zeta
(s)}{\zeta (2s)}x^ss^{-1}\mathrm{d}s,\\
&amp;&amp;I_3=\frac {1}{2\pi i}\int_{\frac 12+\delta
-ix^2}^{1+\varepsilon-ix^2}g_y(2s)\frac {\zeta (s)}{\zeta
(2s)}x^ss^{-1}\mathrm{d}s.
\end{eqnarray*}
 From \eqref{eq:E37} it is not difficult to see that
\begin{eqnarray}\label{eq:E40}
I_j\ll y^{-\frac 12}x^{\frac 12+8\delta},(j=1,2,3).
 \end{eqnarray}

Combining \eqref{eq:E38}--\eqref{eq:E40}, we get
\begin{eqnarray}\label{eq:E41}
S_2(x)=\frac {x}{\zeta (2)}\sum_{d&gt; y}\frac {\mu(d)}{d^2}+O\left(
y^{-\frac 12}x^{\frac 12+\varepsilon}+x^\varepsilon\right).
\end{eqnarray}

Finally, combining \eqref{eq:E27}, \eqref{eq:E28}, \eqref{eq:E31}
and \eqref{eq:E41} we obtain
\begin{eqnarray}\label{eq:E42}
T_1(x)=\frac {x}{\zeta^2 (2)}+\sum_{d\le y}\mu (d)\Delta\left(\frac
{x}{d^2}\right)+O\left( x^{\frac 12+\varepsilon}y^{-\frac
12}+x^\varepsilon\right).
\end{eqnarray}


In \cite{ji}, Jia proved the  estimate  $\Delta (u)\ll
u^{\frac{17}{54}+\varepsilon}$. Inserting this estimate  into
\eqref{eq:E42} and on taking $y=x^{\frac{10}{57}}$,  we get
\begin{eqnarray}\label{eq:E43}
T_1(x)=\frac {x}{\zeta^2 (2)}+O\left( x^{\frac{37}{94}+\varepsilon}
\right).
\end{eqnarray}

The asymptotic formula \eqref{eq:E5} follows form
\eqref{eq:E25}-\eqref{eq:E27} and \eqref{eq:E43} by the well-known
convolution method.


Now we prove the asymptotic formula \eqref{eq:E7}.  We define the
functions $a_2(n)$ by the following
 identity
\begin{eqnarray}\label{eq:E44}
\frac{\zeta(s)\zeta(2s)}{\zeta(4s)}=\sum_{n=1}^{\infty}\frac{a_2(n)}{n^s}
,\ \Re s&gt;1.
\end{eqnarray}
Hence
$$a_2(n)=\sum_{d^4m=n}\mu (d) d(1,2;m).$$
Thus by the same approach  as \eqref{eq:E42}, we easily get that for
$1\le y\le x^{\frac 14}$
\begin{eqnarray}
&amp;&amp;T_2(x):=\sum_{n\le x}a_2(n)\label{eq:E45}\\
&amp;&amp;=\frac {\zeta (2)}{\zeta (4)}x+\frac{\zeta(\frac 12)}{\zeta
(2)}x^{\frac 12}+\sum_{d\le y}\mu(d)\Delta\left(1,2;\frac
{x}{d^4}\right)+O\left(x^{\frac 12+\varepsilon}y^{-\frac
32}+x^\varepsilon\right).\nonumber
\end{eqnarray}

Graham and Kolesnik \cite{gk} proved that $\Delta(1,2;u)\ll u^{\frac
{1057}{4785}+\varepsilon}$. Inserting this estimate into
\eqref{eq:E45} and on taking $y=^{\frac {2671}{15469}}$, we get

\begin{eqnarray}\label{eq:E46}
T_2(x)=\sum_{n\le x}a_2(n)=\frac {\zeta (2)}{\zeta
(4)}x+\frac{\zeta(\frac 12)}{\zeta (2)}x^{\frac 12}+O\left(x^{\frac
{3728}{15469}+\varepsilon}\right).
\end{eqnarray}

The asymptotic formula  \eqref{eq:E7} follows from \eqref{eq:E22},
\eqref{eq:E44} and \eqref{eq:E46} by the convolution method.


%*************************\section 5****************************************************************

\section{\bf Proof of \eqref{eq:E6}}

Throughout this section we assume RH.

 Define the function $a_3(n)$ by
\begin{equation}\label{eq:E47}
\frac{\zeta(s)\zeta(5s)}{\zeta(4s)}=\sum_{n=1}^{\infty}\frac{a_3(n)}{n^s},
\ \Re s&gt;1. \end{equation}


Similar to (45) we have for some $1\le y\le x^{\frac 14}$ that
\begin{eqnarray*}
&amp;&amp;T_3(x):=\sum_{n\le x}a_3(n)\\
&amp;&amp;=\frac {\zeta (5)}{\zeta (4)}x+\frac{\zeta(\frac 15)}{\zeta (\frac
45)}x^{\frac 15}+\sum_{d\le y}\mu(d)\Delta\left(1,5;\frac
{x}{d^4}\right)+O\left(x^{\frac 12+\varepsilon}y^{-\frac
32}+x^{\frac 15+\varepsilon}y^{-\frac
{3}{10}}+x^\varepsilon\right).\nonumber
\end{eqnarray*}
Unfortunately, we can not improve T\'{o}th's exponent $\frac 15$ in
\eqref{eq:E3} by the above formula even if we use the conjectural
bound $\Delta(1,5;t)\ll t^{1/12+\varepsilon}.$ So we use a different
approach to prove \eqref{eq:E6}.


Let $q_4(n)$ denote the characteristic function of  the set of
$4$-free numbers, then
$$
\sum_{n=1}^{\infty}\frac{q_4(n)}{n^s}=\frac
{\zeta(s)}{\zeta(4s)},\Re s&gt;1.
$$
We write
\begin{eqnarray}\label{eq:E48}
\Delta_4 (x):=\sum_{n\le 1}q_4(n)-\frac {x}{\zeta(4)}.
\end{eqnarray}

From  \eqref{eq:E47} and \eqref{eq:E48}, it follows from the
Dirichlet hyperbolic approach that for some $1\le y\le x^{\frac
15},$
\begin{eqnarray}\label{eq:E49}
&amp;&amp;B_3(x)=\sum_{n\le x}a_3(n)=\sum_{md^5\le x}q_4(m)\\
&amp;&amp;=\sum_{d\le y}\sum_{m\le \frac {x}{d^5}}q_4(m)+\sum_{m\le \frac
{x}{y^5}}q_4(m)\sum_{y&lt;d\le (\frac xm)^{\frac 15}}1\nonumber\\
&amp;&amp;=\frac {x}{\zeta(4)}\sum_{d\le y}\frac {1}{d^5}+x^{\frac
15}\sum_{m\le \frac {x}{y^5}}\frac {q_4(m)}{m^{\frac
15}}-y\sum_{m\le \frac
{x}{y^5}}q_4(m)\nonumber\\
&amp;&amp;+\sum_{d\le y}\Delta_4\left(\frac {x}{d^5}\right)- \sum_{m\le
\frac {x}{y^5}}q_4(m)\psi\left((\frac x m)^{\frac 15}\right)+ \psi
(y)\sum_{m\le \frac {x}{y^5}}q_4(m).\nonumber
\end{eqnarray}

Applying partial summation formula, we get
\begin{eqnarray}\label{eq:E50}
&amp;&amp;\sum_{m\le \frac {x}{y^5}}\frac {q_4(m)}{m^{\frac 15}}\\
=&amp;&amp;\frac {\sum_{m\le \frac {x}{y^5}}q_4(m)}{(\frac{x}{y^5})^{\frac
15 }}+\frac 15 \int_{1}^{\frac {x}{y^5}}\frac {\sum_{m\le
t}q_4(m)}{t^{\frac 65 }}\mathrm{d}t\nonumber\\
=&amp;&amp;\frac {y}{x^{\frac 15}}\sum_{m\le \frac {x}{y^5}}q_4(m)+\frac
15 \int_{1}^{\frac {x}{y^5}}\frac {\frac {t}{\zeta(4)}+\Delta_4(t)}{t^{\frac 65 }}\mathrm{d}t\nonumber\\
=&amp;&amp;\frac {y}{x^{\frac 15}}\sum_{m\le \frac {x}{y^5}}q_4(m)+\frac
{1}{4\zeta(4)}\left(\frac{x^{\frac 45}}{y^4}-1\right)+\frac 15
\int_{1}^{\frac {x}{y^5}}\frac {\Delta_4(t)}{t^{\frac 65
}}\mathrm{d}t.\nonumber
\end{eqnarray}

In addition, we have by Euler-Maclaurin formula
\begin{eqnarray}\label{eq:E51}
\sum_{d\le y}\frac {1}{d^5}=\zeta (5)-\frac
{1}{4y^4}-\psi(y)y^{-5}+O(y^{-6}).
\end{eqnarray}

If RH is true, Graham and Pintz \cite{gp} showed that

\begin{eqnarray}\label{eq:E52}
\Delta_4 (x)=O(x^{\frac {7}{38}+\varepsilon}),
\end{eqnarray}
 and so
\begin{eqnarray}\label{eq:E53}
\int_{1}^{\frac {x}{y^5}}\frac {\Delta_4(t)}{t^{\frac 65
}}dt&amp;&amp;=\int_{1}^{\infty}\frac {\Delta_4(t)}{t^{\frac 65
}}\mathrm{d}t-\int_{\frac {x}{y^5}}^{\infty}\frac
{\Delta_4(t)}{t^{\frac 65
}}\mathrm{d}t\\
&amp;&amp;=\int_{1}^{\infty}\frac {\Delta_4(t)}{t^{\frac 65
}}\mathrm{d}t+O\left(x^{-\frac{3}{190}+\varepsilon}y^{\frac{3}{38}}\right).\nonumber
\end{eqnarray}

Combining \eqref{eq:E49}--\eqref{eq:E53} and taking on
$y=x^{31/193}$, we obtain
\begin{eqnarray}\label{eq:E54}
T_3(x)&amp;&amp;=\sum_{md^5\le x}q_4(m)\\
&amp;&amp;=\frac {\zeta (5)}{\zeta(4)}x+\left(\frac 15\int_{1}^{\infty}\frac
{\Delta_4(t)}{t^{\frac 65
}}\mathrm{d}t-\frac{1}{4\zeta(4)}\right)x^{\frac 15}\nonumber\\
&amp;&amp;\ \ \ - \sum_{m\le \frac {x}{y^5}}q_4(m)\psi\left((\frac x
m)^{\frac
15}\right)+O\left(x^{\frac{7}{38}+\varepsilon}y^{\frac{3}{38}}+xy^{-6}\right)\nonumber\\
&amp;&amp;=\frac {\zeta
(5)}{\zeta(4)}x+C_3x^{1/5}+O(x^{\frac{7}{38}+\varepsilon}y^{\frac{3}{38}}+xy^{-5}),\nonumber\\
&amp;&amp;=\frac {\zeta (5)}{\zeta(4)}x+C_3x^{1/5}+O(x^{\frac{38}{193}
+\varepsilon} ),\nonumber
\end{eqnarray}
where   we used the trivial estimate  $$\sum_{m\le \frac
{x}{y^5}}q_4(m)\psi\left((\frac x m)^{\frac 15}\right)\ll
\frac{x}{y^5}$$ and where $$C_3:= \frac 15\int_{1}^{\infty}\frac
{\Delta_4(t)}{t^{\frac 65 }}\mathrm{d}t-\frac{1}{4\zeta(4)}. $$


Now the asymptotic formula \eqref{eq:E6} follows from
\eqref{eq:E18}, \eqref{eq:E47}, \eqref{eq:E49} and \eqref{eq:E54} by
the convolution approach.




\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.

This work is supported by National Natural Science Foundation of China
(Grant No.\ 10771127) and  Research Award Foundation for  Young
Scientists of Shandong Province (No.\ BS2009SF018).

\begin{thebibliography}{99}


\bibitem{brs} R. Balasubramanian, K. Ramachandra and M. V. Subbarao,
On the error function in the asymptotic formula for the counting
function of $k$-full numbers, {\it Acta Arith.\/} {\bf 50} (1988),
107--118.

\bibitem{gk}  S. W. Graham and G. Kolesnik,  On the difference between consecutive
 squarefree integers, {\it Acta Arith.\/}  {\bf 49} (1988), 435--447.

\bibitem{gp}S. W. Graham and J. Pintz, The distribution of $r$-free
numbers, {\it Acta Math. Hungar.\/} {\bf 53} (1989), 213--236.

\bibitem{ha}P. Jr. Hagis, Some results concerning exponential
divisors, {\it Internat. J. Math. Math. Sci.\/} {\bf 11} (1988),
343--349.



\bibitem{ji} Chaohua Jia, The distribution of square-free numbers,
{\it Sci.  China Ser. A\/} {\bf 36} (1993), 154--169.

\bibitem{ks1}I. K\'{a}tai and M. V. Subbarao, On the iterates of the sum of
exponential divisors, {\it Math. Pannon.\/} {\bf 10} (1999),
153--158.

\bibitem{ks2}I. K\'{a}tai and M. V. Subbarao, On the distribution of exponential
divisors,  {\it Ann. Univ. Sci. Budapest. Sect. Comput.\/} {\bf
22} (2003), 161--180.


\bibitem{kn}M. K\"{u}leitner and W. G. Nowak, An omega theorem for a
class of arithmetic functions, {\it Math. Nachr.\/} {\bf 165} (1994),
79--98.

\bibitem{lu}L. Lucht, On the sum of exponential divisors and its
iterates, {\it Arch. Math. (Basel)\/}, {\bf 27} (1976), 383--386.

\bibitem{mv} H. L. Montgomery and R. C. Vaughan, On the distribution
of square-free numbers, in {\it Recent Progress in Analytic Number
Theory}, Vol.\ 1, Academic Press,
1981, pp.\ 247--256.

\bibitem{ns} W. G. Nowak and M. Schmeier, Conditional asymptotic
fomula for a class of arithmetic functions, {\it Proc. Amer. Math.
Soc.\/} {\bf 103} (1988), 713--717.

\bibitem{pw}Y.-F. S. P\'{e}termann and J. Wu, On the sum of exponential divisors of
an integer, {\it Acta Math. Hungar.\/} {\bf 77} (1997), 159--175.

\bibitem{pe}Y.-F. S. P\'{e}termann, Arithmetical functions
involving exponential divisors: Note on two paper by L. T\'{o}th,
{\it Ann. Univ. Sci. Budapest. Sect. Comput.\/}, to appear.

\bibitem{sa} J. S\'{a}ndor, A note on exponential divisors and related
arithmetic functions, {\it Sci. Magna\/} {\bf 1} (2005), 97--101.

\bibitem{sw} A. Smati and J. Wu, On the exponential divisor function, {\it Publ. Inst.
Math. (Beograd) (N.S.)\/}, {\bf 61} (1997), 21--32.

\bibitem{ss}E. G. Straus and M. V. Subbarao,  On exponential divisors, {\it Duke Math.
J.\/} {\bf 41} (1974), 465--471.

\bibitem{su} M. V. Subbarao, On some arithmetic convolutions, in
{\it The Theory of Arithmetic Functions}, Lecture Notes in
Mathematics, Vol.\ 251, Springer, 1972, pp.\ 247--271.

\bibitem{ti} E. C. Titchmarsh, {\it The Theory of the Riemann
Zeta-Function} (revised by D. R. Heath-Brown), Oxford University
Press, 1986.

\bibitem{to1} L. T\'{o}th,  On certain arithmetic functions
involving exponential divisors, {\it Ann. Univ. Sci. Budapest. Sect.
Comput.\/} {\bf 24} (2004), 285--294.

\bibitem{to2}  L. T\'{o}th, On certain arithmetical functions involving exponential
divisors II, {\it Ann. Univ. Sci. Budapest. Sect. Comput.\/} {\bf
27} (2007), 155--166.

\bibitem{to3}L. T\'{o}th, An order result for the exponential divisor
function, {\it Publ. Math. Debrecen\/} {\bf 71} (2007), 165--171.

\bibitem{wu} Jie Wu,  Probl\`{e}me de diviseurs
exponentiels et entiers exponentiellement sans facteur carr\'{e},
{\it J. Th\'{e}or. Nombres Bordeaux\/} {\bf 7} (1995), 133--141.



\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}: Primary
11N37.

\noindent \emph{Keywords: }  exponential divisor, error term,
asymptotic lower bound.

\bigskip
\hrule
\bigskip

\vspace*{+.1in} \noindent Received January 29 2010; revised version
received March 10 2010. Published in {\it Journal of Integer
Sequences}, March 12 2010.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}
</pre></body></html>