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\begin{center}
\vskip 1cm{\LARGE\bf Reciprocals of the Gcd-Sum Functions} \vskip
1cm
\large Shiqin Chen \\
Experimental Center \\
Linyi University\\
Linyi, 276000, Shandong\\
China\\
\href{mailto:shiqinchen2010@163.com}{\tt shiqinchen2010@163.com }\\

\ \\

Wenguang Zhai \footnote{This work is supported by  the Natural Science
Foundation of Beijing (Grant No.\ 1112010).} \\
Department of Mathematics\\
China University of Mining and Technology \\
Beijing, 100083 \\
China\\
\href{mailto:zhaiwg@hotmail.com}{\tt zhaiwg@hotmail.com} \\
\end{center}

\vskip .2 in


\begin{abstract}
In this paper we study the reciprocals of the $\gcd$-sum function
and some related functions and improve some results of T\'oth.  The
harmonic mean of the $\gcd$ function is also studied.
\end{abstract}

\newcommand{\bvec}[1]{\mbox{\boldmath$#1$}}
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\section{Introduction }

Recently, L. T\'oth published  a paper  \cite{t1} about the
$\gcd$-sum function $P(n)$, called also Pillai's function 
\cite{P} defined by
$$P(n)=\sum_{k=1}^{n}\gcd(k,n).$$ 
In this well-written paper, T\'oth
not only listed many classical results about this function, its
analogues and generalizations, but also proved several new results.
The aim of this short note is to improve some results in T\'oth's
paper.

Let $H(n)$ denote the harmonic mean of $\gcd(1,n), \gcd(2,n),\ldots,
\gcd(n,n),$ namely,
\begin{equation}\label{Eq:1}
H(n):=n\left(\sum_{j=1}^n\frac{1}{\gcd(j,n)}\right)^{-1}=\frac{n^2}{\sum_{d|n}d\varphi(d)},
\end{equation}
where $\varphi(n)$ denotes  Euler's function. Theorem 4 of T\'oth
\cite{t1} states that
\begin{equation}\label{eq:2}
\sum_{n\leq x}\frac{H(n)}{n}=C_1\log x+C_2+O(x^{-1+\varepsilon}),
\end{equation}
where $C_1$ and $C_2$ are computable constants, $\varepsilon>0$ is a
small positive constant. As a corollary, he deduced that
\begin{equation}\label{Eq:3}
\sum_{n\leq x}H(n)=C_1x+O(x^\varepsilon).
\end{equation}

In this note we shall prove first the following

  \begin{theorem}\label{thm:1}
  {\it We have the asymptotic formula
\begin{equation}\label{Eq:4}
\sum_{n\leq x}H(n)=C_1x+C_3\log x+O((\log x)^{2/3}),
\end{equation}
where $C_1$ and $C_3$ are constants. }\end{theorem}

As a corollary of Theorem~\ref{thm:1}, we have

\begin{corollary}{\it
\begin{equation}\label{eq:5}
\sum_{n\leq x}\frac{H(n)}{n}=C_1\log x+C_2+O(x^{-1}(\log x)^{2/3}).
\end{equation}
}\end{corollary}

 \begin{remark} T\'oth \cite{t1} studied the average order of
$\sum_{n\leq x}H(n)/n$ first and then deduced an asymptotic formula
of $\sum_{n\leq x}H(n)$ through it. However in this note we study
$\sum_{n\leq x}H(n)$ first, then deduce an asymptotic formula for
$\sum_{n\leq x}H(n)/n.$\end{remark}


Now we study the reciprocals of $P(n)$ and some other related
functions. We first recall the definitions of some functions.


An integer $d$ is called a unitary divisor of $n$ if $d|n$ and
$(d,n/d)=1,$ notation $d\Vert n.$ The unitary analogue of the
function $P$ is defined by
$$P^*(n):=\sum_{j=1}^n(j,n)_*,$$
where $(j,n)_*=\max\{d\in {\Bbb N}: d|j, d\Vert n\},$ which was
first introduced in T\'oth \cite{t2}. This function $P^*$ (\seqnum{A145388})
is multiplicative and for every prime power $p^\alpha(\alpha\geq 1)$
we have $P^*(p^\alpha)=2p^\alpha-1.$

Let $n>1$ be an integer of canonical form
$n=\prod_{j=1}^rp_j^{a_j}.$ An integer $d$ is called an exponential
divisor of $n$ if $d=\prod_{j=1}^rp_j^{b_j}$ such that
$b_j|a_j(1\leq j\leq r), $ notation $d|_e n.$ By convention $1|_e
1.$ The   kernel of $n$ is denoted by $\kappa(n)=\prod_{j=1}^rp_j.$

Two positive integers $m>1, n>1$ have common exponential divisors
iff they have the canonical forms $n=\prod_{j=1}^rp_j^{a_j}$ and
$m=\prod_{j=1}^rp_j^{b_j}$ with $a_j\geq 1, b_j\geq 1(1\leq j\leq
r).$ The greatest common exponential divisor of $n$ and $m$ is
$(n,m)_e=\prod_{j=1}^rp_j^{(a_j,b_j)}.$ By convention define
$(1,1)_e=1$ and note that $(1,m)_e$ does not exist for $m>1.$ Now
define the function $P^{(e)}(n)$ by the relation
$$P^{(e)}(n)=\sum_{\stackrel{1\leq j\leq n}{\kappa(j)=\kappa(n)}}(j,n)_e.$$
The function  $P^{(e)}$(see T\'oth \cite{t4}) is multiplicative and
for any prime power $p^\alpha\ (\alpha\geq 1)$ we have
\begin{equation}\label{Eq:6}
P^{(e)}(p^\alpha)=\sum_{1\leq j\leq
\alpha}p^{(j,\alpha)}=\sum_{d|\alpha}p^d\varphi(\alpha/d).
\end{equation}

Now we recall another analogue of $P.$ Let $n>1$ be an integer and
we say an integer $j$ is regular (mod \ $n$) if there exists an
integer $x$ such that $j^2x\equiv j$ (mod \ $n$). T\'oth \cite{t3}
introduced the function (\seqnum{A176345})
$$\tilde{P}(n):=\sum_{j\in \Reg_n}\gcd(j,n),$$
where $\Reg_n=\{1\leq j\leq n: j \ \mbox{is regular} ({\rm mod}\ n)\}.$
T\'oth\cite{t3} showed that $\tilde{P}$ is multiplicative and
\begin{equation}\label{Eq:7}
\tilde{P}(n)=n\prod_{p|n}(2-1/p).
\end{equation}

The above mentioned functions have been extensively studied and many
papers about them have been published. See T\'oth \cite{t1} and
references therein.

Theorem 6 of T\'oth \cite{t1} states that
\begin{eqnarray}
\sum_{n\leq x}\frac{1}{P(n)}&&=K(\log x)^{1/2}+O((\log x)^{-1/2}),\label{Eq:8}\\
\sum_{n\leq x}\frac{1}{P^*(n)}&&=K^*(\log x)^{1/2}+O((\log x)^{-1/2}),\label{Eq:9}\\
\sum_{n\leq x}\frac{1}{\tilde{P}(n)}&&=\tilde{K}(\log x)^{1/2}+O((\log x)^{-1/2}),\label{Eq:10}\\
\sum_{n\leq x}\frac{1}{P^{(e)}(n)}&&=K^{(e)}\log
x+O(1),\label{Eq:11}
\end{eqnarray}
where $K, K^*, \tilde{K}, K^{(e)}$ are computable constants.

In this note we shall prove the following result.

\begin{theorem} \label{thm:4}{\it Suppose $N\geq 1$ is a fixed integer, then
\begin{eqnarray}
\sum_{n\leq x}\frac{1}{P(n)}&&=\sum_{j=0}^NK_j(\log x)^{1/2-j}+O((\log x)^{-1/2-N}),\label{Eq:12}\\
\sum_{n\leq x}\frac{1}{P^*(n)}&&=\sum_{j=0}^NK^*_j(\log x)^{1/2-j}+O((\log x)^{-1/2-N}),\label{Eq:13}\\
\sum_{n\leq x}\frac{1}{\tilde{P}(n)}&&=\sum_{j=0}^N\tilde{K}_j(\log x)^{1/2-j}+O((\log x)^{-1/2-N}),\label{Eq:14}\\
\sum_{n\leq x}\frac{1}{P^{(e)}(n)}&&=K^{(e)}_0\log
x+K^{(e)}_1+O(x^{-1}\log^2 x),\label{Eq:15}
\end{eqnarray}
where $K_j, K^*_j, \tilde{K}_j, K^{(e)}_j(j\geq 0)$ are computable
constants. }\end{theorem}

{\bf Notation.} Throughout this paper, $\varepsilon$ denotes a
sufficiently small positive constant. $\zeta(s)$ denotes the Riemann
zeta-function. For any real number $t,$ $[t]$ denotes the greatest
integer not exceeding $t,$ $\{t\}=t-[t],$ and $\psi(t)=\{t\}-1/2.$
For any  complex number $z$, $\sigma_z(n)=\sum_{d|n}d^z$ and
$d_z(n)$ denotes the generalized divisor function, $\mu(n)$ denotes
the M\"obius function.




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

Define for $\Re s>1$ that
\begin{equation}\label{Eq:16}
D_H(s):=\sum_{n=1}^\infty H(n)n^{-s}.
\end{equation}

Since $H(n)$ is multiplicative, by Euler's product we get
\begin{equation}\label{Eq:17}
D_H(s)=\prod_{p}\left(1+\sum_{\alpha=1}^\infty H(p^\alpha)p^{-\alpha
s}\right).
\end{equation}

We evaluate $H(p^\alpha)$ first for any prime $p$ and $\alpha\geq
1.$ By \eqref{Eq:1} we have
\begin{eqnarray}\label{Eq:18}
H(p^\alpha)&&=p^{2\alpha}\times (1+\sum_{j=1}^\alpha
p^{2j-1}(p-1))^{-1}\\
&&=p^{2\alpha}\times (1+\frac{p-1}{p}\sum_{j=1}^\alpha
p^{2j})^{-1}\nonumber\\
&&=p^{2\alpha}\times (1+\frac{p^{2\alpha+2}-p^2}{p(p+1)}
)^{-1}\nonumber\\
&&= p^{2\alpha}\times \left(\frac{p^{2\alpha+1}+1}{p+1}\right)^{-1}\nonumber\\
&&=\frac{p^{2\alpha}(p+1)}{p^{2\alpha+1}+1}=\frac{1+p^{-1}}{1+p^{-2\alpha-1}}.\nonumber
\end{eqnarray}
Hence we have
\begin{eqnarray}\label{Eq:19}
&&\ \ \ \ \ 1+\sum_{\alpha=1}^\infty H(p^\alpha)p^{-\alpha s}\\
&&=1+\sum_{\alpha=1}^\infty \frac{1+p^{-1}}{1+p^{-2\alpha-1}}
p^{-\alpha s}\nonumber\\
&&=1+(1+p^{-1})\sum_{\alpha=1}^\infty p^{-\alpha s}\sum_{m=0}^\infty
\left(-p^{-2\alpha-1}\right)^m\nonumber\\
&&=1+(1+p^{-1})\sum_{m=0}^\infty (-1)^mp^{-m}\sum_{\alpha=1}^\infty
p^{-\alpha s-2\alpha m}\nonumber\\
&&=1+(1+p^{-1})\sum_{m=0}^\infty
(-1)^mp^{-m}\frac{p^{-s-2m}}{1-p^{-s-2m}}\nonumber\\
&&=1+(1+p^{-1})\frac{p^{-s}}{1-p^{-s}}+(1+p^{-1})\sum_{m=1}^\infty
(-1)^mp^{-m}\frac{p^{-s-2m}}{1-p^{-s-2m}}\nonumber\\
&&= \frac{1+p^{-1-s}}{1-p^{-s}}+(1+p^{-1})\sum_{m=1}^\infty
(-1)^mp^{-m}\frac{p^{-s-2m}}{1-p^{-s-2m}}\nonumber,
\end{eqnarray}
which implies that
\begin{eqnarray}\label{Eq:20}
&&\ \ \ \ \ \ (1-p^{-s})(1-p^{-s-1})( 1+\sum_{\alpha=1}^\infty H(p^\alpha)p^{-\alpha s})\\
&& =1-p^{-2s-2}+(1-p^{-s})(1-p^{-s-1})(1+p^{-1})\sum_{m=1}^\infty
(-1)^mp^{-m}\frac{p^{-s-2m}}{1-p^{-s-2m}}\nonumber\\
&&=1+O(p^{-2\sigma-2}+(1+p^{-\sigma})(1+p^{-\sigma-1})p^{-\sigma-3})\nonumber\\
&&=1+O(p^{-2\sigma-2}+p^{-\sigma-3}+p^{-2\sigma-3}+p^{-3\sigma-4}),\nonumber
\end{eqnarray}
where $\sigma=\Re s.$




From \eqref{Eq:17}, \eqref{Eq:20} and noting
\begin{equation}\label{Eq:21}
\zeta(s)=\prod_{p}\left(1-p^{-s}\right)^{-1}\ (\Re s>1)
\end{equation}
we get
\begin{equation}\label{Eq:22}
D_H(s)=\zeta(s)\zeta(s+1)G(s)\ (\Re s>1),
\end{equation}
where
\begin{equation}\label{Eq:23}
G(s)=\prod_{p}(1-p^{-s})(1-p^{-s-1})\left(1+\sum_{\alpha=1}^\infty
H(p^\alpha)p^{-\alpha s}\right)
\end{equation}
such that if we expand $G(s)$ into a Dirichlet series
\begin{eqnarray}\label{Eq:24}
G(s)=\sum_{n=1}^\infty g(n)n^{-s},
\end{eqnarray}
 then this Dirichlet series is
absolutely convergent for $\Re s>-1/2.$

From \eqref{Eq:22} and \eqref{Eq:24} we have
\begin{equation}\label{Eq:25}
H(n)=\sum_{d|n}\sigma_{-1}(d)g(n/d).
\end{equation}
For $\sigma_{-1}(n)$ we have
\begin{eqnarray}\label{Eq:26}
\sum_{n\leq x}\sigma_{-1}(n)&&=\sum_{nm\leq x}\frac 1n=\sum_{n\leq
x}\frac 1n\sum_{m\leq x/n}1\\
&&=\sum_{n\leq x}\frac 1n\left[\frac xn\right]=\sum_{n\leq x}\frac
1n\left(\frac xn-\frac 12-\psi(\frac xn)\right)\nonumber\\
&&=x\sum_{n\leq x}n^{-2}-\frac 12\sum_{n\leq x}n^{-1}-\sum_{n\leq
x}\frac 1n\psi(\frac xn)\nonumber\\
&&=\frac{\pi^2x}{6}-\frac{\log x}{2} +O((\log
x)^{2/3}),\nonumber\end{eqnarray} where in the last step we used the
well-known bound(see Walfisz \cite{w})
\begin{equation}\label{Eq:27}
\sum_{n\leq x}\frac 1n\psi(\frac xn)\ll (\log x)^{2/3}.
\end{equation}

From \eqref{Eq:25}-\eqref{Eq:27} we have
\begin{eqnarray}\label{Eq:28}
\sum_{n\leq x}H(n)&&=\sum_{m\leq x}g(m)\sum_{n\leq
x/m}\sigma_{-1}(n)\\
&&= \frac{\pi^2 x}{6}\sum_{m\leq x}\frac{g(m)}{m}-\frac{\log
x}{2}\sum_{m\leq x}g(m)+\frac{1}{2}\sum_{m\leq x}g(m)\log
m\nonumber\\
&&\ \ \ \ +O\left(\log^{2/3} x\sum_{m\leq x}|g(m)|\right).\nonumber
\end{eqnarray}
Recall that the Dirichlet series $\sum_{n=1}^\infty g(n)n^{-s}$ is
absolutely convergent for $\Re s>-1/2.$ Hence for any $U>1$ we have
$$\sum_{U<n\leq 2U}|g(n)|\ll U^{-1/2+\varepsilon}.$$
It follows that the infinite series $\sum_{m\geq 1}g(m)m^{-1},
\sum_{m\geq 1}|g(m)|, \sum_{m\geq 1}g(m)\log m$ are all convergent
and that
\begin{eqnarray}\label{Eq:29}
&&\sum_{m\leq
x}\frac{g(m)}{m}=\sum_{m=1}^\infty\frac{g(m)}{m}+O(x^{-3/2+\varepsilon}),\\
&&\sum_{m\leq x}g(m)=\sum_{m=1}^\infty
g(m)+O(x^{-1/2+\varepsilon})\nonumber\\
&&\sum_{m\leq x}g(m)\log m\ll 1, \ \sum_{m\leq x}|g(m)|\ll
1.\nonumber
\end{eqnarray}

Now Theorem \ref{thm:1} follows from \eqref{Eq:28} and
\eqref{Eq:29}.







\section{\bf Proof of Theorem \ref{thm:4} }


In this section we shall prove Theorem \ref{thm:4}. We only prove
\eqref{Eq:12} and \eqref{Eq:15} since the proofs of \eqref{Eq:12},
\eqref{Eq:13} and \eqref{Eq:14} are the same.

\subsection{\bf The generalized divisor problem}

Suppose $k\geq 2\ (k\in {\Bbb N})$ is a fixed integer.  The divisor
function $d_k(n)$ denotes the number of ways $n$ can be written as a
product of $k$ natural-number factors. It is an important problem in
the analytic number theory to study the mean value of the divisor
function $d_k(n)$.  It is well-known that $d_k(n)$ are the
coefficients of the Dirichlet series
$$\zeta^k(s)=\sum_{n=1}^\infty d_k(n)n^{-s}\ (\Re s>1).$$


Now suppose $z$ is a fixed complex number and  let $d_z(n)$ denote
the coefficients of the Dirichlet series
\begin{equation}
\label{Eq:30} \zeta^z(s):=\sum_{n=1}^\infty d_z(n)n^{-s}\ (\Re s>1),
\end{equation}
where $\zeta^z(s)=e^{z\log \zeta(s)}$ such that for $\log s$ we take
the main value $\log 1=0.$ The function $d_z(n)$ is called the
generalized divisor function.

Suppose $A>0$ is an arbitrary but fixed real number and $N\geq 1$ is
an arbitrary but fixed integer. Then uniformly for $|z|\leq A$ we
have
\begin{equation}\label{Eq:31}
\sum_{n\leq x}d_z(n)=x\sum_{j=1}^N c_j(z)(\log x)^{z-j}+O(x(\log
x)^{\Re z-N-1}),
\end{equation}
where the functions $  c_1(z), \cdots, c_N(z)$ are regular  in the
region $|z|\leq A$.

The above result is Theorem 14.9 of Ivi\'c \cite{I}.


\subsection{\bf Proof of \eqref{Eq:12}}\

Define $P_1(n)=n/P(n).$ Then by Euler's product we have for $\Re
s>1$ that
\begin{equation}\label{Eq:32}
\sum_{n=1}^\infty P_1(n)n^{-s}=\prod_p\left(1+\sum_{\alpha=1}^\infty
P_1(p^\alpha)p^{-\alpha s}\right).
\end{equation}

We evaluate $P_1(p^\alpha)(\alpha\geq 1)$ first. The formula (14) of
 \cite{t1} reads $$P(p^\alpha)=(\alpha+1)p^\alpha-\alpha
p^{\alpha-1},$$ which implies that
\begin{equation}\label{Eq:33}
P_1(p^\alpha)=\frac{p^\alpha}{(\alpha+1)p^\alpha-\alpha
p^{\alpha-1}}=\frac{1}{\alpha+1}+O(p^{-1}).
\end{equation}

Inserting \eqref{Eq:33} into  \eqref{Eq:32} and recalling
\eqref{Eq:21} we get that
\begin{equation}\label{Eq:34}
\sum_{n=1}^\infty P_1(n)n^{-s}=\zeta^{1/2}(s)G_1(s),\ {\Re s>1},
\end{equation}
where
\begin{equation}\label{Eq:35}
G_1(s)=\prod_{p}(1-p^{-s})^{1/2} \left(1+\sum_{\alpha=1}^\infty
P_1(p^\alpha)p^{-\alpha s}\right)
\end{equation}
such that if we expand $G_1(s)$ into a Dirichlet series
\begin{eqnarray}\label{Eq:36}
G_1(s)=\sum_{n=1}^\infty g_1(n)n^{-s},
\end{eqnarray}
 then this Dirichlet series is
absolutely convergent for $\Re s>1/2.$ For the function $g_1$ we
have the trivial  estimate
\begin{equation}\label{Eq:37}
\sum_{m\leq y}|g_1(m)|\leq y^{1/2+\varepsilon}.
\end{equation}

From \eqref{Eq:37} we get by partial summation that
\begin{equation}\label{Eq:38}
\sum_{m\leq y}|g_1(m)|m^{-1}\ll 1, \ \ \sum_{m> y}|g_1(m)|m^{-1}\ll
y^{-1/2+\varepsilon}
\end{equation}
and for any fixed constant $C$ that
\begin{equation}\label{Eq:39}
\sum_{m\leq y}\frac{g_1(m)\log^C m}{m}=\sum_{m=1}^\infty
\frac{g_1(m)\log^C m}{m}+O(y^{-1/2+\varepsilon}).
\end{equation}
Let $e_C$ denote the value of the infinite series in \eqref{Eq:39}.

Suppose $\beta$ is a real number which is not a non-negative
integer. By Taylor's expansion we have
\begin{equation}\label{Eq:40}
(1-u)^\beta=\sum_{\ell=0}^Nd_\ell^{(\beta)}u^{\ell}+O(|u|^{N+1}),\
|u|\leq 1/2,
\end{equation}
where $d_\ell^{(\beta)}=(-1)^l \beta(\beta-1)\cdots
(\beta-\ell+1)/\ell!.$



By the hyperbolic approach, \eqref{Eq:31} with $A=z=1/2$,
\eqref{Eq:38} and \eqref{Eq:39} with $y=\sqrt x$ and \eqref{Eq:40}
we get for any fixed $N\geq 1$ that
\begin{eqnarray}\label{Eq:41}
\sum_{n\leq x}P_1(n)&&=\sum_{nm\leq x}d_{1/2}(n)g_1(m)\\
&&=\sum_{m\leq \sqrt x}g_1(m)\sum_{n\leq x/m}d_{1/2}(n)+ \sum_{n\leq
\sqrt x}d_{1/2}(n)\sum_{\sqrt x<m\leq
x/n}g_1(m)\nonumber\\
&&=\sum_{m\leq \sqrt x}g_1(m)\sum_{n\leq
x/m}d_{1/2}(n)+O(x^{3/4+\varepsilon})\nonumber\\
&&=\sum_{m\leq \sqrt x}g_1(m)\left(\frac
xm\sum_{j=1}^Nc_j(1/2)(\log\frac xm)^{1/2-j}+O(\frac xm (\log
x)^{-1/2-N})\right)\nonumber\\&&\ \ \ \ \
+O(x^{3/4+\varepsilon})\nonumber\\
&&=x\sum_{j=1}^Nc_j(1/2)(\log x)^{1/2-j}\sum_{m\leq \sqrt
x}\frac{g_1(m)}{m}\left(1-\frac{\log m}{\log
x}\right)^{1/2-j}\nonumber\\&&\ \ \ \ \ \ +O(x(\log
x)^{-1/2-N})\nonumber\\
&&=x\sum_{j=1}^Nc_j(1/2)(\log
x)^{1/2-j}\sum_{\ell=0}^Nd_\ell^{(1/2-j)}\sum_{m\leq \sqrt
x}\frac{g_1(m)}{m}\frac{\log^\ell m}{\log^\ell x} \nonumber\\
&&\ \ \ \ \ +O\left(x\sum_{j=1}^N (\log x)^{1/2-j} \sum_{m\leq \sqrt
x}\frac{|g_1(m)|}{m}\frac{\log^{N+1} m}{\log^{N+1}
x}\right)\nonumber\\
&&\ \ \ \ \ \ +O(x(\log x)^{-1/2-N})\nonumber\\&&
=x\sum_{j=1}^Nc_j(1/2)(\log
x)^{1/2-j}\sum_{\ell=0}^Nd_\ell^{(1/2-j)}e_\ell \log^{-\ell} x \nonumber\\
&&\ \ \ \ \ +O(x(\log x)^{-1/2-N}) \nonumber\\
&&=x\sum_{j=1}^N{\cal K}_j(\log x)^{1/2-j}  +O(x(\log
x)^{-1/2-N}),\nonumber
\end{eqnarray}
where
$${\cal K}_j=\sum_{\stackrel{j=j_1+\ell}{j_1\geq 1,\ell\geq 0}}c_{j_1}(1/2)d_\ell^{(1/2-j_1)}e_\ell
=\sum_{   \ell=0}^{j-1} c_{j-\ell}(1/2)d_\ell^{(1/2-j+\ell)}e_\ell \
\ (1\leq j\leq N).$$

From \eqref{Eq:41} we get  \eqref{Eq:12} immediately by partial
summation and some easy calculations.


\subsection{\bf Proof of \eqref{Eq:15}}\

Now we prove \eqref{Eq:15}. Define $P_1^{(e)}(n)=n/P^{(e)}(n).$ By
Euler's product we have   for $\Re s>1$ that
\begin{equation}\label{Eq:42}
\sum_{n=1}^\infty
P_1^{(e)}(n)n^{-s}=\prod_p\left(1+\sum_{\alpha=1}^\infty
P_1^{(e)}(p^\alpha)p^{-\alpha s}\right).
\end{equation}
Suppose $p$ is a prime. From \eqref{Eq:6} it is easy to see that
$$P^{(e)}(p)=p, P^{(e)}(p^2)=p^2+p,P^{(e)}(p^3)=p^3+2p, P^{(e)}(p^\alpha)=p^\alpha+O(p^{\alpha/2})\ (\alpha\geq 4).$$
Hence
\begin{eqnarray}\label{Eq:43}
&&P_1^{(e)}(p)=1,P_1^{(e)}(p^2)=\frac{1}{1+p^{-1}}, P_1^{(e)}(p^3)=\frac{1}{1+2p^{-2}}, \\
&&P_1^{(e)}(p^\alpha)=\frac{p^\alpha}{p^\alpha+O(p^{\alpha/2})}=
\frac{1}{1+O(p^{-\alpha/2})}=1+O(p^{-\alpha/2})\ (\alpha\geq
4).\nonumber
\end{eqnarray}

From \eqref{Eq:43} we get
\begin{eqnarray}\label{Eq:44}
&&\ \ \ \ \ \ \ 1+\sum_{\alpha=1}^\infty
P_1^{(e)}(p^\alpha)p^{-\alpha s}\\
&&=1+\sum_{\alpha=1}^\infty  p^{-\alpha s}+\sum_{\alpha=1}^\infty
(P_1^{(e)}(p^\alpha)-1)p^{-\alpha s}\nonumber\\
&&=1+\sum_{\alpha=1}^\infty  p^{-\alpha s}+\sum_{\alpha=2}^\infty
(P_1^{(e)}(p^\alpha)-1)p^{-\alpha s}\nonumber\\
&&=\frac{1}{1-p^{-s}}+(\frac{1}{1+p^{-1}}-1)p^{-2s}+
(\frac{1}{1+2p^{-2}}-1)p^{-3s}+\sum_{\alpha=4}^\infty
(P_1^{(e)}(p^\alpha)-1)p^{-\alpha s}\nonumber\\
&&=\frac{1}{1-p^{-s}}+(\frac{1}{1+p^{-1}}-1)p^{-2s}+O\left(p^{-3\sigma-2}+\sum_{\alpha=4}^\infty
p^{-\alpha/2-\alpha\sigma}\right)\nonumber\\
&&=\frac{1}{1-p^{-s}}+(\frac{1}{1+p^{-1}}-1)p^{-2s}+O\left(p^{-3\sigma-2}+p^{-4\sigma-2}
\right)\nonumber\\
&&=\frac{1}{1-p^{-s}}-p^{-1-2s}+O\left(p^{-2\sigma-2}+p^{-3\sigma-2}+p^{-4\sigma-2}
\right)\nonumber\\
&&=\frac{1}{1-p^{-s}}-p^{-1-2s}+O\left(p^{-2\sigma-2}
+p^{-4\sigma-2} \right).\nonumber
\end{eqnarray}
 Hence  we get
\begin{eqnarray}\label{Eq:45}
&&\ \ \ \ \ \ \ (1-p^{-s})(1+\sum_{\alpha=1}^\infty
P_1^{(e)}(p^\alpha)p^{-\alpha s})\\
&&=1-p^{-1-2s}(1-p^{-s}) +O\left(p^{-2\sigma-2}(1
+p^{-2\sigma})(1+p^{-\sigma}) \right)\nonumber\\
&&=1-p^{-1-2s}+p^{-1-3s}+O\left(p^{-2\sigma-2}(1 +p^{-3\sigma})
\right)\nonumber
\end{eqnarray}
and
\begin{eqnarray}\label{Eq:46}
&&\ \ \ \ \ \ \ (1-p^{-s})(1+p^{-1-2s})(1+\sum_{\alpha=1}^\infty
P_1^{(e)}(p^\alpha)p^{-\alpha s})\\
&& =1-p^{-2-4s}+p^{-1-3s}+p^{-2-5s}+O\left(p^{-2\sigma-2}(1
+p^{-3\sigma})(1+p^{-1-2\sigma}) \right)\nonumber\\
&&=1 +p^{-1-3s} +O\left(p^{-2\sigma-2}(1
+p^{-3\sigma})(1+p^{-1-2\sigma}) \right)\nonumber
\end{eqnarray}
and
\begin{eqnarray}\label{Eq:47}
&&\ \ \ \ \ \ \ (1-p^{-s})(1+p^{-1-2s})(1
-p^{-1-3s})(1+\sum_{\alpha=1}^\infty
P_1^{(e)}(p^\alpha)p^{-\alpha s})\\
&&  =1 -p^{-2-6s} +O\left(p^{-2\sigma-2}(1
+p^{-3\sigma})(1+p^{-1-2\sigma})(1+p^{-1-3\sigma})
\right)\nonumber\\
&&=1+O\left(p^{-2-6\sigma}+p^{-2\sigma-2}(1
+p^{-3\sigma})(1+p^{-1-2\sigma})(1+p^{-1-3\sigma})\right).\nonumber
\end{eqnarray}

We  write for $\sigma>1$
\begin{eqnarray}\label{Eq:48}
&&\ \ \ \ \ 1+\sum_{\alpha=1}^\infty P_1^{(e)}(p^\alpha)p^{-\alpha
s}\\
&&=\frac{1}{1-p^{-s}}\times \frac{1}{1+p^{-1-2s}}\times
\frac{1}{1-p^{-1-3s}}\nonumber\\&&\ \ \times \left((1-p^{-s})\times
(1+p^{-1-2s})\times (1-p^{-1-3s})\times (1+\sum_{\alpha=1}^\infty
P_1^{(e)}(p^\alpha)p^{-\alpha s})\right)\nonumber\\
&&=\frac{1}{1-p^{-s}}\times \frac{1-p^{-1-2s}}{1-p^{-2-4s}}\times
\frac{1}{1-p^{-1-3s}}\nonumber\\&&\ \ \times \left((1-p^{-s})\times
(1+p^{-1-2s})\times (1-p^{-1-3s})\times (1+\sum_{\alpha=1}^\infty
P_1^{(e)}(p^\alpha)p^{-\alpha s})\right).\nonumber
\end{eqnarray}
From \eqref{Eq:48} we may write for $\sigma>1$ that
 \begin{equation}\label{Eq:49}
 \sum_{n=1}^\infty
P_1^{(e)}(n)n^{-s}=\frac{\zeta(s)\zeta(3s+1)}{\zeta(2s+1)}G_{P_1^{(e)}}(s),
\end{equation}
where
 \begin{equation}\label{Eq:50}
 G_{P_1^{(e)}}(s)=\zeta(4s+2)\prod_p\left((1-p^{-s})
(1+p^{-1-2s}) (1-p^{-1-3s}) (1+\sum_{\alpha=1}^\infty
P_1^{(e)}(p^\alpha)p^{-\alpha s})\right).
\end{equation}

 From (47) we see that if we write the function $G_{P_1^{(e)}}(s)$
 into a Dirichlet series
\begin{equation}\label{Eq:51}
 G_{P_1^{(e)}}(s)=\sum_{n=1}^\infty g_{P_1^{(e)}}(n)n^{-s},
 \end{equation}
then this Derichlet series is absolutely for $\sigma>-1/6.$ This
fact implies that
\begin{equation}\label{Eq:52}
\sum_{n\leq x}|g_{P_1^{(e)}}(n)|\ll 1,\ \sum_{n\leq
x}\frac{g_{P_1^{(e)}}(n)}{n}=\sum_{n=1}^\infty\frac{g_{P_1^{(e)}}(n)}{n}+O(x^{-7/6+\varepsilon}).
\end{equation}

From \eqref{Eq:49}, \eqref{Eq:51} and \eqref{Eq:52} we get
\begin{eqnarray}\label{Eq:53}
&&\ \ \ \ \sum_{n\leq x}P_1^{(e)}(n)=\sum_{n_1n_2^2n_3^3n_4\leq
x}\frac{\mu(n_2)}{n_2n_3}g_{P_1^{(e)}}(n_4)\\
&&=\sum_{n_4\leq x} g_{P_1^{(e)}}(n_4) \sum_{n_2^2n_3^3\leq
\frac{x}{n_4}}\frac{\mu(n_2)}{n_2n_3} \sum_{n_1\leq
\frac{x}{n_2^2n_3^3n_4}}1\nonumber\\
&&=\sum_{n_4\leq x} g_{P_1^{(e)}}(n_4) \sum_{n_2^2n_3^3\leq
\frac{x}{n_4}}\frac{\mu(n_2)}{n_2n_3}\left(\frac{x}{n_2^2n_3^3n_4}+O(1)\right)\nonumber\\
&&=x\sum_{n_4\leq x}\frac{ g_{P_1^{(e)}}(n_4)}{n_4}
\sum_{n_2^2n_3^3\leq
\frac{x}{n_4}}\frac{\mu(n_2)}{n_2^3n_3^4}+O(\sum_{n_4\leq x}
|g_{P_1^{(e)}}(n_4)|\times \log^2 x)\nonumber \\
&&=x\sum_{n_4\leq x}\frac{ g_{P_1^{(e)}}(n_4)}{n_4} \sum_{n_2 \leq
\sqrt\frac{x}{n_4}}\frac{\mu(n_2)}{n_2^3}\sum_{n_3\leq
(\frac{x}{n_4n_2^2})^{1/3}}\frac{1}{n_3^4}+O( \log^2 x)\nonumber\\
&&=x\sum_{n_4\leq x}\frac{ g_{P_1^{(e)}}(n_4)}{n_4} \sum_{n_2 \leq
\sqrt\frac{x}{n_4}}\frac{\mu(n_2)}{n_2^3}\left(\zeta(4)+O(\frac{n_4n_2^2}{x})\right)
+O( \log^2 x)\nonumber\\
&&=x\zeta(4)\sum_{n_4\leq x}\frac{ g_{P_1^{(e)}}(n_4)}{n_4}
\sum_{n_2 \leq \sqrt\frac{x}{n_4}}\frac{\mu(n_2)}{n_2^3}  +O( \log^2
x)\nonumber\\
&&=x\zeta(4)\sum_{n_4\leq x}\frac{ g_{P_1^{(e)}}(n_4)}{n_4}
 \left(\frac{1}{\zeta(3)}+O(\frac{n_4}{x})\right)  +O( \log^2
x)\nonumber\\
&&=x\frac{\zeta(4)}{\zeta(3)} \sum_{n_4\leq x}\frac{
g_{P_1^{(e)}}(n_4)}{n_4}  +O( \log^2 x)\nonumber\\
&&=x\frac{\zeta(4)}{\zeta(3)} \sum_{n_4=1}^\infty\frac{
g_{P_1^{(e)}}(n_4)}{n_4}  +O( \log^2 x),\nonumber
\end{eqnarray}
where we used the following easy estimates($y\geq 2$)
\begin{eqnarray}
&&\sum_{n\leq y}n^{-4}=\zeta(4)+O(y^{-3}),\\
&&\sum_{n\leq y}\mu(n)n^{-3}=\frac{1}{\zeta(3)}+O(y^{-2}),\\
&&\sum_{n\leq y}n^{-1}\ll \log y.
\end{eqnarray}

 Now \eqref{Eq:15} follows from
\eqref{Eq:53} by partial summation.











\section{\bf 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.











\begin{thebibliography}{99}



\bibitem{I} A. Ivi\'c, \emph{The Riemann Zeta-Function}, John Wiley \& Sons,
1985.

\bibitem{P} S.\ S.\ Pillai, On an arithmetic function, \emph{J. Annamalai
Univ.} \textbf{2} (1933), 243--248.

\bibitem{t1} L. T\'oth,  A survey of gcd-sum functions, \emph{J.
Integer Sequences,}  \textbf{13} (2010), 
\href{http://www.cs.uwaterloo.ca/journals/JIS/VOL13/Toth/toth10.html}{Article 10.8.1}.

\bibitem{t2}L. T\'oth, The unitary analogue of Pillai's arithmetical
function, \emph{Collect. Math.} \textbf{40} (1989), 19--30.

\bibitem{t4}L. T\'oth, On certain arithmetic functions involving exponential
divisors, \emph{Annales Univ. Sci. Budapest., Sect. Comp.} \textbf{24}
(2004), 285--294.

\bibitem{t3}L. T\'oth, A gcd-sum function over regular integers
(mod\ $n$), \emph{J. Integer Sequences} \textbf{12} (2009), 
\href{http://www.cs.uwaterloo.ca/journals/JIS/VOL12/Toth/toth3.html}{Article 09.2.5}.

\bibitem{w}A. Walfisz, \emph{Weylsche Exponentialsummen in der neueren
Zahlentheorie,} Berlin, 1963.




\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}: Primary
11A25; Secondary 11N37.

\noindent \emph{Keywords: } gcd-sum function, regular integers,
harmonic mean, unitary divisor.

\bigskip\hrule\bigskip

\noindent (Concerned with sequences \seqnum{A018804},
\seqnum{A145388} and \seqnum{A176345}.)


\bigskip
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\vspace*{+.1in}
\noindent
Received February 13 2011;
revised versions received  May 15 2011; June 28 2011; August 12 2011.
Published in {\it Journal of Integer Sequences}, September 25 2011.
 
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\noindent
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