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\begin{center}
\vskip 1cm{\LARGE\bf 
Some Remarks on a Paper of L.\,Toth
}
\vskip 1cm
\large
Jean-Marie De Koninck \\
D\'ep. de math\'ematiques \\
Universit\'e Laval \\
Qu\'ebec, PQ G1V 0A6 \\
Canada \\
\href{mailto:jmdk@mat.ulaval.ca}{\tt jmdk@mat.ulaval.ca} \\
\ \\
Imre K\'atai \\
Computer Algebra Department \\
E\"otv\"os Lor\'and University\\
P\'azm\'any P\'eter S\'et\'any I/C \\
1117 Budapest \\
Hungary  \\
\href{mailto:katai@compalg.inf.elte.hu}{\tt katai@compalg.inf.elte.hu} \\
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\vskip .2 in

\begin{abstract}
Consider the functions $P(n):=\sum_{k=1}^n \gcd(k,n)$ (studied by
Pillai in 1933) and $\widetilde{P}(n):=n \prod_{p|n}(2-1/p)$
(studied by Toth in 2009). From their results, one can obtain
asymptotic expansions for $\sum_{n\le x} P(n)/n$ and $\sum_{n\le x}
\widetilde{P}(n)/n$. We consider two wide classes of functions
${\mathcal R}$ and ${\mathcal U}$ of arithmetical functions which
include $P(n)/n$ and $\widetilde{P}(n)/n$ respectively. For any
given $R\in {\mathcal R}$ and $U\in {\mathcal U}$, we obtain
asymptotic expansions for $\sum_{n\le x} R(n)$, $\sum_{n\le x}
U(n)$, $\sum_{p\le x} R(p-1)$ and $\sum_{p\le x} U(p-1)$.
\end{abstract}

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\section{Introduction}


In 1933, Pillai \cite{kn:pillai} introduced the function
$$P(n):=\sum_{k=1}^n \mbox{gcd}(k,n)$$
and showed that
$$P(n) = \sum_{d|n} d\varphi(n/d) \quad \mbox{ and } \sum_{d|n}P(d) =
 n\tau(n) = \sum_{d|n} \sigma(d) \varphi(n/d),$$
where $\varphi$ stands for the Euler function and where $\tau(n)$
and $\sigma(n)$ stand for the number of divisors of $n$ and the sum
of the divisors of $n$ respectively.

It is easily shown that
$$P(n)=n \tau(n) \prod_{p^a\| n} \left( 1 - \frac{ a/(a+1)}p
\right).$$

In 1985, Chidambaraswamy and Sitaramachandrarao \cite{kn:chi} showed that, given an arbitrary
$\varepsilon&gt;0$,
\begin{equation} \label{eq:asymP}
\sum_{n\le x} P(n) = e_1 x^2\log x + e_2 x^2 + O\left(
x^{1+\theta+\varepsilon} \right) ,
\end{equation}
where $\displaystyle{e_1=\frac 1{2\zeta(2)}}$ and
$\displaystyle{e_2= \frac 1{2\zeta(2)} \left( 2\gamma - \frac 12 -
\frac{\zeta'(2)}{\zeta(2)} \right)}$, and where $\theta$ is the
constant appearing below in Lemma \ref{lem:diviseur}, $\zeta$ stands
for the Riemann Zeta Function and $\gamma$ stands for Euler's
constant.


Using partial summation, one easily deduces from (\ref{eq:asymP})
that
\begin{equation} \label{eq:asym2}
\sum_{n\le x} \frac{P(n)}n = e_1 x\log x + (2e_2-e_1) x + O\left(
x^{\theta+\varepsilon} \right).
\end{equation}


In \cite{kn:toth}, Toth introduced the function
$$
\widetilde{P}(n)  =  n \prod_{p|n} \left( 2 - \frac 1p \right) =
n\cdot 2^{\omega(n)} \prod_{p|n} \left( 1 - \frac 1{2p} \right),$$
where $\omega(n)$  stands  for the number of distinct prime factors
of $n$.
Toth obtained an estimate for $\sum_{n\le x}
\widetilde{P}(n)$, analogous to (\ref{eq:asymP}) and from which one can easily derive an asymptotic
expansion for $\sum_{n\le x} \widetilde{P}(n)/n$.


In this paper, we consider two wide classes of arithmetical
functions ${\cal R}$ and ${\cal U}$, the first of which includes the
function $P(n)/n$, and the second of which includes
$\widetilde{P}(n)/n$. Given $R\in {\cal R}$, we obtain an asymptotic
expansion for $\sum_{n\le x} R(n)$; similarly for $U\in {\cal U}$.
We then examine the behavior of $\sum_{p\le x} R(p-1)$ and
$\sum_{p\le x} U(p-1)$.


More precisely, the class ${\cal R}$ is made of the following
functions $R$. First, let $\gamma(n)$ stand for the kernel of $n\ge
2$, that is $\gamma(n)=\prod_{p|n}p$ (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
$\displaystyle{|\kappa(n)|\le \frac c{\gamma(n)^\alpha}}$ for all
$n\ge 2$, let $R\in {\cal R}$ be defined by
\begin{equation} \label{eq:defR}
R(n)= R_{\kappa,c,\alpha}(n)  :=\tau(n) \sum_{d\|n} \kappa(d) =
\tau(n) \prod_{p^a\|n} (1+ \kappa(p^a)).
\end{equation}
Here $d\|n$
means that the sum runs over the unitary divisors of $n$, that is
over all divisors $d$ of $n$ for which $(d,n/d)=1$.

It is easily seen that if we let $\displaystyle{\kappa(p^a)= -
\frac{a/(a+1)}p}$, then the corresponding function $R(n)$ is
precisely $P(n)/n$.

As for the class of functions ${\cal U}$, it is made of the
functions
\begin{equation} \label{eq:defU}
U(n)=U_{h,c,\alpha}(n):= 2^{\omega(n)} \sum_{d|n} h(d),
\end{equation}
 where $h$ is a
multiplicative function satisfying $\displaystyle{|h(d)|\le \frac
c{\gamma(d)^\alpha}}$ for each integer $d\ge 2$, where $\alpha&gt;0$ is a given number. It is easily seen
that by taking $h(p)=-\frac1{2p}$ and $h(p^a)=0$ for $a\ge 2$, we
obtain the particular case $U(n)=\widetilde{P}(n)/n$.

Throughout this paper, $c_1,c_2,\ldots$ denote absolute positive
constants.

\section{Main results}


\begin{theorem} \label{th:1}
Let $R$ be as in (\ref{eq:defR}). For any arbitrary $\varepsilon&gt;0$,
as $x\to \infty$,
\begin{equation}  \nonumber
T(x):=\sum_{n\le x} R(n) = A_0x \log x + B_0 x + O\left( x^{\beta}
\right),
\end{equation}
with
$$\beta= \left\{ \begin{array}{cl}  \theta+\varepsilon, &amp; \mbox{if } \alpha \ge 1-\theta;\\
1-\alpha +\varepsilon, &amp; \mbox{if } \alpha &lt; 1 - \theta; \end{array}
\right.$$ where $\theta$ is the number mentioned in Lemma
\ref{lem:diviseur} below and where
\begin{equation}    \label{eq:1.9}
A_0 = \sum_{d\ge 1} \frac{\lambda(d)}d \qquad \mbox{and} \qquad B_0=
\sum_{d\ge 1} \frac{\lambda(d)}d (2\gamma-1 - \log d),
\end{equation}
the function $\lambda$  being defined below in
(\ref{eq:def-lambda1}) and (\ref{eq:def-lambda2}).
\end{theorem}

\bigskip

\begin{theorem} \label{th:2}
Let $U$ be as in (\ref{eq:defU}). As $x\to \infty$,
$$S(x):= \sum_{n\le x} U(n) = t_1 x \log x + t_2 x + O\left( \frac x{\log x}   \right),$$
where
\begin{eqnarray*}
t_1 &amp; = &amp; \sum_{\delta=1}^\infty \frac{h(\delta)
2^{\omega(\delta)}B_1(\delta) \eta_\delta}{\delta} \sum_{n=1 \atop
\gamma(n)|\delta}^\infty \frac 1n,\\
t_2 &amp; = &amp; \sum_{\delta=1 \atop b \in
 G_\delta}^\infty
 \frac{h(\delta) 2^{\omega(\delta)}}{\delta b}
  \big(B_2(\delta)\eta_\delta-2B_1(\delta)\mu_\delta - B_1(\delta) \eta_\delta \log(\delta b) \big),
  \end{eqnarray*}
where $B_1(\delta)$ and $B_2(\delta)$  are defined below in Lemma
\ref{lem:diviseur-phi}, while $\eta_\delta$ and $\mu_\delta$ are
defined respectively in (\ref{eq:def-eta}) and (\ref{eq:def-mu}).
\end{theorem}

\bigskip

\begin{theorem} \label{th:3}
Let $R$ be as in (\ref{eq:defR}). As $x\to \infty$,
\begin{equation} \label{eq:th3}
M(x) := \sum_{p\le x} R(p-1) = K_1 x + O \left( \frac x{\log \log x}
\right),
\end{equation}
where $\displaystyle{K_1=\frac 12 \sum_{d=1}^\infty
\frac{\kappa(d)\tau(d) c_d}d }$, where $c_d$ is itself defined in
Lemma \ref{lem:1-sur-phi}.
\end{theorem}

\bigskip

\begin{theorem} \label{th:4}
Let $U$ be as in (\ref{eq:defU}). As $x\to \infty$,
\begin{equation} \label{eq:th4}
N(x):=\sum_{p\le x} U(p-1) =  K_2 x + O \left( \frac x{\log \log x}
\right),
\end{equation}
where $K_2$  is a positive constant which may depend on the function
$\kappa$.
\end{theorem}

\section{Preliminary results}


\begin{lemma} \label{lem:diviseur}
As $x\to \infty$,
\begin{equation} \label{eq:1.7}
D(x):=\sum_{n\le x} \tau(n) = x \log x + (2\gamma-1) x + O(x^\theta)
\qquad (x\to \infty),
\end{equation}
for some positive constant $\theta &lt; 1/3$.
\end{lemma}

\begin{proof}
A proof can be found in the book of Ivi\'c \cite{kn:ivic}, where one
can also find a history of the improvements concerning the size of
$\theta$.
\end{proof}

\bigskip

\begin{lemma} \label{lem:diviseur-modulo}
Given $1\le \ell &lt;k$ with $\gcd(\ell,k)=1$,
$$\sum_{n\le x \atop n\equiv \ell \pmod k} \tau(n) = A_1(k) x \log x
+ A_2(k) x + O\left( k^{7/3} x^{1/3} \log x \right),$$
where
$$A_1(k) =  \frac{\varphi(k)}{k^2}, \qquad A_2(k)= (2\gamma-1)
\frac{\varphi(k)}{k^2} - \frac 2k \sum_{d|k} \frac{\mu(d) \log
d}d.$$
\end{lemma}

\begin{proof}
This result is due to Tolev \cite{kn:tolev}.
\end{proof}

\bigskip

\begin{lemma} \label{lem:diviseur-phi}
Given a positive integer $k$,
$$\sum_{n\le x \atop (n,k)=1} \tau(n) = B_1(k)  x\log x + B_2(k) x + O\left( k^{10/3} x^{1/3}
\log x \right),$$ where $\displaystyle{ B_1(k) = \left(
\frac{\varphi(k)}k \right)^2}$ \quad and \quad
$\displaystyle{B_2(k)=\varphi(k) A_2(k)}$.
\end{lemma}

\begin{proof}
Observing that
$$ \sum_{n\le x \atop \gcd(n,k)=1} \tau(n) = \sum_{\ell=1 \atop \gcd(\ell,k)=1}^k
\sum_{n\le x \atop n\equiv \ell \pmod k} \tau(n),$$ and using the
trivial fact that $\varphi(k)\le k$,  the result follows immediately
from Lemma \ref{lem:diviseur-modulo}.
\end{proof}

\bigskip



\begin{lemma} \label{lem:1surnoyau}
Given an arbitrary positive real number $\alpha&lt;1$,
$$\sum_{n&gt;x} \frac 1{n\gamma(n)^\alpha} \ll \frac 1{x^\alpha}.$$
\end{lemma}


\begin{proof}
Writing each number $n$ as $n=rm$, where $r$ and $m$ are the
square-full and square-free parts of $n$ respectively with
$(r,m)=1$, we may write
\begin{eqnarray*}
\sum_{n&gt;x} \frac 1{n\gamma(n)^\alpha} &amp; = &amp; \sum_{rm&gt;x \atop{
\gcd(r,m)=1 \atop r\ \mbox{\tiny square-full}}}
\frac{\mu^2(m)}{r\gamma(r)^\alpha m^{1+\alpha}} = \sum_{r\ge 1 \atop
r\ \mbox{\tiny square-full}} \frac 1{r\gamma(r)^\alpha} \sum_{m&gt;x/r
\atop \gcd(m,r)=1}
\frac{\mu^2(m)}{m^{1+\alpha}} \\
&amp; \ll &amp; \sum_{r\ge 1 \atop r\ \mbox{\tiny square-full}} \frac
1{r\gamma(r)^\alpha} \frac{\varphi(r)}r \int_{x/r}^\infty
\frac{dt}{t^{1+\alpha}} = \frac 1{x^{\alpha}} \sum_{r\ge 1 \atop r\
\mbox{\tiny square-full}} \frac{\varphi(r)}{r^2} \left( \frac
r{\gamma(r)}
\right)^\alpha \\
&amp; = &amp; \frac 1{x^{\alpha}} \prod_p \left( 1+
\frac{p-1}{p^2(p^{1-\alpha}-1)} \right) \ll \frac 1{x^{\alpha}}.
\end{eqnarray*}
\end{proof}

\bigskip

\begin{lemma} \label{lem:selberg}
Given any fixed number $z&gt;0$,
\begin{equation} \label{eq:sel}
\sum_{n\le x}z^{\omega(n)} \ll x\ \log^{z-1}x.
\end{equation}
\end{lemma}

\begin{remark}
This result is a weak form of the well known Selberg-Sathe formula
$$\sum_{n\le x}z^{\omega(n)}= C(t) x \log^{t-1}x +O(x\log^{t-2}x),$$
an estimate which holds uniformly for $x\ge 2$ and  $|t|\le 1$, where $C(t)$ is a constant depending only on $t$
(see Selberg \cite{kn:selberg}). Here we give a simple direct proof of (\ref{eq:sel}).
\end{remark}

\begin{proof}
For each positive integer $k$, let
$$\pi_k(x):= \#\{n\le x: \omega(n)=k\}.$$
From a classical result of Hardy and Ramanujan \cite{kn:hardy}, we
know that
$$\pi_k(x) \le \frac{c_1 x}{(k-1)!\log x} (\log \log x +
c_2)^{k-1} \qquad (x\ge 3),$$ for all $k \ge 1$, where $c_1$ and
$c_2$ are some absolute constants. Using this estimate, it follows
that
\begin{eqnarray*}
\sum_{n\le x}z^{\omega(n)} &amp; = &amp; \sum_{k\ge 1} \pi_k(x) z^k &lt; c_1z
\frac x{\log x}  \sum_{k\ge 1} \frac{(z(\log \log x +
c_2))^{k-1}}{(k-1)!} \\
&amp; = &amp; c_1z \frac x{\log x} e^{z\log \log x + zc_2} \ll \frac x{\log
x} e^{z\log \log x} = x \log^{z-1} x,
\end{eqnarray*}
as required.
\end{proof}

\bigskip

\begin{lemma} \label{lem:pi} ({\sc Brun-Titchmarsh Theorem})
Let $\pi(y;k,\ell):=\#\{p\le y: p\equiv \ell \pmod k\}$. Given a
fixed positive number $\beta&lt;1$, then, uniformly for $k\in
[1,x^\beta]$, there exists a positive constant $\xi_1$ such that
$$  \pi(x;k,\ell) &lt; \xi_1 \frac x{\varphi(k) \log(x/k)}.$$
\end{lemma}

\begin{proof}
For a proof, see Titchmarsh \cite{kn:titchmarsh}.
\end{proof}

\bigskip

\begin{lemma} \label{lem:bombieri}
 ({\sc Bombieri-Vinogradov Theorem}).  Given
an arbitrary positive constant $A$, and let $B=\frac 32 A+17$. Then
$$ \sum_{k\le \frac{\sqrt x}{\log^B x}}  \max_{\gcd(k,\ell)=1} \max_{y\le x}
\left| \pi(y;k,\ell) - \frac{\mbox{li}(x)}{\varphi(k)} \right| \ll
\frac x{\log^A x}.$$
\end{lemma}

\begin{proof}
For a proof, see Cheng-Dong \cite{kn:pan}.
\end{proof}

\bigskip

\begin{lemma} \label{lem:p-1}
Given a fixed positive number $\beta&lt;1$, then, uniformly for $d\in
[1,x^{1-\beta}]$, there exists a positive constant $\xi_3$ such that
$$ \sum_{p\le x \atop p\equiv 1 \mod d} \tau(p-1) \le \xi_3
x\frac{\tau(d)}{\varphi(d)}.$$
\end{lemma}

\begin{proof}
Since $\displaystyle{\tau(n)\le 2 \sum_{d|n \atop d\le \sqrt n}1 }$
and $\tau(mn)\le \tau(m)\tau(n)$ for all positive integers $m,n$, it
follows that
\begin{eqnarray} \label{eq:tt} \nonumber
E:&amp; = &amp; \sum_{p\le x \atop p\equiv 1 \pmod d} \tau(p-1) \\ \nonumber
&amp; \le &amp; \tau(d) \sum_{p\le x \atop p\equiv 1 \pmod d}
\tau(\frac{p-1}d) \\
&amp; \le &amp; 2\tau(d) \sum_{u\le \sqrt{x/d}} \pi(x;du,1).
\end{eqnarray}
On the other hand, since $\displaystyle{du=\sqrt d \cdot u\sqrt d
\le \sqrt d \cdot \sqrt x \le x^{\frac 12 - \frac{\beta}2} \cdot
x^{\frac 12} = x^{1-\frac{\beta}2}}$, we have, in light of Lemma
\ref{lem:pi}, that
\begin{equation} \label{eq:ttpi}
\pi(x;du,-1)\le \xi_1 \frac x{\varphi(du) \log(x/du)} \le \xi_2
\frac x{\varphi(du)\log x},
\end{equation}
with $\xi_2=2\xi_1/\beta$. Using (\ref{eq:ttpi}) in (\ref{eq:tt})
and keeping in mind that $\varphi(d)\varphi(u)\le \varphi(du)$ for
positive integers $d,u$, we obtain that
\begin{eqnarray*}
E &amp; \le &amp; 2\xi_2  \frac x{\log x} \tau(d) \sum_{u\le \sqrt{x/d}}
\frac 1{\varphi(du)} \\
&amp; \le &amp; 2\xi_2  \frac x{\log x} \frac{\tau(d)}{\varphi(d)}
\sum_{u\le \sqrt{x/d}}
\frac 1{\varphi(u)} \\
&amp; \le &amp; \xi_3 x \frac{\tau(d)}{\varphi(d)},
\end{eqnarray*}
for some positive constant $\xi_3$, where we used the fact that
$\displaystyle{\sum_{n\le y}\frac 1{\varphi(n)} \ll \log y}$, thus
completing the proof of the lemma.
\end{proof}

\bigskip

\begin{lemma} \label{lem:tau-gamma}
Given fixed numbers $A&gt;0$ and $\kappa&lt;\alpha$, then
$$ \sum_{d \ge \log^A x}
\frac{\tau(d)}{d \gamma(d)^\alpha} \ll \frac 1{\log^{A\kappa}x}.
$$
\end{lemma}



\begin{proof}
Clearly we have
\begin{eqnarray*}
\sum_{d \ge \log^A x} \frac{\tau(d)}{d \gamma(d)^\alpha} &amp; \le &amp;
\sum_{d \ge \log^A x} \frac{\tau(d)}{d \gamma(d)^\alpha} \left(
\frac d{\log^A x} \right)^\kappa \\
&amp; = &amp; \frac 1{\log^{A\kappa}x} \sum_{d\ge \log^A x}
\frac{\tau(d)}{d^{1-\kappa} \gamma(d)^\alpha} \\
&amp; &lt; &amp; \frac 1{\log^{A\kappa}x} \prod_p \left( 1+ \frac
2{p^{1-\kappa} p^\alpha} +  \frac 2{p^{2(1-\kappa)} p^\alpha} +
\ldots \right)  \\
&amp; \ll &amp; \frac 1{\log^{A\kappa}x},
\end{eqnarray*}
since the above infinite product converges in light of the fact that
$\kappa&lt;\alpha$.
\end{proof}

\bigskip

\begin{lemma} \label{lem:1-sur-phi}
Given a fixed positive integer $D$, then
$$\sum_{n\le x \atop \gcd(n,D)=1} \frac 1{\varphi(n)} = c_D \log x +
O(1),$$ where $$c_D= \prod_p \left( 1+ \frac 1{p(p-1)} \right) \cdot
\prod_{p|D} \left(1 + \frac p{(p-1)^2} \right)^{-1}.$$
\end{lemma}

\begin{remark}
Note that the result of Lemma \ref{lem:1-sur-phi} is known in a more
precise form (see the book of Montgomery and Vaughan
\cite[pp.\ 42--43]{kn:mon}).
\end{remark}


\begin{proof} We first compute the generating series of
$n/\varphi(n)$. We have
\begin{eqnarray*}
\sum_{n=1 \atop \mbox{\tiny gcd}(n,D)=1}^\infty \frac{n/\varphi(n)}{n^s} &amp; = &amp; \prod_{p\not|D}
\left(1 + \frac{(1-1/p)^{-1}}{p^s} +  \frac{(1-1/p)^{-1}}{p^{2s}} +
\ldots \right) \\
&amp; = &amp; \frac{ \prod_{p} \left(1 + \frac{(1-1/p)^{-1}}{p^s} +
\frac{(1-1/p)^{-1}}{p^{2s}} + \ldots \right) }{ \prod_{p|D} \left(1
+ \frac{(1-1/p)^{-1}}{p^s} +  \frac{(1-1/p)^{-1}}{p^{2s}} + \ldots
\right)} \\
&amp; = &amp; \zeta(s) \prod_p \left(1- \frac 1{p^s} \right) \frac{
\prod_{p} \left(1 + \frac{(1-1/p)^{-1}}{p^s} +
\frac{(1-1/p)^{-1}}{p^{2s}} + \ldots \right) }{ \prod_{p|D} \left(1
+ \frac{(1-1/p)^{-1}}{p^s} +  \frac{(1-1/p)^{-1}}{p^{2s}} + \ldots
\right)} \\
&amp; = &amp; \zeta(s) \prod_p \left(1+ \frac 1{p^s(p-1)} \right)
\prod_{p|D} \left( 1+ \frac p{(p-1)(p^s-1)}  \right)^{-1},
\end{eqnarray*}
which by Wintner's Theorem yields
\begin{equation}
\sum_{n \le x \atop \gcd(n,D)=1} \frac n{\varphi(n)} = c_D x +
O(x^{1/2} \log x),
\end{equation}
where
$$c_D = \prod_p \left( 1+ \frac 1{p(p-1)} \right) \cdot \prod_{p|D}
\left(1 + \frac p{(p-1)^2} \right)^{-1}.$$ Then, using partial
summation, we get that
$$\sum_{n \le x \atop \gcd(n,D)=1} \frac 1{\varphi(n)} = c_D + O\left(\frac{\log x}{x^{1/2}}\right) +
\int_1^x  c_D \frac{dt}t + O(1)= c_D \log x + O(1),$$ as required.
\end{proof}

\bigskip

\begin{lemma} \label{lem:noyau}
For each integer $\delta\ge 2$, let $G_\delta$ be the semigroup
generated by the prime factors of $\delta$, i.e. for
$\delta=q_1^{\alpha_1} \ldots q_r^{\alpha_r}$, let $G_\delta=\{
q_1^{\beta_1} \ldots q_r^{\beta_r}: \beta_i \ge 0\}$. Then
$$\sum_{n\in G_\delta} \frac 1n = \sum_{n=1 \atop \gamma(n)|\delta}^\infty \frac
1n \ll \log \log \delta.$$
\end{lemma}

\begin{proof}
Using the well known result of Landau
$$\limsup_{n\to \infty} \frac n{\varphi(n) \log \log n} =e^\gamma,$$
we certainly have that there exists  some constant $C&gt;0$ such that
$$\frac n{\varphi(n)} &lt; C \log \log n  \qquad (n\ge
3),$$ from which is follows easily that
\begin{eqnarray*}
\sum_{n=1 \atop \gamma(n)|\delta}^\infty \frac 1n &amp; = &amp;
\prod_{p|\delta} \left( 1+ \frac 1p + \frac 1{p^2} + \ldots \right)
= \prod_{p|\delta} \left(1- \frac 1p \right)^{-1} \\
&amp;=&amp; \frac{\delta}{\varphi(\delta)} &lt; C \log \log \delta,
\end{eqnarray*}
which proves our result.
\end{proof}

\bigskip

\begin{lemma} \label{lem:noyau2}
Let $G_\delta$ be as in Lemma \ref{lem:noyau}. Then
$$\sum_{\delta=1}^\infty \frac{|h(\delta)| 2^{\omega(\delta)}}\delta
\sum_{b\in G_{\delta}} \frac 1b &lt; + \infty.$$
\end{lemma}

\begin{proof}
In light of Lemma \ref{lem:noyau}, the fact that
$\displaystyle{|h(\delta)|\le \frac{c}{\gamma(\delta)^\alpha}}$ and
since given any $\varepsilon&gt;0$, $\log \delta &lt; \delta^\varepsilon$
provided $\delta\ge \delta_0(\varepsilon)$, we have
\begin{eqnarray*}
\sum_{\delta=1}^\infty \frac{|h(\delta)| 2^{\omega(\delta)}}\delta
\sum_{b\in G_{\delta}} \frac 1b
&amp; \ll &amp; \sum_{\delta=1}^\infty
\frac{|h(\delta)| 2^{\omega(\delta)} \log\log \delta}\delta \\
&amp;  \ll &amp; \sum_{\delta=1}^\infty \frac{ 2^{\omega(\delta)} \log \log
\delta}{\delta \gamma(\delta)^\alpha}\\
&amp; \ll &amp;  \sum_{\delta=\delta_0}^\infty \frac{ 2^{\omega(\delta)}
}{\delta^{1-\varepsilon} \gamma(\delta)^\alpha} \\
&amp; &lt; &amp; \prod_p \left(1 + \frac{2}{p^{1-\varepsilon} p^\alpha} +
\frac{2}{p^{2-2\varepsilon} p^\alpha} + \ldots \right)  \\
&amp; = &amp; \prod_p \left( 1 + \frac{2}{p^\alpha(p^{1-\varepsilon}-1)}
\right) &lt;+\infty,
\end{eqnarray*}
provided  $\varepsilon &lt; \alpha$.
\end{proof}



\bigskip

\begin{lemma} \label{lem:sum-h-delta}
Let $G_\delta$ be as in Lemma \ref{lem:noyau} and let $\rho(x)$ be a
real function which tends to $+\infty$ as $x\to \infty$. Given any
fixed positive constant $\kappa&lt;\alpha$, we have
$$Z(x):=\sum_{\rho(x) &lt;\delta b \le x \atop b \in G_\delta}
\frac{h(\delta) 2^{\omega(\delta)} }{\delta b} \ll \frac
1{\rho(x)^\kappa}.$$
\end{lemma}

\begin{proof} We have
\begin{equation} \label{eq:z1z2}
|Z(x)| \le \sum_{\delta b &gt; \rho(x) \atop b\in G_\delta}
\frac{|h(\delta)|\cdot 2^{\omega(\delta)}}{\delta b} = \sum_{\delta
b &gt; \rho(x) \atop \delta&lt;\rho(x)} + \sum_{\delta b &gt; \rho(x) \atop
\delta \ge \rho(x)} = Z_1+Z_2,
\end{equation}
say. We first estimate $Z_2$. Recalling that $|h(\delta)|\le
c/\gamma(\delta)^\alpha$,
\begin{eqnarray} \label{eq:z2} \nonumber
Z_2 &amp; \le &amp; \sum_{\delta \ge \rho(x)} \frac{|h(\delta)|\cdot
2^{\omega(\delta)}}{\delta} \prod_{p|\delta} \left( 1+ \frac 1p +
\frac 1{p^2} + \ldots \right) \\  \nonumber
&amp; \le &amp;
\sum_{\delta=1}^\infty \frac{|h(\delta)|\cdot
2^{\omega(\delta)}}{\delta} \prod_{p|\delta} \frac 1{1-1/p} \cdot
\left( \frac{\delta}{\rho(x)} \right)^\kappa \\
&amp; \le &amp; \frac c{\rho(x)^\kappa} \sum_{\delta=1}^\infty \frac
{2^{\omega(\delta)}}{\delta^{1-\kappa} \gamma(\delta)^\alpha}
\prod_{p|\delta} \frac 1{1-1/p}.
\end{eqnarray}
Define $$f(\delta):=\frac{2^{\omega(\delta)}}{\delta^{1-\kappa}
\gamma(\delta)^\alpha} \prod_{p|\delta} \frac 1{1-1/p}.$$ Clearly
$f$ is a multiplicative function with the following values on prime
powers:
$$f(q^a) = \frac 2{q^\alpha\cdot q^{a(1-\kappa)} (1-1/q)}.$$
Moreover, $f$ is such that $\sum_{\delta=1}^\infty f(\delta)$ is
bounded provided $\kappa&lt;\alpha$. Hence, taking this into account in
(\ref{eq:z2}), we obtain that
\begin{equation} \label{eq:z2a}
Z_2 \ll \frac 1{\rho(x)^\kappa}.
\end{equation}

We now move on to estimate $Z_1$. We have
\begin{eqnarray} \label{eq:z1} \nonumber
Z_1 &amp; \le &amp; \sum_{\delta &lt; \rho(x)} \frac{|h(\delta)|
2^{\omega(\delta)}}{\delta} \sum_{b&gt; \rho(x)/\delta \atop b\in
G_\delta} \frac 1b \le \sum_{\delta &lt; \rho(x)} \frac{|h(\delta)|
2^{\omega(\delta)}}{\delta} \sum_{b\in G_\delta} \frac 1b \left(
\frac{b\delta}{\rho(x)} \right)^\kappa \\
&amp; = &amp; \frac 1{\rho(x)^\kappa} \sum_{\delta &lt; \rho(x)}
\frac{|h(\delta)| 2^{\omega(\delta)}}{\delta^{1-\kappa}} \sum_{b\in
G_\delta} b^{\kappa-1}.
\end{eqnarray}
Since
$$\sum_{b\in
G_\delta} b^{\kappa-1} \le \prod_{p|\delta} \left(1+ \frac
1{p^{1-\kappa}} + \frac 1{p^{2(1-\kappa})} + \ldots \right)=
\prod_{p|\delta} \left( 1- \frac 1{p^{1-\kappa}} \right)^{-1} ,$$ it
follows from (\ref{eq:z1}) and the fact that $|h(\delta)| \le
c/\gamma(\delta)^\alpha$, that
\begin{equation} \label{eq:zia}
Z_1 \le \frac c{\rho(x)^\kappa} \sum_{\delta&lt; \rho(x)}
\frac{2^{\omega(\delta)}}{\gamma(\delta)^\alpha \delta^{1-\kappa}}
\prod_{p|\delta} \left( 1- \frac 1{p^{1-\kappa}} \right)^{-1} =
\frac c{\rho(x)^\kappa} \sum_{\delta&lt;\rho(x)} g(\delta),
\end{equation}
say, where $g$ is clearly a multiplicative function whose values at
the prime powers are given by
$$g(q^a) =  \frac 2{q^\alpha\cdot q^{a(1-\kappa)}} \left(1 + \frac
1{q^{1-\kappa}} + \frac 1{q^{2(1-\kappa})} + \ldots\right).$$ Since
we assumed that $\kappa&lt;\alpha$and since
$\displaystyle{\sum_{a=1}^\infty g(q^a) &lt; \frac
{c_3}{q^{\alpha+1-\kappa}}}$, with a suitable constant $c_3&gt;0$, it
follows that
$$\sum_{\delta=1}^\infty g(\delta) = \prod_p (1+ g(p) + g(p^2) +
\ldots )&lt; +\infty.$$ Using this information, (\ref{eq:zia}) yields
\begin{equation} \label{eq:z1b}
Z_1 \ll \frac 1{\rho(x)^\kappa}.
\end{equation}
Substituting (\ref{eq:z2a}) and (\ref{eq:z1b}) in (\ref{eq:z1z2}),
the lemma is proved.


\end{proof}


\section{Proof of Theorem~\ref{th:1}}

One can easily see that
$$F(s):= \sum_{n=1}^\infty \frac{R(n)}{n^s}$$
can be written as
\begin{equation} \label{eq:e1}
F(s) = A(s) \zeta^2(s),
\end{equation}
where
$$
A(s) = \prod_p \left( 1+ \frac{\lambda(p)}{p^s} +
\frac{\lambda(p^2)}{p^{2s}} + \ldots \right),
$$
with
\begin{equation} \label{eq:def-lambda1}
\lambda(p) = 2 \kappa(p), \qquad \lambda(p^2) = 3 \kappa(p^2)-4
\kappa(p),
\end{equation}
and more generally, for each $\beta\ge 3$, by
\begin{equation} \label{eq:def-lambda2}
 \lambda(p^\beta) = (\beta+1) \kappa(p^\beta)-2\beta
\kappa(p^{\beta-1}) + (\beta-1) \kappa(p^{\beta-2}).
\end{equation}
Hence,
$$|\lambda(p^\beta)| \le \frac{4\beta c}{p^\alpha} \qquad (\beta\ge 1).$$
Consequently,
\begin{equation} \label{eq:1.6}
|\lambda(d)| \le \frac{\tau(d)^{c_4}}{\gamma(d)^\alpha},
\end{equation}
where $c_4$ is a  suitable constant ($c_4=4c$ will do).

Now observe that, in light of (\ref{eq:e1}) and of Lemma
\ref{lem:diviseur},
\begin{eqnarray} \label{eq:p1} \nonumber
T(x) &amp; = &amp; \sum_{d\le x} \lambda(d) D(x /d) \\
&amp; = &amp; \sum_{d\le x} \lambda(d) \left( \frac xd \log(\frac xd) +
(2\gamma-1)\frac xd\right) + O \left( x^\theta \sum_{d\le x}
\frac{|\lambda(d)|}{d^\theta}   \right).
\end{eqnarray}
It follows from (\ref{eq:1.6}) and Lemma \ref{lem:1surnoyau} that
 \begin{equation} \label{eq:p2}
\sum_{d&gt;x} \frac{|\lambda(d)|}d  \le  \sum_{d&gt;x} \frac{c
\tau(d)^{c_4}}{d\gamma(d)^\alpha} \ll x^{\varepsilon-\alpha}.
\end{equation}
Indeed, we have
$$S_U:= \sum_{d\in [U,2U]} \frac{\tau(d)^{c_4}}{d\gamma(d)}  &lt; U^\varepsilon \sum_{d\in [U,2U]} \frac 1{d\gamma(d)}  = U^\varepsilon K_U,$$
say, provided $U$ is sufficiently large. Now, from Lemma \ref{lem:1surnoyau}, it follows that $K_U\le \frac c{U^\alpha}$. Hence, if we set $U_j=2^j x$ for $j=0,1,\ldots$, then, if $x$ is large enough,
$$\sum_{d&gt;x} \frac{|\lambda(d)|}d  \le c \sum_{j=0}^\infty \frac{U_j^\varepsilon}{U_j^\alpha} = c\, x^{\varepsilon-\alpha} \sum_{j=0}^\infty \frac 1{2^{j(\alpha-\varepsilon)}} \ll x^{\varepsilon-\alpha}.$$
Similarly, observing that $\tau(d)^{c_4} (c+\log d) \le d^\varepsilon$ and arguing in a similar way, one can easily proves that
\begin{equation}\label{eq:p3}
\sum_{d&gt;x} \frac{|\lambda(d)|\cdot |c+\log d|}d \ll x^{\varepsilon-\alpha}. \end{equation} Substituting
(\ref{eq:p2}) and (\ref{eq:p3}) in (\ref{eq:p1}), one obtains
\begin{equation}    \label{eq:1.8}
T(x)= A_0x \log x + B_0 x + O\left( x^{1-\alpha+\varepsilon} \right)
+ O \left( x^\theta \sum_{d\le x} \frac{|\lambda(d)|}{d^\theta}
\right),
\end{equation}
where $\displaystyle{A_0= \sum_{d=1}^\infty \frac{\lambda(d)}d}$ and
$\displaystyle{B_0=\sum_{d=1}^\infty \frac{\lambda(d)}d(2\gamma -1
-\log d)}$.

 Since
$$\sum_{d\le x} \frac{|\lambda(d)|}{d^\theta}
\le \sum_{d\le x} \frac{c\tau(d)^{c_4}}{d^\theta \gamma(d)^\alpha} 
\le \left\{ \begin{array}{cl} x^{\varepsilon}, &amp; \mbox{if } \
\theta + \alpha \ge 1;
\\
x^{1-(\theta+\alpha)}, &amp; \mbox{if }\
\theta + \alpha &lt; 1; \end{array} \right.$$
it follows that
$$x^\theta \sum_{d\le x} \frac{|\lambda(d)|}{d^\theta}  \le \left\{ \begin{array}{cl} x^{\theta + \varepsilon}, &amp; \mbox{if }\
\theta + \alpha \ge 1;
\\
 x^{ 1  - \alpha}, &amp; \mbox{if }\ \theta + \alpha &lt; 1.
\end{array} \right.$$ Using this last estimate in (\ref{eq:1.8})
completes the proof of Theorem~\ref{th:1}.

\section{Proof of Theorem \ref{th:2}}

Since
$$ \sum_{n=1}^\infty \frac{2^{\omega(n)}}{n^s} =
\frac{\zeta^2(s)}{\zeta(2s)} \qquad (\Re(s)&gt;1),$$ we have that
\begin{equation} \label{eq:1.12}
2^{\omega(n)} = \sum_{d^2e =n} \mu(d)\tau(e).
\end{equation}
Let $G_\delta$ be as in the statement of Lemma \ref{lem:noyau}. If
$\delta|n$, then let $m$ be the largest divisor of $n/\delta$ which
is co-prime with $\delta$, and let $b=b(\delta)\in G_\delta$ be
defined implicitly by $n=\delta\cdot b \cdot m$. Note that with this
setup, the numbers $m$ and $b$ are uniquely determined, and we
therefore have  $2^{\omega(n)} = 2^{\omega(\delta)} \cdot
2^{\omega(m)}$. Hence,
\begin{eqnarray} \label{eq:1.13} \nonumber
S(x) &amp; = &amp; \sum_{\delta \le x} h(\delta) 2^{\omega(\delta)}
\sum_{b\le x/\delta \atop b\in G_\delta} \sum_{m\le  \frac x{\delta
b}  \atop \gcd(m,\delta)=1} 2^{\omega(m)} \\
&amp; = &amp; \sum_{\delta \le x} h(\delta) 2^{\omega(\delta)} \sum_{b\le
x/\delta \atop b\in G_\delta} E_\delta\left( \frac x{\delta b}
\right),
\end{eqnarray}
say. We therefore need to estimate
$$E_k(X):= \sum_{n\le X \atop \gcd(n,k)=1} 2^{\omega(n)}.$$
Observe that it follows from (\ref{eq:1.12}) that
\begin{equation} \label{eq:1.14}
E_k(X)= \sum_{d\le \sqrt X \atop \gcd(d,k)=1} \mu(d)  \sum_{d^2e \le
X \atop \gcd(e,k)=1} \tau(e) = \sum_{d\le \sqrt X \atop \gcd(d,k)=1}
\mu(d) V_k\left( \frac X{d^2} \right),
\end{equation}
where
$$V_k(y):= \sum_{n\le y \atop \gcd(n,k)=1} \tau(n),$$
while we trivially have
\begin{equation} \label{eq:1.14a}
E_k(X) \ll X \log X.
\end{equation}


We shall now make use of a function $\rho(x)$ satisfying
\begin{equation} \label{eq:rho}
\exp\left\{ \sqrt{\log x} \right\} \le \rho(x) \le \sqrt  x
\end{equation}
and which we will later determine more precisely. 

\bigskip

We first write (\ref{eq:1.14}) as
\begin{eqnarray} \label{eq:evaS} \nonumber
E_k(X) &amp; = &amp; \sum_{d\le \rho(X) \atop \gcd(d,k)=1} \mu(d) V_k\left(
\frac X{d^2} \right) + \sum_{\rho(X) &lt; d \le \sqrt X \atop
\gcd(d,k)=1}
\mu(d) V_k\left( \frac X{d^2} \right) \\
&amp; = &amp; W_1(X) + W_2(X),
\end{eqnarray}
say.

It follows from Lemma \ref{lem:diviseur-phi} that
\begin{eqnarray}\label{eq:evaS2} \nonumber
W_2(X) &amp; \ll &amp; \sum_{\rho(X) &lt; d \le \sqrt X} \frac X{d^2}
\log(X/d^2) \le X \log X \sum_{d&gt;\rho(X)} \frac 1{d^2} \\
&amp; \ll &amp; \frac X{\rho(X)} \log X \ll \frac X{\log X}.
\end{eqnarray}

On the other hand, again using Lemma \ref{lem:diviseur-phi}, we have
\begin{eqnarray} \label{eq:s1temp} \nonumber
W_1 (X) &amp; = &amp; \sum_{d\le \rho(X) \atop \gcd(d,k)=1} \mu(d) \left\{
B_1(k) \frac X{d^2} \log \left( \frac X{d^2} \right) \right.
\\ \nonumber
&amp; &amp; \qquad  \left. +  B_2(k) \frac X{d^2} + O \left( k^{10/3} \left(
\frac X{d^2} \right)^{1/3}
 \log \left( \frac X{d^2} \right) \right) \right\}      \\ \nonumber
 &amp; = &amp;
 X \log X B_1(k) \sum_{d \le \rho(X) \atop \gcd(d,k)=1} \frac{\mu(d)}{d^2}
 - 2 X B_1(k) \sum_{d\le \rho(X) \atop \gcd(d,k)=1} \frac{\mu(d) \log
 d}{d^2}\\
 &amp; &amp; + \quad X  B_2(k) \sum_{d\le \rho(X) \atop \gcd(d,k)=1}
 \frac{\mu(d)}{d^2}  + O \left( k^{10/3} X^{1/3} \log X \right).
\end{eqnarray}
Since
\begin{eqnarray} \label{eq:def-eta} \nonumber
\sum_{d\le \rho(X) \atop \gcd(d,k)=1}
 \frac{\mu(d)}{d^2} &amp; = &amp; \sum_{d=1 \atop \gcd(d,k)=1}^\infty
 \frac{\mu(d)}{d^2} + O\left( \frac 1{\rho(X)} \right)
 = \frac 6{\pi^2} \prod_{p|k} \left(1 - \frac 1{p^2} \right)^{-1} + O\left( \frac 1{\rho(X)} \right)
 \\
 &amp; = &amp; \eta_k +  O\left( \frac 1{\rho(X)} \right)
 \end{eqnarray}
say,  and since
\begin{equation} \label{eq:def-mu}
\sum_{d\le \rho(X) \atop \gcd(d,k)=1} \frac{\mu(d) \log
 d}{d^2} = \sum_{d=1 \atop \gcd(d,k)=1}^\infty \frac{\mu(d) \log
 d}{d^2} +  O\left( \frac 1{\rho(X)} \right) = \mu_k +  O\left( \frac{\log X}{\rho(X)} \right),
 \end{equation}
say, it follows that (\ref{eq:s1temp}) can be written as
\begin{equation} \label{eq:evaS1}
W_1(X) = X \log X B_1(k) \eta_k -2X B_1(k) \mu_k  + X B_2(k) \eta_k
+ O\left( \frac X{\log X} \right) +  O \left( k^{10/3} X^{1/3} \log
X \right).
\end{equation}

Let us then write (\ref{eq:1.13}) as
\begin{eqnarray} \label{eq:1.13a} \nonumber
S(x) &amp; = &amp; \sum_{\delta b \le \rho(x) \atop b\in G_\delta} h(\delta)
2^{\omega(\delta)} E_\delta \left( \frac x{\delta b} \right) +
\sum_{\rho(x) &lt;\delta b \le x \atop b\in G_\delta} h(\delta)
2^{\omega(\delta)} E_\delta \left( \frac x{\delta b}
\right)\\
&amp; = &amp; Z_1(x)+Z_2(x),
\end{eqnarray}
say.

First of all, we have using (\ref{eq:1.14a}) and Lemma
\ref{lem:sum-h-delta},
\begin{equation} \label{eq:zz2}
Z_2(x) \ll \sum_{\rho(x)&lt;\delta b \le x \atop b\in G_\delta}
\frac{h(\delta) 2^{\omega(\delta)}}{\delta b} x \log(x/\delta b) \ll
x \log x \frac 1{\rho(x)^\kappa} \ll \frac x{\log x}.
\end{equation}

On the other hand, substituting (\ref{eq:evaS2}) and
(\ref{eq:evaS1}) in (\ref{eq:evaS}), we get that $Z_1(x)$ from
(\ref{eq:1.13a}) can be written as
\begin{eqnarray} \label{eq:Z1} \nonumber
Z_1(x) &amp; = &amp; \sum_{\delta b \le \rho(x) \atop b \in G_\delta}
h(\delta) 2^{\omega(\delta)} \left\{  \frac x{\delta b} \log \left(
\frac x{\delta b} \right)
 B_1(\delta) \eta_\delta +(B_2(\delta)\eta_\delta-2B_1(\delta)\mu_\delta)\frac x{\delta b} \right.
 \\  \nonumber
 &amp; &amp; \qquad \left.
  + O \left( \frac{\frac x{\delta b}}{\log(\frac x{\delta b})}\right) +
  \left( \delta^{10/3} \left( \frac x{\delta b} \right)^{1/3} \log \left( \frac x{\delta b}
 \right) \right)
 \right\} \\  \nonumber
&amp; = &amp; x\log x \, \sum_{\delta b \le \rho(x) \atop b \in G_\delta}
\frac{h(\delta) 2^{\omega(\delta)}}{\delta b}
 B_1(\delta) \eta_\delta \\  \nonumber
 &amp; &amp;  + x \sum_{\delta b \le \rho(x) \atop b \in G_\delta}
 \frac{h(\delta) 2^{\omega(\delta)}}{\delta b}
  \big(B_2(\delta)\eta_\delta-2B_1(\delta)\mu_\delta - B_1(\delta) \eta_\delta \log(\delta b) \big)
  \\  \nonumber
 &amp; &amp;
   \qquad + O \left( x\sum_{\delta b \le \rho(x) \atop b \in G_\delta} \frac{h(\delta) 2^{\omega(\delta)}}
  {\delta b\log(x/(\delta b))}\right) \\ \nonumber
 &amp; &amp; \qquad  +  O\left( \sum_{\delta b \le \rho(x) \atop b \in G_\delta}h(\delta) 2^{\omega(\delta)} \delta^{10/3} \left( \frac x{\delta b} \right)^{1/3} \log \left( \frac x{\delta b}
 \right) \right) \\
&amp;  = &amp; (x\log x) T_1(x) + x\,T_2(x) + O(T_3(x)) + O(T_4(x)),
\end{eqnarray}
say.

First, it follows from Lemma \ref{lem:noyau2} that
\begin{equation} \label{eq:T3}
T_3(x) \ll \frac x{\log x}
\end{equation}
and that
\begin{eqnarray} \label{eq:T4} \nonumber
T_4(x) &amp; \ll &amp; x^{1/3} \log x \sum_{\delta b\le \rho(x)} \sum_{b\in
G_\delta} \frac{h(\delta) 2^{\omega(\delta)}}{\delta b} \delta^4
b^{2/3} \\ \nonumber &amp; \ll &amp; x^{1/3} \log x (\rho(x))^4
\sum_{\delta=1}^\infty \sum_{b\in
G_\delta} \frac{h(\delta) 2^{\omega(\delta)}}{\delta b} \\
&amp; \ll &amp; x^{1/3} \log x (\rho(x))^4 \ll \frac x{\log x}.
\end{eqnarray}

On the other hand, using Lemma \ref{lem:sum-h-delta}, we have
\begin{eqnarray} \label{eq:T1} \nonumber
T_1(x) &amp; = &amp;  \sum_{\delta=1  \atop b \in G_\delta}^\infty
\frac{h(\delta) 2^{\omega(\delta)}  B_1(\delta) \eta_\delta}{\delta
b} - \sum_{\rho(x) &lt;\delta b \le x \atop b \in G_\delta}
\frac{h(\delta)
2^{\omega(\delta)}  B_1(\delta) \eta_\delta}{\delta b} \\
&amp; = &amp; t_1 + O \left( \frac 1{\rho(x)^\kappa} \right),
\end{eqnarray}
say.  Similarly, one can prove that
\begin{equation} \label{eq:T2}
T_2(x)  =   t_2 + O \left( \frac 1{\rho(x)^\kappa} \right),
\end{equation}
say.

Substituting (\ref{eq:T3}), (\ref{eq:T4}), (\ref{eq:T1}) and
(\ref{eq:T2}) in (\ref{eq:Z1}), we get
\begin{equation} \label{eq:zz1}
Z_1(x) \ll \frac{x \log x}{\rho(x)^\kappa}.
\end{equation}
Gathering (\ref{eq:zz2}) and (\ref{eq:zz1}) in (\ref{eq:1.13a}), we
obtain
$$Z(x) \ll \frac{x \log x}{\rho(x)^\kappa}.$$
Choosing $\rho(x)$, already introduced in (\ref{eq:rho}), in such a
way that
$$\frac{\log x}{\rho(x)^\kappa} \ll \frac 1{\log x}$$
completes the proof of Theorem~\ref{th:2}.

\section{Proofs of Theorems \ref{th:3} and \ref{th:4}}

Defining
$$S_d(x):= \sum_{p\le x \atop{ p\equiv 1 \mod d \atop
\gcd(\frac{p-1}d,d)=1}} \tau(p-1),$$ it is clear that
\begin{equation} \label{eq:defM}
 M(x) = \sum_{d\le x-1} \kappa(d) S_d(x).
\end{equation}
  Moreover, let
$$E_d(x):= \sum_{p\le x \atop p\equiv 1 \pmod d}
2^{\omega(p-1)}\qquad \mbox{ and } \qquad H_{d,\delta} (x):=
\sum_{p\le x \atop{p\equiv 1\pmod d \atop p\equiv 1
\pmod{\delta^2}}} \tau \left( \frac{p-1}{\delta^2}  \right).$$ Since
$$U(n) = 2^{\omega(n)} \sum_{d|n} h(d),$$
it follows that
\begin{equation} \label{eq:defN}
N(x) = \sum_{d\le x-1} h(d) E_d(x). \end{equation}
 On the other hand,
since
$$2^{\omega(n)} = \sum_{\delta^2 e=n} \mu(\delta) \tau(e),$$
we have
$$E_d(x)= \sum_{\delta \le \sqrt {x-1}} \mu(\delta) H_{d,\delta}(x).$$

It follows from Lemma \ref{lem:p-1} that
\begin{equation} \label{eq:6.8}
E_d(x), S_d(x) \le c_5 \frac{\tau(d)}{\varphi(d)} x \qquad
\mbox{say, for } d\le \sqrt x, \end{equation} while it is clear that
\begin{equation} \label{eq:6.8dd} \nonumber
\sum_{p\le x \atop p\equiv 1 \pmod d} \tau(p-1) \le \tau(d)
\sum_{n\le x/d} \tau(n) \le \frac{ c_6 x \log x \cdot \tau(d)}d
\qquad \mbox{for }1\le d\le x,
\end{equation}
and therefore,
\begin{equation} \label{eq:6.9}
E_d(x), S_d(x) \le c_7 \frac{\tau(d)}d x \log x.
\end{equation}

Let $A$ be a large constant. We shall now estimate
$$T_1:= \sum_{\log^A x \le d \le x-1} |\kappa(d)|S_d(x) \qquad
\mbox{and} \qquad  T_2:= \sum_{\log^A x \le d \le x-1} |h(d)|
E_d(x).$$

From (\ref{eq:6.8}) and (\ref{eq:6.9}), we have
\begin{equation} \label{eq:t1}
T_1 \ll x\log x \sum_{d\ge \sqrt x}
\frac{\tau(d)}{d\gamma(d)^\alpha} + x \sum_{d\ge \log^A
x}\frac{\tau(d)}{d\gamma(d)^\alpha} = x\log x\cdot U_1 + x\cdot U_2,
\end{equation}
say. Given a positive constant $\kappa&lt;\alpha$, we have that
\begin{eqnarray} \label{eq:u1} \nonumber
U_1 &amp; \le &amp; \sum_{d=1}^\infty \frac{\tau(d)}{d\gamma(d)^\alpha}
\left( \frac d{\sqrt x} \right)^\kappa = \frac 1{x^{\kappa/2}}
\sum_{d=1}^\infty
\frac{\tau(d)}{\gamma(d)^\alpha d^{1-\kappa}} \\
&amp; \le &amp; \frac 1{x^{\kappa/2}} \prod_p \left( 1+ \sum_{a=1}^\infty
\frac{\tau(p^a)}{p^\alpha \cdot p^{a(1-\kappa)}} \right),
\end{eqnarray}
where this last product, which we denote by $Q$, is convergent
provided $\kappa&lt;\alpha$.

In a similar way, we obtain
\begin{equation}  \label{eq:u2}
U_2 \le \sum_{d=1}^\infty \frac{\tau(d)}{d\gamma(d)^\alpha} \left(
\frac d{\log^A x} \right)^\kappa \le \frac 1{\log ^{A\kappa} x} Q.
\end{equation}
Substituting (\ref{eq:u1}) and  (\ref{eq:u2}) in  (\ref{eq:t1}), we
get that
\begin{equation}  \label{eq:t1a}
T_1 \ll \frac x{\log^{\kappa A-1} x} \qquad (0&lt;\kappa&lt;\alpha).
\end{equation}
In a similar manner, we obtain
\begin{equation}  \label{eq:t2a}
T_2 \ll \frac x{\log^{\kappa A-1} x} \qquad (0&lt;\kappa&lt;\alpha).
\end{equation}

We further define
\begin{eqnarray} \label{eq:def1}
M_1(x) &amp; = &amp; \sum_{d\le \log^A x} \kappa(d) S_d(x), \\
 N_1(x) &amp; = &amp; \sum_{d\le \log^A x} h(d) E_d(x).
\end{eqnarray}

Let us first fix $d\le \log^A x$ and move on to estimate $S_d(x)$
and $E_d(x)$, by using the Bombieri-Vinogradov Theorem.

It turns out that it is more convenient to estimate
$$S_d(x) - S_d(x/2) \qquad \mbox{ and } \qquad  E_d(x) - E_d(x/2)$$
and similarly for $x/2$, $x/2^2$, \ldots in place of $x$. This is
why, for any fixed integer $r\ge 1$, we write
\begin{equation} \label{eq:M1aa}
M_1(x)  =  \sum_{d\le \log^A x} \kappa(d) \sum_{j=0}^r A_d(x/2^j) +
O\left( \frac x{2^r} \right) + O \left(  \frac x{\log^{A\kappa} x}
\right),
\end{equation}
where
$$A_d(x/2^j) = S_d(x/2^j) - S_d(x/2^{j+1}).$$

Similarly, for any fixed integer $r\ge 1$, we may write
\begin{equation} \label{eq:N1a}
N_1(x)  =  \sum_{d\le \log^A x} h(d) \sum_{j=0}^r B_d(x/2^j) +
O\left( \frac x{2^r} \right) + O \left(  \frac x{\log^{A\kappa} x}
\right),
\end{equation}
where
$$B_d(x/2^j) = E_d(x/2^j) - E_d(x/2^{j+1}).$$


For now, fix $x$ and set
\begin{equation} \label{eq:defrr}
r:= \left\lfloor \frac{A\kappa \log \log x}{\log 2} \right\rfloor
\quad \mbox{ so that } \quad  2^r \approx \log^{A\kappa} x.
\end{equation}

We now proceed to estimate $A_d(x)$ and $B_d(x)$.

Clearly we have
\begin{equation} \label{eq:Ad}
A_d(x)= \sum_{\frac x2 &lt; p \le x \atop {\gcd(\frac{p-1}d,d)=1 \atop
p\equiv 1 \pmod d} } \tau(p-1) = \tau(d) \sum_{\frac x2 &lt; p \le x
\atop {\gcd(\frac{p-1}d,d)=1 \atop p\equiv 1 \pmod d} }
\tau(\frac{p-1}d).
\end{equation}

Observe that
$$\tau(n) = 2 \#\{(u,v): u&lt;v \mbox{ and } uv=n\} + \theta_n,$$
where
$$\theta_n = \left\{ \begin{array}{ll}
1, &amp; \mbox{if } n= \mbox{square};\\
0, &amp; \mbox{otherwise}.
\end{array} \right.$$


Thus, in light of (\ref{eq:Ad}),
\begin{equation} \label{eq:uv}
A_d(x)  =  \tau(d) \sum_{\frac x2 &lt; p \le x \atop p\equiv 1 \pmod d}
\#\{ (u,v): u&lt;v, \ \gcd(uv,d)=1,\ uv = \frac{p-1}d  \}  + O \left(
\sqrt x  \right),
\end{equation}
where the error term is there to account for those $p$ for which
$\frac{p-1}d = u^2$.

It is clear that in the sum appearing in (\ref{eq:uv}), we have
$u\le \sqrt x$. On the other hand, the contribution of those $u$'s
for which $u&gt; \sqrt x/(\log^B x)$ can be bounded above, using Lemma
\ref{lem:pi}, by
\begin{eqnarray} \label{eq:uvd} \nonumber
 \tau(d) \sum_{\sqrt x/(\log^B x)  &lt; u \le \sqrt x \atop
\gcd(u,d)=1} \pi(x;du,1) &amp; \le &amp; c_8 \tau(d)
\frac{\mbox{li}(x)}{\varphi(d)} \sum_{\sqrt x/(\log^B x) &lt; u \le
\sqrt x \atop \gcd(u,d)=1} \frac
1{\varphi(u)} \\
&amp; \le &amp;  c_9 \tau(d) \frac{\mbox{li}(x)}{\varphi(d)} B \log \log x,
\end{eqnarray}
where we also used Lemma \ref{lem:1-sur-phi}.

Concerning the equation $\frac{p-1}d=uv$, the condition
$\gcd(uv,d)=1$ is satisfied if and only if $\gcd(u,d)=1$ and
$v\equiv \ell \pmod d$ for some positive integer $\ell$ co-prime to
$d$, meaning that $v=\ell + dt$ and $p-1=du(\ell+dt)=du\ell +d^2 ut$
for some integer $t$.

In light of these observations and of (\ref{eq:uvd}), relation
(\ref{eq:uv}) can be replaced by
\begin{eqnarray} \label{eq:uv2} \nonumber
A_d(x) &amp; = &amp; \tau(d) \sum_{u \le \sqrt x/(\log^B x) \atop
\gcd(u,d)=1} \sum_{\gcd(\ell,d)=1} \left( \pi(x;d^2u,1+du\ell)-
\pi(\frac x2;
d^2u,1+du\ell) \right) \\
&amp; &amp; \qquad \qquad + O \left(   \frac{\tau(d)}{\varphi(d)}
(\mbox{li}(x)) \log \log x \right).
\end{eqnarray}

We shall assume that $B$ is sufficiently large to insure that $d^2u
\le \sqrt x/(\log^\Delta x)$, for some large number $\Delta$.

Using Lemma \ref{lem:bombieri}, (\ref{eq:uv2}) becomes
\begin{eqnarray} \label{eq:uv3} \nonumber
A_d(x) &amp; = &amp; \tau(d) \left( \mbox{li}(x) - \mbox{li}(x/2) \right)
\sum_{u \le \sqrt x/(\log^B x) \atop \gcd(u,d)=1}
\sum_{\gcd(\ell,d)=1} \frac 1{\varphi(d^2u)} \\ \nonumber &amp; &amp; \qquad
\qquad + O \left( \frac{\tau(d)}{\varphi(d)} (\mbox{li}(x))\log \log
x \right).
\end{eqnarray}

But observe that in the above sum, we have
$$ \sum_{\gcd(\ell,d)=1}\frac
1{\varphi(d^2u)} = \frac 1{\varphi(u)} \cdot
\frac{\varphi(d)}{d\varphi(d)}= \frac 1{\varphi(u)}\cdot \frac 1d
,$$ so that
\begin{eqnarray} \label{eq:uv4} \nonumber
A_d(x) &amp; = &amp; \frac{\tau(d)}{d}  \left( \mbox{li}(x) - \mbox{li}(x/2)
\right) \sum_{u \le \sqrt x/(\log^B x) \atop \gcd(u,d)=1} \frac
1{\varphi(u)} + O \left( \frac{\tau(d)}{\varphi(d)}
(\mbox{li}(x))\log \log x \right) \\
&amp;= &amp; \frac 14 \frac{\tau(d)c_d}{d} x + O\left( \frac{x \log \log
x}{\log x} \right) ,
\end{eqnarray}
($c_d$ being the constant appearing in the statement of Lemma
\ref{lem:1-sur-phi}) where we used Lemma \ref{lem:1-sur-phi} and the
fact that
$$(\mbox{li}(x) - \mbox{li}(x/2)) \cdot \left( \frac 12 \log x -B\log \log x \right) = \frac 14 x
+ O \left( \frac{ x \log \log x}{\log x} \right)$$ and that
$$\frac{\tau(d)}{\varphi(d)}\mbox{li}(x)\log \log x =O \left( \frac{ x \log \log x}{\log x} \right).$$


Substituting (\ref{eq:uv4}) in (\ref{eq:M1aa}) and taking into
account the choice of $r$ made in (\ref{eq:defrr}), we get
\begin{eqnarray} \label{eq:M1zz} \nonumber
M_1(x) &amp;  = &amp;  \sum_{d\le \log^A x} \kappa(d) \sum_{j=0}^r
A_d(x/2^j) +
 O \left(  \frac x{\log^{A\kappa} x} \right)\\     \nonumber
&amp; = &amp; \frac 14 \sum_{d\le \log^A x} \frac{\kappa(d)\tau(d)c_d}{d}
\sum_{j=0}^r
 \frac{x}{2^j}  + O\left( \frac{x (\log
\log x)^2}{\log x} \right) \\
&amp; = &amp; \frac 12 x \sum_{d\le \log^A x} \frac{\kappa(d)\tau(d)c_d}{d}
+ O\left( \frac{x (\log \log x)^2}{\log x} \right)
\end{eqnarray}
provided $A$ is chosen so that $A\kappa&gt;1$. Finally, using Lemma
\ref{lem:tau-gamma}, recalling the initial formulation of $M(x)$
given in (\ref{eq:defM}) and using (\ref{eq:M1zz}), we get
$$M(x) = x \frac 12 \sum_{d=1}^\infty \frac{\kappa(d)\tau(d)c_d}{d}
+ O\left( \frac{x (\log \log x)^2}{\log x} \right),$$ thus
completing the proof of Theorem~\ref{th:3}.


We now move to complete the proof of Theorem~\ref{th:4}. For this, we shall
use essentially the same kind of technique to obtain an estimate for
$E_d(x)$ for $d\le \log^A x$.

Since $$Q_{d,\delta}(x) := \sum_{n\le x/\delta^2} \tau(n) \le
\frac{c_{10}x \log x}{\delta^2},$$ it follows that
$$E_d(x) =\sum_{\delta\le \log^B x} \mu(\delta) Q_{d,\delta}(x) + O\left(
\frac xd \frac 1{\log ^2 x} \right),$$ provided $B$ is sufficiently
large.

We shall now estimate $Q_{d,\delta}(x)$ assuming that $d\le \log^A
x$ and $\delta\le \log^B x$.

We proceed to estimate
$$B(x)=B_{d,\delta}(x):=Q_{d,\delta}(x) - Q_{d,\delta}(x/2).$$

We have
$$B(x)=2 \sum_{\frac x2 &lt; p \le x \atop p \equiv 1
\pmod{\delta^2}} \#\{(u,v):u&lt;v,\ uv=\frac{p-1}{\delta^2},\
d|\delta^2uv\} +O(\sqrt x).$$

As in the proof of Theorem~\ref{th:3}, we can drop in the above solution
count those pairs $u,v$ for which $u &gt; \sqrt x/(\log^{\Delta} x)$
for any fixed $\Delta&gt;0$, arbitrarily large.

Now let $K,L\in G_d$. Given $u,v$ such that
$uv=\frac{p-1}{\delta^2}$, write $u=K \widetilde{u}$ and $v=L
\widetilde{v}$, where
$\gcd(\widetilde{u},d)=\gcd(\widetilde{v},d)=1$. We then have
\begin{equation} \label{eq:uv1}
\frac{p-1}{\delta^2} = KL \widetilde{u} \widetilde{v}.
\end{equation}
And in order to guarantee that $\gcd(\widetilde{v},d)=1$, we seek
$\widetilde{v}$ from the arithmetical progression
$\widetilde{v}=\ell + td$ with $\gcd(\ell,d)=1$. It follows from
(\ref{eq:uv1}) that
$$p-1= \delta^2 K L \widetilde{u} (\ell +td) = \delta^2 KL
\widetilde{u} \ell + \delta^2 KL d \widetilde{u} t.$$ With this set
up, we have
\begin{eqnarray} \label{eq:bx} \nonumber
B(x) &amp; = &amp; 2 \sum_{K,L\in G_d} \sum_{\gcd(\ell,d)=1}
\sum_{\widetilde{u} \le (\sqrt x/K)/\log^{\Delta} x \atop
\gcd(\widetilde{u},d)=1} \Big( \pi(x; \delta^2 KL d
\widetilde{u}, \delta^2 KL \widetilde{u}\ell +1) \\
&amp; &amp; \qquad \qquad - \quad \pi(x/2; \delta^2 KL d \widetilde{u}, \delta^2
KL \widetilde{u}\ell +1) \Big) \quad + \quad O(\sqrt x).
\end{eqnarray}
First we drop all pairs $K,L$ for which $KL &gt; \log^D x$, where $D$
is a fixed large number.
Indeed observe that
\begin{eqnarray} \label{eq:DD} \nonumber
&amp; &amp; \sum_{\log^D x &lt; KL \le x^{1/10} \atop K,L\in G_d}
\sum_{\widetilde{u} \le \sqrt x}\quad \sum_{\ell \pmod d}
\pi(x;\delta^2 KL d\widetilde{u},\delta^2 KL \widetilde{u}\ell+1) \\
\nonumber &amp; &amp; \qquad \qquad \qquad \qquad  + \sum_{x^{1/10}&lt;KL \le x
\atop K,L\in G_d} \sum_{\widetilde{u} \le \sqrt x}
\frac{x\varphi(d)}{\delta^2 KL d \widetilde{u}} \\ \nonumber &amp; &amp;
\qquad \le \mbox{li}(x) \sum_{KL&gt;\log^D x \atop K,L\in G_d}
\sum_{\widetilde{u} \le \sqrt x} \frac{\varphi(d)}{\varphi(\delta^2
KL d \widetilde{u})} + \sum_{x^{1/10} &lt; KL \le x \atop K&lt;L\in G_d}
\frac x{KL} \frac{\log x}{\varphi(\delta^2)} \\  \nonumber&amp; &amp; \qquad
\le \frac{\mbox{li}(x)}{\varphi(\delta^2)}\cdot \log x \cdot
\sum_{KL &gt; \log^D x \atop K,L\in G_d} \frac 1{\varphi(KL)} +
\frac{x\log x}{\varphi(\delta^2)} \sum_{KL &gt; x^{1/10} \atop K,L \in
G_d} \frac 1{KL} \\
&amp;  &amp; \qquad = \frac{\mbox{li}(x)}{\varphi(\delta^2)}\cdot \log x
\cdot {\cal H}_\delta(x) + \frac{x\log x}{\varphi(\delta^2)} \cdot
{\cal J}_\delta(x),
\end{eqnarray}
say.

First of all, using the fact that $\displaystyle{ \frac
1{\varphi(n)} \ll \frac{\log \log n}n}$, we have, for a fixed
$\kappa&gt;0$,
\begin{eqnarray} \label{eq:ch} \nonumber
{\cal H}_\delta(x) &amp; \le &amp; c_{11} \log\log  \log x \sum_{KL &gt; \log^D
x \atop K,L \in G_d} \frac 1{KL}  \le c_{11}\log \log \log x
\sum_{KL
&gt; \log^D x \atop K,L \in G_d} \frac 1{KL}  \left(
\frac{KL}{\log^D x}
\right)^\kappa \\
&amp; \le &amp; c_{11}\frac{\log \log \log x}{\log^{D\kappa} x} \prod_{p|d}
\left( \frac 1{1- \frac 1{p^{1-\kappa}}} \right)^2.
\end{eqnarray}
Choosing $\kappa=1/2$ and observing that for $d&lt; \log^A x$, we have
that  $\omega(d) \ll \frac{\log d}{\log \log d}$. Therefore for such
a number $d$,
$$\prod_{p|d} \left(
\frac 1{1- \frac 1{\sqrt p}} \right)^2
 &lt; c_{12} 4^{\omega(d)} &lt; (\log x)^{\varepsilon},$$
 for any $\varepsilon&gt;0$ arbitrarily small. Using this observation,
 we may conclude from (\ref{eq:ch}) that
\begin{equation} \label{eq:ch2}
{\cal H}_\delta(x) \le \frac{c_{13}}{(\log x)^{D/3}},
\end{equation}
say.  Proceeding in a somewhat similar manner, we get that
\begin{eqnarray} \label{eq:cj} \nonumber
{\cal J}_\delta(x) &amp; = &amp; \sum_{KL &gt; x^{1/10} \atop K,L\in G_d} \frac
1{KL} \le \sum_{KL &gt; x^{1/10} \atop K,L\in G_d} \frac
1{KL} \left( \frac{KL}{x^{1/10}} \right)^{1/2}\\
&amp;  \le &amp;  \frac 1{x^{1/20}} \prod_{p|d} \frac 1{\left(1- \frac
1{\sqrt p}\right)^2} &lt;
 \frac{(\log x)^{\varepsilon}}{x^{1/20}}.
\end{eqnarray}

Estimates (\ref{eq:DD}), (\ref{eq:ch2}) and (\ref{eq:cj}) therefore
establish that we can drop from the sum in (\ref{eq:bx}) those
$K,L\in G_d$ for which $KL &gt; \log^D x$ for a large $D$, the error
thus created being no larger than $\displaystyle{O \left( \frac{
\mbox{li}(x)}{(\log x)^{D/3}}\right)}$. In light of these
observations and using the fact that $\varphi(\delta^2 KL d
\widetilde{u})= KL \varphi(\delta^2 d \widetilde{u})$ and that
$$\sum_{K,L\in G_d \atop KL &lt; (\log x)^D} \frac 1{KL} = \sum_{K,L\in
G_d} \frac 1{KL}  - \sum_{K,L\in G_d \atop KL \ge  (\log x)^D} \frac
1{KL}  = \prod_{p|G_d} \left( 1-\frac 1p \right)^{-2}  + O \left(
\frac{ \mbox{li}(x)}{(\log x)^{D/3}}  \right),$$ we may write
(\ref{eq:bx}) as
\begin{eqnarray} \label{eq:bx2} \nonumber
B(x) &amp;  = &amp; 2(\mbox{li}(x)-\mbox{li}(x/2))  \sum_{KL&lt;(\log x)^D
\atop K,L \in G_d} \sum_{\widetilde{u} &lt; \sqrt x/(\log x)^\Delta
\atop \gcd(\widetilde{u},d)=1} \frac{\varphi(d)}{\varphi(\delta^2 KL
d \widetilde{u})} + O \left(
\frac{ \mbox{li}(x)}{(\log x)^{D/3}}\right) \\
&amp; = &amp;2(\mbox{li}(x)-\mbox{li}(x/2)) \prod_{p\in G_d} \left( 1 -
\frac 1p \right)^{-2} \sum_{\widetilde{u} &lt; \sqrt x/(\log x)^\Delta
\atop \gcd(\widetilde{u},d)=1} \frac{\varphi(d)}{\varphi(\delta^2 d
\widetilde{u})} + O \left( \frac{ \mbox{li}(x)}{(\log
x)^{D/3}}\right).
\end{eqnarray}
Set
$$Y_d(x):=
\sum_{\widetilde{u} &lt; \sqrt x/(\log x)^\Delta \atop
\gcd(\widetilde{u},d)=1} \frac 1{\varphi(\delta^2 d
\widetilde{u})}.$$ In order to evaluate $Y_d(x)$, we let $\delta=
\delta_1 \delta_2$, where $\delta_1 \in G_d$, $\gcd(\delta_2,d)=1$,
and $\widetilde{u} = \widetilde{u_1} \cdot \widetilde{u_2}$, where
$\widetilde{u_1} \in G_{\delta_2}$, $\gcd(\widetilde{u_2}, d
\delta_2)= \gcd(\widetilde{u_2}, d\delta)=1$. Then,
$$\varphi(\delta^2 d \widetilde{u}) =  \varphi(\delta_1^2 \delta_2^2  \widetilde{u_1})
\varphi( \widetilde{u_2}) = \varphi(\delta_1^2 d )
\varphi(\delta_2^2  \widetilde{u_1}) \varphi(\widetilde{u_2}).$$

It follows from Lemma \ref{lem:1-sur-phi} that
\begin{equation} \label{eq:6.28}
\sum_{\gcd(\widetilde{u_2},d\delta)=1 \atop \widetilde{u_2} \le y}
\frac 1{\varphi(\widetilde{u_2})} = c_{d\delta} \log y +O(1).
\end{equation}
Now, the contribution of those $\widetilde{u_1}$ in $Y_d(x)$ for
which $\widetilde{u_1}\in G_{\delta_2}$ and $\widetilde{u_1}&gt; (\log
x)^{\Delta_0}$ is, using (\ref{eq:6.28}),
\begin{eqnarray*}
&amp;&amp; \ll \sum_{\widetilde{u_1}\in G_{\delta_2} \atop
\widetilde{u_1}&gt;(\log x)^{\Delta_0}} \frac 1{\varphi(\delta_1^2 d)}
\frac 1{\varphi(\delta_2^2) \widetilde{u_1}} \sum_{ } \frac
1{\varphi(\widetilde{u_2})} \\
&amp; &amp; \ll (\log x) \sum \frac 1{\varphi(\delta_1^2 d)} \frac
1{\varphi(\delta_2^2) }  \sum_{\widetilde{u_1}\in G_{\delta_2} \atop
\widetilde{u_1}&gt;(\log x)^{\Delta_0}} \frac 1{\widetilde{u_1}} \\
&amp; &amp; \ll \frac{\log x}{\varphi(\delta_1^2 d) \varphi(\delta_2^2)}
\frac 1{(\log x)^{\Delta_0/2}} \sum_{\widetilde{u_1}\in
G_{\delta_2}}
\frac 1{\sqrt{\widetilde{u_1}}} \\
&amp; &amp; \ll (\log x)^{1-\Delta_0/2} \frac 1{\delta_1^2 \varphi(d)
\varphi(\delta_2^2)} \prod_{p|\delta_2} \left( 1- \frac 1{\sqrt p}
\right)^{-1} \\
&amp; &amp; \ll (\log x)^{1-\Delta_0/2} \frac 1{\varphi(d)}
\frac{2^{\omega(\delta_2)}}{\delta_1^2 \delta_2^2}.
\end{eqnarray*}

On the other hand, the contribution of those $\widetilde{u_1} \le
(\log x)^{\Delta_0}$ in $Y_d(x)$, as $\widetilde{u_2}$ runs up to
$y_0:=\sqrt x/((\log x)^C \widetilde{u_1})$, and also using Lemma
\ref{lem:1-sur-phi} so that
\begin{equation} \label{eq:7.28} \nonumber
\sum_{\gcd(\widetilde{u_2},d\delta)=1 \atop \widetilde{u_2} \le y_0}
\frac 1{\varphi(\widetilde{u_2})} = \frac 12 c_{d\delta} \log x
+O(\log \log x),
\end{equation}
we get
\begin{eqnarray} \label{eq:mm} \nonumber
Y_d(x) &amp; = &amp; \frac 12 c_{d\delta} (\log x + \log \log x) \frac
1{\varphi(\delta_1^2 d)} \frac 1{\varphi(\delta_2^2)}
\sum_{\widetilde{u_1} \in G_{\delta_2}} \frac 1{\widetilde{u_1}} \\
&amp; &amp; \qquad  + \ O \left( \frac{2^{\omega(\delta_2)}}{\delta^2
\varphi(d)} (\log x)^{1-C_0/2} \right).
\end{eqnarray}
Observing that $\varphi(\delta_1^2 d) = \delta_1^2 \varphi(d)$ and
that
$$\sum_{\widetilde{u}\in G_{\delta_2}} \frac 1{\widetilde{u}} =
\prod_{p|G_{\delta_2}} \left( 1- \frac 1p \right)^{-1} =
\frac{\delta_2}{\varphi(\delta_2)},$$ (\ref{eq:mm}) becomes
\begin{eqnarray} \label{eq:mm1} \nonumber
Y_d(x) &amp; = &amp; \frac 12 \frac{c_{d\delta}}{\varphi(d)} \frac
1{\delta_1^2 \delta_2^2} \frac{\delta_2}{\varphi(\delta_2)} (\log x
+ \log \log x) \\
&amp; &amp; \qquad  + O \ \left( \frac{2^{\omega(\delta_2)}}{\delta^2
\varphi(d)} (\log x)^{1-C_0/2} \right).
\end{eqnarray}
Using (\ref{eq:mm1}) in (\ref{eq:bx2}), we obtain
\begin{eqnarray*}
B(x) &amp; = &amp; (\mbox{li}(x) - \mbox{li}(x/2))
\frac{c{_d\delta}}{\delta_1^2 \delta_2 \varphi(\delta_2)} (\log x +
\log \log x) \\
&amp; &amp; +O \left( \frac{2^{\omega(\delta_2)}}{\delta^2 \varphi(d)} (\log
x)^{1-C_0/2} \right).
\end{eqnarray*}
Finally, summing over $d\le (\log x)^A$ and $\delta \le (\log x)^B$,
the theorem follows.

\section{Final remarks}

It is interesting to inquire about the solutions to the equation
\begin{equation} \label{eq:f6}
\widetilde{P}(n)=\widetilde{P}(n+1),
\end{equation}
or equivalently
\begin{equation} \label{eq:f7}
\frac n{\gamma(n)} \prod_{p|n} (2p-1) =  \frac{n+1}{\gamma(n+1)}
\prod_{q|n+1} (2q-1).
\end{equation}
The first 13  solutions are 45, 225, 1125, 2025, 3645, 140\,625,
164\,025, 257\,174, 703\,125, 820\,125,  1\,265\,625, 2\,657\,205
and 3\,515\,625.

 One can easily check that if for some positive
integers $a$ and $b$, the number $p=\frac{3^a\cdot 5^b+1}2$ is a
prime number, then $n=3^a\cdot 5^b$ is a solution of (\ref{eq:f6}),
with $\widetilde{P}(n)=3^{a+1}\cdot 5^b$. If one could prove that
there exist infinitely many primes of this form, it would follow
that equation (\ref{eq:f6}) has infinitely many solutions.
Interestingly, a computer search establishes that (\ref{eq:f6}) has
37 solutions $n&lt;10^{10}$ and that 21 of them are of the form $n=3^a
\cdot 5^b$.

Moreover, all 37 solutions $n_0$ are such that $3^3\cdot
5|\widetilde{P}(n_0)$. To see that $3|n_0$, proceed as follows.
Suppose that $3\not|\,\widetilde{P}(n_0)$, then, in light of
(\ref{eq:f7}), $3\not|\,(2p-1)$ for each prime divisor $p$ of $n$
and $3\not|\,(2q-1)$ for each prime divisor $q$ of $n+1$. This would
imply that $p\equiv 1 \pmod 3$ for all $p|n$ (implying that $n\equiv
1 \pmod 3$) and $q\equiv 1 \pmod 3$ for all $q|n+1$ (implying that
$n+1\equiv 1 \pmod 3$), a contradiction.

The fact that $3|\widetilde{P}(n_0)$ clearly implies that
$5|\widetilde{P}(n_0)$.

\section{Acknowledgement}

The authors would like to thank the
referee for some very helpful suggestions which improved the quality
of this paper.

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\bibitem{kn:ivic} A.\,Ivi\'c, {\sl The Riemann Zeta Function},
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\bibitem{kn:pillai} S. S.\,Pillai, On an arithmetic function,
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\bibitem{kn:selberg} A. Selberg, Note on a paper by L. G. Sathe, 
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\bibitem{kn:titchmarsh} E. C.\,Titchmarsh, A divisor problem,
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\bibitem{kn:tolev} D. I.\,Tolev,  On the divisor problem in
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\bibitem{kn:toth} L.\,Toth, A gcd-sum function over regular
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\end{thebibliography}

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\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11A25; Secondary 11N37.

\noindent \emph{Keywords: } 
gcd-sum function, Dirichlet divisor problem, shifted primes.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  August 31 2009;
revised version received December 29 2009. 
Published in {\it Journal of Integer Sequences}, December 31 2009.

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\noindent
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