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\begin{center}
\vskip 1cm{\LARGE\bf 
On the Powerful and Squarefree Parts of an  \\
\vskip .1in
Integer
}
\vskip 1cm
\large
Maurice-\'Etienne Cloutier, Jean-Marie De Koninck, and Nicolas Doyon \\
D\'epartement de math\'ematiques et de statistique\\
Universit\'e Laval \\
Qu\'ebec G1V 0A6 \\
Canada \\
\href{mailto:maurice-etienne.cloutier.1@ulaval.ca}{\tt maurice-etienne.cloutier.1@ulaval.ca} \\
\href{mailto:jmdk@mat.ulaval.ca}{\tt jmdk@mat.ulaval.ca} \\
\href{mailto:nicolas.doyon@mat.ulaval.ca}{\tt nicolas.doyon@mat.ulaval.ca} \\
\end{center}

\vskip .2 in


\begin{abstract}
Any integer $n\ge 2$ can be written in a unique way as the product of
its powerful part and its squarefree part, that is, as $n=mr$ where $m$
is a powerful number and $r$ a squarefree number, with $\gcd(m,r)=1$. We
denote these two parts of an integer $n$ by $\pow(n)$ and $\sq(n)$
respectively, setting for convenience $\pow(1)=\sq(1)=1$.  We first
examine the behavior of the counting functions $\sum_{n\le x,\,
{\scriptsize \sq}(n)\le y} 1$ and $\sum_{n\le x,\, {\scriptsize
\pow}(n)\le y} 1$. Letting $P(n)$ stand for the largest prime factor of
$n$, we then provide asymptotic values of $A_y(x):=\sum_{n\le x,\,
P(n)\le y} \pow(n)$ and $B_y(x) :=\sum_{n\le x,\, P(n)\le y} \sq(n)$
when $y=x^{1/u}$ with $u\ge 1$ fixed. We also examine the size of
$A_y(x)$ and $B_y(x)$ when $y=(\log x)^\eta$ for some $\eta&gt;1$.
Finally, we prove that $A_y(x)$ will coincide with $B_y(x)$ in the
sense that $\log(A_y(x)/x) = (1+o(1))\log(B_y(x)/x)$ as $x\to \infty$
if we choose $y=2\log x$.
\end{abstract}




\section{Introduction and notation}

A {\it powerful number} (or {\it square-full number}) is a positive integer $n$ such that $p^2\mid n$ for every prime factor $p$ of $n$.
In 1934, Erd\H{os} and Szekeres \cite{kn:erdos} were the first to study the distribution of these numbers.
In 1958, Bateman and Grosswald \cite{kn:bateman} established a very accurate estimate for the number $N(x)$ of powerful numbers not exceeding $x$, namely
$$
N(x) = C_2 x^{1/2} +C_3 x^{1/3} + o(x^{1/6}) \qquad (x\to \infty),
$$
where $\displaystyle{C_2= \frac{\zeta(3/2)}{\zeta(3)} \doteq 2.1732}$ and $\displaystyle{C_3=\frac{\zeta(2/3)}{\zeta(2)} \doteq -1.4879}$. Here $\zeta$ stands for the Riemann Zeta Function.

Further estimates and generalizations concerning powerful numbers were later obtained by Ivi\'c and Shiu \cite{kn:ivic}.

On the other hand, a {\it squarefree number} is a positive integer $n$ such that $p^2\nmid n$ for every prime factor $p$ of $n$. It is well known 
(see, for instance, Hardy and Wright \cite[pp.\ 169--170]{kn:hardy})  that
the number $Z(x)$ of squarefree numbers not exceeding $x$ can be
written as
$$
Z(x) =\sum_{n\le x} \mu^2(n)  = \frac 6{\pi^2} x + O\left( x^{1/2} \right),
$$
where $\mu$ stands for the Moebius function.

Interestingly, any integer $n\ge 2$ can be written in a unique way as the product of its powerful part and its squarefree part, that is, as $n=mr$ where $m$ is a {\it powerful number}  and $r$ a squarefree number, with $\gcd(m,r)=1$. We denote these two parts of an integer $n$ by $\pow(n)$ and $\sq(n)$ respectively, setting for convenience $\pow(1)=\sq(1)=1$.

In his master's thesis,
Cloutier \cite{kn:MEC} obtained asymptotic values for $\sum_{n\le x} \pow(n)^a \sq(n)^b$, where $a$ and $b$ are fixed integers. Particular cases of his results yield the following estimates.
\begin{eqnarray} \label{eq:me1}
\sum_{n\le x} \pow(n) &amp; = &amp; \frac{d_1}{3} x^{3/2} +O(x^{4/3}),\\ \label{eq:me2}
\sum_{n\le x} \sq(n) &amp; = &amp; \frac{d_2}{2} x^2 +O(x^{3/2}),\\ \label{eq:me3}
\sum_{n\le x} \frac 1{\pow(n)} &amp; = &amp; d_2 x +O(x^{1/2}),\\ \label{eq:me4}
\sum_{n\le x} \frac 1{\sq(n)} &amp; = &amp; d_1 x^{1/2} +O(x^{1/3}),
\end{eqnarray}
where
\begin{eqnarray} \label{eq:d1}
d_1 &amp; = &amp; \prod_p \left( 1+ \frac 2{p^{3/2}} - \frac 1{p^{5/2}} \right)\doteq 3.52,\\ \label{eq:d2}
d_2 &amp; = &amp; \prod_p \left( 1-  \frac 1{p^2}  + \frac 1{p^3(p+1)}  \right) \doteq 0.65.
\end{eqnarray}



In particular, it follows from (\ref{eq:me1}) and (\ref{eq:me2}) that $B(x):=\sum_{n\le x} \sq(n)$ is much larger than $A(x):=\sum_{n\le x} \pow(n)$ for large values of $x$. However, letting $P(n)$ stand for the largest prime factor of $n$, with $P(1)=1$, one can easily check that if we set
\begin{equation} \label{eq:defAB}
A_y(x):= \sum_{\substack{n\le x \\ P(n)\le y}} \pow(n) \qquad \mbox{ and } \qquad B_y(x)= \sum_{\substack{n\le x\\ P(n)\le y}} \sq(n),
\end{equation}
then, for small values of $y=y(x)$, we have $B_y(x) = o(A_y(x))$ as $x\to \infty$, while for large values of  $y=y(x)$, we have $A_y(x) = o(B_y(x))$.


Here, after examining
the comparative sizes of $\#\{n\le x: \sq(n)\le y\}$ and $\#\{n\le x: \pow(n)\le y\}$,
we show that $\#\{n\le x: \pow(n) &gt; \sq(n)\}$ is asymptotic to $cx^{3/4}$ for some computable constant $c&gt;0$.
We then provide asymptotic values for $A_y(x)$ and $B_y(x)$ when $y=x^{1/u}$ with $u\ge 1$ fixed. We also examine the size of $A_y(x)$ and $B_y(x)$ when $y=(\log x)^\eta$ for some $\eta&gt;1$.
Finally, we establish that if $y=2\log x$, then
$\log(A_y(x)/x) = (1+o(1))\log(B_y(x)/x)$ as $x\to \infty$.

Throughout this paper, we let $\phi$ stand for the Euler totient function, $\omega(n)$ for the number of distinct prime factors of $n$ (with $\omega(1)=0$), $\Omega(n)$ for the number of prime factors of $n$ counting their multiplicity (with $\Omega(1)=0$) and $p(n)$ for the smallest prime factor of $n$ (with $p(1)=1$). 
We shall also be using the function  $\Psi(x,y):=\#\{n\le x: P(n)\le y\}$ defined for $2\le y\le x$.

Denoting by $\pi(x)$ the number of prime numbers not exceeding $x$, we shall be using the prime number theorem in its various forms, in particular as follows:
\begin{eqnarray} \label{eq:pnt1}
\pi(x) &amp; = &amp; (1+o(1)) \frac x{\log x} \qquad (x\to \infty),\\ \label{eq:pnt2}
\prod_{p\le z} p &amp; = &amp; \exp\{ (1+o(1)) z \} \qquad (z\to \infty), \\ \label{eq:pnt3}
\sum_{p\le x} \log p &amp; = &amp; (1+o(1)) x \qquad (x\to \infty), \\ \label{eq:pnt4}
p_k &amp; = &amp; (1+o(1)) k \log k \qquad (k\to \infty),
\end{eqnarray}
where $p_k$ stands for the $k$-th prime number.

\section{The distribution of the values of $\pow(n)$ and of $\sq(n)$}

We examine the distribution of the values of $\pow(n)$ and of $\sq(n)$ separately in the following two theorems.


\begin{theorem} \label{thm:sqy}
For $2\le y\le x$,
\begin{equation} \label{eq:sqy}
\sum_{\substack{n\le x \\ {\scriptsize \sq(n)}\le y}} 1 =  2C_0\frac{\zeta(3/2)}{\zeta(3)} \sqrt{xy} + O\left( x^{1/3} y^{2/3} \log y  \right),
\end{equation}
where
\begin{equation} \label{eq:defC0}
C_0= \prod_p \left( 1+ \frac{1-1/p}{p(1+1/p^{3/2})} \right) \left(1 - \frac 1p \right) \doteq 0.38.
\end{equation}
\end{theorem}

\begin{remark}
It is clear that estimate (\ref{eq:sqy}) has some interest only if the error term is of smaller order than the main term, that is, when $y\log^6 y =o(x)$ as $x\to \infty$.
\end{remark}


\begin{theorem} \label{thm:powy}
For $2\le y\le x$,
\begin{equation} \label{eq:powy}
\sum_{\substack{n\le x \\ {\scriptsize \pow}(n)\le y}} 1 =  x \frac 6{\pi^2} \sum_{m\le y \atop m\ \mbox{\tiny powerful}} \frac 1{m \prod_{p\mid m} \left( 1 + \frac 1p \right)} + O\left( \sqrt{x} \log y  \right).
\end{equation}
\end{theorem}

\begin{remark}
Observe that the main term on the right hand side of (\ref{eq:powy}) is asymptotic to $x$ as $y\to \infty$. Indeed, this follows from the fact that
\begin{eqnarray*}
\frac 6{\pi^2} \lim_{y\to \infty} \sum_{m\le y \atop m\ \mbox{\tiny powerful}} \frac 1{m \prod_{p\mid m} \left( 1 + \frac 1p \right)} &amp; = &amp; \prod_p \left( 1 - \frac 1{p^2} \right) \prod_p \left( 1+ \frac 1{p^2(1+\frac 1p)} +  \frac 1{p^3(1+\frac 1p)} + \cdots \right)\\
&amp; = &amp; \prod_p \left( 1 - \frac 1{p^2} \right) \prod_p \left( 1+ \frac 1{p^2-1} \right) =1.
\end{eqnarray*}
\end{remark}


In order to prove these two theorems, we first establish some preliminary results.

\begin{lemma} \label{lem:1}
Given an arbitrary positive integer $r$,
\begin{equation} \label{eq:pui-r}
\#\{n\le x: n \mbox{ powerful}, \ (n,r)=1\} =  \frac{\phi(r)}r \frac{\zeta(3/2)}{\zeta(3)} \prod_{p|r} \left( 1+ \frac 1{p^{3/2}} \right)^{-1} \sqrt x  + O\left( 2^{\omega(r)} x^{1/3} \right),
\end{equation}
where the constant implied by the Landau symbol is absolute.
\end{lemma}

\begin{proof}
Using the estimate
\begin{equation} \label{eq:uni}
\sum_{n\le X \atop (n,r)=1} 1 = \frac{\phi(r)}r X + O\left(2^{\omega(r)}\right) \qquad \mbox{ uniform for } r\ge 1,
\end{equation}
we have, using the fact that any powerful number $n$ can be written in a unique way as $n=a^2b^3$ with $\mu^2(b)=1$,
\begin{eqnarray} \label{eq:tempo1} \nonumber
\sum_{\substack{n\le x \\n\text{ powerful} \\ (n,r)=1}} 1 &amp; = &amp;
\sum_{\substack{a^2b^3 \le x \\ (a,r)=1\\ (b,r)=1}} \mu^2(b)
 =  \sum_{\substack{b^3\le x \\ (b,r)=1}} \mu^2(b) \sum_{\substack{a^2 \le x/b^3 \\ (a,r)=1}} 1 \\ \nonumber
&amp; = &amp; \sum_{\substack{b^3\le x \\ (b,r)=1}} \mu^2(b) \left( \frac{\phi(r)}r  \frac{x^{1/2}}{b^{3/2}} +O\left( 2^{\omega(r)} \right) \right) \\
&amp; = &amp;  x^{1/2} \frac{\phi(r)}r \sum_{\substack{b\le x^{1/3} \\ (b,r)=1}} \frac{\mu^2(b)}{b^{3/2}}
+ O\left( 2^{\omega(r)} x^{1/3} \right).
\end{eqnarray}
Using the fact that
\begin{eqnarray*}
\sum_{b\le x^{1/3} \atop (b,r)=1} \frac{\mu^2(b)}{b^{3/2}}
&amp; = &amp;
\sum_{b=1 \atop (b,r)=1}^\infty \frac{\mu^2(b)}{b^{3/2}} + O\left( \sum_{b&gt; x^{1/3}} \frac 1{b^{3/2}} \right)\\
&amp; = &amp; \prod_{p\nmid r} \left( 1 + \frac 1{p^{3/2}} \right) +  O\left( \int_{x^{1/3}}^\infty \frac 1{t^{3/2}}\,dt  \right)\\
&amp; = &amp; \prod_{p\nmid r} \left( 1 + \frac 1{p^{3/2}} \right) +
O\left( \frac 1{x^{1/6}} \right),
\end{eqnarray*}
we obtain from (\ref{eq:tempo1}) that
\begin{eqnarray*}
 \sum_{\substack{n\le x \\n\text{ powerful} \\ (n,r)=1}} 1
&amp; = &amp; x^{1/2} \frac{\phi(r)}r  \left(
\prod_{p\nmid r} \left( 1 + \frac 1{p^{3/2}} \right) +
O\left( \frac 1{x^{1/6}} \right) \right) +  O\left( 2^{\omega(r)} x^{1/3} \right) \\
&amp; = &amp; x^{1/2} \frac{\phi(r)}r
\prod_{p\nmid r} \left( 1 + \frac 1{p^{3/2}} \right)
+ O\left( 2^{\omega(r)} x^{1/3} \right)\\
&amp; = &amp; x^{1/2} \frac{\phi(r)}r
\frac{\prod_p \left( 1 + \frac 1{p^{3/2}} \right)}{\prod_{p\mid r} \left( 1 + \frac 1{p^{3/2}} \right)}
+ O\left( 2^{\omega(r)} x^{1/3} \right),
\end{eqnarray*}
which proves (\ref{eq:pui-r}).
\end{proof}


\begin{lemma} \label{lem:2}
Given an arbitrarily small $\delta&gt;0$, as $x\to \infty$,
\begin{equation} \label{eq:pui-r2}
\sum_{r\le x} \frac{\mu^2(r)\phi(r)}{r \prod_{p|r} \left(1+\frac 1{p^{3/2}} \right)} = C_0 x + O\left( x^{\frac 12 + \delta}\right),
\end{equation}
where $C_0$ is the constant defined in (\ref{eq:defC0}).
\end{lemma}



\begin{proof}
For $s&gt;1$, we have
\begin{eqnarray*}
\sum_{n=1}^\infty \frac{ \mu^2(n) \phi(n)/(n \prod_{p|n} (1 + \frac 1{p^{3/2}}))}{n^s} &amp; = &amp;
\prod_p \left( 1 + \frac{1-1/p}{1+ \frac 1{p^{3/2}}}\, \frac 1{p^s} \right) \\
&amp; = &amp;
\zeta(s) \prod_p \left(1 - \frac 1{p^s} \right) \prod_p \left( 1 + \frac{1-1/p}{1+ \frac 1{p^{3/2}}}\, \frac 1{p^s} \right)\\
&amp; = &amp;
\zeta(s)  H(s),
\end{eqnarray*}
say. Observing that, for any fixed $s&gt; \frac 12$, the infinite product
$H(s)$ is absolutely convergent, then (\ref{eq:pui-r2}) follows by
using Wintner's theorem (see, for instance, 
\cite[Theorem 6.13 and Problem 6.3]{kn:DL}).
\end{proof}

As an immediate consequence of this lemma, we obtain the following.


\begin{lemma} \label{lem:3}
Given an arbitrarily small $\delta&gt;0$,
$$
\sum_{r\le y} \frac{\mu^2(r)\phi(r)}{r^{3/2} \prod_{p|r} \left(1+\frac 1{p^{3/2}} \right)} =
2C_0 \sqrt y +O \left( y^{\frac 14 +\delta} \right).
$$
\end{lemma}

\begin{lemma} \label{lem:katai}
Given an arbitrary powerful number $m$,
$$\sum_{r\le x/m \atop (r,m)=1 } \mu^2(r) = \alpha(m) x + O \left( \sqrt{\frac xm} \beta(m) \right),$$
where
$$\alpha(m) = \frac 6{\pi^2 m} \prod_{p\mid m} \left( 1 + \frac 1p \right)^{-1} \qquad \mbox{ and }
\qquad \beta(m) = \prod_{p\mid m} \left( 1 - \frac 1{\sqrt p} \right)^{-1}.$$
\end{lemma}

\begin{proof}
This is relation (1) in the paper of De Koninck, K\'atai and Subbarao \cite{kn:DKS} with a correction by Cloutier \cite{kn:MEC}.
\end{proof}

We now have the necessary tools to prove Theorems \ref{thm:sqy} and \ref{thm:powy}.



\begin{proof}[Proof of Theorem \ref{thm:sqy}]
As a consequence of Lemma \ref{lem:1}, we have, as $x\to \infty$,
\begin{eqnarray*}
\sum_{\substack{n\le x \\ {\scriptsize \sq}(n)\le y}} 1 &amp; = &amp; \sum_{\substack{mr \le x \\m\text{ powerful} \\ r\le y\\ (m,r)=1}} \mu^2(r)
 =  \sum_{r\le y} \mu^2(r) \sum_{\substack{m\le x/r \\m \text{ powerful} \\  (m,r)=1}} 1\\
&amp; = &amp; \sum_{r\le y} \mu^2(r) \left( \frac{\phi(r)}r \frac{\zeta(3/2)}{\zeta(3)} \prod_{p|r} \left( 1+ \frac 1{p^{3/2}} \right)^{-1} \sqrt{\frac xr} + O \left( 2^{\omega(r)} \left( \frac xr \right)^{1/3}  \right) \right)\\
&amp; = &amp;  \sqrt x \frac{\zeta(3/2)}{\zeta(3)} \sum_{r\le y} \frac{\mu^2(r) \phi(r)}{r^{3/2} \prod_{p|r} \left( 1+ \frac 1{p^{3/2}} \right)} + O \left(  x^{1/3} \sum_{r\le y} \frac{2^{\omega(r)}}{r^{1/3}}  \right) \\
&amp; = &amp;  \sqrt x \frac{\zeta(3/2)}{\zeta(3)} \sum_{r\le y} \frac{\mu^2(r) \phi(r)}{r^{3/2} \prod_{p|r} \left( 1+ \frac 1{p^{3/2}} \right)} + O \left(  x^{1/3} y^{2/3} \log y \right),
\end{eqnarray*}
where we used the fact that, since $\displaystyle{\sum_{r=1}^\infty \frac{2^{\omega(r)}}{r^s}=\frac{\zeta^2(s)}{\zeta(2s)}}$ say for $s&gt;1$, it follows that
\begin{equation} \label{eq:omega}
\sum_{r\le y} 2^{\omega(r)} = \frac 1{\zeta(2)} y \log y + O(y),
\end{equation}
implying by partial summation that
$$\sum_{r\le y} \frac{2^{\omega(r)}}{r^{1/3}} \ll y^{2/3}\log y.$$
Using Lemma \ref{lem:3}, the proof of Theorem \ref{thm:sqy}  is thus complete.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm:powy}]
Using Lemma \ref{lem:katai}, we may write
\begin{eqnarray} \label{eq:x11} \nonumber
\sum_{n\le x \atop {\scriptsize \pow}(n)\le y }  1 &amp; = &amp; \sum_{m\le y \atop m\ \mbox{\tiny powerful}} \sum_{r\le x/m \atop (r,m)=1}\mu^2(r)
 =  x\sum_{m\le y \atop m\ \mbox{\tiny powerful}}  \alpha(m)  + O \left( \sum_{m\le y \atop m\ \mbox{\tiny powerful}} \sqrt{\frac xm} \beta(m) \right)\\
&amp; = &amp; x\frac 6{\pi^2} \sum_{m\le y \atop m\ \mbox{\tiny powerful}} \frac 1{m \prod_{p\mid m}(1+\frac 1p)}  + O \left( \sqrt x \sum_{m\le y \atop m\ \mbox{\tiny powerful}} \frac{ \beta(m)}{\sqrt m} \right).
\end{eqnarray}
Using the estimate
$$\sum_{m\le y \atop m\ \mbox{\tiny powerful}} \frac{ \beta(m)}{\sqrt m}
\le \prod_{p\le y} \left( 1+ \frac{\beta(p)}{p^{2/2}} + \frac{\beta(p)}{p^{3/2}} +\cdots  \right)\le \prod_{p\le y} \left( 1 + \frac 1p + O\left( \frac 1{p^{3/2}} \right) \right) \ll \log y$$
in (\ref{eq:x11}) completes the proof of Theorem \ref{thm:powy}.
\end{proof}


\section{How often is $\pow(n)$ larger than $\sq(n)$}

In this section, we show that the set of positive integers  $n$ such that $\pow(n)&gt;\sq(n)$ is of zero density. In fact, we show much more.

\begin{theorem} \label{thm:pow-sq} As $x$ becomes large,
\begin{equation} \label{eq:star}
\#\{n\le x: \pow(n) &gt; \sq(n)\} = D_0 x^{3/4} + O\left( x^{2/3} \log x \right),
\end{equation}
where $\displaystyle{D_0 = \frac 43 C_0 \frac{\zeta(3/2)}{\zeta(3)} \doteq 1.10}$.
\end{theorem}

\begin{proof}
We clearly have
\begin{eqnarray} \label{eq:x1} \nonumber
\#\{n\le x: \pow(n) &gt; \sq(n)\} &amp;=&amp; \sum_{r\le \sqrt x} \mu^2(r) \sum_{r&lt;m \le x/r \atop{ m \ \mbox{\tiny powerful} \atop (m,r)=1}} 1 \\ \nonumber
&amp; =&amp; \sum_{r\le \sqrt x} \mu^2(r) \sum_{m \le x/r \atop{ m \ \mbox{\tiny powerful} \atop (m,r)=1}} 1
- \sum_{r\le \sqrt x} \mu^2(r) \sum_{m \le r \atop{ m \ \mbox{\tiny powerful} \atop (m,r)=1}} 1 \\
&amp; = &amp; T_1(x) - T_2(x),
\end{eqnarray}
say. On the one hand, using Lemma \ref{lem:1} and (\ref{eq:omega}), we have
\begin{eqnarray} \label{eq:x2} \nonumber
T_1(x)  &amp; = &amp; \sqrt x \frac{\zeta(3/2)}{\zeta(3)} \sum_{r\le \sqrt x} \frac{\mu^2(r) \phi(r)}{r^{3/2} \prod_{p\mid r} (1+1/p^{3/2})} + O(x^{1/3} \log x) \\
&amp; = &amp; 2C_0 \frac{\zeta(3/2)}{\zeta(3)}  x^{3/4}  + O(x^{1/3} \log x).
\end{eqnarray}
On the other hand,
using (\ref{eq:uni}) and following the same reasoning as in the proof of Lemma \ref{lem:2}, we obtain
\begin{eqnarray*}
\sum_{m \le r \atop{ m \ \mbox{\tiny powerful} \atop (m,r)=1}} 1 &amp; = &amp; \sum_{a^2b^3\le r \atop{(a,r)=1 \atop (b,r)=1}} \mu^2(b) = \sum_{b^3\le r \atop (b,r)=1} \mu^2(b) \sum_{a^2\le r/b^3 \atop (a,r)=1} 1\\
&amp; = &amp; \sum_{b\le r^{1/3} \atop (b,r)=1} \mu^2(b) \left( \frac{\phi(r)}r \frac{r^{1/2}}{b^{3/2}} + O(2^{\omega(r)})  \right)\\
&amp; = &amp; \frac{\phi(r)}{r^{1/2}}\sum_{b\le r^{1/3} \atop (b,r)=1} \frac{\mu^2(b)}{b^{3/2}} + O\left( 2^{\omega(r)} r^{1/3} \right) \\
&amp; = &amp; \frac{\phi(r)}{r^{1/2}}\frac{\zeta(3/2)}{\zeta(3)} \prod_{p\mid r}\left( 1 + \frac 1{p^{3/2}} \right)^{-1} + O\left( 2^{\omega(r)} r^{1/3} \right).
\end{eqnarray*}
Using this last estimate, we obtain that
\begin{eqnarray} \label{eq:x3} \nonumber
T_2(x)  &amp; = &amp; \frac{\zeta(3/2)}{\zeta(3)}  \sum_{r\le \sqrt x} \frac{\mu^2(r) \phi(r)}{r^{1/2} \prod_{p\mid r}\left( 1 + \frac 1{p^{3/2}} \right)} +O\left(\sum_{r\le \sqrt x}  2^{\omega(r)} r^{1/3} \right)\\
&amp;=&amp; \frac 23 C_0 \frac{\zeta(3/2)}{\zeta(3)} x^{3/4} + O\left( x^{2/3} \log x \right).
\end{eqnarray}
Using (\ref{eq:x2}) and (\ref{eq:x3}) in (\ref{eq:x1}), estimate (\ref{eq:star}) follows thus completing the proof of Theorem\,\ref{thm:pow-sq}.
\end{proof}

\section{The sums $A_{y}(x) \text{ and } B_{y}(x)$}

Recall the definitions of $A_{y}(x)$ and $B_{y}(x)$ given in (\ref{eq:defAB}) as well as the following.
\begin{definition}
The Dickman function $\rho(u)$ is the continuous function satisfying $\rho(u)=1$ for $0\leq u\leq 1$ and $u\rho'(u) =-\rho(u-1)$ for $u&gt;1$.
\end{definition}
It follows from this definition that
\begin{equation} \label{eq:rhou}
\rho(u) = \rho(k) - \int_{k}^{u}\frac{\rho(z-1)}{z}dz.
\end{equation}

\begin{theorem} \label{asympfriable}
Let $y = x^{1/u}$, with $u\ge 1$ fixed. Then, as $x\to \infty$,
\begin{equation} \label{eq:t4-a}
\sum_{\substack{n\leq x\\ P(n)\leq y}}\sq(n)=(1+o(1))\rho(u)\frac{d_{2}}{2}x^2
\end{equation}
and
\begin{equation} \label{eq:t4-b}
\sum_{\substack{n\leq x\\ P(n)\leq y}}\pow(n)=(1+o(1))\rho(u/2)\frac{d_{1}}{3}x^{3/2}.
\end{equation}
\end{theorem}


\begin{proof}
The proof of Theorem \ref{asympfriable} is very similar to the proof of the estimate
$$\Psi(x,x^{1/u}) =(1+o(1)) \rho(u)x \qquad (x\to \infty),$$
as performed, for instance, in the book of Tenenbaum \cite{kn:Tenenbaum1}. We start by studying the expression
\begin{eqnarray} \label{eq:r1} \nonumber
\sum_{\substack{n\leq x\\ y&lt;P(n)\leq x}}\sq(n) &amp; = &amp; \sum_{y&lt; p\le x}\sum_{\substack{n\leq x\\P(n)=p}}\sq(n)\\ \nonumber
          &amp; = &amp; \sum_{y &lt; p\le x}\sum_{\substack{m\leq x/p\\P(m)&lt;p}}\sq(mp) +
          \sum_{y&lt; p\le x}\sum_{\substack{m\leq x/p\\P(m)=p}}\sq(mp)\\
          &amp; = &amp; \Sigma_1(x) + \Sigma_2(x),
\end{eqnarray}
say.

We first show that $\Sigma_2(x)$ is relatively small and can thus be neglected. Writing $m=r\cdot p$, we obtain
\begin{equation}\label{eq:ss2}
\Sigma_2(x) = \sum_{y&lt; p\le x}\sum_{\substack{rp\leq x/p\\P(r)\le p}}\sq(rp^{2}) \leq  \sum_{y &lt; p\le x}\sum_{r\leq x/p^2}\sq(r) \leq \sum_{y &lt; p\le x}\sum_{r\leq x/p^2}r\ll x^2\sum_{y &lt; p\le x}p^{-4}\ll\frac{x^2}{y^3}.
\end{equation}
We now turn our attention to the estimation of $\Sigma_1(x)$, which we can write as
$$\Sigma_1(x)=\sum_{y&lt; p\le x}p\sum_{\substack{m\leq x/p\\P(m)&lt;p}}\sq(m).$$
 In the range $1\le u\le 2$, the condition $m\leq x/p$ implies $P(m)\leq x^{1/2} &lt; p$, so we can drop the condition $P(m)&lt;p$. Hence, using equation (\ref{eq:me2}), we obtain, as $x\to \infty$,
\begin{eqnarray} \label{eq:ss1} \nonumber
\Sigma_1(x) &amp; = &amp; \sum_{y &lt; p\le x}p\sum_{m\leq x/p}\sq(m)=(1+o(1))\sum_{y &lt; p\le x}p \frac{d_{2}}{2}\left(\frac xp\right)^2\\ \nonumber
&amp; = &amp; (1+o(1))\frac{d_{2}}{2}x^2\sum_{y &lt; p\le x}\frac{1}{p}=(1+o(1))\frac{d_{2}}{2}x^2\int_{y}^{x}\frac{1}{t\log t}dt\\
&amp; =  &amp; (1+o(1))\frac{d_{2}}{2}(\log u) x^2.
\end{eqnarray}
Again using (\ref{eq:me2}) and after substituting  (\ref{eq:ss1}) and (\ref{eq:ss2}) in (\ref{eq:r1}), we obtain that, in the range $1\le u\le 2$,
\begin{eqnarray} \label{eq:ss3} \nonumber
\sum_{n\le x \atop P(n)\le y} \sq(n) &amp; = &amp; \sum_{n\le x} \sq(n) - \sum_{n\le x \atop y&lt;P(n) \le x}\sq(n) \\ \nonumber
&amp; = &amp; (1+o(1)) \frac{d_2}{2} (1-\log u) x^2 + O\left( \frac{x^2}{y^3} \right) \\
&amp; = &amp; (1+o(1)) \frac{d_2}{2} \rho(u) x^2 ,
\end{eqnarray}
where we used the fact that $\rho(u)=1-\log u$ when $u\in [1,2]$.

In order to extend the range of validity of (\ref{eq:ss3}) for $u&gt;2$, we proceed recursively. Let $M\ge 2$ be a positive integer and assume that the asymptotic estimate
\begin{equation} \label{eq:assume}
\sum_{\substack{n\leq x\\P(n)\leq x^{1/u}}}\sq(n)=(1+o(1))\frac{d_{2}}{2}\rho(u)x^2 \quad \mbox{holds for any } u\in[1,M] \mbox{ as } x\to \infty.
\end{equation}
We will then show that it also holds for  $u\in [M,M+1]$.

First observe that, in light of (\ref{eq:ss2}), we have
\begin{eqnarray} \label{eq:s4} \nonumber
\sum_{\substack{n\leq x\\x^{1/u}&lt;P(n)\leq x^{1/M}}}\sq(n) &amp; = &amp; \sum_{x^{1/u}&lt;p\leq x^{1/M}}\sum_{\substack{m\leq \frac{x}{p}\\P(m)\leq p}}\sq(mp)\\
\nonumber &amp;=&amp; \sum_{x^{1/u}&lt;p\leq x^{1/M}}p\sum_{\substack{m\leq \frac{x}{p}\\P(m)&lt;p}}\sq(m)+O\left(x^{2-3/u}\right)\\
\nonumber &amp; = &amp; \sum_{x^{1/u}&lt;p\leq x^{1/M}}p\left(\sum_{\substack{m\leq \frac{x}{p}\\P(m)\leq p}}\sq(m)-\sum_{\substack{m\leq \frac{x}{p}\\P(m)= p}}\sq(m)\right)+O\left(x^{2-3/u}\right)\\
&amp; = &amp; \sum_{x^{1/u}&lt;p\leq x^{1/M}}p\sum_{\substack{m\leq \frac{x}{p}\\P(m)\leq p}}\sq(m) + O\left(x^{2-1/u}\right).
\end{eqnarray}

Using our induction hypothesis (\ref{eq:assume}), we obtain, as $x\to \infty$,
\begin{eqnarray} \label{eq:4.10} \nonumber
\sum_{x^{1/u}&lt;p\leq x^{1/M}}p\sum_{\substack{m\leq \frac{x}{p}\\ P(m)\leq p}}\sq(m)&amp;=&amp;(1+o(1))\frac{d_{2}}{2}\sum_{x^{1/u}&lt;p\leq x^{1/M}}p\left(\frac{x}{p}\right)^2\rho\left(\frac{\log x}{\log p}-1\right)\\ \nonumber
&amp;=&amp;(1+o(1))\frac{d_{2}}{2}x^2\sum_{x^{1/u}&lt;p\leq x^{1/M}}\frac{1}{p}\rho\left(\frac{\log x}{\log p}-1\right)\\
&amp;=&amp;(1+o(1))\frac{d_{2}}{2}x^2\int_{x^{1/u}}^{x^{1/M}}\frac{1}{t\log t}\rho\left(\frac{\log x}{\log t}-1\right)dt.
\end{eqnarray}
Setting $z=\log x/\log t$, we obtain $dz=\frac{-\log x}{t(\log t)^2}dt$, so that $\frac{1}{t\log t}dt=-\frac{dz}{z}$ and
\begin{equation} \label{eq:4.11}
\frac{d_{2}}{2}x^2\int_{x^{1/u}}^{x^{1/M}}\frac{1}{t\log t}\rho\left(\frac{\log x}{\log t}-1\right)dt=(1+o(1))\frac{d_{2}}{2}x^2\int_{M}^{u}\frac{\rho(z-1)}{z}dz.
\end{equation}
Gathering (\ref{eq:s4}), (\ref{eq:4.10})  and (\ref{eq:4.11}), we obtain
\begin{equation}\label{sqind}
\sum_{\substack{n\leq x\\x^{1/u}&lt;P(n)\leq x^{1/M}}}\sq(n)=(1+o(1))\frac{d_{2}}{2}x^2\int_{M}^{u}\frac{\rho(z-1)}{z}dz.
\end{equation}
From equation (\ref{sqind}) and our induction hypothesis (\ref{eq:assume}), we then obtain
\begin{eqnarray*}
\sum_{\substack{n\leq x\\P(n)\leq x^{1/u}}}\sq(n) &amp; = &amp;
\sum_{n\le x \atop P(n)\le x^{1/M}} \sq(n) -  \sum_{n\le x \atop  x^{1/u} &lt; P(n) \le x^{1/M}} \sq(n)  \\
&amp; = &amp; (1+o(1))\frac{d_{2}}{2}x^2\left(\rho(M)-\int_{M}^{u}\frac{\rho(z-1)}{z}dz\right),
\end{eqnarray*}
which, in light of (\ref{eq:rhou}), can be replaced by
$$\sum_{\substack{n\leq x\\P(n)\leq x^{1/u}}}\sq(n)= (1+o(1))\frac{d_{2}}{2}x^2\rho(u),$$
thus completing the proof of (\ref{eq:t4-a}).


We now turn to the proof of (\ref{eq:t4-b}). We begin by writing
\begin{equation} \label{eq:vide}
\sum_{\substack{n\leq x\\x^{1/u}&lt;P(n)\leq x}}\pow(n)=\Sigma_1(x)+\Sigma_2(x)+\Sigma_3(x),
\end{equation}
where $\Sigma_1(x),\Sigma_2(x),\Sigma_3(x)$ stand respectively for the three expressions
$$\sum_{\substack{n\leq x\\x^{1/u}&lt;P(n)\leq x\\P(n)\|n}}\pow(n),\qquad
 \sum_{\substack{n\leq x\\x^{1/u}&lt;P(n)\leq x\\P(n)^2\|n}}\pow(n),\qquad
 \sum_{\substack{n\leq x\\x^{1/u}&lt;P(n)\leq x\\P(n)^3|n}}\pow(n).$$
First observe that, in light of (\ref{eq:me1}),
\begin{eqnarray*}
\Sigma_1(x)&amp;=&amp;\sum_{x^{1/u}&lt;p\leq x}\sum_{\substack{m\leq\frac{x}{p}\\P(m)&lt; p}}\pow(mp)=\sum_{x^{1/u}&lt;p\leq x}\sum_{\substack{m\leq \frac{x}{p}\\ P(m)&lt; p}}\pow( m)\\
&amp;\ll&amp;\sum_{x^{1/u}&lt;p\leq x}\left(\frac{x}{p}\right)^{3/2}\ll  x^{3/2}\sum_{t&gt;x^{1/u}}\frac{1} {t^{3/2}}=O\left(x^{3/2-1/2u}\right).
\end{eqnarray*}
On the other hand, again using (\ref{eq:me1}), we have
\begin{eqnarray*}
\Sigma_3(x)&amp;=&amp;\sum_{a\geq 3}\sum_{x^{1/u}&lt;p\leq x}\sum_{\substack{m\leq\frac{x}{p^a}\\ P(m)&lt; p}}\pow(mp^a)\le\sum_{a\geq 3}\sum_{x^{1/u}&lt;p\leq x}p^a\sum_{m\leq\frac{x}{p^a}}\pow(m)\\
&amp;\ll&amp; x^{3/2}\sum_{a\geq 3}\sum_{x^{1/u}&lt;p\leq x}\frac{1}{p^{a/2}}\ll \sum_{a\geq 3}x^{\frac{3}{2}-\frac{1}{u}\left(\frac{a}{2}-1\right)}=O\left(x^{3/2-1/2u}\right).
\end{eqnarray*}
It follows from these two estimates that
\begin{equation} \label{eq:zt1}
\Sigma_1(x)+\Sigma_3(x)=O\left(x^{3/2-1/2u}\right).
\end{equation}
Now estimating $\Sigma_2(x)$ in the range $1\le u\le 2$, we have, since $P(n)^2&gt;x$,
\begin{equation} \label{eq:zt2}
\Sigma_2(x)=\sum_{\substack{n\leq x\\x^{1/u}&lt;P(n)\leq x\\P(n)^2\|n}}\pow(n)=0.
\end{equation}

Recalling (\ref{eq:me1}) and using both (\ref{eq:zt1}) and (\ref{eq:zt2}) in (\ref{eq:vide}), it follows that, in the range $u\in [1,2]$,
\begin{eqnarray} \label{eq:powu12} \nonumber
\sum_{\substack{n\leq x\\P(n)\leq x^{1/u}}} \pow(n) &amp; = &amp; \sum_{n\le x} \pow(n) -  \sum_{\substack{n\leq x\\x^{1/u}&lt;P(n)\le x}} \pow(n) \\ \nonumber
 &amp; = &amp; (1+o(1))\frac{d_{1}}{3}x^{3/2} - \left(\Sigma_1(x) + \Sigma_2(x) + \Sigma_3(x)\right)\\
  &amp; = &amp; (1+o(1))\frac{d_{1}}{3}\rho(u/2)x^{3/2},
\end{eqnarray}
where we used the fact that $\rho(u/2) = 1$ for $u\in [1,2]$.
\vskip 5pt

In order to extend the range of validity of (\ref{eq:powu12}) for $u&gt;2$, we will proceed recursively, namely by assuming that the asymptotic estimate
\begin{equation} \label{eq:assume2}
\sum_{\substack{n\leq x\\ P(n)\leq x^{1/v}}}\pow(n)=(1+o(1))\frac{d_{1}}{3}\rho(v/2)x^{3/2} \quad \mbox{holds for any $v\in [1,2M]$ as $x\to \infty$}
\end{equation}
and then showing that it also holds for $u\in [2M,2M+2]$. Indeed we have
\begin{eqnarray} \label{eq:sumpowind} \nonumber
\sum_{\substack{n\leq x\\ x^{1/u}&lt;P(n)\leq x^{1/2M}}}\pow(n) &amp; = &amp; \sum_{x^{1/u}&lt;p\leq x^{1/2M}}\sum_{\substack{m\leq \frac{x}{p^2}\\{P(m)&lt;p}}}\pow(mp^2)+O\left(x^{3/2-1/2u}\right)\\ \nonumber
 &amp; = &amp; \sum_{x^{1/u}&lt;p\leq x^{1/2M}}p^2\sum_{\substack{m\leq \frac{x}{p^2}\\P(m)&lt;p}}\pow(m)
+O\left(x^{3/2-1/2u}\right)\\
&amp; = &amp; \sum_{x^{1/u}&lt;p\leq x^{1/2M}}p^2\sum_{\substack{m\leq \frac{x}{p^2}\\P(m)\leq
p}}\pow(m)
-S_{1}(x)+ O\left(x^{3/2-1/2u}\right)
\end{eqnarray}
where
\begin{eqnarray} \label{eq:SS} \nonumber
S_{1}(x) &amp; := &amp; \sum_{x^{1/u}&lt;p\leq x^{1/2M}}p^2\sum_{\substack{j\leq \frac{x}{p^3}\\P(j)\leq
p}}\pow(jp)\\ \nonumber
 &amp; = &amp; \sum_{x^{1/u}&lt;p\leq x^{1/2M}}p^2\left(\sum_{\substack{j\leq \frac{x}{p^3}\\P(j)&lt; p}}\pow(j)+\sum_{\substack{j\leq \frac{x}{p^3}\\P(j)= p}}\pow(jp) \right)\\ \nonumber
&amp; \ll &amp;  \sum_{x^{1/u}&lt;p\leq x^{1/2M}}p^2\left(\sum_{j\leq \frac{x}{p^3}}\pow(j) + \sum_{a\geq 1}\sum_{p^{a}k\leq \frac{x}{p^3}}p^{a+1}\pow(k)\right)\\ \nonumber
&amp;\ll &amp; \sum_{x^{1/u}&lt;p\leq x^{1/2M}}p^2\left(\left(\frac{x}{p^{3}}\right)^{3/2} + \sum_{a\geq 1}p^{a+1}\left(\frac{x}{p^{a+3}}\right)^{3/2}\right)\\
&amp;\ll&amp; x^{3/2-1/u},
\end{eqnarray}
where again we used estimate (\ref{eq:me1}). Combining (\ref{eq:sumpowind}) and (\ref{eq:SS}) along with our induction hypothesis (\ref{eq:assume2}), it follows that
$$
\sum_{\substack{n\leq x\\x^{1/u}&lt;P(n)\leq x^{1/2M}}}\pow(n)=(1+o(1))\frac{d_{1}}{3}x^{3/2}\sum_{x^{1/u}&lt;p\leq x^{1/2M}}\frac{1}{p}\rho\left(\frac{\log x}{2\log p}-1\right),
$$
which we can rewrite in its integral form as
\begin{equation}\label{powdickman}
\sum_{\substack{n\leq x\\x^{1/u}&lt;P(n)\leq x^{1/2M}}}\pow(n)=(1+o(1))\frac{d_{1}}{3}x^{3/2}\int_{x^{1/u}}^{x^{1/2M}}\frac{1}{t\log t}\rho\left(\frac{\log x}{2\log t}-1\right)dt.
\end{equation}
Setting $z=\frac{\log x}{2\log t}$ in (\ref{powdickman}), we obtain
$$
\sum_{\substack{n\leq x\\x^{1/u}&lt;P(n)\le x^{1/2M}}}\pow(n)=(1+o(1))\frac{d_{1}}{3}x^{3/2}\int_{M}^{u/2}\frac{1}{z}\rho\left(z-1\right)dz.
$$
From this we get
$$
\sum_{\substack{n\leq x\\P(n)\leq x^{1/u}}}\pow(n)=(1+o(1))\frac{d_{1}}{3}x^{3/2}\left(\rho(M)-\int_{M}^{u/2}\frac{\rho(z-1)}{z}dz\right),
$$
which, in light of (\ref{eq:rhou}), yields
$$
\sum_{\substack{n\leq x\\P(n)\leq x^{1/u}}}\pow(n)=(1+o(1))\frac{d_{1}}{3}x^{3/2}\rho(u/2),
$$
thus proving (\ref{eq:t4-b}) and completing the proof of Theorem \ref{asympfriable}.
\end{proof}


\section{The sizes of $A_y(x)$ and $B_y(x)$ for small $y$}

In this section, we establish the size of $A_y(x)$ and $B_y(x)$ when $y=(\log x)^\eta$ for some $\eta&gt;1$. We prove the following.

\begin{theorem}\label{other1}
Let $\eta&gt;1$ be fixed. Then, as $x\to \infty$,
\begin{equation} \label{eq:th-1}
\sum_{\substack{n\leq x\\P(n)\leq(\log x)^{\eta}}} \pow(n)= x^{\frac{3}{2}-\frac{1}{2\eta}+o(1)}
\end{equation}
and
\begin{equation} \label{eq:th-2}
\sum_{\substack{n\leq x\\P(n)\leq(\log x)^{\eta}}} \sq(n)= x^{2-\frac{1}{\eta}+o(1)}.
\end{equation}
\end{theorem}

As we will soon see, Theorem \ref{other1} is a consequence of the following lemmas.
\vskip 5pt

\begin{lemma} \label{lem:g1}
Given a real number $A&gt;1$,
$$
\Psi(x, \log^{A}x) = x^{1-1/A + o(1)} \qquad (x\rightarrow\infty).
$$
\end{lemma}

\begin{proof}
This is relation (1.14) in the paper of Granville \cite{kn:Granville1}.
\end{proof}



\begin{lemma}\label{lem:7}
Let $\eta&gt;1$ be fixed. Then, as $x\to \infty$,
\begin{equation} \label{eq:def-F-eta}
F_\eta(x):=\#\{n\le x: P(n)\le (\log x)^{\eta},\ n\ \mbox{\rm powerful}\}=x^{\frac{1}{2}\left(1-\frac{1}{\eta}\right)+o(1)}.
\end{equation}
\end{lemma}


\begin{proof}
Since all perfect squares are powerful numbers, we have, using Lemma \ref{lem:g1},
\begin{equation} \label{eq:p-s-1}
F_\eta(x) \ge
\#\{n\le x^{1/2}: P(n)\le (\log x)^{\eta}\} =  x^{\frac 12\left(1-\frac  1{\eta}\right)+o(1)}.
\end{equation}


On the other hand, using Lemma \ref{lem:g1} and writing each powerful number $n\le x$ as $n=a^2b^3$, where $b$ is squarefree, we have for any $\varepsilon&gt;0$,
\begin{eqnarray}
\nonumber F_\eta(x)  &amp;\le&amp; \sum_{\substack{a\leq \sqrt{x}\\P(a)\le (\log x)^{\eta}}}\#\left\{b\le \frac{x^{1/3}}{a^{2/3}}: P(b)\le (\log x)^{\eta}\right\} \\
\nonumber &amp; \leq &amp;  \sum_{\substack{a\leq x^{\frac{1}{2}-\varepsilon}\\P(a)\le (\log x)^{\eta}}}\#\left\{b\le \frac{x^{1/3}}{a^{2/3}}: P(b)\le (\log x)^{\eta}\right\} +
\sum_{\substack{x^{\frac{1}{2}-\varepsilon}&lt;a\leq \sqrt{x}\\P(a)\le (\log x)^{\eta}}}x^{\frac{2\varepsilon}{3}}\\
\nonumber &amp; \leq &amp;  \sum_{\substack{a\leq x^{\frac{1}{2}-\varepsilon}\\P(a)\le (\log x)^{\eta}}}\#\left\{b\le \frac{x^{1/3}}{a^{2/3}}: P(b)\le (\log x)^{\eta}\right\}
+ x^{\frac{2\varepsilon}{3}}\Psi(\sqrt{x},(\log x)^{\eta})\\
\nonumber &amp; = &amp;  \sum_{\substack{a\leq x^{\frac{1}{2}-\varepsilon}\\P(a)\le (\log x)^{\eta}}}\#\left\{b\le \frac{x^{1/3}}{a^{2/3}}: P(b)\le (\log x)^{\eta}\right\}
+ x^{\frac{2\varepsilon}{3}}\cdot x^{\frac{1}{2}(1-\frac{1}{\eta})+o(1)}\\
\label{S1}&amp; = &amp; S_{1}(x) + x^{\frac{1}{2}(1-\frac{1}{\eta})+O(\varepsilon)},
\end{eqnarray}
say. Then, using Lemma \ref{lem:g1} again, we obtain
\begin{eqnarray}
\nonumber S_{1}(x) &amp; \le &amp; \sum_{\substack{a\leq x^{\frac{1}{2}-\varepsilon}\\P(a)\leq(\log x)^{\eta}}}\left(\frac{x}{a^2}\right)^{\frac{1}{3}(1-\frac 1{\eta})+o(1)}
 \leq  x^{\frac 13 (1-\frac 1{\eta}) +o(1)} \sum_{a\le \sqrt{x} \atop P(a) \le (\log x)^\eta} a^{-\frac 23(1-\frac 1{\eta})}\\
\nonumber &amp; = &amp; x^{\frac 13 (1-\frac 1{\eta})+o(1)} \int_{1}^{\sqrt x} t^{-\frac 23(1-\frac 1{\eta})} d\Psi(t,(\log x)^\eta)\\
\nonumber &amp; = &amp; x^{\frac 13 (1-\frac 1{\eta}) +o(1)}
\Bigg( \left. t^{-\frac 23(1-\frac 1{\eta})} \Psi(t,(\log x)^\eta)\right|_1^{\sqrt x} \\
\nonumber &amp; &amp; \qquad \qquad + \quad
\frac 23 \left(1- \frac 1{\eta}\right) \int_1^{\sqrt x} t^{-\frac 23 (1-\frac 1{\eta})-1} \Psi(t,(\log x)^\eta)\,dt \Bigg)\\
\label{sep_int}&amp; = &amp; x^{\frac 13 (1-\frac 1{\eta})+o(1)}\left(\left. t^{-\frac 23(1-\frac 1{\eta})} \Psi(t,(\log x)^\eta)\right|_1^{\sqrt x}
+\frac 23 \left(1- \frac 1{\eta}\right)\Bigg(I_{1}(x)+I_{2}(x)\Bigg)\right)
\end{eqnarray}
say, where
$$I_{1}(x) = \int_{1}^{x^{\varepsilon}} t^{-\frac 23 (1-\frac 1{\eta})-1} \Psi(t,(\log x)^\eta)\,dt \quad
\text{and} \quad I_{2}(x)= \int_{x^{\varepsilon}}^{\sqrt{x}} t^{-\frac 23 (1-\frac 1{\eta})-1} \Psi(t,(\log x)^\eta)\,dt.$$

On the one hand,
\begin{equation}\label{I1}
I_{1}(x) \leq \int_{1}^{x^{\varepsilon}}t^{-\frac 23 (1-\frac 1{\eta})}\,dt \ll \left(x^{\varepsilon}\right)^{-\frac{2}{3}(1-\frac{1}{\eta})} = x^{O(\varepsilon)}.
\end{equation}
On the other hand, using Lemma \ref{lem:g1},
\begin{equation}\label{I2}
I_{2}(x) =  \int_{x^{\varepsilon}}^{\sqrt{x}} t^{-\frac 23 (1-\frac 1{\eta})-1}\cdot t^{1-\frac{1}{\eta}+o(1)}\,dt\ll \left(\sqrt{x}\right)^{\frac{1}{3}(1-\frac{1}{\eta})+o(1)}.
\end{equation}

Then, combining (\ref{I1}), (\ref{I2}), (\ref{sep_int}) and (\ref{S1}) as well as using Lemma \ref{lem:g1} one more time, we obtain
\begin{eqnarray*}
F_\eta(x) &amp; \ll &amp; x^{\frac 13 (1-\frac 1{\eta}) +o(1)}
\left( (\sqrt{x})^{\frac 13(1-\frac 1{\eta})+o(1)}+\frac 23 \left(1-\frac 1{\eta} \right)\left(\sqrt{x}\right)^{\frac{1}{3}(1-\frac{1}{\eta})+o(1)}\right)
+x^{\frac{1}{2}(1-\frac{1}{\eta})+O(\varepsilon)}\\
&amp; \ll &amp; x^{\frac 13 (1-\frac 1{\eta}) +o(1)} \cdot x^{\frac 16 (1-\frac 1{\eta})+o(1)}+x^{\frac{1}{2}(1-\frac{1}{\eta})+O(\varepsilon)}
 \ll  x^{\frac 12 (1-\frac 1{\eta}) +O(\varepsilon)},
\end{eqnarray*}
an estimate which in combination with estimate (\ref{eq:p-s-1}) completes the proof of Lemma\,\ref{lem:7}.
\end{proof}

\begin{lemma}\label{lem:8}
Let $\eta&gt;1$ be fixed. Then, as $x\to \infty$,
$$G_\eta(x,y):=\#\{n\le x:\pow(n)&gt;y, P(n)\leq (\log x)^{\eta}\}\ll\left(\frac{x}{y^{1/2}}\right)^{1-\frac{1}{\eta}+o(1)}.$$
\end{lemma}

\begin{proof}
Writing each integer $n\le x$ as $n=m\cdot r$,
where $m$ and $r$ represent their powerful part and squarefree part, respectively, and using Lemma \ref{lem:g1}, we get for any $\varepsilon &gt;0$
\begin{eqnarray*}
G_\eta(x,y) &amp; \le &amp; \sum_{\substack{y&lt;m\leq x\\m \text{ powerful}\\ P(m)\le (\log x)^{\eta}}}
\sum_{r\le x/m \atop P(r)\le (\log x)^\eta} \mu^2(r)
 \le \sum_{\substack{y&lt;m\leq x\\m \text{ powerful}\\ P(m)\le (\log x)^{\eta}}}\Psi\left(\frac{x}{m},(\log x)^{\eta}\right) \\
&amp; = &amp; \sum_{\substack{y&lt;m\leq x^{1-\varepsilon}\\m \text{ powerful}\\ P(m)\le (\log x)^{\eta}}}\Psi\left(\frac{x}{m},(\log x)^{\eta}\right) +
\sum_{\substack{x^{1-\varepsilon} &lt;m\leq x\\m \text{ powerful}\\ P(m)\le (\log x)^{\eta}}}\Psi\left(\frac{x}{m},(\log x)^{\eta}\right)\\
 &amp; \le &amp; x^{1-\frac 1{\eta}+o(1)}\sum_{\substack{y&lt;m\le x^{1-\varepsilon}\\m \text{ powerful } \\ P(m)\le (\log x)^{\eta}}} \frac 1{m^{1-\frac{1}{\eta}}}
 + x^{\varepsilon}\sum_{\substack{x^{1-\varepsilon} &lt;m\leq x\\m \text{ powerful}\\ P(m)\le (\log x)^{\eta}}}1.
\end{eqnarray*}
Recalling the definition of $F_\eta(x)$ given in (\ref{eq:def-F-eta}), we then have, using Lemma \ref{lem:7},
\begin{eqnarray*}
G_\eta(x,y) &amp; \ll &amp;
x^{1-\frac 1{\eta}+o(1)} \int_y^x t^{-1+\frac 1{\eta}}\, dF_\eta(t) +x^{\varepsilon}F_{\eta}(x)\\
&amp; = &amp; x^{1-\frac 1{\eta}+o(1)}
\left( \left. t^{-1+\frac 1{\eta}} F_\eta(t)  \right|_y^x +\left(1-\frac 1{\eta} \right) \int_y^x t^{-2 +\frac 1{\eta}} F_\eta(t)\, dt   \right)
+x^{\frac{1}{2}(1-\frac{1}{\eta})+O(\varepsilon)}
\\
&amp; = &amp; x^{1-\frac 1{\eta}+o(1)} \left( \left. t^{-\frac 12 (1-\frac 1{\eta})}   \right|_y^x +
\left(1-\frac 1{\eta}\right) \int_y^x t^{-\frac 32 +\frac 1{2\eta}}dt \right)+x^{\frac{1}{2}(1-\frac{1}{\eta})+O(\varepsilon)}  \\
&amp; \leq &amp; x^{1-\frac 1{\eta}+o(1)} \left(- y^{-\frac 12 (1-\frac 1{\eta})} + 2 y^{-\frac 12 +\frac 1{2\eta}}\right)+x^{\frac{1}{2}(1-\frac{1}{\eta})+O(\varepsilon)}\\
&amp; \ll &amp; \left( \frac x{y^{1/2}}  \right)^{1-\frac 1{\eta} +o(1)},
\end{eqnarray*}
thus completing the proof of Lemma \ref{lem:8}.
\end{proof}


We now have the necessary tools to prove Theorem \ref{other1}.

\begin{proof}[Proof of Theorem \ref{other1}]
First observe that, using Lemma \ref{lem:7},
\begin{eqnarray} \label{eq:o1} \nonumber
\sum_{\substack{n\leq x\\P(n)\le (\log x)^{\eta}}} \pow(n) &amp; \ge &amp;
\sum_{\substack{n\leq x\\P(n)\le (\log x)^{\eta} \\ n\ \text{powerful}}} n
 =  \int_1^x t\, d F_\eta(t)\,dt
 =  \left. t\, F_\eta(t) \right|_1^x - \int_1^x F_\eta(t)\,dt \\
&amp; = &amp; \left. t^{1+ \frac 12(1-\frac 1{\eta}) +o(1)}\right|_{1}^{x} - \int_1^x t^{\frac 12(1- \frac 1{\eta})+o(1)}\,dt
 \gg  x^{\frac 32 - \frac 1{2\eta}+o(1)}.
\end{eqnarray}
Similarly, using Lemmas \ref{lem:g1} and \ref{lem:7}, we have
\begin{eqnarray} \label{eq:o2} \nonumber
\sum_{\substack{n\leq x\\P(n)\le (\log x)^{\eta}}} \pow(n) &amp; \le &amp;
\sum_{m\le x \atop{ m\ \mbox{\tiny powerful} \atop P(m)\le (\log x)^\eta}} m \sum_{r\le x/m \atop P(r)\le (\log x)^\eta} \mu^2(r)
\leq  \sum_{m\le x \atop{ m\ \mbox{\tiny powerful} \atop P(m)\le (\log x)^\eta}} m\Psi\left(\frac{x}{m},(\log x)^{\eta}\right)\\
\nonumber &amp; = &amp; \sum_{m\le x^{1-\varepsilon} \atop{ m\ \mbox{\tiny powerful} \atop P(m)\le (\log x)^\eta}} m \left( \frac xm \right)^{1-\frac 1{\eta} +o(1)}
+ \sum_{x^{1-\varepsilon}&lt;m\leq x \atop{ m\ \mbox{\tiny powerful} \atop P(m)\le (\log x)^\eta}} m\Psi\left(\frac{x}{m},(\log x)^{\eta}\right) \\ \nonumber
&amp; \leq &amp; x^{1-\frac 1{\eta}+o(1)}  \sum_{m\le x \atop{ m\ \mbox{\tiny powerful} \atop P(m)\le (\log x)^\eta}}
m^{\frac 1{\eta}}+ x^{\varepsilon}\sum_{m\leq x \atop{ m\ \mbox{\tiny powerful} \atop P(m)\le (\log x)^\eta}} m\\ \nonumber
&amp; = &amp; x^{1-\frac 1{\eta}+o(1)} \int_1^x t^{\frac 1{\eta}}\, d\,F_\eta(t)+x^{\varepsilon}\int_{1}^{x}t\,dF_{\eta}(x) \\
&amp; \ll &amp; x^{1-\frac 1{\eta}+o(1)}\cdot x^{\frac 1{\eta} + \frac 12 - \frac 1{2\eta}} + x^{\varepsilon}\cdot x^{\frac{3}{2}-\frac{1}{2\eta}+o(1)}
 \ll  x^{\frac 32 - \frac 1{2\eta}+O(\varepsilon)}.
\end{eqnarray}
Combining (\ref{eq:o1}) and (\ref{eq:o2}) and taking $\varepsilon$ arbitrarily small, estimate (\ref{eq:th-1}) of Theorem \ref{other1} follows.


It remains to prove estimate (\ref{eq:th-2}).
First, using Lemma \ref{lem:g1}, we have
\begin{eqnarray} \label{eq:p1} \nonumber
\sum_{\substack{n\leq x\\P(n)\leq (\log x)^{\eta}}} \sq(n) &amp; \le &amp; \sum_{\substack{n\leq x\\P(n)\leq (\log x)^{\eta}}} n
 =  \int_1^x t d\Psi(t,(\log x)^\eta) \\ \nonumber
&amp; = &amp; \left. t\, \Psi(t,(\log x)^\eta) \right|_1^x - \int_1^x  \Psi(t,(\log x)^\eta)\,dt \\
\nonumber &amp; \leq &amp; x\cdot x^{1-\frac{1}{\eta}+o(1)} - \int_{x^{\varepsilon}}^{x}\Psi(t,(\log x)^\eta)\, dt\\
&amp; \ll &amp; x^{2 - \frac 1{\eta} + o(1)}.
\end{eqnarray}
On the other hand, given an arbitrary small number $\varepsilon&gt;0$, we have, using Lemma \ref{lem:8},
\begin{eqnarray} \label{eq:p2}\nonumber
\sum_{\substack{n\leq x\\P(n)\leq (\log x)^{\eta}}} \sq(n) &amp; = &amp;
\sum_{\substack{n\leq x\\P(n)\leq (\log x)^{\eta}}} \frac n{\pow(n)}
 \ge  x^{-\varepsilon} \sum_{n\le x \atop{ P(n) \le (\log x)^\eta \atop \pow(n)\leq x^\varepsilon}} n
 \ge  x^{-\varepsilon} \sum_{x^{1-\varepsilon}&lt; n\le x \atop{ P(n) \le (\log x)^\eta \atop \pow(n)\leq x^\varepsilon}} x^{1-\varepsilon}\\ \nonumber
&amp; = &amp; x^{1-2\varepsilon}\left(\Psi(x,(\log x)^\eta)-G_\eta(x,x^{\varepsilon})-\Psi(x^{1-\varepsilon},(\log x)^\eta)+G_\eta(x^{1-\varepsilon},x^{\varepsilon})\right)\\
&amp; \gg &amp; x^{1-2\varepsilon}x^{1-\frac{1}{\eta}+o(1)}=x^{2-\frac{1}{\eta}-2\varepsilon+o(1)}.
\end{eqnarray}


Since $\varepsilon$ can be chosen arbitrarily small, combining (\ref{eq:p1}) and (\ref{eq:p2}), estimate (\ref{eq:th-2}) follows, thus completing the proof of Theorem \ref{other1}.
\end{proof}


\section{Determining $y$ such that $A_y(x)$ and $B_y(x)$ are of the same order}

In this section, the functions
$$g(x):=x\log x-(x-1)\log (x-1)$$
and
$$h(x):=\max_{0&lt;b&lt;1/2}\left(x\log x-(x-b)\log(x-b) - 2b\log b + (1-b)\log (1-b) - (1-2b)\log (1-2b)\right),$$
both defined for real $x&gt;1$,
play an important role.

Our goal is to show that when $y=C\log x$ the expressions $\log(A_y(x)/x)$ and $\log(B_y(x)/x)$ are equal to
$(1+o(1))h(C)\frac{\log x}{\log \log x}$  and $(1+o(1))g(C)\frac{\log x}{\log \log x}$ respectively, so that by choosing $C$ as the solution of $h(C)=g(C)$, both expressions will be asymptotic as $x\to \infty$.

\begin{theorem}\label{main}
The estimate
$$\log(A_y(x)/x)=(1+o(1))\log(B_y(x)/x)
 \qquad (x\rightarrow \infty)$$
holds for $y=y(x)=2\log x$.
\end{theorem}

The proof of Theorem \ref{main} will follow from a series of lemmas.

\begin{lemma}\label{lem:88}
The only solution $x&gt;1$ to $g(x)=h(x)$ is $x=2$, in which case $g(2)=h(2) = 2\log 2$.
\end{lemma}

\begin{proof}
Clearly, $g(2)=2\log 2$. We will prove that $g(2)=h(2)$ and then show that the solution is unique. First observe that
\begin{eqnarray*}
h(2)-g(2) &amp; = &amp; \max_{0&lt;b&lt;1/2}\left(-(2-b)\log(2-b) - 2b\log b + (1-b)\log (1-b) - (1-2b)\log (1-2b)\right)\\
&amp; = &amp; \max_{0&lt;b&lt;1/2} f(b),
\end{eqnarray*}
say. Differentiating $f(b)$ with respect to $b$ yields
\begin{eqnarray*}
f'(b) &amp;=&amp;  \log(2-b) - 2\log b-\log(1-b) + 2\log(1-2b).
\end{eqnarray*}
We then have
$$
 f'(b)=0 \Longleftrightarrow \frac{(2-b)(1-2b)^{2}}{b^{2}(1-b)}=1
 \Longleftrightarrow (3b-2)(b^2-3b+1)=0,
$$
from which we get $b=b_0:=\frac{3-\sqrt{5}}{2}$ as the only solution to $f'(b)=0$ in $(0,1/2)$.

It remains to show that $f(b_0)=0$. This follows directly from the fact that $$f(b) = b f'(b) -2\log(2-b) + \log(1-b) - \log (1-2b),$$
which implies that
\begin{eqnarray*}
 f(b_0) = 0 &amp; \Longleftrightarrow &amp;-2\log(2-b_0) + \log(1-b_0) - \log (1-2b_0) = 0\\
 &amp; \Longleftrightarrow &amp;\frac{1-b_0}{(2-b_{0})^2(1-2b_{0})}=1
 \Longleftrightarrow (2b_{0}-3)(b_{0}^2-3b_{0}+1)=0,
\end{eqnarray*}
completing the first part of the proof. We also have
$$g'(x) - h'(x) \geq \left(\log x - \log (x-1)\right) - \max_{0&lt;b&lt;\frac{1}{2}}\left(\log x - \log(x-b)\right)
&gt; \log\left(x-\frac{1}{2}\right) - \log (x-1) &gt; 0,$$
so that $g(x)$ is increasing at a faster rate than $h(x)$ and thus proving the solution $x=2$ to be unique.
\end{proof}



\begin{lemma} \label{lem:g2}
Given any real number $\kappa&gt;0$,
\begin{equation}\label{thmB}
\Psi(x,\kappa\log x) =
\exp \left( g(\kappa+1) \frac{\log x}{\log\log x}\left(1+O\left(\frac{1}{\log\log x}\right)\right)\right) .
\end{equation}
\end{lemma}


\begin{proof}
This is the first relation in Granville \cite[p.~271]{kn:Granville1}.
\end{proof}

We now introduce the function
$$L(x):= \frac{\log x}{\log \log x}.$$

\begin{lemma}\label{appstirling}
Given arbitrary numbers $C&gt;D&gt;0$ and integer valued functions $a(x)$ and $b(x)$ satisfying $a(x)=(1+o(1))C\, L(x)$ and $b(x)=(1+o(1))D\,L(x)$, then, as $x\to \infty$,
$${a(x) \choose  b(x)} =
\exp\left\{(1+o(1))(C\log C-D\log D-(C-D)\log (C-D))L(x)\right\}.$$
\end{lemma}

\begin{proof}
Using the well known Stirling formula $n!=(1+o(1))n^n e^{-n} \sqrt{2\pi n}$ as $n\to \infty$, we have, as $x\to \infty$,
\begin{eqnarray*}
 {a(x) \choose b(x)}&amp; =&amp; \frac{ a(x)!}{b(x)!\left(a(x) - b(x)\right)!}\\
 &amp; = &amp;(1+o(1)) \frac{\sqrt{2\pi a(x)} a(x)^{ a(x)}}
 {\sqrt{2\pi b(x)}b(x)^{b(x)}\cdot
 \sqrt{2\pi (a(x)-b(x))}\left(a(x)-b(x)\right)^{a(x)-b(x)}}\\
 &amp; &amp; \\
 &amp;= &amp;(1+o(1)) \frac{1}{\sqrt{2\pi}}\frac{\sqrt{C}}{\sqrt{D}\sqrt{(C-D)L(x)}}\frac{((1+o(1))C)^{a(x)}}{((1+o(1))D)^{b(x)}((1+o(1))(C-D))^{a(x)-b(x)}}\\
 %&amp; = &amp; \exp\left\{O(\log L(x))
 % + a(x)\left(o(1) + \log C\right) - b(x)\left(o(1)+\log D\right) -(a(x)-b(x))\left(o(1) + \log(C-D)\right)\right\}\\
 &amp; = &amp; \exp\left\{(1+o(1))(C\log C-D\log D-(C-D)\log (C-D))L(x)\right\}.
 \end{eqnarray*}
\end{proof}


\begin{lemma} \label{lem:5}
As $x\rightarrow \infty$,
$$
\sum_{\substack{n\le x\\P(n)\le C\log x}}\sq(n) = xe^{(1+o(1))g(C)L(x)}.
$$
\end{lemma}

\begin{proof}
To prove this result, we proceed in two steps, establishing upper and lower bounds.
We start with the upper bound.
Let $\varepsilon&gt;0$ be an arbitrarily small number.
It is clear that
\begin{equation} \label{eq:w1}
\sum_{\substack{n\le x\\P(n)\le C\log x}}\sq(n)\le x\#\{n\leq x: P(n)\le C\log x,\;\sq(n)&gt;x^{1-\varepsilon}\}
+ x^{1-\varepsilon}\Psi(x,C\log x).
\end{equation}
It follows from Lemma \ref{lem:g2} that
\begin{equation} \label{eq:w2}
x^{1-\varepsilon}\Psi(x,C\log x) = o(x).
\end{equation}

Since $n=\sq(n)\pow(n)$, we have
\begin{equation} \label{eq:e1e2}
\#\{n\leq x: P(n)\leq C\log x,\;\sq(n)&gt;x^{1-\varepsilon}\}\leq E_{1}(x)\cdot E_{2}(x),
\end{equation}
where
\begin{eqnarray*}
E_{1}(x)&amp; := &amp; \#\{m\le x^{\varepsilon} : m \text{ powerful}, P(m)\le C\log x\},\\
E_{2}(x) &amp; := &amp; \#\{ x^{1-\varepsilon} &lt; m \le x :m\text{ squarefree},P(m)\le C\log x\}.
\end{eqnarray*}

Using Lemma \ref{lem:g2}, we get
\begin{equation} \label{eq:upper1}
E_{1}(x)\leq \Psi(x^{\varepsilon},C\log x) = e^{O\left(\varepsilon L(x) \right)}.
\end{equation}
To estimate $E_{2}(x)$, we will need information on the size of $\omega(m)$ for those integers $m$ counted by $E_2(x)$. On the one hand,
$$x^{1-\varepsilon} &lt; \sq(m) = \prod_{p|\sq(m)}p \leq \left(C\log x\right)^{\omega(\sq(m))},$$
which leads to
\begin{equation}\label{omegasqninf}
\omega(\sq(m))\ge \left(1-\varepsilon\right)\left(1+o\left(1\right)\right)L(x).
\end{equation}

On the other hand, it was proved by Robin \cite{kn:Robin1} that, for $n&gt;e^e$,
$$\omega(n) \le L(n) + 1.45743 \frac{L(n)}{\log \log n},$$
with equality when $\displaystyle n=\prod_{i=1}^{47}p_i$. Hence it follows that for all $n\le x$,
\begin{equation}\label{omegasqnsup}
\omega(\sq(n))\le \omega(n)\le (1+o(1)) L(x) \qquad (x\to \infty).
\end{equation}

Combining equations (\ref{omegasqninf}) and (\ref{omegasqnsup}), we obtain
\begin{equation}\label{omegasqn}
 \omega(\sq(n))=\left(1+O\left(\varepsilon\right)\right)L(x).
\end{equation}

We can now get an upper bound for $E_{2}(x)$. Equation (\ref{omegasqn}) tells us that there exists $\delta=\delta(\varepsilon)&gt;0$ tending to $0$ with $\varepsilon$ such that
\begin{eqnarray} \label{eq:upper2} \nonumber
E_{2}(x)&amp;\le &amp; \sum_{(1-\delta)L(x) &lt; r &lt; (1+\delta)L(x)}
\#\{n\le x: P(n)\leq C\log x,\omega(n)=r,\mu^{2}(n)=1\} \\
&amp; \le &amp; \sum_{(1-\delta)L(x) &lt; r &lt; (1+\delta)L(x)}\left(\substack{\pi\left(C\log x\right)\\\text{ }\\r}\right)
= \exp\left\{(g(C)+o(1))L(x)\right\},
\end{eqnarray}
where the last equality follows as in the proof of Lemma \ref{appstirling} after observing that, in light of  (\ref{eq:pnt1}),
$$\pi(C\log x) = (1+o(1))\frac{C\log x}{\log \log x} =  (1+o(1))C\,L(x).$$

Using (\ref{eq:upper1}) and (\ref{eq:upper2}) in (\ref{eq:e1e2}),  and then taking into account (\ref{eq:w2}) and (\ref{eq:w1}), we
have proven the upper bound implied in Lemma \ref{lem:5}.
\vskip 5pt

It remains to prove the lower bound.  Let us first choose a squarefree number $t&lt;x$ such that
$\varepsilon\log x &lt; p(t) &lt; P(t)\leq C\log x$
and also satisfying
$$
 \omega(t)=\ell_1(x):=\left\lfloor\left(1-\frac{\varepsilon}{2}\right)L(x)\right\rfloor.
$$
For such an integer $t$, using (\ref{eq:pnt2}) and then  (\ref{eq:pnt4}), we have
\begin{equation}\label{eq:5.11}
t=\prod_{p|t}p\ge \prod_{p\leq p_{\omega(t)}} p\ge e^{\omega(t)\log(\omega(t))(1+o(1))}\ge x^{1-\frac{\varepsilon}{2}+o(1)}
\end{equation}
and
\begin{equation}\label{eq:5.12}
t\le (C\log x)^{\omega(t)}\le x^{1-\frac{\varepsilon}{2}+o(1)}, \quad \mbox{so that } t=x^{1-\frac{\varepsilon}{2}+o(1)}.
\end{equation}
Now define $p_t$ as the largest prime number such that
\begin{equation} \label{eq:ppt}
t\cdot\prod_{p\leq p_t}p\le x.
\end{equation}
Note that such a prime $p_t$ exists because of (\ref{eq:5.12}). It follows from (\ref{eq:ppt}), using  (\ref{eq:pnt3}), that, as $x$ (and thus $t$) tends to infinity,
\begin{eqnarray*}
&amp; &amp; \log t + \sum_{p\le p_t} \log p \le \log x,\\
&amp; &amp; \log t + (1+o(1)) p_t \le \log x,\\
&amp; &amp; (1+o(1)) p_t  \le \log x - (1-\varepsilon/2+o(1)) \log x=(\varepsilon/2+o(1))\log x,
\end{eqnarray*}
where we used (\ref{eq:5.11}), thus implying that, if $x$ is sufficiently large,
$$
p_t\le \left(\frac{\varepsilon}2 +o(1) \right) \log x &lt; \varepsilon \log x.
$$
This allows us to choose an integer  $s_t$ defined by
$\displaystyle{s_t:=\prod_{p\leq p_t}p}$
such that, in light of  (\ref{eq:ppt}), $n=s_{t}\cdot t\le x$.
Observe that we also have
\begin{equation} \label{eq:6.15}
\frac{x}{\log x} \leq n=s_{t}\cdot t
\end{equation}
because if we had $ s_{t}\cdot t&lt; \frac{x}{\log x}$, then multiplying both side by $\varepsilon \log x $ would give us
$\varepsilon\log x\cdot s_{t}\cdot t &lt; \varepsilon x &lt; x$, which would contradict the definition of $p_{t}$. Using (\ref{eq:6.15}), we therefore have
\begin{eqnarray} \label{choose1} \nonumber
\sum_{\substack{n\leq x\\P(n)  \le  C\log x}}\sq(n) &amp; \ge &amp; \frac{x}{\log x}\#\left\{t\leq x:\mu^2(t)=1, \varepsilon\log x&lt;p(t)&lt; P(t)\leq C\log x, \omega(t)=\ell_{1}(x)\right\}\\
&amp; \ge &amp;
 \frac{x}{\log x}{\pi(C\log x)-\pi(\varepsilon\log x)) \choose \ell_{1}(x)}.
\end{eqnarray}
Finally, observing that as a consequence of  (\ref{eq:pnt1}),
$$\pi(C\log x)-\pi(\varepsilon\log x)=(1+o(1))C L(x)$$
and that $\ell_{1}(x) = (1+o(1))L(x)$,
it follows from (\ref{choose1}) and Lemma \ref{appstirling} that
$$
\sum_{\substack{n\leq x\\P(n)\leq C\log x}}\sq(n)\ge xe^{(g(C)+o(1))L(x)},
$$
thus
completing the proof of the lower bound and thereby that of Lemma  \ref{lem:5}.
\end{proof}

\begin{lemma} \label{lem:6}
As $x\rightarrow \infty$, we have
\begin{equation} \label{eq:pow-clogx}
\sum_{\substack{n\le x\\P(n)\le C\log x}}\pow(n) = xe^{(1+o(1))h(C)L(x)}.
\end{equation}
\end{lemma}

\begin{proof}
We begin by proving the implied lower bound by using the trivial inequality
\begin{equation}\label{powgex2}
\sum_{\substack{n\le x\\P(n)\le C\log x}}\pow(n)\ge\frac{x}{2}\#\left\{\frac{x}{2}&lt;n\leq x: P(n)\leq C\log x, n \mbox{ powerful}\right\}.
\end{equation}
In this last set, if we focus only on those $n=2^a\cdot m$ with $m$ odd and $a\ge 2$, we may write
\begin{eqnarray*}
&amp;&amp; \#\left\{\frac{x}{2}&lt;n\leq x: P(n)\leq C\log x, n \mbox{ powerful}\right\}\\
&amp; &amp; \qquad \ge
\#\left\{m\leq\frac{x}{8}: P(m)\leq C\log x, m \mbox{ odd, powerful}\right\}.
\end{eqnarray*}
Letting $\varepsilon&gt;0$ be a small number, we further focus our attention on those integers $m\le x/8$ such that
$\Omega(m)=\ell_2(x):= \left\lfloor (1-\varepsilon) L(x)   \right\rfloor$,
implying that we may write
\begin{eqnarray} \label{eq:88} \nonumber
&amp;&amp; \#\left\{m\leq \frac{x}{8}: P(m)\leq C\log x, m \mbox{ odd and powerful}\right\} \\
&amp;&amp; \quad \ge
\#\{m\leq \frac{x}{8}: P(m)\leq C\log x, \Omega(m)=\ell_{2}(x), m \mbox{ odd and powerful}\}.
\end{eqnarray}

Note that the two conditions $ \Omega(m)=\ell_{2}(x)$ and $ P(m)\leq C\log x$ imply that
$$ m\leq \left(C\log x\right)^{\ell_{2}(x)} \leq x^{(1-\varepsilon)(1+o(1))}, $$
thus allowing us to replace $m\leq \frac{x}{8}$ by $m\ge 1$ in the last set appearing in (\ref{eq:88}).
\vskip 5pt

We now write each integer $m$ in that set as $m=s^2\cdot t$ where $s=\prod_{p|m}p$ and $t=m/s^2$.
Observe that for each such integer $m$, we have
$$\Omega(t)=\Omega(m)-\Omega(s^2) =\Omega(m)-2\omega(s)= \Omega(m)-2b \Omega(m) = (1-2b)\ell_2(x)$$
for some $b\in (0,1/2)$.
It follows from these observations that
\begin{eqnarray}\label{ge:maxb} \nonumber
&amp;&amp; \#\{m\geq 1: P(m)\leq C\log x, \Omega(m)=\ell_{2}(x), m \mbox{ odd and powerful}\}\\
&amp;&amp; \qquad \ge
\max_{0&lt;b&lt;1/2}\sum_{\substack{s\geq 1\\ \mu^2(s)=1\\P(s)\le C\log x\\ \omega(s)=b \ell_{2}(x)\\ s\scriptstyle{\text{ odd}}}}\#\{t\ge 1:\Omega(t)=(1-2b)\ell_{2}(x),p|t\Rightarrow p|s\}.
\end{eqnarray}
Since $\omega(s)=b \ell_{2}(x)$, we obtain
\begin{eqnarray}\label{cardt} \nonumber
\#\{t\geq 1:\Omega(t)=(1-2b)\ell_{2}(x),p|t\Rightarrow p|s\}
 &amp; = &amp; { \Omega(t) + \omega(s) -1 \choose \Omega(t) } \\ \nonumber
 &amp; = &amp; {(1-2b) \ell_{2}(x)+ b\ell_{2}(x)-1 \choose (1-2b) \ell_{2}(x)}\\
&amp; = &amp; {(1-b) \ell_{2}(x)-1 \choose (1-2b) \ell_{2}(x)}
\end{eqnarray}
and we also have
\begin{equation}\label{cards}
\#\{s\geq 1: \mu^2(s)=1,P(s)\le C\log x,\omega(s)=b \ell_{2}(x), s\text{ odd}\}={\pi(C\log x)-1 \choose b\ell_{2}(x)}.
\end{equation}
From estimates (\ref{ge:maxb}), (\ref{cardt}) and (\ref{cards}), we obtain
\begin{eqnarray}\label{exph} \nonumber
&amp;&amp;
\#\{m\geq 1: P(m)\leq C\log x, \Omega(m)=\ell_{2}(x),m \mbox{ odd and powerful }\}\\ \nonumber
&amp; &amp; \qquad \ge
\max_{0&lt;b&lt;1/2}{(1-b) \ell_{2}(x) -1 \choose (1-2b) \ell_{2}(x)}{\pi(C\log x)-1 \choose b\ell_{2}(x)} \\
&amp;&amp; \qquad =\exp\left((1+o(1))h(C) L(x)\right),
\end{eqnarray}
where the last equality follows from Lemma \ref{appstirling}. The proof of the lower bound in (\ref{eq:pow-clogx}) then follows from (\ref{powgex2}) and (\ref{exph}).


We now prove the upper bound implied in Lemma \ref{lem:6}. Let $\varepsilon&gt;0$ be small. We start with the trivial inequality
\begin{equation} \label{eq:xx1}
\sum_{\substack{n\leq x\\P(n)\leq C\log x}}\pow(n)\le x\#\{n\leq x:\pow(n)&gt;x^{1-\varepsilon},P(n)\leq C\log x\}+x^{1-\varepsilon}\Psi\left(x,C\log x\right).
\end{equation}
First observe that,  using Lemma \ref{lem:g2} and provided that $x$ is large enough, we  have
\begin{equation} \label{eq:xx2}
x^{1-\varepsilon}\Psi\left(x,C\log x\right)\le x^{1-\varepsilon}x^{\varepsilon/2}=x^{1-\varepsilon/2}.
\end{equation}
On the other hand, writing each integer $n\in \{n\le x : \pow(n)&gt;x^{1-\varepsilon}, P(n)\leq C\log x\}$ as $n=\pow(n)\cdot \sq(n)=m\cdot r$, we have
\begin{equation} \label{eq:xx3}
\#\{n\leq x:\pow(n)&gt;x^{1-\varepsilon},P(n)\leq C\log x\}\le A(x) \times B(x),
\end{equation}
where
$$
A(x):=\#\{r\le x^{\varepsilon}, \mu^2(r)=1, P(r)\leq C\log x\}
$$
and
$$B(x): = \#\{m\le x:m \mbox{ powerful},P(m)\leq C\log x\}.$$
Using Lemma \ref{lem:g2}, we obtain
\begin{equation}  \label{eq:xx4}
A(x) \le\Psi(x^{\varepsilon},C\log x) = \exp\left(O(\varepsilon) L(x)\right).
\end{equation}
Concerning $B(x)$, we have
\begin{equation} \label{eq:v1}
B(x)  \le  \#B_1(x) \times \#B_2(x),
\end{equation}
where
\begin{eqnarray*}
B_1(x) &amp; := &amp; \{m\le x:m\mbox{ powerful},P(m)\le \varepsilon\log x\}, \\
B_2(x) &amp; := &amp; \{m\le x:m\mbox{ powerful},\varepsilon \log x&lt;p(m) \le P(m) \le C\log x\}.
\end{eqnarray*}
On the one hand, using Lemma \ref{lem:g2}, we have
\begin{equation} \label{eq:v2}
\#B_1(x)   \le  \Psi(x,\varepsilon\log x) \le  \left(\frac{1}{\varepsilon}\right)^{\pi(\varepsilon\log x)}=\exp\left(O(\varepsilon)\frac{\log x}{\log \log x}\right).
\end{equation}
On the other hand, for each $m\in B_2(x)$, we have
$\displaystyle{(\varepsilon \log x)^{\Omega(m)}&lt;m\le x}$,
and therefore, for some small $\delta&gt;0$,
$$
\Omega(m)\le \frac{\log x}{\log\log x+\log \varepsilon}=(1+\delta) L(x),
$$
provided $x$ is large enough.
It follows that
\begin{eqnarray*}
\#B_2(x) &amp; \le &amp; \#\left\{m\le x:P(m)\le C\log x,\Omega(m)&lt; (1+\delta) L(x)\right\}\\
&amp; \le &amp; (1+\delta) L(x) \times \#\left\{m\leq x:P(m)\le C\log x,\Omega(m)=\left\lfloor (1+\delta)L(x) \right\rfloor\right\} \\
&amp; \le &amp;  L(x)^2\\
 &amp; &amp; \times \max_{0&lt;b&lt;1/2}\#\left\{m\leq x:P(m)\leq C\log x,\Omega(m)=\left\lfloor (1+\delta) L(x) \right\rfloor ,\omega(m)=\left\lfloor b L(x) \right\rfloor\right\}.
\end{eqnarray*}
As we did in the proof of the lower bound, namely when establishing (\ref{exph}), it follows from this last series  of  inequalities that
\begin{eqnarray} \label{eq:v4} \nonumber
&amp;&amp; \max_{0&lt;b&lt;1/2}\#\left\{n\leq x:P(n)\leq C\log x,\Omega(n)=\left\lfloor (1+\delta)L(x) \right\rfloor,\omega(n)=\left\lfloor b L(x) \right\rfloor\right\}\\
&amp; &amp; \qquad = \quad \exp\left((1+o(1))h(C) L(x) \right).
\end{eqnarray}
Gathering estimates (\ref{eq:v1}) through (\ref{eq:v4}), we obtain that
\begin{equation} \label{eq:v5}
B(x) \le\exp\left((1+O(\varepsilon))h(C)L(x)\right).
\end{equation}
Collecting estimates (\ref{eq:xx1}) through (\ref{eq:xx4})    and also (\ref{eq:v5}), the proof of the upper bound implied in Lemma \ref{lem:6} then follows, thus completing the proof of Lemma \ref{lem:6}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{main}]
The proof of Theorem \ref{main} is an immediate consequence of Lemmas \ref{lem:5} and \ref{lem:6}.
\end{proof}


\begin{remark}
While Lemma \ref{lem:6} is valid only if $y=C\log x$ with $C\ge 1$, the domain of validity of Lemma \ref{lem:5} can be extended to $y=C\log x$ for any $C&gt;0$.
\end{remark}

\begin{remark}
As is shown in \cite{kn:MEC}, considering, for instance, only powers of 2, for any $y&gt;2$, we have $A_y(x)&gt;x/2$.
\end{remark}

\begin{remark}
For any $\varepsilon&gt;0$, if $C&lt;1-\varepsilon$ and $y\le C\log x$, then $$B_y(x)&lt;\left(\prod_{p&lt;y}p\right)\Psi(x,y)&lt;e^{C\log x+o(1)}=x^{C+o(1)}.$$
\end{remark}

\section{Final remarks}

Collecting our results from Sections 4, 5 and 6, it is interesting to observe that the behaviors of
$A_y(x)$ and $B_y(x)$ vary greatly depending on the relative size of $y=y(x)$. Indeed, we obtained asymptotic estimates for $A_y(x)$ and $B_y(x)$ when $y=x^{1/u}$, as $x\to \infty$, namely
$$A_y(x)= (1+o(1))\rho(u/2)\frac{d_{1}}{3}x^{3/2} \quad \mbox{and} \quad
B_y(x) =(1+o(1))\rho(u)\frac{d_{2}}{2}x^2.$$
In other ranges, we obtained less precise but nevertheless interesting estimates. In particular, when $y=(\log x)^\eta$ for some $\eta&gt;1$, we obtained asymptotic formulas for $\log(A_y(x))/\log x$ and $\log(B_y(x))/\log x$, that is,
$$A_y(x)=x^{\frac{3}{2}-\frac{1}{2\eta}+o(1)}
\quad \mbox{and} \quad
B_y(x) =x^{2-\frac{1}{\eta}+o(1)} \qquad (x\to \infty).$$
For even smaller values of $y$, that is, when $y=y(x)=2\log x$, we obtained
 $$\log(A_y(x)/x) = (1+o(1))\log(B_y(x)/x) \qquad (x\to \infty),$$
since we proved that both $A_y(x)$ and $B_y(x)$ are equal to $x\exp\left( (1+o(1))2\log 2\frac{\log x}{\log\log x}\right)$ as $x\to \infty$.

Gathering our results, we obtain that given any $\varepsilon&gt;0$, if $y&lt;(2-\varepsilon)\log x$, then $B_y(x)&gt;A_y(x)$.  On the other hand, although we did not prove it, we believe that $A_y(x)&gt;B_y(x)$ when $y&gt;(2+\varepsilon)\log x$.  While this is the case in every range investigated in the paper, a rigorous proof would require studying $A_y(x)$ and $B_y(x)$ in the ranges not covered by the present work, an interesting challenge.

\section{Acknowledgment}
The work of the second author was supported in part by a grant from NSERC.


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

\noindent \emph{Keywords: } 
powerful numbers, arithmetic function.

\bigskip
\hrule
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\vspace*{+.1in}
\noindent
Received July 8 2014;
revised version received  July 31 2014.
Published in {\it Journal of Integer Sequences}, August 5 2014.

\bigskip
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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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\end{document}

                                                                                

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