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\begin{center}
\vskip 1cm{\LARGE\bf On the Truncated Kernel Function
}
\vskip 1cm
\large
Jean-Marie De Koninck \\
D\'epartement de Math\'ematiques et de Statistique  \\
Universit\'e Laval  \\
Qu\'ebec G1V 0A6 \\
Canada \\
\href{mailto:jmdk@mat.ulaval.ca}{\tt jmdk@mat.ulaval.ca}  \\
\ \\
Isma\"{\i}la Diouf \\
D\'epartement de Math\'ematiques et d'Informatique \\
FST - Universit\'e Cheikh Anta DIOP  \\
BP 5005, Dakar-Fann  \\
Senegal  \\
\href{mailto:isma.diouf@gmail.com}{\tt isma.diouf@gmail.com}   \\
\ \\
Nicolas Doyon \\
D\'epartement de Math\'ematiques et de Statistique  \\
Universit\'e Laval \\
Qu\'ebec G1V 0A6 \\
Canada \\
\href{mailto:nicodoyon77@hotmail.com}{\tt nicodoyon77@hotmail.com}\\
\end{center}

\vskip .2 in

\begin{abstract}
We study properties of the {\it truncated kernel function} $\gamma_2$ defined on integers $n\ge 2$ by $\gamma_2(n)=\gamma(n)/P(n)$, where $\gamma(n)=\prod_{p|n}p$ is the well-known {\it kernel function} and $P(n)$ is the largest prime factor of $n$. In particular, we show that the maximal order of $\gamma_2(n)$ for $n\le x$ is $(1+o(1))x/\log x$ as $x\to \infty$ and that $\sum_{n\le x} 1/\gamma_2(n)= (1+o(1)) \eta x/\log x$, where $\eta=\zeta(2)\zeta(3)/\zeta(6)$. We further show that, given any positive real number $u<1$,
$\lim_{x\to \infty} \frac 1x \#\{n\le x: \gamma_2(n)<x^u\}=
\lim_{x\to \infty} \frac 1x \#\{n\le x: n/P(n) < x^u\}
 = 1-\rho(1/(1-u))$, where $\rho$ is the Dickman function.
We also show that $n/P(n)$ can very often be much larger than $\gamma_2(n)$, namely by proving that, given any $c\in [1,\xi)$, where $\xi$ is the unique solution to $\xi\log 2 = \log(1+\xi)+\xi \log(1+1/\xi)$, then
$$\#\{n\le x: \gamma_2(n) \ge n/(c\log n)\} = o\left( \#\{n\le x: n/P(n) \ge n/(c\log n)\}  \right) \qquad (x\to \infty).$$
\end{abstract}


\section{Introduction}

Let $\gamma(n)=\prod_{p|n}p$ for $n\ge 2$ and $\gamma(1)=1$ be the traditional {\it kernel function}, also at times called the {\it largest squarefree divisor function}. This arithmetic function has been extensively studied.
For instance, it was shown by Cohen \cite{kn:cohen} that
\begin{equation} \label{eq:cohen}
\sum_{n\le x} \gamma(n) = c_0\, x^2 + O(x^{3/2}\log x ),
\end{equation}
where
\begin{equation} \label{eq:defc0}
c_0= \frac 12 \prod_p \left(1- \frac 1{p(p+1)} \right)\approx 0.352.
\end{equation}
It was later shown by De Koninck and Sitaramachandrarao \cite{kn:dk-si} that, given any positive integer\,$k$,
$$
\sum_{n\le x} \frac 1{n\log \gamma(n)} = \log \log x + \sum_{m=0}^{k-1} \frac{b_m}{\log^m x} + O \left(  \frac 1{\log^k x} \right),$$
where the $b_m$'s are computable constants.

The kernel function is at the heart of several problems in number theory, one of which is the famous $abc-$conjecture (stated below). Although simple in appearance, the kernel function remains intriguing and in some aspects very mysterious.

As an example of its intriguing character, let us mention the search for an asymptotic formula for the sum of its reciprocal values up to a given number $x$, namely the sum
$$
I(x):= \sum_{n\leqslant x} \frac {1}{\gamma(n)}.
$$
At first, since $\gamma(n)=n$ when $n$ is squarefree and since this occurs for approximately $6x/\pi^2$ of all integers up to $x$, one might think that $I(x)$ would be near $\displaystyle\sum_{n\leqslant x} 1/n~\sim~\log x$. But, in 1962, N. G. de Bruijn \cite{kn:ngb} proved that, as $x\to \infty$,
$$
\log I(x) = (1+o(1)) \sqrt{\frac{8\log x}{\log\!\log x}},
$$
the asymptotic behavior of $I(x)$ itself remaining at that time a total  mystery.
Three years later, W. Schwarz \cite{kn:ws} came up with the unexpected formula
$$
I(x) =(1+o(1))\frac 1{2^{1/4}\sqrt{4\pi}} \left( \frac{\log\!\log x}{\log x} \right)^{1/4} Q(x)\qquad (x\to \infty),
$$
where
$$
Q(x)= \min_{0<\sigma < \infty} x^\sigma \kappa(\sigma) \qquad \mbox{with} \quad \kappa(\sigma) := \sum_{n=1}^\infty \frac 1{n^\sigma \gamma(n)}= \prod_p \left( 1+ \frac 1{p(p^\sigma -1)} \right),
$$
and was thus able to prove the asymptotic formula
$$I(x) =(1+o(1))\frac 1{2^{1/4}\sqrt{4\pi}} \left( \frac{\log\!\log x}{\log x} \right)^{1/4} \exp\{-R(\log x)\}\qquad (x\to \infty),$$
where, by setting $\varphi(\sigma):=\log \kappa(\sigma)$ and $\Phi(\sigma):=\varphi'(\sigma)$, the function $R(u)$ is defined by the relation
$$-R(u)=\varphi(\Phi^{-1}(-u)) + u\Phi^{-1}(-u).$$

The study of the kernel function can be seen as a way to investigate the multiplicative structure of integers. Now, the quantity  $\gamma(n)$ is easy to grasp when it is fairly large compared with $n$, say in the neighborhood of  $n/k$ for a fixed $k\ge 1$. However, when $\gamma(n)$ is very small compared with $n$, say when $\gamma(n)\le \sqrt n$ or even worst when $\gamma(n)\le \log n$, then its behavior depends on the small prime factors of $n$ (since $P(n)$ is small)  and is therefore very hard to grasp, thus for instance explaining  the difficulty in obtaining the above asymptotic value of $I(x)$.
This motivates us to study the {\it truncated kernel function} $\gamma_2$ defined by $\gamma_2(1)=1$ and for each integer $n\ge 2$ by $\gamma_2(n)=\gamma(n)/P(n)$, where
 $P(n)$ stands for the largest prime factor of $n$ (with $P(1)=1$).
This new function is very similar to $\gamma(n)$ in many respects but it also has the advantage of being simpler to investigate, essentially because the distribution of its small values no longer depends on $P(n)$ and is therefore much easier to understand.

We will study the global and local behavior of $\gamma_2(n)$ and also compare it with that of $\gamma(n)$ and of $n/P(n)$.
Although $\gamma_2(n)$ is not a multiplicative function, in general its behavior is more easily understood than that of $\gamma(n)$, as is the case when comparing the corresponding sums of their reciprocal values. However, as we will see, it turns out to be quite the opposite in many aspects.


\section{Main results}

Our first two theorems provide information about the global behavior of $\gamma_2(n)$.


\begin{theorem} \label{thm:moyenne}
With $c_0$ defined as in (\ref{eq:defc0}),
we have
\begin{equation} \label{eq:th1}
\sum_{n\le x} \gamma_2(n) = (1+o(1)) 2c_0 x^2 \delta(x)  \qquad (x\to \infty),
\end{equation}
where
\begin{equation} \label{eq:defdelta}
\delta(x)= \int_2^x \frac 1{t^2}\rho\left( \frac{\log x}{\log t} \right)\,{\rm d}t
\end{equation}
and $\rho$ is the well-known {\it Dickman function} defined as the (unique) continuous solution to the differential-difference equation
$$u\rho'(u)=-\rho(u-1) \qquad (u>1)$$
satisfying the initial condition $\rho(u)=1$ for $0\le u \le 1$.
\end{theorem}


\begin{theorem} \label{thm:reciproque}
For all $x\ge 2$,
\begin{equation} \label{eq:th2}
\sum_{n\le x} \frac 1{\gamma_2(n)}=  c_2 \frac x{\log x} + O \left( \frac{x\log \log x}{\log^2 x} \right),
\end{equation}
where $c_2=\displaystyle{\frac{\zeta(2)\zeta(3)}{\zeta(6)}\approx 1.9436}$ (here $\zeta$ stands for the Riemann Zeta Function).
\end{theorem}

The next theorem allows us to study the distribution of $\gamma_2(n)$, in particular by showing that it behaves like the function $n/P(n)$ almost everywhere.


\begin{theorem}\label{thm:distribution2}
Given any positive real number $u<1$,
$$\lim_{x\to \infty} \frac 1x \#\{n\le x: \gamma_2(n) < x^u\}
=\lim_{x\to \infty} \frac 1x \#\{n\le x: n/P(n) < x^u\}
= 1 -\rho\left(\frac 1{1-u}\right).$$
\end{theorem}



Given a real number $h\ge 1$, let us set
$$\Psi_h(x,y) = \#\{n\le x: P(n)\le y,\ \gamma(n)<n/h\}.$$



Observe that $\gamma_2(n)$ can be written as the product of two functions, namely $\displaystyle{\gamma_2(n)= \frac{\gamma(n)}n \cdot \frac n{P(n)}}$. The next result shows that $\gamma_2(n)$ is the product of two basic functions which are statistically independent almost everywhere.


\begin{theorem}\label{thm:distribution3}
Given real numbers $u>0$ and $h\ge 1$, then
\begin{equation} \label{eq:th7}
\Psi_h(x,x^{1/u}) = (1+o(1)) x\,D(h)\,\rho(u) \qquad (x\to \infty),
\end{equation}
where
$$D(h) := \lim_{x\to \infty} \frac 1x \#\{n\le x: \gamma(n)<n/h\}, $$
a number whose existence is established in Lemma \ref{lem:Dh}.
\end{theorem}

While $\gamma_2(n)$ and $n/P(n)$ are independent almost everywhere, this is not true for extremal values. Indeed, our next theorem shows that the maximal value of $\gamma_2(n)$ is smaller than the maximal value of $n/P(n)$, while Theorem \ref{thm:unique} shows that $n/P(n)$ takes on large values more often than $\gamma_2(n)$.


\begin{theorem} \label{thm:max}
As $x\to \infty$,
\begin{equation} \label{eq:max}
\max_{n\le x} \gamma_2(n) = (1+o(1)) \frac x{\log x}.
\end{equation}
\end{theorem}




\begin{theorem} \label{thm:unique}
Let $\xi$ be the unique solution of the equation
$$ \xi\, \log 2 = \log(1+\xi) + \xi\, \log(1+1/\xi),$$
so that $\xi \approx 3.403$. Fix any number $c\in [1,\xi)$ and set
\begin{eqnarray*}
A(x) &: = & \#\left\{n\le x: \gamma_2(n) \ge \frac n{c\log n}\right\},\\
B(x) & := & \#\left\{n\le x : \frac n{P(n)} \ge \frac n{c\log n}\right\}.
\end{eqnarray*}
Then,
$$A(x) = o(B(x))  \qquad (x\to \infty).$$
\end{theorem}

The following theorem is essentially the counterpart to the last two theorems, that is that there are many more integers $n$ with a small $\gamma_2(n)$ than there are integers $n$ with a small $P(n)$.


\begin{theorem}\label{thm:distribution1}
Given a fixed positive integer $k$,
$$\#\{n\le x: n/P(n) \le k\} =C(k) \frac x{\log x}+ O\left( \frac x{\log^2 x} \right),$$
where $\displaystyle{C(k):= \sum_{j=1}^k \frac 1j}$, and
$$\#\{n\le x: \gamma_2(n) \le k\} =E(k) \frac x{\log x} + O\left( \frac x{\log^2 x} \right) ,$$
where $\displaystyle{E(k):= \sum_{m\ge 1\atop \gamma(m)\le k} \frac 1m= \sum_{j=1}^k \frac{\mu^2(j)}j \prod_{p|j} \left(1- \frac 1p \right)^{-1}=\sum_{j=1}^k \frac{\mu^2(j)}{\phi(j)}}$.
\end{theorem}







\section{Preliminary results and notations}

We say that a function $\lambda:[1,+\infty[\to [0,+\infty[$ is {\it slowly increasing} if, for each constant $c>0$,
$$
\lim_{x\to \infty}\lambda(cx)/\lambda(x)=1.
$$ 

\begin{proposition} \label{prop:1}
Let $f$ be a non negative multiplicative function for which there exist a positive real number $k$ and a slowly increasing function $\lambda$ such that, as $x\to \infty$,
$$\sum_{n\le x} f(n) = (1+o(1))\,x\,(\log^{k-1} x)\,\lambda(\log x),$$
and such that for all real numbers $u>1$, as $y\to \infty$,
$$
\sum_{y<p<y^u}\frac{f(p)}{p} = (1+o(1))\, k\, \log u.
$$
Then, as $x=y^u\to \infty$,
$$
\sum_{n\leqslant x \atop P(n)\leqslant y} f(n) = (1+o(1)) \,\Gamma(k)\, u^{1-k} \rho_k(u) \sum_{n\leqslant x} f(n),
$$
uniformly for $u$ bounded, where $\rho_k(u)$ is the continuous solution to the differential difference equation with delayed argument, defined by
$$
\begin{array}{rcll}
\rho_k(u) & = & 0 & (u\leqslant 0),\\
\rho_k(u) & = & \dfrac{u^{k-1}}{\Gamma(k)} & (0<u\leqslant 1), \\
u\rho_k'(u) & = & (k-1)\rho_k(u) -k \rho_k(u-1) & (u>1), \\
\end{array}
$$
where $\Gamma$ stands for the Gamma function.
\end{proposition}


\begin{proof} This result is due to
de Bruijn and van Lint \cite{kn:lint}.
\end{proof}




\begin{proposition} \label{prop:2}
As $x\to \infty$,
\begin{equation} \label{eq:iv}
\sum_{n\leqslant x} \frac 1{P(n)} = \left(1+O \left( \sqrt{ \frac{\log\!\log x}{\log x}}   \right) \right)x\delta(x),
\end{equation}
\begin{equation}
\sum_{n\leqslant x} \frac{\mu^2(n)}{P(n)} = \left(\frac 6{\pi^2}+ O \left(\sqrt{\frac{\log\!\log x}{\log x}} \right)   \right) \sum_{n\leqslant x} \frac 1{P(n)}, \label{eq:iv2}
\end{equation}
\begin{equation}
\sum_{n\leqslant x \atop P(n)^2|n} \frac 1{P(n)} = x \exp\left\{-\sqrt{4\log x \log\!\log x} \left(1+ O \left( \frac{\log\!\log\!\log x}{\log\!\log x} \right) \right) \right\} =
x\delta(x)^{\sqrt 2 +o(1)}, \label{eq:iv3}
\end{equation}
where $\delta(x)$ is the function defined in (\ref{eq:defdelta}).
\end{proposition}

\begin{proof} A proof of (\ref{eq:iv}) was established by
Erd\H{o}s, Ivi\'c, and Pomerance \cite{kn:eip}, while (\ref{eq:iv2}) can be found in Ivi\'c \cite{kn:ivic}, and (\ref{eq:iv3}) in Ivi\'c and Pomerance \cite{kn:ip}.
\end{proof}

\begin{remark}
Using the estimate
\begin{equation}\label{eq:rhou}
\rho(u) = \exp\{ -u(\log u + \log\!\log u -1 +o(1))   \} \qquad (u\to \infty)
\end{equation}
(see the book of Tenenbaum \cite{kn:ten}),
one can show that the function $\delta(x)$ defined in (\ref{eq:defdelta}) is slowly increasing and satisfies
\begin{equation}\label{eq:gL}
\delta(x)=\exp\left\{-(1+o(1))\sqrt{2\log x \log\!\log x}\right\}\quad (x\to \infty),
\end{equation}
so that
$$
\delta(x)=L_0(x)^{-1+o(1)} \quad (x\to \infty),
$$
where
\begin{equation} \label{eq:defL0}
L_0(x):= \exp\{ \sqrt{2\log x\log \log x}\},
\end{equation}
\end{remark}


\begin{lemma} \label{lem:1a}
As $x\to \infty$,
$$
\int_1^x t\, \rho \left( \frac{\log t}{\log y} \right) \, {\rm d}t = \int_y^x t\, \rho \left( \frac{\log t}{\log y}
\right) \, {\rm d}t+O(y) = O(x),
$$
uniformly for $2\leqslant y \leqslant x$.
\end{lemma}


\begin{proof}
 We first estimate the maximum value of
\begin{equation}
g(t):=t\,\rho \left( \frac{\log t}{\log y} \right)  \label{eq:defg}
\end{equation}
for $y\leqslant t \leqslant x$ and fixed $y\in [2,x]$.
For this, consider $h(t):=\log g(t)$ and solve $h'(t)=0$ for $t$.

In view of (\ref{eq:rhou}), we have
\begin{eqnarray*}
h'(t)  =  \frac {\rm d}{{\rm d}t}\left(- \frac{\log t}{\log y}( \log\log t - \log \log y+\log\log\log t -\log\log\log y-1)+ \log t\right)
\end{eqnarray*}
so that
$$\log \log t - \log \log y = \log y,$$
in which case,
$$ t=y^y.$$
Substituting this value of $t$ in (\ref{eq:defg}), we get in view of (\ref{eq:rhou}),
$$g(t) = \rho(y) y^y \ll e^{-y\log y} y^y = 1,$$
which completes the proof of the lemma.
\end{proof}



\begin{lemma} \label{lem:psi}
Uniformly for $x\ge y \ge 2$,
$$\log \Psi(x,y) = Z \left\{ 1 + O\left( \frac 1{\log y} + \frac 1{\log \log x} \right) \right\},$$
where
$$Z= \frac{\log x}{\log y} \log \left( 1+ \frac y{\log x} \right) + \frac y{\log y} \log \left(1+ \frac{\log x}y \right).$$
\end{lemma}

\begin{proof}
This result is due to de Bruijn \cite{kn:bruijn}.
\end{proof}



\begin{lemma} \label{lem:Dh}
Given any real number $h\ge 1$, the limit
\begin{equation} \label{eq:defDh}
D(h):= \lim_{x\to \infty} \frac 1x \#\{n\le x: \gamma(n)<n/h\}
\end{equation}
exists and moreover, as $h$ becomes large,
\begin{equation} \label{eq:borneDh}
D(h) \ll \frac 1{\sqrt h}.
\end{equation}
\end{lemma}

\begin{proof}
Writing each positive integer $n$ as $n=st$, where $s$ is powerful and $t$ squarefree with $(s,t)=1$, we have that the condition $\displaystyle{\frac{\gamma(n)}n < \frac 1h}$ is equivalent to $\displaystyle{\frac{\gamma(s)}s < \frac 1h}$.  In light of these facts, we have that
\begin{equation} \label{eq:fa}
\sum_{n\le x \atop \frac{\gamma(n)}n < \frac 1h} 1
 =  \sum_{s\le x \atop{ \frac{\gamma(s)}s < \frac 1h \atop s\ \mbox{\tiny powerful}}} \sum_{t\le x/s \atop (t,s)=1}\mu^2(t)  = (1+o(1)) \frac 6{\pi^2} x \sum_{s\le x \atop{ \frac{\gamma(s)}s < \frac 1h \atop s\ \mbox{\tiny powerful}}} \frac 1{s \prod_{p|s} \left(1+ \frac 1p \right)} .
\end{equation}
Since
$\displaystyle{\sum_{s\ \mbox{\tiny powerful}\atop \frac{\gamma(s)}s < \frac 1h}  \frac 1{s \prod_{p|s} \left(1+ \frac 1p \right)}}$ is clearly a convergent sequence,
we can define $D(h)$ as
$$D(h) = \sum_{s\ \mbox{\tiny powerful}\atop \frac{\gamma(s)}s < \frac 1h}  \frac 1{s \prod_{p|s} \left(1+ \frac 1p \right)}.$$
Now, recall the following result proved by S. Golomb \cite{kn:golomb} in 1970:
$$S(y):=\#\{n\le y: n \mbox{ powerful}\} = (c_3+o(1))\sqrt y \qquad (y\to \infty),$$
where $c_3=\zeta(3/2)/\zeta(2)\approx 2.173$.
Finally, in order to prove (\ref{eq:borneDh}), we only need to observe that, since the condition $\frac{\gamma(s)}s<\frac 1h$ certainly implies that $s>\frac s{\gamma(s)} >h$, we have
$$
\sum_{s>x \atop{s\ \mbox{\tiny powerful} \atop \frac{\gamma(s)}s<\frac 1h}} \frac 1s <
\sum_{s> h \atop s\ \mbox{\tiny powerful}} \frac 1s
 \ll  \int_h^\infty \frac 1t \mbox{d}\,S(t)
  =  \int_h^\infty \frac{S(t)}t \,{\rm d}t + \int_h^\infty \frac{S(t)}{t^2} \,{\rm d}t\ll \frac 1{\sqrt h}.$$
\end{proof}


\section{The proof of the theorems}


\subsection{The proof of Theorem \ref{thm:moyenne}}

We first write
$$
S_2(x):=\sum_{n\le x} \gamma_2(n) =\sum_{n\le x \atop P(n)\|n} \frac{\gamma(n)}{P(n)} +
\sum_{n\le x \atop P(n)^2|n } \frac{\gamma(n)}{P(n)} =\Sigma_1 + \Sigma_2,
$$
say.
Since it follows from (\ref{eq:iv3}) of Proposition \ref{prop:2} that
\begin{equation}\label{eq:s1s2}
\Sigma_2 \leqslant \sum_{n\leqslant x \atop P(n)^2|n } \frac n{P(n)}
\leqslant x \sum_{n\leqslant x \atop P(n)^2|n } \frac 1{P(n)} \ll x^2\, \delta(x)^{\sqrt 2 + o(1)} \qquad (x\to \infty),
\end{equation}
we only need to estimate $\Sigma_1$.

We first observe that the true order of $\Sigma_1$ is $x^2\,\delta(x)$ and in fact that
$$
\frac{\Sigma_1}{(x^2\delta(x))/2}\; \in\; \left[\frac 6{\pi^2},1\right]
$$
since it is easily shown that, as $x\to \infty$,
\begin{equation}\label{eq:approx}
(1+o(1))\frac 6{\pi^2} \frac{x^2}2\, \delta(x) \leqslant \sum_{n\leqslant x \atop P(n)\|n}
\frac{\gamma(n)}{P(n)} \leqslant (1+o(1)) \frac{x^2}2\, \delta(x).
\end{equation}
Indeed, on the one hand,
$$
\sum_{n\le x \atop P(n)\|n} \frac{\gamma(n)}{P(n)} \leqslant \sum_{n\leqslant x \atop P(n)\|n} \frac n{P(n)}  = (1+o(1))\frac{x^2}2\, \delta(x) \qquad (x\to \infty),
$$
by way of (\ref{eq:iv}) and partial summation. On the other hand, using the trivial observation
$\gamma(n)\geqslant \mu^2(n)\,n$ valid for all $n\geqslant 1$, we have
$$
\sum_{n\le x \atop P(n)\|n} \frac{\gamma(n)}{P(n)} \geqslant \sum_{n\leqslant x \atop P(n)\|n}
\frac{\mu^2(n)\,n}{P(n)} = \sum_{n\leqslant x}\frac{\mu^2(n)n}{P(n)}=(1+o(1))\frac 6{\pi^2}\frac{x^2}2\, \delta(x) \qquad (x\to \infty),
$$
where first we used (\ref{eq:iv2}) and partial summation and thereafter estimate (\ref{eq:iv}) of Proposition\,\ref{prop:2}.

In order to estimate $\Sigma_1$, we shall first prove that
\begin{equation}\label{eq:g1}
G(x,y):=\sum_{n\le x \atop P(n)\leqslant y} \gamma(n) = (1+o(1))\,c_1 x^2 \ \rho(u)\qquad (x\to \infty),
\end{equation}
where $\ u=\dfrac{\log x}{\log y}$ and $\rho$ is the Dickman function.

Let $f(n):=\gamma(n)/n$. First, it is an easy matter to derive from (\ref{eq:cohen}) that
\begin{equation} \label{eq:gn}
\sum_{n\le x} \frac{\gamma(n)}n = c_1 x +O(x^{1/2} \log x),
\end{equation}
where $c_1=2c_0$.

On the other hand, using Mertens' formula, we have that
\begin{equation}\label{eq:pnt}
\sum_{y< p<y^u} \frac{f(p)}p = \sum_{y< p<y^u} \frac 1p  = \log u + O(1) \qquad(y \to \infty).
\end{equation}
Hence, it follows from (\ref{eq:pnt}) and (\ref{eq:gn}) that
$f$ satisfies the conditions of Proposition \ref{prop:1} with $k=1$, yielding
\begin{equation} \label{eq:g3}
\sum_{n\le x \atop P(n)\le y} \frac{\gamma(n)}{n} = (1+o(1))\,c_1   \; \rho(u)\;x \qquad (x\to\infty).
\end{equation}
Using partial summation and Lemma \ref{lem:1a}, we get, as $x\to \infty$,
\begin{eqnarray*}
\sum_{n\le x \atop P(n)\le y} \gamma(n) & = & \sum_{n\leqslant x \atop P(n)\le y}
\frac{\gamma(n)}{n} n \\
& = & (1+o(1))\,c_1 \rho(u) x^2 - c_1(1+o(1))\int_1^x \rho \left( \frac{\log t}{\log y} \right) t\, {\rm d}t \\
& = &(1+o(1))\, c_1 \rho(u) x^2 +O(x),
\end{eqnarray*}
which proves (\ref{eq:g1}).

Getting back to the definition of $\Sigma_1$, we obtain
\begin{eqnarray} \label{eq:t1t2}  \nonumber
\Sigma_1 & = &  \sum_{p\leqslant x} \sum_{mp\leqslant x \atop P(m)<p} \frac{\gamma(m)\,p}p = \sum_{p\leqslant x} \sum_{m\leqslant x/p \atop P(m)<p} \gamma(m)
  = \sum_{p\leqslant x} \sum_{m\leqslant x/p \atop P(m)\leqslant p} \gamma(m) - \sum_{p\leqslant x}
 \sum_{m\leqslant x/p \atop P(m)=p} \gamma(m) \\ \nonumber
& = & \sum_{p\leqslant x} G\left(\frac{x}{p},p\right)-\sum_{mp\leqslant x \atop P(mp)^2 | mp}\frac{\gamma(mp)}{P(mp)}  \\
& = & T_1 - T_2,
\end{eqnarray}
where, by (\ref{eq:g1}),
$$T_1 = (1+o(1)) c_1 x^2  \sum_{p\leqslant x}\frac 1{p^2} \rho\left( \frac{\log x}{\log p} -1 \right) \qquad (x\to \infty)$$
while, by (\ref{eq:s1s2}),
$$T_2 \ll x^2 \delta(x)^{\sqrt{2}+o(1)} \qquad (x\to \infty) .$$
Following an argument used by Ivi\'c and Pomerance \cite{kn:ip}, we obtain that
\begin{equation} \label{eq:t1}
T_1  =  c_1 (1+o(1))\,x^2 \int_2^x \frac 1{t^2\log t }  \rho\left( \frac{\log x}{\log t} -1 \right)\,{\rm d}t
 =   c_1 (1+o(1))\,x^2 \delta(x) \qquad(x\to \infty),
\end{equation}
while clearly
\begin{equation}\label{eq:t2}
T_2 = o\left(x^2\delta(x)\right) \qquad (x\to \infty).
\end{equation}

Substituting (\ref{eq:t1}) and (\ref{eq:t2}) in (\ref{eq:t1t2}),  (\ref{eq:th1}) follows.



\subsection{The proof of Theorem \ref{thm:reciproque}}

Let $K$ be a fixed large integer. Then,
\begin{equation} \label{eq:y0}
\sum_{n\le x} \frac 1{\gamma_2(n)} = \sum_{k\le K} \frac{\mu^2(k)}k \sum_{n\le x \atop \gamma_2(n)=k} 1 + \sum_{K<k\le x} \frac{\mu^2(k)}k \sum_{n\le x \atop \gamma_2(n)=k} 1
= S_1(x;K)+S_2(x;K),
\end{equation}
say.

Using the fact that, for any squarefree integer $k$,
$$\sum_{n\ge 1 \atop \gamma_2(n)=k} \frac 1n= \frac 1k \prod_{p|k} \left(1+ \frac 1p + \frac 1{p^2} + \ldots \right) = \frac 1{\phi(k)},$$
we easily obtain that
\begin{equation} \label{eq:y1}
S_2(x;K) \le \sum_{K<k\le x} \frac{\mu^2(k)}k \sum_{n\le x \atop \gamma_2(n)=k} \frac xn \le x \sum_{k>K} \frac{\mu^2(k)}{k\phi(k)} < x \sum_{k>K} \frac 1{k^{3/2}} < \frac x{K^{1/2}},
\end{equation}
where we used the trivial inequality $\phi(k)>k^{1/2}$ valid for all $k\ge 7$.

On the other hand, in light of estimate (\ref{eq:ajout}), which is proved in section 4.7, we get
\begin{eqnarray} \label{eq:y2} \nonumber
S_1(x;K) & = & \sum_{k\le K} \frac{\mu^2(k)}k \left( \sum_{n\le x \atop \gamma_2(n)\le k} 1 - \sum_{n\le x \atop \gamma_2(n)\le k-1} 1 \right) \\ \nonumber
& = & \sum_{k\le K} \frac{\mu^2(k)}k \left( (E(k)-E(k-1)) \frac x{\log x} + O\left(
\frac x{\log^2 x} \right) +O(\pi(k)) \right) \\
& = & \left(  \sum_{k\le K} \frac{\mu^2(k)}{k\phi(k)} \right) \frac x{\log x}
+ O\left(
\frac{x\log K}{\log^2 x} \right)+ O\left( \frac K{\log K} \right).
\end{eqnarray}
Again using the fact that $\phi(k)< k^{1/2}$ for all $k\ge 7$, we have
\begin{equation} \label{eq:y3}
\sum_{k\le K} \frac{\mu^2(k)}{k\phi(k)} = \sum_{k=1}^\infty \frac{\mu^2(k)}{k\phi(k)}
- \sum_{k> K} \frac{\mu^2(k)}{k\phi(k)} = c_2 + O\left( \frac 1{\sqrt K} \right).
\end{equation}
Choosing $K=\log^4 x$ and using (\ref{eq:y1}), (\ref{eq:y2}) and (\ref{eq:y3}) in (\ref{eq:y0}), estimate (\ref{eq:th2}) follows, thereby completing the proof of Theorem \ref{thm:reciproque}.



\subsection{The proof of Theorem \ref{thm:distribution2}}


Since, for each integer $n\ge 2$, we have
$\displaystyle{\gamma_2(n)=\frac{\gamma(n)}{P(n)} \le \frac n{P(n)}}$, it follows that
\begin{eqnarray}\label{eq:w1} \nonumber
\sum_{n\le x \atop \gamma_2(n)<x^u} 1 & \ge &
\sum_{n\le x \atop \frac n{P(n)} <x^u} 1 =  \sum_{n\le x \atop \frac{P(n)}n >\frac 1{x^u}} 1 \\ \nonumber
& = & [x] - \sum_{n\le x \atop P(n)\le n/x^u} 1
 \ge  [x] - \sum_{n\le x \atop P(n)\le x^{1-u}} 1 \\
& = & x \left( 1- \rho\left(\frac 1{1-u}\right)+o(1) \right) \qquad (x\to \infty).
\end{eqnarray}

On the other hand, let $\varepsilon>0$ be an arbitrarily small number and choose $k$ large enough so that, using estimate (\ref{eq:borneDh}) of Lemma \ref{lem:Dh}, we can claim that
\begin{equation} \label{eq:v1}
\lim_{x\to \infty} \frac 1x \#\{n\le x: \gamma(n)/n < 1/k\} < \varepsilon.
\end{equation}

Then, using (\ref{eq:v1}), provided $x$ is large enough, we have
\begin{equation}\label{eq:w2}
\sum_{n\le x \atop \gamma_2(n)<x^u} 1  =
\sum_{n\le x \atop{\gamma_2(n)<x^u \atop \frac{\gamma(n)}n < \frac 1k}} 1
+
\sum_{n\le x \atop{\gamma_2(n)<x^u \atop \frac{\gamma(n)}n \ge \frac 1k}} 1
 <  \varepsilon x + \sum_{n\le x \atop \frac n{P(n)} < kx^u} 1,
\end{equation}
where the last sum was obtained using the fact that
$$ \frac n{kP(n)}\le  \frac{\gamma(n)}{P(n)} =\gamma_2(n) < x^u.$$
Hence, from (\ref{eq:w2}), we conclude that
\begin{equation}\label{eq:w3}
\sum_{n\le x \atop \gamma_2(n)<x^u} 1  <  \varepsilon x + [x] - \sum_{n\le x \atop P(n) \le x^{1-u}} 1 =  x\left(1 - \rho \left( \frac 1{1-u}\right)+o(1)\right) \qquad (x\to \infty).
\end{equation}
Combining estimates(\ref{eq:w1}) and (\ref{eq:w3}) completes the proof of Theorem \ref{thm:distribution2}.


\subsection{The proof of Theorem \ref{thm:distribution3}}

First, we can assume that $P(n)\|n$ since we know from Proposition \ref{prop:2} that the number of positive integers $n\le x$ such that $P^2(n)|n$ is no larger than $x\delta(x)^{\sqrt 2+1}=o(x)$.

We will use the same pattern of proof as the one that Granville \cite{kn:granville} used to prove that $\psi(x,x^{1/u}) =(1+o(1))x\rho(u)$ as $x\to \infty$.

First, in light of the definition of $D(h)$ given in (\ref{eq:defDh}), it is clear that
\begin{equation} \label{eq:100}
\Psi_h(x,x^{1/u}) =(1+o(1)) xD(h)\rho(u)\quad \mbox{ for all } 0<u\le 1 \qquad (x\to \infty).
\end{equation}

Let us now consider the case where $u\in [1,2]$. Then, with the assumption $P(n)\|n$, we have

\begin{equation} \label{eq:tot1}
\sum_{n\le x \atop{P(n)\le x^{1/u} \atop \frac{\gamma(n)}n < \frac 1h}} 1 =
\sum_{n\le x\atop \frac{\gamma(n)}n < \frac 1h} 1-
\sum_{n\le x \atop{x^{1/u}<P(n)\le x \atop \frac{\gamma(n)}n < \frac 1h}} 1
= \Sigma_1 - \Sigma_2,
\end{equation}
say.

Again, it follows from the definition of $D(h)$  that
\begin{equation} \label{eq:ss1}
\Sigma_1 = (1+o(1)) D(h) x \qquad (x\to \infty).
\end{equation}

On the other hand,  writing $n=mp$ with $P(m)<p$, then, because $u\in [1,2]$, the condition $P(m)<p$ is automatically satisfied, while the assumption $P(n)\|n$ guarantees that the condition $\frac{\gamma(n)}n<\frac 1h$ is equivalent to the condition $\frac{\gamma(m)}m<\frac 1h$,  implying that
\begin{eqnarray} \label{eq:ss2} \nonumber
\Sigma_2 & = & \sum_{x^{1/u}< p\le x} \sum_{m\le \frac xp \atop{P(m)\le p \atop \frac{\gamma(m)}m< \frac 1h}} 1 =
\sum_{x^{1/u}< p\le x} \sum_{m\le \frac xp \atop \frac{\gamma(m)}m< \frac 1h} 1 \\
& = &
\sum_{x^{1/u} < p \le x} \frac xp D(h)
=(1+o(1)) x D(h) \log u \qquad (x\to \infty).
\end{eqnarray}
Gathering (\ref{eq:ss1}) and (\ref{eq:ss2}) in (\ref{eq:tot1}), we obtain
that, in the case $u\in [1,2]$,
\begin{equation} \label{eq:tot2}
\Psi_h(x,x^{1/u})   =
(1+o(1)) x\, D(h)\, (1-\log u) \qquad (x\to \infty).
\end{equation}

Hence, in light of (\ref{eq:100}) and (\ref{eq:tot2}), we have thus proved that
\begin{equation} \label{eq:resum}
\Psi_h(x,x^{1/u}) = (1+o(1)) x\, D(h)\, \rho(u) \qquad (x\to \infty)
\end{equation}
holds for $0\le u \le 2$ with
$$\rho(u)= \begin{cases}  1, & \text{if $0<u\le 1$}, \\
  1-\log u, & \text{if $1\le u \le 2$}.   
\end{cases}
$$

We now use induction. Assuming that (\ref{eq:resum}) holds for all $u\in (0,N]$, we will now prove that it must also hold for all $u\in [N,N+1]$. For this, we will need the Buchstab identity (trivially generalized to $\Psi_h(x,y)$), that is
\begin{equation} \label{eq:B}
\Psi_h(x,y) =  1+ \sum_{p\le y} \sum_{mp\le x \atop{P(m)<p \atop \frac{\gamma(m)}m <\frac 1h}} 1
= 1+ \sum_{p\le y} \Psi_h \left( \frac xp,p \right).
\end{equation}

Using (\ref{eq:B}), we have
\begin{eqnarray*}
\Psi_h(x,x^{1/N}) & = & 1+ \sum_{p\le x^{1/N}} \Psi\left( \frac xp, p \right), \\
\Psi_h(x,x^{1/u}) & = & 1+ \sum_{p\le x^{1/u}} \Psi\left( \frac xp, p \right).
\end{eqnarray*}
Subtracting these two equations, we obtain
\begin{equation} \label{eq:dif-2}
\Psi_h(x,x^{1/u}) = \Psi_h(x,x^{1/N}) - \sum_{x^{1/u}<p \le x^{1/N}} \Psi_h\left( \frac xp, p \right).
\end{equation}
Now, since $p>x^{1/u}$ and $u\in [N,N+1]$, we have
$$\frac{\log(x/p)}{\log p} = \frac{\log x}{\log p} -1 < \frac{\log x}{\log(x^{1/u})} - 1
=u-1 \le N,$$
implying that (\ref{eq:resum}) holds, say with $u'=\frac{\log(x/p)}{\log p}$ instead of $u$, thus allowing us to replace (\ref{eq:dif-2}) by
\begin{eqnarray*}
\Psi_h(x,x^{1/u}) & = &  (1+o(1)) x D(h) \rho(N) - (1+o(1)) D(h) \sum_{x^{1/u} < p \le x^{1/N}} \frac xp\, \rho \left( \frac{\log x}{\log p} -1 \right) \\
& = & (1+o(1)) x D(h) \rho(N) - (1+o(1))x D(h)\int_{x^{1/u}}^{x^{1/N}} \rho \left( \frac{\log x}{\log v}-1 \right) \frac{{\rm d}\,\theta(v)}{v\log v} \\
& = & (1+o(1))x D(h) \left( \rho(N) - \int_N^u \frac{\rho(t-1)}t\,dt \right) \\
& = & (1+o(1))x D(h) \rho(u) \qquad (x\to \infty),
\end{eqnarray*}
(where we used the prime number theorem in the form $\theta(v)=\sum_{p\le v} \log p = (1+o(1))v$ as $v\to \infty$), thus showing that (\ref{eq:resum}) also holds for $u\in [N,N+1]$ and thus completing the induction argument.





\subsection{The proof of Theorem \ref{thm:max}}


We first show that the bound is achieved for $n=\prod_{p\le \log x} p$. Indeed, it follows from the prime number theorem that, as $x\to \infty$,
$$
\gamma_2(n)= \frac 1{\max_{p\le \log x} p} \times \prod_{p\le \log x} p
=\frac 1{\max_{p\le \log x} p} \times e^{(1+o(1))\log x}
= (1+o(1)) \frac x{\log x}.
$$
On the other hand, this last expression is indeed an upper bound for $\gamma_2(n)$. To prove this, first assume that $P(n)>\log(n/\log n)$. Then, in this case, as $x\to \infty$,
$$
\gamma_2(n) \le \frac n{P(n)} \le \frac n{\log n - \log \log n} = \frac n{\log n} \left(1 + O\left( \frac{\log \log n}{\log n} \right) \right)\le (1+o(1)) \frac x{\log x}.
$$
If, on the contrary, $P(n)\le \log(n/\log n)$, we have, by the prime number theorem,
$$
\gamma_2(n) < \prod_{p\le P(n)}p \le \prod_{p\le \log n - \log\log n}p = (1+o(1)) \frac n{\log n}  \le (1+o(1)) \frac x{\log x} \qquad (x\to \infty).
$$
This completes the proof of Theorem \ref{thm:max}.


\subsection{The proof of Theorem \ref{thm:unique}}

Let us write each integer $n\ge 2$ as $n=st$, where $s$ is squarefull and $t$ is squarefree with $(s,t)=1$.  On the one hand,
$$\frac{\gamma_2(n)}n \ge \frac 1{c\log n}
\Longleftrightarrow \frac{s P(st)}{\gamma(s)} \le c\log n.$$
But this last inequality implies that
\begin{equation} \label{eq:a1}
P(st) \le \frac{sP(st)}{\gamma(s)} \le c\log n \le c \log x.
\end{equation}
Since $\gamma_2(n)\le \gamma(n) \le n/\sqrt s$, we get that $s\le (k\log x)^2$, which combined with (\ref{eq:a1}) implies that $P(t)\le c \log x$.

Therefore, as $x\to \infty$,
\begin{eqnarray} \label{eq:t11} \nonumber
A(x) & \le &  \#\{t\le x: \mu^2(t)=1, \ P(t)\le c \log x\}\\
  & \le & 2^{\pi(c\log x)} =  \exp\left\{ (1+o(1)) c \log 2 \frac{\log x}{\log \log x}    \right\},
\end{eqnarray}
where again we made use of the prime number theorem.

We will now obtain a lower bound for $B(x)$. Given any small $\delta>0$, we have, using Lemma \ref{lem:psi}, that as $x\to \infty$, 
\begin{eqnarray} 
B(x) & = & \#\{n\le x: P(n) \le c \log n\}  \nonumber \\
    & \ge & \#\{x^{1-\delta} < n\le x: P(n) \le c \log n\}  \nonumber \\
    & \ge & \#\{x^{1-\delta} < n\le x: P(n) \le (1-\delta) c \log x\}  \nonumber \\
    & = & \Psi(x,(1-\delta)c\log x) - \Psi(x^{1-\delta},(1-\delta)c\log x) \nonumber \\
    & = & (1+o(1))\Psi(x,(1-\delta)c\log x) \label{eq:t21} \\
    & = & (1+o(1)) \exp Z \nonumber \\
    & = &  \exp\left\{ (1+o(1))\frac{\log x}{\log \log x}
    \left(  \log(1+(1-\delta)c) + (1-\delta) c \log \left( 1+ \frac 1{c(1-\delta)} \right)    \right)  \right\} \nonumber .
\end{eqnarray}

Since $\delta$ can be taken arbitrarily small, it follows from (\ref{eq:t21}) that, as $x\to \infty$,
\begin{equation} \label{eq:t3}
B(x) \ge
\exp\left\{ (1+o(1))\frac{\log x}{\log \log x}
    \left(  \log(1+c) +  c \log \left( 1+ \frac 1c \right)    \right)  \right\}.
\end{equation}
Finally, by comparing (\ref{eq:t11}) with (\ref{eq:t3}) and observing that for $c<\xi$, we have
$$c \log 2 < \log(1+c) +  c \log \left( 1+ \frac 1c \right),$$
the proof of Theorem \ref{thm:unique} is complete.


\subsection{The proof of Theorem \ref{thm:distribution1}}



We first evaluate $S_1=\#\{n\le x: n/P(n) \le k\}$. Writing each positive integer $n\le x$ as $n=mp$ with $P(m)\le p$, we have
\begin{eqnarray*}
S_1 & = & \sum_{m\le k } \sum_{k<p\le x/m} 1 \\
& = &    \sum_{m\le k} \sum_{p\le x/m} 1 +O(1) \\
& = & \sum_{m\le k} \pi(x/m)  +O(1) \\
& = &  \sum_{m\le k} \left(\frac x{m \log (x/m)} + O \left( \frac{x/m}{\log^2(x/m)} \right) \right) \\
& = &  \frac x{\log x} \sum_{m=1}^k \frac 1m +  O\left( \frac x{\log^2 x} \right),
\end{eqnarray*}
where we used the prime number theorem,
thus proving our first assertion.

Now let $S_2=\#\{n\le x: \gamma_2(n)\le k\}$. Again, writing each positive integer $n\le x$ as $n=mp$ with $P(m)\le p$, we have, using Proposition \ref{prop:2},
\begin{eqnarray*} 
S_2 & = & \sum_{n\le x \atop {P(n)^2|n \atop \gamma_2(n)\le k}} 1 +
\sum_{m\le x  \atop{p \le x/m \atop P(m)< p,\ \gamma(m)\le k}} 1 \\ \nonumber
& = & O\left( \frac x{\exp\{(1+o(1))2\sqrt{\log x \log \log x} \}} \right)
 + \sum_{m\le x \atop \gamma(m)\le k} \sum_{P(m)<p\le x/m} 1 \\ \nonumber
& = & O\left( \frac x{\log^2x} \right)+ \sum_{m\le x \atop \gamma(m)\le k} \pi(x/m) +  O\left( \pi(k) \right)\\ \nonumber
 & = &    \sum_{m\le x \atop \gamma(m) \le k} \frac{x/m}{\log(x/m)} +  O\left( \frac x{\log^2 x} \right) +O(\pi(k))\\ \nonumber
 \end{eqnarray*}
 \begin{eqnarray}
\label{eq:ajout}
 \hphantom{S_2} 
& = & \frac x{\log x} \sum_{m\ge 1 \atop \gamma(m) \le k} \frac 1m +  O\left( \frac x{\log^2 x} \right) +O(\pi(k))  \nonumber \\
& = &  E(k) \frac x{\log x} +  O\left( \frac x{\log^2 x} \right) +O(\pi(k)) ,
\end{eqnarray}
as we wanted to prove.

\section{Final remarks}

One could certainly extend the above results on $\gamma_2(n)$ to the more general case $\gamma_k(n)$, with $k\ge 3$, defined by $\gamma_k(n)= \gamma_{k-1}(n)/P(\gamma_{k-1}(n))$. For instance, using the above arguments, one can easily show that
$$\sum_{n\le x} \frac 1{\gamma_k(n)} = (c_k+o(1)) x \frac{(\log \log x)^{k-2}}{\log x} \qquad (x\to \infty)$$
and that
$$\max_{n\le x} \gamma_k(n) = (1+o(1)) \frac x{\log^{k-1}x} \qquad (x\to \infty).$$

On the other hand, an interesting problem would be to investigate what kind of results one could obtain if $k$ is replaced by a function of $n$, for instance, by choosing $k=\lfloor \frac{\omega(n)}2 \rfloor$.



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\bibitem{kn:ws} W. Schwarz, Einige Anwendungen tauberscher S\"atz in
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\end{thebibliography}

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

\noindent \emph{Keywords: } kernel function, arithmetic function.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  November 24 2011;
revised version received  January 28 2012.
Published in {\it Journal of Integer Sequences}, February 5 2012.

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\noindent
Return to
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