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\begin{center}
\vskip 1cm{\LARGE\bf 
A Sum of Gcd's over Friable Numbers 
}
\vskip 1cm
\large
Julian Ransford\footnote{Research supported by an NSERC USRA scholarship.}\\
D\'epartement de Math\'ematiques et de Statistique\\
Universit\'e Laval \\
Qu\'ebec, PQ G1V 0A6 \\
Canada \\
\href{mailto:julian.ransford.1@ulaval.ca}{\tt julian.ransford.1@ulaval.ca} \\
\end{center}

\bigskip

\begin{abstract}
We study the function 
$$g(n,y):=\sum_{\substack{i \leq n\\ P(i)\leq y}}\gcd(i,n),$$
where $P(n)$ denotes the largest prime factor of $n$,
and we derive some estimates for its summatory function.
\end{abstract}

\section{Introduction}
Let $P(n)$ denote the largest prime factor of $n$, with the convention that $P(1)=1$. We say that $n$ is $y$\textit{-friable} if $P(n)\leq y$. Friable numbers have been studied in great detail by many authors, and a lot of attention has been devoted to estimates and properties of the function
\[\Psi(x,y):=\sum_{\substack{n\leq x \\ P(n)\leq y}}1.\]
The article of Hildebrand--Tenenbaum \cite{hildebrand+tenenbaum93} contains a survey on the subject, as well as numerous references. One of the most interesting results concerning this function is due to Dickman. He showed that, for any fixed $u$, we have
\[\Psi(x,x^{1/u}) \sim x\rho(u) \qquad (x \rightarrow \infty),\] 
where $\rho(u)$ is the \textit{Dickman function}, defined as the solution to the delayed differential equation
\[\begin{cases}
\rho(u)=1, &amp; \text{if } 0 \leq u \leq 1;\\ 
u\rho'(u)+\rho(u-1)=0, &amp; \text{if } u&gt;1.\\ 
\end{cases}\]
For more details, see for example De Koninck--Luca \cite[Thm.\ 9.3, p.\ 134]{dekoninck+luca12}. Hildebrand \cite{hildebrand86} showed that, in fact, the following stronger estimate holds.
\begin{theorem}
Given any $\epsilon &gt;0$, uniformly for $\exp((\log\log x)^{\frac{5}{3}+\epsilon})\leq y \leq x,$
\begin{equation}\label{estime psi 1} 
\Psi(x,y)=x\rho(u)\left(1+O\left(\frac{\log(u+1)}{\log y}\right)\right).
\end{equation}
\end{theorem}
 It turns out that improving the region of validity for this estimate improves our knowledge of the distribution of the zeros of the Riemann zeta function, and that \eqref{estime psi 1} holds uniformly for $(\log x)^{2+\epsilon}\leq y \leq x$ if and only if the Riemann hypothesis is true; see Hildebrand \cite{hildebrand84}.

 On the other hand, the function $\Psi(x,y)$ behaves quite differently for very small values of $y$. Indeed, Ennola showed \cite[Thm.\ 1.5]{hildebrand+tenenbaum93} the following estimate.
\begin{theorem}Uniformly for $2 \leq y \leq \sqrt{\log x}$,
\[\Psi(x,y)=\frac{1}{\pi(y)!}\prod_{p \leq y}\left(\frac{\log x}{\log p}\right)\left(1+O\left(\frac{y^2}{\log x \log y}\right)\right).\]
\end{theorem}
 In particular, if $y$ is fixed, this yields
\begin{equation}\label{Ennola 2} \Psi(x,y)=\frac{1}{\pi(y)!}\prod_{p \leq y}\left(\frac{1}{\log p}\right)(\log x)^{\pi(y)}+O((\log x)^{\pi(y)-1}).
\end{equation}
 In this paper, we introduce and study the function
\begin{equation}\label{Definition 1} 
g(n,y):=\sum_{\substack{i \leq n\\ P(i)\leq y}}\gcd(i,n).
\end{equation}
This can be seen as a generalization of the function
\[g(n):=\sum_{i \leq n}\gcd(i,n).\]
The latter was studied by Broughan \cite{broughan01}, who showed that $g(n)$ is a multiplicative function. Indeed, we have that \cite[Thm.\ 2.3]{broughan01}
\begin{equation}\label{dirichlet produit 1} 
g(n)=\sum_{d \mid n}\phi(d)\frac{n}{d},
\end{equation}
where $\phi(n)$ is Euler's totient function. Using well-known estimates for summatory functions related to $\phi(n)$, Broughan has shown that
\[\sum_{n \leq x}g(n)=\frac{3x^2 \log x}{\pi^2}+O(x^2) \qquad (x\rightarrow \infty).\]
Our main goal is to derive estimates for
\[G(x,y):=\sum_{n \leq x}g(n,y)\]
in various ranges of $y \leq x$.

\section{Main results}
 Let $g(n,y)$ be defined as in \eqref{Definition 1}. Also, here and in what follows, we set $u:=\log x/\log y$.
\begin{theorem}\label{resultat majeur 1} For $y$ fixed, we have, as $x\rightarrow \infty$
\[G(x,y)=\left(\frac{1}{(2\pi(y))!}\prod_{p \leq y}\frac{p-1}{p(\log p)^2}\right)x(\log x)^{2\pi(y)}+O(x(\log x)^{2\pi(y)-1}),\]
where $\pi(y)$ denotes the prime counting function.
\end{theorem}

\begin{theorem}\label{resultat majeur 2}We have
\[G(x,y)=\frac{3x^2\rho_2(u)\log y}{\pi^2}\left(1+O\left(\frac{\log(u+1)}{\log y}\right)\right),\]
uniformly for
\[\exp((\log\log x)^{\frac{5}{3}+\epsilon})\leq y \leq x,\]
where $\rho_2(u)$ is the continuous solution to the delayed differential equation
\[\begin{cases}
\rho_2(u)=u, &amp; \text{if } 0 \leq u \leq 1;\\ 
u\rho_2'(u)-\rho_2(u)+2\rho_2(u-1)=0, &amp; \text{if } u&gt;1.\\ 
\end{cases}
\]
\end{theorem}

\section{Preliminary results}
An important fact that we will need later is the value of $g(n)$ when $n$ is a prime power.
\begin{lemma}\label{g puissances premiers} 
If $p$ is a prime, then
\[g(p^k)=(k+1)p^k-kp^{k-1}.\]
\end{lemma}
\begin{proof}
Using \eqref{dirichlet produit 1}, we have
\begin{align*}
g(p^k)&amp;=\sum_{d \mid p^k}d\phi\left(\frac{p^k}{d}\right)=\left(\sum_{j=0}^{k-1} p^j\cdot p^{k-j-1}(p-1)\right)+p^k\\
&amp;=p^{k-1}(p-1)\sum_{j=0}^{k-1}1 + p^k =p^{k-1}(p-1)k+p^k=(k+1)p^k-kp^{k-1}.\qedhere
\end{align*}
\end{proof}

 The next lemma is the analog of \eqref{dirichlet produit 1} for friable numbers.
\begin{lemma}\label{dirichlet produit 2} 
The function $g(n,y)$ satisfies the relation
\begin{equation}\label{formule dirichlet produit 2}
g(n,y)=\sum_{\substack{d \mid n \\ P(d)\leq y}}d \phi\left(\frac{n}{d},y\right),
\end{equation}
where we set
\[\phi(n,y):=\sum_{\substack{k \leq n \\ (n,k)=1 \\ P(k) \leq y}}1.\]
\end{lemma}  
\begin{proof}
The proof is very similar to the proof of \eqref{dirichlet produit 1} given in Broughan \cite{broughan01}. It is clear that the gcd's in the sum  are divisors of $n$ and must be $y$-friable. How many times does a $y$-friable divisor of $n$ appear in the sum \eqref{formule dirichlet produit 2}? We have that 
\[\gcd(i,n)=d \Longleftrightarrow \gcd\left( \frac{n}{d}, \frac{i}{d} \right)=1.\]
Since $i/d$ is always $y$-friable we conclude that $d$ is in the sum as often as there are integers less than or equal to $n/d$ that are coprime with $n/d$ and $y$-friable. This is precisely the same thing as $\phi(n/d,y)$, and therefore
\[g(n,y)=\sum_{\substack{d \mid n \\ P(d) \leq y} } d  \phi\left(\frac{n}{d},y\right).\qedhere\] 
\end{proof}
 Using Lemma \ref{dirichlet produit 2}, we obtain that
\begin{align*}
\sum_{n \leq x}g(n,y)&amp;=\sum_{n \leq x}\sum_{\substack{d \mid n \\ P(d)\leq y}}d \phi\left(\frac{n}{d},y\right)=\sum_{\substack{d \leq x\\ P(d)\leq y}}\sum_{\substack{n \leq x \\ d\mid n}}d \phi\left(\frac{n}{d},y\right)\\
&amp;=\sum_{\substack{d \leq x\\ P(d)\leq y}}d\sum_{kd \leq x}\phi\left(\frac{kd}{d},y\right)=\sum_{\substack{d \leq x\\ P(d)\leq y}}d\sum_{k \leq \frac{x}{d}}\phi(k,y)\\
&amp;=\sum_{\substack{d \leq x\\ P(d)\leq y}}d\Phi\left(\frac{x}{d},y\right),\\
\end{align*}
where we have set
\[\Phi(x,y):=\sum_{n \leq x}\phi(n,y).\]

\begin{lemma}\label{lemme Phi}For fixed $y$, we have
\[\Phi(x,y)=\frac{1}{\pi(y)!}\prod_{p \leq y}\left(\frac{p-1}{p\log p}\right)x(\log x)^{\pi(y)}+O(x(\log x)^{\pi(y)-1}) \qquad (x \rightarrow \infty).\]
\end{lemma}
\begin{proof}
Let $P_y:=\prod_{p \leq y}p$. Thus
\[\sum_{n \leq x}\phi(n,y)=\sum_{d \mid P_y}\sum_{\substack{n \leq x\\ (n,P_y)=d}}\phi(n,y).\]
If $(n,P_y)=d$, then $\phi(n,y)$ will only count integers less than or equal to $n$ that contain only prime factors of $P_y$ that do not divide $d$. It is therefore clear that the smaller $d$ is, the bigger $\phi(n,y)$ will be. Hence, it suffices to evaluate the sum
\[\sum_{\substack{n \leq x\\ (n,P_y)=1}}\phi(n,y),\]
since all the other terms will be absorbed in its error term. But $\phi(n,y)=\Psi(n,y)$ when $(n,P_y)=1$. Therefore
\begin{align*}
\sum_{\substack{n \leq x\\ (n,P_y)=1}}\phi(n,y)&amp;=\sum_{\substack{n \leq x\\ (n,P_y)=1}}\Psi(n,y)\\
&amp;=\sum_{n \leq x}\Psi(n,y)-\sum_{\substack{n \leq x\\ 2 \mid n}}\Psi(n,y)-\sum_{\substack{n \leq x\\ 3 \mid n}}\Psi(n,y)-\cdots\\
&amp;\phantom{=}+\sum_{\substack{n \leq x\\ 2\cdot3 \mid n}}\Psi(n,y)+\sum_{\substack{n \leq x\\ 2\cdot5 \mid n}}\Psi(n,y)+\sum_{\substack{n \leq x\\ 2\cdot7 \mid n}}\Psi(n,y)+\cdots\\
&amp;\phantom{=}+(-1)^{\omega(P_y)}\sum_{\substack{n \leq x\\ P_y \mid n}}\Psi(n,y)\\
&amp;=\left(1-\frac{1}{2}-\cdots+\frac{1}{2\cdot3}+\cdots+\frac{(-1)^{\omega(P_y)}}{P_y}\right)\frac{x(\log x)^{\pi(y)}}{\pi(y)!\prod_{p \leq y}\log p}\\
&amp;\phantom{=}+O(x(\log x)^{\pi(y)-1})\\
&amp;=\left(\sum_{d \mid P_y}\frac{\mu(d)}{d}\right)\frac{x(\log x)^{\pi(y)}}{\pi(y)!\prod_{p \leq y}\log p}+O(x(\log x)^{\pi(y)-1})\\
&amp;=\frac{1}{\pi(y)!}\prod_{p \leq y}\left(\frac{p-1}{p\log p}\right)x(\log x)^{\pi(y)}+O(x(\log x)^{\pi(y)-1}),
\end{align*}
where we used \eqref{Ennola 2}, the fact that for every integer $k \geq 1$
\[\sum_{n \leq x}(\log n)^k=x(\log x)^k+O(x(\log x)^{k-1}),\]
and the fact that
\[\sum_{d \mid P_y}\frac{\mu(d)}{d}=\frac{\phi(P_y)}{P_y}=\prod_{p \leq y}\frac{p-1}{p}.\qedhere\]
\end{proof}

\begin{lemma}\label{somme binomes}For each integer $n \geq 1$,  we have
\[\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{n+k}=\frac{1}{n\binom{2n}{n}}.\]
\end{lemma}
\begin{proof}
We will show that, for every $n \geq 1$ and for all $x$ that do not cause a division by 0,
\begin{equation}\label{induction} 
\frac{n!}{x(x+1)(x+2)\cdots(x+n)}=\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{x+k}.
\end{equation}
It is easy to see that \eqref{induction} is true if $n=1$. Assume that \eqref{induction} is true for some $n\geq 1$. Then,
\begin{align*}
\frac{(n+1)!}{x(x+1)\cdots(x+n)(x+n+1)}&amp;=\left(\frac{n+1}{x+n+1}\right)\frac{n!}{x(x+1)(x+2)\cdots(x+n)}\\
&amp;=(n+1)\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{(x+n+1)(x+k)}.
\end{align*}
Using partial fractions, we find that
\[\frac{1}{(x+n+1)(x+k)}=\frac{1}{(n+1-k)(x+k)}-\frac{1}{(n+1-k)(x+n+1)}.\]
Since
\[\frac{n+1}{n+1-k}\binom{n}{k}=\binom{n+1}{k},\]
we have
\begin{align*}
&amp;\frac{(n+1)!}{x(x+1)\cdots(x+n)(x+n+1)}=(n+1)\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{(x+n+1)(x+k)}\\
&amp;=(n+1)\sum_{k=0}^n \binom{n}{k}(-1)^k\left(\frac{1}{(n+1-k)(x+k)}-\frac{1}{(n+1-k)(x+n+1)}\right)\\
&amp;=\sum_{k=0}^n \binom{n+1}{k}\frac{(-1)^k}{x+k}-\frac{1}{x+n+1}\sum_{k=0}^n \binom{n+1}{k}(-1)^k\\
&amp;=\sum_{k=0}^n \binom{n+1}{k}\frac{(-1)^k}{x+k}-\frac{1}{x+n+1}\left(\sum_{k=0}^{n+1} \binom{n+1}{k}(-1)^k-(-1)^{n+1}\right)\\
&amp;=\sum_{k=0}^n \binom{n+1}{k}\frac{(-1)^k}{x+k}+\frac{(-1)^{n+1}}{x+n+1}\\
&amp;=\sum_{k=0}^{n+1} \binom{n+1}{k}\frac{(-1)^k}{x+k},
\end{align*}
the simplification on the penultimate line because, by the binomial theorem,
\[\sum_{k=0}^{n+1} \binom{n+1}{k}(-1)^k=(1-1)^{n+1}=0.\]
Hence, by induction, \eqref{induction} is true for all $n \geq 1$. Setting $x=n$, we obtain the desired result.
\end{proof}

\begin{lemma}\label{lemme ordre sommation}
For all $2 \leq y \leq x$, we have
\[G(x,y)=\sum_{\substack{i \leq x\\ P(i)\leq y}}\left(x\frac{g(i)}{i}-g(i)\right)+O(\Psi(x,y)).\]
\end{lemma}
\begin{proof}
Using the fact that, for every $k$, $\gcd(i,n)=\gcd(i,n+ki)$, we easily find that
\[\sum_{n \leq x}(i,n)=x\frac{g(i)}{i}+O(1) \qquad (x\rightarrow \infty).\]
Thus, changing the order of summation gives
\begin{align*}
G(x,y)&amp;=\sum_{n \leq x}g(n,y)=\sum_{n \leq x}\sum_{\substack{i \leq n \\ P(i)\leq y}}(i,n)=\sum_{\substack{i \leq x\\ P(i)\leq y}}\sum_{i\leq n \leq x}(i,n)\\
&amp;=\sum_{\substack{i \leq x\\ P(i)\leq y}}\left(x\frac{g(i)}{i}-g(i)\right)+O(\Psi(x,y)).\qedhere
\end{align*}
\end{proof}


 In order to evaluate sums of multiplicative functions over friable numbers, we will require a result due to Tenenbaum and Wu \cite[Cor.\ 2.3]{tenenbaum+wu03}, which is a generalization of a theorem of Song \cite[Main Theorem]{song02}.
\begin{theorem}\label{Tenenbaum} 
Let $\epsilon &gt; 0$ and let $f(n)$ be an arithmetic function which satisfies the two conditions
\begin{equation}\label{condition 1} 
\sum_{p \leq x}f(p)\log p=\kappa x+O\left(x\exp(-(\log x)^{\frac{3}{5}-\epsilon})\right)
\end{equation}
\begin{equation}\label{condition 2} 
\sum_p \sum_{k \geq 2}\frac{f(p^k)}{p^{(1-\eta)k}}\leq A,
\end{equation}
for some constants $\kappa &gt;0, 0&lt;\eta&lt;\frac{1}{2}, A&gt;0$. Then
\[\sum_{\substack{n \leq x \\ P(n)\leq y}}f(n)=C_{\kappa}(f)x\rho_{\kappa}(u)(\log y)^{\kappa-1}\left(1+O\left(\frac{\log(u+1)}{\log y}+\frac{1}{(\log y)^{\kappa}}\right)\right),\]
uniformly for
\[\exp((\log\log x)^{\frac{5}{3}+\epsilon})\leq y \leq x,\]
where
\[C_{\kappa}(f):=\prod_p \left(1-\frac{1}{p}\right)^{\kappa}\sum_{j=0}^{\infty}\frac{f(p^j)}{p^j},\]
and $\rho_{\kappa}(u)$ is the continuous solution to the delayed differential equation
\begin{equation}\label{rho_kappa def}
\begin{cases}
\rho_{\kappa}(u)=u^{\kappa-1}/\Gamma(\kappa), &amp; \text{if } 0 \leq u \leq 1;\\ 
u\rho_{\kappa}'(u)+(1-\kappa)\rho_{\kappa}(u)+\kappa\rho_{\kappa}(u-1)=0, &amp; \text{if } u&gt;1.\\ 
\end{cases}
\end{equation}
\end{theorem}

 In our case, it is the function $\rho_2(u)$ that will be of interest. We will need the following lemma that summarizes some basic properties of $\rho_{\kappa}(u)$.
\begin{lemma}\label{lemme rho_kappa}
Let $\rho_{\kappa}(u)$ be defined as in \eqref{rho_kappa def}, for some $\kappa&gt;1$. Then
\begin{enumerate}
\item[(i)] As $u \rightarrow \infty$,
\[\rho_{\kappa}'(u)\asymp \rho_{\kappa}(u)\log(u+1).\]
\item[(ii)] As $x \rightarrow \infty$,
\[\int_1^x \rho_{\kappa}\left(\frac{\log t}{\log y}\right) dt \ll x\rho_{\kappa}(u).\]
\end{enumerate}
\end{lemma}
\begin{proof}
For (i), see Tenenbaum--Wu \cite{tenenbaum+wu03}; for (ii), see Song \cite[Lem.\ 8]{song02}.
\end{proof}


 Finally, we will need the following sharp form of the prime number theorem; see for example Ellison \cite[Thm.\ 11.3, p.\ 425]{ellison+mendes75}.

\begin{theorem}[Prime Number Theorem]\label{Prime Number Theorem}Let $\vartheta(x):=\sum_{p \leq x}\log p$, and $\epsilon &gt;0$. Then
\[\vartheta(x)=x+O(x\exp(-(\log x)^{\frac{3}{5}-\epsilon})) \qquad (x \rightarrow \infty).\]
\end{theorem}

\section{Proof of Theorem \ref{resultat majeur 1}}

 Let
\[C=C(y):=\frac{1}{\pi(y)!}\prod_{p \leq y}\left(\frac{p-1}{p\log p}\right).\]
We have, by Lemma \ref{lemme Phi},
\begin{align*}
G(x,y)&amp;=\sum_{\substack{n \leq x \\ P(n)\leq y}}n\Phi\left(\frac{x}{n},y\right)\\
&amp;=\sum_{\substack{n \leq x\\ P(n)\leq y}}n\left(C\frac{x}{n}\left(\log\left(\frac{x}{n}\right)\right)^{\pi(y)}+O\left(\frac{x}{n}\left(\log\left(\frac{x}{n}\right)\right)^{\pi(y)-1}\right)\right)\\
&amp;=Cx\sum_{\substack{n \leq x\\ P(n)\leq y}}(\log x-\log n)^{\pi(y)}+O(x(\log x)^{\pi(y)-1}\Psi(x,y)).
\end{align*}
Therefore, by the binomial theorem,
\begin{align}\label{expression intermediaire}
G(x,y)&amp;=Cx\sum_{\substack{n \leq x\\ P(n)\leq y}}\left(\sum_{j=0}^{\pi(y)}\binom{\pi(y)}{j}(\log x)^{\pi(y)-j} (-1)^j (\log n)^j \right)\nonumber\\ 
&amp;\quad +O(x(\log x)^{2\pi(y)-1})\nonumber \\ 
&amp;=Cx(\log x)^{\pi(y)}\sum_{j=0}^{\pi(y)}\binom{\pi(y)}{j}(\log x)^{-j}(-1)^j\Bigg(\sum_{\substack{n \leq x\\ P(n)\leq y}}(\log n)^j \Bigg)\nonumber\\
&amp;\quad+O(x(\log x)^{2\pi(y)-1}). 
\end{align}
However, for each $j \geq 0$,
\begin{align*}
\sum_{\substack{n \leq x\\ P(n)\leq y}}(\log n)^j &amp;=\Psi(x,y)(\log x)^j-\int_1^x\frac{j\Psi(t,y)(\log t)^{j-1}}{t}dt\\
&amp;=\frac{(\log x)^{\pi(y)+j}}{\pi(y)!\prod_{p \leq y}\log p}-\int_1^x\frac{j(\log t)^{\pi(y)+j-1}}{t\pi(y)!\prod_{p \leq y}\log p}dt\\
&amp;\quad+O((\log x)^{\pi(y)+j-1})+\int_1^x O\left(\frac{(\log t)^{\pi(y)+j-2}}{t}\right)dt\\
&amp;=\frac{(\log x)^{\pi(y)+j}}{\pi(y)!\prod_{p \leq y}\log p}-\frac{j(\log x)^{\pi(y)+j}}{(\pi(y)+j)\pi(y)!\prod_{p \leq y}\log p}\\
&amp;\quad+O((\log x)^{\pi(y)+j-1})\\
&amp;=\frac{(\log x)^{\pi(y)+j}}{(\pi(y)+j)(\pi(y)-1)!\prod_{p \leq y}\log p}+O((\log x)^{\pi(y)+j-1}).
\end{align*}
Substituting this in \eqref{expression intermediaire} yields
\begin{align*}
&amp;G(x,y)\\
&amp;=Cx(\log x)^{\pi(y)}\sum_{j=0}^{\pi(y)}\binom{\pi(y)}{j}(-\log x)^{-j}\left(\frac{(\log x)^{\pi(y)+j}}{(\pi(y)+j)(\pi(y)-1)!\prod_{p \leq y}\log p}\right)\\
&amp;\quad+O(x(\log x)^{2\pi(y)-1})\\
&amp;=Dx(\log x)^{2\pi(y)}+O(x(\log x)^{2\pi(y)-1}),
\end{align*}
where
\[D=D(y):=\frac{C}{(\pi(y)-1)!\prod_{p \leq y}\log p}\sum_{j=0}^{\pi(y)}\binom{\pi(y)}{j}\frac{(-1)^j}{\pi(y)+j}.\]
But by Lemma \ref{somme binomes}, we have
\begin{align*}
D&amp;=\frac{1}{\pi(y)!(\pi(y)-1)!}\left(\prod_{p \leq y}\frac{p-1}{p\log^2p}\right)\frac{1}{\pi(y)\binom{2\pi(y)}{\pi(y)}}\\
&amp;=\left(\prod_{p \leq y}\frac{p-1}{p\log^2p}\right)\frac{1}{(\pi(y)!)^2\binom{2\pi(y)}{\pi(y)}}=\left(\prod_{p \leq y}\frac{p-1}{p\log^2p}\right)\frac{1}{(2\pi(y))!}.
\end{align*}
Substituting $C$ and $D$ by their appropriate expressions gives the required estimate for $G(x,y)$.
\qed

\section{Proof of Theorem \ref{resultat majeur 2}}

 Using Lemma \ref{g puissances premiers}, we have that
\begin{align*}\sum_{p \leq x}\frac{g(p)}{p}\log p &amp;=\sum_{p \leq x}\left(2-\frac{1}{p}\right)\log p=2\sum_{p \leq x}\log p-\sum_{p \leq x}\frac{\log p}{p}\\
&amp;=2x+O(x\exp(-(\log x)^{\frac{3}{5}-\epsilon})),
\end{align*}
where we used Theorem \ref{Prime Number Theorem} and the fact that
\[\sum_{p \leq x}\frac{\log p}{p}\ll \log x \qquad (x \rightarrow \infty).\]
Hence, setting $f(n):=g(n)/n$, condition \eqref{condition 1} of Theorem \ref{Tenenbaum} is satisfied with $\kappa=2$. Also, we have
\begin{align*}
\sum_p \sum_{k \geq 2}\frac{f(p^k)}{p^{(1-\eta)k}}&amp;=\sum_p \sum_{k\geq 2}\frac{(k+1)p^k-kp^{k-1}}{p^{(2-\eta)k}}\leq \sum_p\sum_{k\geq 2}\frac{2kp^k}{p^{(2-\eta)k}}\\
&amp;=\sum_p \frac{4-2p^{\eta-1}}{(1-p^{1-\eta})^2}\leq\sum_p \frac{4}{(p^{1-\eta}-1)^2}&lt;\infty,
\end{align*}
for any $0&lt;\eta&lt;\frac{1}{2}$. It follows that condition \eqref{condition 2} of Theorem \ref{Tenenbaum} is also satisfied. Note that for this particular $f$,
\begin{align*}
C_{2}(f)&amp;=\prod_p\left(1-\frac{1}{p}\right)^{2}\sum_{j=0}^{\infty}\frac{g(p^j)}{p^{2j}}=\prod_p\left(\frac{p-1}{p}\right)^{2}\left(\frac{p^2}{(p-1)^2}-\frac{1}{(p-1)^2}\right)\\
&amp;=\prod_p\left(1-\frac{1}{p^2}\right)=\frac{1}{\zeta(2)}=\frac{6}{\pi^2}.
\end{align*}
This gives
\begin{equation}\label{somme g(n)/n}
\sum_{\substack{n \leq x\\ P(n)\leq y}}\frac{g(n)}{n}=\frac{6x\rho_2(u)\log y}{\pi^2}\left(1+O\left(\frac{\log (u+1)}{\log y}\right)\right).
\end{equation}
Using Abel's summation formula, we get
\begin{align}\label{somme g_y sur friables}
\begin{split}
\sum_{\substack{n \leq x\\ P(n)\leq y}}g(n)&amp;=\sum_{\substack{n \leq x\\ P(n)\leq y}}n\frac{g(n)}{n}=\frac{6x^2\rho_2(u)\log y}{\pi^2}\left(1+O\left(\frac{\log(u+1)}{\log y}\right)\right)\\
&amp;\quad-\int_1^x \frac{6t\rho_2\left(\frac{\log t}{\log y}\right)\log y}{\pi^2}dt\\
&amp;\quad+\int_1^x O\left(t\rho_2\left(\frac{\log t}{\log y}\right)\log\left(\frac{\log t}{\log y}+1\right)\right)dt.
\end{split}
\end{align}
Using integration by parts on the first integral gives
\begin{equation}\label{integrale 1}
\int_1^x \frac{6t\rho_2\left(\frac{\log t}{\log y}\right)\log y}{\pi^2}dt=\frac{3x^2 \rho_2(u)\log y}{\pi^2}-\int_1^x \frac{3t\rho_2'\left(\frac{\log t}{\log y}\right)}{\pi^2} dt.
\end{equation}
From Lemma \ref{lemme rho_kappa}(i), we can see that the remaining integral from \eqref{integrale 1} is of the same order as the second integral in \eqref{somme g_y sur friables}. But using Lemma \ref{lemme rho_kappa}(ii), we have that
\begin{align*}
\int_1^x t\rho_2\left(\frac{\log t}{\log y}\right)\log\left(\frac{\log t}{\log y}+1\right) dt &amp;\leq x\log(u+1)\int_1^x \rho_2\left(\frac{\log t}{\log y}\right) dt\\
&amp;\ll x^2\rho_2(u)\log(u+1),
\end{align*}
as $x \rightarrow \infty$. Gathering these estimates in \eqref{somme g_y sur friables}, we obtain
\begin{equation}\label{somme g_y sur friables 2}
\sum_{\substack{n \leq x\\ P(n)\leq y}}g(n)=\frac{3x^2\rho_2(u)\log y}{\pi^2}\left(1+O\left(\frac{\log(u+1)}{\log y}\right)\right). 
\end{equation}
Finally, using \eqref{somme g_y sur friables 2} and \eqref{somme g(n)/n} in Lemma \ref{lemme ordre sommation}, we obtain the desired estimate.
\qed

\section{Acknowledgments}
This project was supervised by Jean-Marie De Koninck, to whom I am grateful for his advice and interest.


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\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11N25.

\noindent \emph{Keywords: } 
friable number, largest prime factor, summatory function, Dickman function.

\bigskip
\hrule
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\noindent (Concerned with sequence
\seqnum{A018804}.)

\bigskip
\hrule
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\vspace*{+.1in}
\noindent
Received September 14 2015;
revised version received  February 8 2016.
Published in {\it Journal of Integer Sequences}, April 5 2016.

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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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