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\begin{document}
\vspace*{-40pt}
\centerline{\smalltt INTEGERS: \smallrm
ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 7
(2007), \#A13}
\vskip 40pt
\begin{center}
{\bf ON THE AVERAGE ORDERS OF A CLASS OF DIVISOR FUNCTIONS} \vskip
20pt
{\bf Roger Woodford\footnote{The author is operating under an NSERC CGS D research grant}}\\
{\smallit Department of Mathematics, University of British Columbia, Vancouver, British Columbia V6T 1Z2, Canada}\\
{\tt rogerw@math.ubc.ca}\\ \vskip 10pt
\end{center}
\vskip 30pt \centerline{\smallit Received: 8/14/06,
Revised: 1/23/07, Accepted: 3/11/07,
Published: 3/20/07}
\vskip 30pt

\centerline{\bf Abstract}

\noindent We compute the average orders and study the distribution
of values of a class of divisor functions defined by symmetric
polynomials on the multi-set of prime factors of a number. These
generalize those we have previously defined.
The
simplest case of these functions is the sum of prime factors with
repetition function, whose average order has been computed in
various ways by Alladi and Erd\H{o}s, %~\cite{kA77}
 LeVan,%~\cite{mJ70}
and Kerawala.%~\cite{sK69}.

\pagestyle{myheadings} \markright{\smalltt INTEGERS: \smallrm
ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 7 (2007),
\#A13\hfill}

\thispagestyle{empty} \baselineskip=15pt \vskip 30pt












\section*{\normalsize 1. Introduction}

Let us begin by consolidating some notation. Let $\mathbb{N}_0$
denote the set of nonnegative integers. In~\cite{rW04} we considered
the functions $s_k$ defined as follows:

\ni
{\bf Definition 1} Let $k\in\mathbb{N}_0$. Define $s_{k}:
\mathbb{N}_0 \rightarrow \mathbb{N}_0$ as follows: if $n=0$, then
$s_k(0)=0$, for all $k$. For $n>0$, if $k = 0$, $s_{k}(n) = 1$. If
$k
> 0$, and $n = p_{1}\cdots p_{r}$, where $r = \Omega(n)$ is the
number of prime factors (with multiplicity) of $n$, then
\begin{equation*}
s_{k}(n) = \sum p_{i_{1}}\cdots p_{i_{k}},
\end{equation*}
where the sum is taken over all products of $k$ prime factors from
the multi-set $\{p_{1}, \ldots , p_{r}\}$.

These functions are mere special cases of a larger class of
functions:

\ni
{\bf Definition 2} Let $k, \ell\in\mathbb{N}_0$. We define
$s_{k,\ell}(0)=0$. If $n=p_1\cdots p_r\in\mathbb{N}$, where the
$p_i$ are primes, not necessarily distinct, then
\begin{equation*}
s_{k,\ell}(n) = \sum_{1\le i_1<\ldots<i_k\le k}(p_{i_1}\cdots
p_{i_k})^{\ell}.
\end{equation*}
We write the function $s_{1,k}$ as $e_k$. The function
$s_{0,k}\equiv1$.

The special case when $k=\ell=1$, i.e. the sum of prime factors with
repetition function, has been studied in other contexts. Denote this
function by $s$. Lal~\cite{mL68} observes that the sequence of
iterates $n, s(n), s(s(n)),\ldots$, for $n\ge5$ always terminates in
a prime number $p\ge5$. Based on empirical evidence, he conjectures
that corresponding to a fixed such prime, the set
\begin{equation*}
\{n\in\mathbb{N}: s^{(i)}(n) = p \text{ for some $i\ge 0$}\}
\end{equation*}
has a positive asymptotic density. Here we are writing $s^{(i)}(n)$
for the $i$th iterate of $n$ by $s$.

Alladi and Erd\H{o}s~\cite{kA77} show that $s(n)$ is uniformly
distributed modulo 2 by proving that
\begin{equation*}
\sum_{n\le x}(-1)^{s(n)} = o(x).
\end{equation*}
Since $s(n)\equiv e_k(n) \text{ (mod 2)}$ for any $k\in\mathbb{N}$,
this result clearly extends to $e_k(n)$.

Gupta~\cite{hG70} shows that for a given $m$,
\begin{equation*}
\max{\{ n\in\mathbb{N}: s(n)=m \}} = t\cdot3^{\lfloor m/3\rfloor},
\end{equation*}
where $t = 1, 4/3$, or 2; according as $m\equiv 0, 1$, or 2 (mod 3).

The size of the set $e_k^{(-1)}(n)$ is equal to the number of
partitions of $n$ into $k$th powers of primes,
$p_{\mathbb{P}^{(k)}}(n)$. Bateman and Erd\H{o}s~\cite{pB55} have
shown that the number of partitions into primes $p_{\mathbb{P}}(n)$
is strictly increasing, and Hardy and Ramanujan~\cite{gH16} first
demonstrated the asymptotic formula
\begin{equation*}
\log{p_{\mathbb{P}^{(k)}}(n)} \sim
(k+1)\left[\Gamma\lt\frac{1}{k}+2\rt\zeta\lt\frac{1}{k}+1\rt\right]^{k/(k+1)}\left[\frac{n}{\log^k{n}}\right]^{1/(k+1)}.
\end{equation*}

From this it is easily seen that the functions $e_k(n)$ attain all
sufficiently high values, for any $k\in\mathbb{N}$. We have
shown~\cite{rW04} that $s_2(n)$ does as well.

In Section 3 of the present paper, we shall prove the asymptotic
formula
\begin{equation}\label{e:1}
\sum_{n\le x}s_{k,\ell}(n) \sim
\frac{\zeta(\ell+1)x^{\ell+1}(\log{\log{x}})^{k-1}}{(\ell+1)(k-1)!\log{x}},
\end{equation}
for $k, \ell\ge1$ fixed, with a precise error term included. Section
2 is devoted to $e_k(n)$, and, in addition to proving~\eqref{e:1} in
this case, we study the distribution of values of these functions.

\vskip 30pt









\section*{\normalsize 2. Statistical Results for $e_k$}



\subsection*{\normalsize 2.1 The Average Order of $e_k$}

\ni
{\bf Theorem 2} For $k\in\mathbb{N}$,
$ \displaystyle \sum_{n\le
x}e_k(n)=\frac{\zeta(k+1)x^{k+1}}{(k+1)\log{x}} +
O\lt\frac{x^{k+1}\log{\log{x}}}{\log^2{x}}\rt.
$

The average order of $s$ itself is studied by Alladi and
Erd\H{o}s~\cite{kA77} by comparing it to the average order of the
largest prime factor dividing $n$. The asymptotic in Theorem 2
(without the error term included) has been advanced by
Kerawala~\cite{sK69} and LeVan~\cite{mJ70}. We shall make precise
LeVan's sketch of the proof. First we require a lemma.

\ni
{\bf Lemma 1}\label{le:4} For $x\ge 2$, and $k, \ell\in\mathbb{N}_0$
we have
$\displaystyle \sum_{p\le x}\frac{p^k}{\log^{\ell}{p}} =
\frac{x^{k+1}}{(k+1)\log^{\ell+1}{x}} +
O\lt\frac{x^{k+1}}{\log^{\ell+2}{x}} \rt.
$

The proof of Lemma 1 follows from a simple application of
Riemann-Stieltjes integration and integration by parts.


\ni
{\it Proof of Theorem 2.} We have
\begin{align}\sum_{n\le x}e_k(n) &= \sum_{p\le
x}\sum_{i=1}^{\infty}p^k\left\lfloor\frac{x}{p^i}\right\rfloor%\notag\\
= \sum_{p\le x}p^k\left\lfloor\frac{x}{p}\right\rfloor +
\sum_{i=2}^{\infty}\sum_{p\le x^{1/i}}p^k
\left\lfloor\frac{x}{p^i}\right\rfloor.\label{ee:5}
\end{align}

As we shall see, the first term contributes the greater portion to
the sum:
\begin{align}\sum_{p\le x}p^k\left\lfloor\frac{x}{p}\right\rfloor
&= \sum_{i\le x/2}\sum_{\frac{x}{i+1}<p\le\frac{x}{i}}ip^k%\notag\\
= \sum_{i\le x/2}\sum_{p\le x/i}p^k\notag\\ &= \sum_{i\le x/2}\lt
\frac{(x/i)^{k+1}}{(k+1)\log{(x/i)}} + O\lt
\frac{(x/i)^{k+1}}{\log^2{(x/i)}} \rt \rt.\label{ee:1}
\end{align}

Let
$ \displaystyle \Sigma_1=\sum_{i\le \log^2{x}}
\frac{1}{i^{k+1}\log{(x/i)}}
$
and
$ \displaystyle\Sigma_2=\sum_{\log^2{x}<i\le x/2}
\frac{1}{i^{k+1}\log{(x/i)}}\,\,\,
$
so that

\begin{align*}\sum_{i\le x/2}
\frac{1}{i^{k+1}\log{(x/i)}} &=  \Sigma_1 + \Sigma_2.
\end{align*}

We have that
\begin{align*}\Sigma_1 &\ge \frac{1}{\log{x}}\sum_{i\le
\log^2{x}}\frac{1}{i^{k+1}}%\\ &
= \frac{1}{\log{x}}\lt \zeta(k+1) -
\sum_{i>\log^2{x}}\frac{1}{i^{k+1}}\rt\\ &= \frac{1}{\log{x}}\lt
\zeta(k+1) + O\lt\frac{1}{\log^{2k}{x}}\rt\rt%\\ &
=
\frac{\zeta(k+1)}{\log{x}} + O\lt\frac{1}{\log^{2k+1}{x}}\rt.
\end{align*}

On the other hand,
\begin{align*}\Sigma_1 &\le \sum_{i\le
\log^2{x}}\frac{1}{i^{k+1}(\log{x}-2\log{\log{x}})}%\\
%&
=\frac{1}{\log{x}}\lt 1 +
O\lt\frac{\log{\log{x}}}{\log{x}}\rt\rt\sum_{i\le\log^2{x}}\frac{1}{i^{k+1}}\\
&= \frac{1}{\log{x}}\lt 1 + O\lt\frac{\log{\log{x}}}{\log{x}}\rt\rt
\lt\zeta(k+1) + O\lt \frac{1}{\log^{2k}{x}} \rt \rt\\
&= \frac{\zeta(k+1)}{\log{x}} +
O\lt\frac{\log{\log{x}}}{\log^2{x}}\rt.
\end{align*}
Combining these results we have that
\begin{equation}\Sigma_1=\frac{\zeta(k+1)}{\log{x}} +
O\lt\frac{\log{\log{x}}}{\log^2{x}}\rt.\label{ee:3}
\end{equation}

The sum $\Sigma_2$ is negligible by comparison:
\begin{align}\Sigma_2
=\sum_{\log^2{x}<i\le x/2} \frac{1}{i^{k+1}\log{(x/i)}}%\notag\\
\ll\sum_{i>\log^2{x}} \frac{1}{i^{k+1}}%\notag\\
\ll\int_{\log^2{x}}^{\infty}\frac{1}{t^{k+1}}\,dt%\notag\\
\ll\frac{1}{\log^{2k}{x}}.\label{ee:2}
\end{align}

Now we need to bound the error term in~\eqref{ee:1}:

\begin{align}\sum_{i\le x/2}\frac{1}{i^{k+1}\log^2{(x/i)}}
&= O\lt \int_1^{x/2}\frac{dt}{t^2\log^2{x/t}} \rt\notag\\
&= O\lt\int_1^{\sqrt{x}}\frac{dt}{t^2\log^2{x/t}} +
\int_{\sqrt{x}}^{x/2}\frac{dt}{t^2\log^2{x/t}}\rt\notag\\
&= O\lt\frac{4}{\log^2{x}}\int_1^{\sqrt{x}}\frac{dt}{t^2} +
\int_{\sqrt{x}}^{x/2}\frac{dt}{t^2}\rt\notag\\
&=O\lt\frac{1}{\log^2{x}}\rt\label{ee:4}
\end{align}

Hence by~\eqref{ee:3},~\eqref{ee:2}, and~\eqref{ee:4}, we have that
\begin{equation*}\sum_{p\le
x}p^k\left\lfloor\frac{x}{p}\right\rfloor =
\frac{\zeta(k+1)x^{k+1}}{(k+1)\log{x}} +
O\lt\frac{x^{k+1}\log{\log{x}}}{\log^2{x}}\rt.
\end{equation*}

To conclude the proof, we need to bound the second term
in~\eqref{ee:5}:
\begin{align*}\sum_{i=2}^{\infty}\sum_{p\le x^{1/i}}p^k
\left\lfloor\frac{x}{p^i}\right\rfloor \le
x\sum_{i=2}^{\infty}\sum_{p\le x^{1/i}}\frac{p^k}{p^i}
\le x\sum_{p\le\sqrt{x}}\frac{p^{k-1}}{p-1}
&\le 2x\sum_{p\le\sqrt{x}}p^{k-2}\\
&= \begin{cases}O(x\log{\log{x}}), \text{ if $k=1$};\\
O\lt \frac{x^{\frac{k+1}{2}}}{\log{x}} \rt, \text{ if $k>1$}.
\end{cases}
\end{align*}

Note that for the $k=1$ case,
$ \displaystyle
x\log{\log{x}} = O\lt\frac{x^2\log{\log{x}}}{\log^2{x}}\rt,
$
and for the $k>1$ case,
$ \displaystyle
\frac{x^{\frac{k+1}{2}}}{\log{x}} =
O\lt\frac{x^{k+1}\log{\log{x}}}{\log^2{x}}\rt.
$
%Hence, these error terms may be absorbed to yield the theorem.
\hfill
$\square$








\subsection*{\normalsize 2.2 The Distribution of Values of $e_k$}

In this section we shall relate $e_k(n)$ to the largest prime factor
dividing $n$, which we denote by $P(n)$. The following trivial
identity turns out to be rather useful in deriving some statistical
properties for $e_k$:
\begin{equation}\label{E:boundP(n)}P(n)^k\leq e_k(n) \leq P(n)^k\Omega(n).
\end{equation}

\ni
{\bf Definition 3} Let
\begin{equation}
b_k(x,y) = \#\{n\leq x : e_k(n)\leq y\},
\end{equation}
\begin{equation}
\Psi(x,y)=\#\{n\leq x : P(n)\leq y\}.
\end{equation}
So $b_k(x,y)$ is the number of partitions into $k$-th powers of
primes whose sum is less than or equal to $y$, and whose product of
parts is less than or equal to $x$. We wish to relate the two
quantities defined above. To do so, we state the following
well-known fact concerning the function $\Omega(n)$, which counts
the number of prime factors with repetition of $n$. The proof is by
analogy with~\cite{hI04}, p.30.
\begin{equation}
\#\{n\leq x: |\Omega(n)-\log\log{n}|> (\log\log{n})^{3/4}\} =
O\left(\frac{x}{\sqrt{\log\log{x}}}\right).\label{r:stddevOm}
\end{equation}

\ni
{\bf Theorem 3}\label{t:relb_kPsi_k} There is an absolute 
constant $c_1>0$ such that the following inequalities hold:
\begin{equation}
\Psi\left(x,\left(\frac{y}{\log\log{x}+(\log\log{x})^{3/4}}\right)^{1/k}\right)
- \frac{c_1x}{\sqrt{\log\log{x}}} \leq b_k(x,y) \leq \Psi(x,y^{1/k}).
\end{equation}


\ni
{\it Proof.} By equation~\eqref{E:boundP(n)}, we have that
\begin{equation*}
\#\{n\leq x : P(n)^k\Omega(n)\leq y \} \leq b_k(x,y) \leq \#\{n\leq
x : P(n)^k\leq y \}.
\end{equation*}
This implies the second inequality. By ~\eqref{r:stddevOm}, there is
an absolute positive constant $c_1$ such that
$ \displaystyle
\#\{n\leq x:|\Omega(n)-\log\log{n}| \leq (\log\log{n})^{3/4}\} \geq
x\left(1-\frac{c_1}{\sqrt{\log\log{x}}}\right).
$
Given this, we have the following chain of inequalities:
\begin{align*}\#\{n\leq x :
P(n)^k\Omega(n)\leq y \} \geq &\#\{n\leq x : P(n)^k\Omega(n)\leq
y,\text{ and }\\ &\Omega(n) \leq \log\log{n} + (\log\log{n})^{3/4} \}\\
\geq &\#[\{n\leq x : P(n)^k(\log\log{n} + (\log\log{n})^{3/4})\leq
y\}\\ &\cap \{n\leq x : \Omega(n) \leq \log\log{n} +
(\log\log{n})^{3/4} \}]\\ \geq &\#\{n\leq x : P(n)^k(\log\log{x} +
(\log\log{x})^{3/4})\leq y\}\\&+ \#\{n\leq x:|\Omega(n)-\log\log{n}|
\leq (\log\log{n})^{3/4}\} - x\\ \geq &
\Psi\left(x,\left(\frac{y}{\log\log{x}+(\log\log{x})^{3/4}}\right)^{1/k}\right)
- \frac{c_1x}{\sqrt{\log\log{x}}},
\end{align*}
and the theorem is proved. \hfill
$\square$

The ``Dickman function" $\rho(u)$ is defined to be the unique
continuous solution to the differential-difference equation
\begin{equation*}
u\rho'(u)=-\rho(u-1)\text{ }\text{ } (u>1),
\end{equation*}
satisfying the initial condition
$
\rho(u)=1\text{ }\text{ } (0\leq u \leq 1).
$
The Dickman function is nonnegative for $u>0$, and decreasing for
$u>1.$ This definition and description is taken from~\cite{aH93}, an
extensive survey of work done on the function $\Psi(x,y)$.

It is also true that $\rho(u)$ is convex on $(1,\infty)$. By the
functional equation, it is apparent that
$
\lim_{u\rightarrow1^+}\rho'(u) = -1.
$
It follows that
\begin{equation}\label{e:rho(a+b)}\rho(a+b)\ge\rho(a)-b,\text{ for } a\ge 1,\text{
}b\ge 0.
\end{equation}

De Bruijin~\cite{nB51} proved that
\begin{equation}\label{e:Psias}\Psi(x,y) = x\rho(u)\left[ 1 + O\left( \frac{\log{(u+1)}}{\log{y}} \right)
\right]
\end{equation}
holds uniformly for $u=\log{x}/\log{y}$ in the range
\begin{equation}y\ge 2,\text{ } 1\le u \le (\log{y})^{3/5-\epsilon}.
\end{equation}
This has since been improved by Hildebrand~\cite{aH86} to the range
\begin{equation}\label{e:rangePsivalid}y\ge 2,\text{ } 1\le u \le \exp{\left((\log{y})^{3/5-\epsilon}\right)}.
\end{equation}

For the next theorem, we are interested in the range
$y=x^{\alpha/k}$, and hence $u=k/\alpha$.

\ni
{\bf Theorem 4} Let $0<\alpha<1$, and let $k\in\mathbb{N}$ be fixed
constants. Then
\begin{equation*}
b_k(x,x^{\alpha}) \sim \Psi(x,x^{\alpha/k}) \sim
\rho\left(\frac{k}{\alpha}\right)x.
\end{equation*}
In fact, we have
\begin{equation*}
b_k(x,x^{\alpha}) = \rho\left(\frac{k}{\alpha}\right)x +
O\lt\frac{x}{\sqrt{\log\log{x}}}\rt.
\end{equation*}

\ni
{\it Proof.} Note that for $x\ge2^{k/\alpha}$, we are within the
range~\eqref{e:rangePsivalid}. Consequently~\eqref{e:Psias} applies:
\begin{equation*} \Psi(x,x^{\alpha/k}) =
x\rho\left(\frac{k}{\alpha}\right)\left[ 1 + O\left(
\frac{1}{\log{x}} \right) \right].
\end{equation*}
Furthermore, the second inequality of Theorem 3 gives us
$
b_k(x,x^{\alpha}) \leq \Psi(x,x^{\alpha/k}).
$
Similarly, the first inequality from Theorem 3 yields:
\begin{equation*} \Psi\left(x,\left(\frac{x^{\alpha}}{2\log\log{x}}\right)^{1/k}\right)
- \frac{c_1x}{\sqrt{\log\log{x}}} \leq b_k(x,x^{\alpha}),
\end{equation*}
since $\Psi(x,y)$ is increasing in $y$.

For $x$ sufficiently large,~\eqref{e:Psias} gives
\begin{align*}
\Psi\left(x,\left(\frac{x^{\alpha}}{2\log\log{x}}\right)^{1/k}\right)
&= x\rho\left( \frac{\log{x}}{(\alpha/k)\log{x} -
(1/k)\log(2\log\log{x})} \right)%\\ &
%\text{ }\text{ }\text{
%}\times
\left[ 1 + O\left( \frac{\log{(u+1)}}{\log{y}} \right)
\right],
\end{align*}
where, in this instance,
\begin{equation*}
y=\left(\frac{x^{\alpha}}{2\log\log{x}}\right)^{1/k}
\end{equation*}
and
\begin{align*}u &= \frac{\log{x}}{\log{y}}= \frac{\log{x}}{(\alpha/k)\log{x}
- (1/k)\log(2\log\log{x})}\\ &=
\left(\frac{k}{\alpha}\right)\frac{1}{1 -
(1/\alpha)\log(2\log\log{x})/\log{x}}.
\end{align*}
The identity
$
\frac{1}{1-a} < 1+2a
$
holds for $0<a<1/2$. So if $x$ is chosen sufficiently large such
that
$ \displaystyle
0<\frac{\log(2\log\log{x})}{\alpha\log{x}}<\frac{1}{2},
$
then
$ \displaystyle u< \frac{k}{\alpha} +
\frac{2k\log(2\log\log{x})}{\alpha^2\log{x}}.
$
Hence, since $\rho(u)$ decreases on $(1,\infty)$, we have
\begin{align*}
\Psi\left(x,\left(\frac{x^{\alpha}}{2\log\log{x}}\right)^{1/k}\right)
&\ge x\rho\left( \frac{k}{\alpha} +
\frac{2k\log(2\log\log{x})}{\alpha^2\log{x}}
\right)\left[ 1 + O\left( \frac{1}{\log{x}} \right) \right]\\
&= x \rho\left( \frac{k}{\alpha} +
\frac{2k\log(2\log\log{x})}{\alpha^2\log{x}} \right) +
O\left(\frac{x}{\log{x}}\right)\\ &\ge x \rho\left(
\frac{k}{\alpha}\right) -
\frac{2kx\log(2\log\log{x})}{\alpha^2\log{x}} +
O\left(\frac{x}{\log{x}}\right)\\ &= x\rho\left(
\frac{k}{\alpha}\right) +
O\left(\frac{x\log\log\log{x}}{\log{x}}\right),
\end{align*}
using equation~\eqref{e:rho(a+b)}. Combining this information with
Theorem 3 we obtain the following inequalities:
\begin{align*}x \rho\left(\frac{k}{\alpha}\right) \!+\!
O\left(\frac{x}{\sqrt{\log\log{x}}}\right) \!\le
\!\Psi\left(x,\left(\frac{x^{\alpha}}{2\log\log{x}}\right)^{1/k}\right)
&\!\le b_k(x,x^{\alpha})\\ &\!\le \Psi(x,x^{\alpha/k}) =
x\rho\left(\frac{k}{\alpha}\right) \!+\!
O\left(\frac{x}{\log{x}}\right),
\end{align*}
which proves the theorem. \hfill
$\square$






\vskip 20pt














\section*{\normalsize 3. Average Order of $s_{k,\ell}$}

We now devote our attention to equation~\eqref{e:1}. We have already
proved the $k=1$ case, so henceforth we shall assume that $k\ge2$.

\ni
{\bf Theorem 5}\label{t:sa:1} The average order of $s_{k,\ell}(n)$
has the following expression:
\begin{align*}
\sum_{n\le x}s_{k,\ell}(n) =
&\sum_{r=1}^k\sum_{p_1<\ldots<p_r}\sum_{\substack{i_1+\cdots +i_r=k\\
i_1,\ldots,i_r>0}} (p_1^{i_1}\cdots p_r^{i_r})^{\ell}\times\\ &\lt
\sum_{j_1,\ldots,j_r=1}^{\infty}\binom{j_1-1}{i_1-1}\cdots\binom{j_r-1}{i_r-1}\left\lfloor
\frac{x}{p_1^{j_1}\cdots p_r^{j_r}}\right\rfloor \rt.
\end{align*}

\ni
{\it Proof.} The terms in the sum $\sum_{n\le x}s_{k,\ell}(n)$ are
products of $k$ $\ell$-th powers of primes, not necessarily
distinct. In other words, an arbitrary term is of the form
$(p_1^{i_1}\cdots p_r^{i_r})^{\ell}$, where $r\le k$,
$p_1<\ldots<p_r$, $i_1,\ldots,i_r>0$, and $i_1+\cdots +i_r=k$. Fix
$(p_1^{i_1}\cdots p_r^{i_r})^{\ell}$. We shall count the number of
times this expression occurs in the sum.

By the inclusion-exclusion principle, the number of $n\le x$ such
that $p_1^{j_1},\ldots,p_r^{j_r}||n$ is
\begin{align*}&\left\lfloor\frac{x}{p_1^{j_1}\cdots
p_r^{j_r}}\right\rfloor%\\
%&
- \lt \left\lfloor\frac{x}{p_1^{j_1+1}p_2^{j_2}\cdots
p_r^{j_r}}\right\rfloor +
\left\lfloor\frac{x}{p_1^{j_1}p_2^{j_2+1}\cdots
p_r^{j_r}}\right\rfloor + \cdots +
\left\lfloor\frac{x}{p_1^{j_1}p_2^{j_2}\cdots
p_r^{j_r+1}}\right\rfloor\rt\\
&\hskip 66pt +\cdots + (-1)^r\left\lfloor\frac{x}{p_1^{j_1+1}\cdots
p_r^{j_r+1}}\right\rfloor,
\end{align*}
which we write as $\beta(j_1,\ldots,j_r)$. Each such $n$ contributes
$\binom{j_1}{i_1}\cdots\binom{j_r}{i_r}$ copies of $(p_1^{i_1}\cdots
p_r^{i_r})^{\ell}$ to the sum $\sum_{n\le x}s_k(n)$. Thus
$(p_1^{i_1}\cdots p_r^{i_r})^{\ell}$ occurs
\begin{equation*}
\sum_{j_1,\ldots,j_r=1}^{\infty}\binom{j_1}{i_1}\cdots\binom{j_r}{i_r}\beta(j_1,\ldots,j_r)
\end{equation*}
times. We make the following claim:
\begin{align}\label{e:sa:1}
\sum_{j_1,\ldots,j_r=1}^{\infty}\binom{j_1}{i_1}\cdots\binom{j_r}{i_r}
\beta(j_1,\ldots,j_r)
=
\sum_{j_1,\ldots,j_r=1}^{\infty}\binom{j_1-1}{i_1-1}\cdots\binom{j_r-1}{i_r-1}\left\lfloor\frac{x}{p_1^{j_1}\cdots
p_r^{j_r}}\right\rfloor.
\end{align}
To prove~\eqref{e:sa:1}, first note that
$\left\lfloor\frac{x}{p_1^{j_1}\cdots p_r^{j_r}}\right\rfloor$
occurs
\begin{align*}
\binom{j_1}{i_1}\cdots\binom{j_r}{i_r} &- \lt
\binom{j_1-1}{i_1}\binom{j_2}{i_2}\cdots\binom{j_r}{i_r} +\cdots +
\binom{j_1}{i_1}\binom{j_2}{i_2}\cdots\binom{j_r-1}{i_r} \rt\\
&+\cdots +
(-1)^r\binom{j_1-1}{i_1}\binom{j_2-1}{i_2}\cdots\binom{j_r-1}{i_r}
\end{align*}
times in the left hand side. But an induction on $r$, with the
identity
\begin{equation}\label{e:sa:3}
\binom{j}{i} - \binom{j-1}{i}= \binom{j-1}{i-1}
\end{equation}
being the $r=1$ case, gives us 
\begin{align}
&\binom{j_1}{i_1}\cdots\binom{j_r}{i_r} - \lt
\binom{j_1-1}{i_1}\binom{j_2}{i_2}\cdots\binom{j_r}{i_r} +\cdots +
\binom{j_1}{i_1}\binom{j_2}{i_2}\cdots\binom{j_r-1}{i_r} \rt\notag\\
&+\cdots +
(-1)^r\binom{j_1-1}{i_1}\binom{j_2-1}{i_2}\cdots\binom{j_r-1}{i_r} =
\binom{j_1-1}{i_1-1}\cdots\binom{j_r-1}{i_r-1}.\label{e:sa:2}
\end{align}
Indeed, suppose that~\eqref{e:sa:2} holds for $r-1$. Denote the
left-hand side of~\eqref{e:sa:2} by $C_r$. We need to show that
$\displaystyle
C_r = \binom{j_1-1}{i_1-1}\cdots\binom{j_r-1}{i_r-1}.
$
But factoring, the identity~\eqref{e:sa:3}, and the induction
hypothesis give us that
\begin{align*}
C_r = \binom{j_r}{i_r}C_{r-1} - \binom{j_r-1}{i_r}C_{r-1}
= C_{r-1}\binom{j_r-1}{i_r-1}
=
\binom{j_1-1}{i_1-1}\cdots\binom{j_{r-1}-1}{i_{r-1}-1}\binom{j_r-1}{i_r-1}.
\end{align*}
Thus, the claim~\eqref{e:sa:1} is proved. The theorem follows by
summing over all values of $r$ from 1 to $k$, all possible
$r$-tuples of primes, and for each such $r$-tuple, all $r$-tuples
$(i_1,\ldots,i_r)$ satisfying $i_j>0$ for $j=1,\ldots,r$, and
$i_1+\cdots+i_r=k$. \hfill $\square$


We will first investigate the part of the sum in Theorem 5
corresponding to $r=k$, and hence $i_1=\ldots = i_k = 1$. To do so,
we require some generalizations of the prime number theorem. These
are taken from Nathanson~\cite{mN00}, pp.313-319, however we also
include precise error terms.

Let
$
\pi_k(x)=\#\{n\le x:\Omega(n)=\omega(n)=k\}\text{ and }
\pi_k^*(x)=\#\{n\le x:\Omega(n)=k\}.
$
That is, $\pi_k(x)$ counts the number of $n\le x$ that are products
of exactly $k$ distinct prime factors, and $\pi_k^*(x)$ counts the
number of $n\le x$ which have $k$ prime factors with repetition.
Note that $\pi_1(x)=\pi_1^*(x)=\pi(x)$. For $k\ge2$, we have that
\begin{equation}
\pi_k(x)=\frac{x(\log{\log{x}})^{k-1}}{(k-1)!\log{x}} +
O\lt\frac{x(\log{\log{x}})^{k-2}}{\log{x}}\rt,
\end{equation}
and
\begin{equation}\label{e:sa:10}0\le\pi_k^*(x)-\pi_k(x)\ll\frac{x(\log{\log{x}})^{k-2}}{\log{x}}.
\end{equation}

The first result can be used to prove the next lemma via
Riemann-Stieltjes integration.

\ni
{\bf Lemma 6}\label{l:sa:1} Let $u\ge 0$, and $k\ge2$. Then
\begin{equation*}
\sum_{\substack{n\le x \\ \omega(n)=\Omega(n)=k}}n^u =
\frac{x^{u+1}(\log{\log{x}})^{k-1}}{(u+1)(k-1)!\log{x}} +
O\lt\frac{x^{u+1}(\log{\log{x}})^{k-2}}{\log{x}}\rt.
\end{equation*}





For $r=k$, we have the following:
\begin{align}\label{e:sa:4}\sum_{p_1<\ldots<p_k}(p_1\cdots
p_k)^{\ell}&\sum_{j_1,\ldots,j_k=1}^{\infty}\left\lfloor
\frac{x}{p_1^{j_1}\cdots p_k^{j_k}}\right\rfloor =
\sum_{p_1<\ldots<p_k}(p_1\cdots p_k)^{\ell}\left\lfloor
\frac{x}{p_1\cdots p_k}\right\rfloor\notag\\ &+
\sum_{p_1<\ldots<p_k}(p_1\cdots
p_k)^{\ell}\sum_{\substack{j_1,\ldots,j_k=1\\
j_1\cdots j_k>1}}^{\infty}\left\lfloor \frac{x}{p_1^{j_1}\cdots
p_k^{j_k}}\right\rfloor.
\end{align}
We further focus by looking at the first term on the right-hand side
of~\eqref{e:sa:4}, that is, the term corresponding to $j_1=\ldots =
j_k = 1$. Making use of Lemma 6, we have
\begin{align}
\sum_{p_1<\ldots<p_k}(p_1\cdots p_k)^{\ell}\left\lfloor
\frac{x}{p_1\cdots p_k}\right\rfloor =& \sum_{m\le
x/2^k}\sum_{\substack{p_1<\ldots< p_k \\ \frac{x}{m+1}< p_1\cdots
p_k\le \frac{x}{m}}}m(p_1\cdots p_k)^{\ell}\notag\\
=&\sum_{m\le x/2^k}\sum_{\substack{p_1<\ldots < p_k \\
p_1\cdots p_k\le x/m}}(p_1\cdots p_k)^{\ell}\notag\\
=& \frac{x^{\ell+1}}{(\ell+1)(k-1)!}\sum_{m\le
x/2^k}\frac{(\log{\log{(x/m)}})^{k-1}}{m^{\ell+1}\log{(x/m)}}\notag\\
&+ O\lt x^{\ell+1}\sum_{m\le
x/2^k}\frac{(\log{\log{(x/m)}})^{k-2}}{m^{\ell+1}\log{(x/m)}}\rt.\label{e:sa:8}
\end{align}
Now
\begin{align} \!\!\!
\sum_{m\le
x/2^k}\frac{(\log{\log{(x/m)}})^{k-1}}{m^{\ell+1}\log{(x/m)}} = \!\!
\sum_{m\le
\log^2{x}}\frac{(\log{\log{(x/m)}})^{k-1}}{m^{\ell+1}\log{(x/m)}}
+ \!\!\sum_{\log^2{x}<m\le
x/2^k}\frac{(\log{\log{(x/m)}})^{k-1}}{m^{\ell+1}\log{(x/m)}}.\label{e:sa:7}
\end{align}
For $m\in[1,\log^2{x}]$,
\begin{align*}
\frac{(\log{\log{(x/m)}})^{k-1}}{m^{\ell+1}\log{(x/m)}} \le
\frac{(\log{\log{x}})^{k-1}}{m^{\ell+1}(\log{x}-2\log{\log{x}})}
=
\frac{(\log{\log{x}})^{k-1}}{m^{\ell+1}\log{x}}\lt1+O\lt\frac{\log{\log{x}}}{\log{x}}\rt\rt,
\end{align*}
which implies that
\begin{align}
\sum_{m\le
\log^2{x}}\frac{(\log{\log{(x/m)}})^{k-1}}{m^{\ell+1}\log{(x/m)}}
\le&
\frac{(\log{\log{x}})^{k-1}}{\log{x}}\lt1+O\lt\frac{\log{\log{x}}}{\log{x}}\rt\rt
\sum_{m\le\log^2{x}}\frac{1}{m^{\ell+1}}\notag\\
=&
\frac{(\log{\log{x}})^{k-1}}{\log{x}}\lt1+O\lt\frac{\log{\log{x}}}{\log{x}}\rt\rt\notag\\
&\times \zeta(\ell+1)\lt1+O\lt\frac{1}{\log^{2\ell}{x}}\rt\rt\notag\\
=& \frac{\zeta(\ell+1)(\log{\log{x}})^{k-1}}{\log{x}} +
O\lt\frac{(\log{\log{x}})^k}{\log^2{x}}\rt.\label{e:sa:5}
\end{align}

On the other hand, for $m\in[1,\log^2{x}]$ we have
\begin{align*}
\frac{(\log{\log{(x/m)}})^{k-1}}{m^{\ell+1}\log{(x/m)}} \ge
\frac{(\log{(\log{x}-2\log{\log{x}})})^{k-1}}{m^{\ell+1}\log{x}}
&= \frac{\lt\log{\log{x}} + \log{\lt 1-
\frac{2\log{\log{x}}}{\log{x}}\rt}\rt^{k-1}}{m^{\ell+1}\log{x}}\\
&= \frac{\lt\log{\log{x}} +
O\lt\frac{\log{\log{x}}}{\log{x}}\rt\rt^{k-1}}{m^{\ell+1}\log{x}}\\
&= \frac{(\log{\log{x}})^{k-1} +
O\lt\frac{(\log{\log{x}})^{k-1}}{\log{x}}\rt}{m^{\ell+1}\log{x}},
\end{align*}
and so
\begin{align}
\sum_{m\le
\log^2{x}}\frac{(\log{\log{(x/m)}})^{k-1}}{m^{\ell+1}\log{(x/m)}}
&\ge \frac{(\log{\log{x}})^{k-1}}{\log{x}}\lt\zeta(\ell+1) +
O\lt\frac{1}{\log^{2\ell}{x}}\rt\rt%\notag\\
+ O\lt\frac{(\log{\log{x}})^{k-1}}{\log^2{x}}\rt\notag\\
&= \frac{\zeta(\ell+1)(\log{\log{x}})^{k-1}}{\log{x}} +
O\lt\frac{(\log{\log{x}})^{k-1}}{\log^2{x}}\rt\label{e:sa:6}.
\end{align}
Combining~\eqref{e:sa:5} and~\eqref{e:sa:6} we have that
\begin{equation}
\sum_{m\le
\log^2{x}}\frac{(\log{\log{(x/m)}})^{k-1}}{m^{\ell+1}\log{(x/m)}} =
\frac{\zeta(\ell+1)(\log{\log{x}})^{k-1}}{\log{x}} +
O\lt\frac{(\log{\log{x}})^k}{\log^2{x}}\rt.
\end{equation}

Now we must bound the second term on the right-hand side
of~\eqref{e:sa:7}.
\begin{align*}
\sum_{\log^2{x}<m\le
x/2^k}\frac{(\log{\log{(x/m)}})^{k-1}}{m^{\ell+1}\log{(x/m)}} \ll
\sum_{\log^2{x}<m\le
x/2^k}\frac{(\log{\log{x}})^{k-1}}{m^{\ell+1}}
\ll \frac{(\log{\log{x}})^{k-1}}{\log^2{x}}.
\end{align*}

A similar argument shows that
\begin{equation}
\sum_{m\le
x/2^k}\frac{(\log{\log{(x/m)}})^{k-2}}{m^{\ell+1}\log{(x/m)}} \ll
\frac{(\log{\log{x}})^{k-2}}{\log{x}}.
\end{equation}
This we use to bound the error term in~\eqref{e:sa:8}.

Applying this information to~\eqref{e:sa:8} we have that
\begin{align}
\sum_{p_1<\ldots<p_k}(p_1\cdots p_k)^{\ell}\left\lfloor
\frac{x}{p_1\cdots p_k}\right\rfloor =&
\frac{x^{\ell+1}}{(\ell+1)(k-1)!}\times\notag\\
&\lt\frac{\zeta(\ell+1)(\log{\log{x}})^{k-1}}{\log{x}}
+ O\lt\frac{(\log{\log{x}})^k}{\log^2{x}}\rt \rt\notag\\
&+O\lt\frac{x^{\ell+1}(\log{\log{x}})^{k-2}}{\log{x}}\rt\notag\\
=&
\frac{\zeta(\ell+1)x^{\ell+1}(\log{\log{x}})^{k-1}}{(\ell+1)(k-1)!
\log{x}}
+
O\lt\frac{x^{\ell+1}(\log{\log{x}})^{k-2}}{\log{x}}\rt.\label{e:sa:12}
\end{align}

This is the main term in $\sum_{n\le x}s_{k,\ell}(n)$. To complete
the computation of the sum, we need only bound all that remains. We
will first complete the case when $r=k$, by bounding the second term
in the right-hand side of~\eqref{e:sa:4}:
\begin{align}
\sum_{p_1<\ldots<p_k}(p_1\cdots
p_k)^{\ell}&\sum_{\substack{j_1,\ldots,j_k=1\\
j_1\cdots j_k>1}}^{\infty}\left\lfloor \frac{x}{p_1^{j_1}\cdots
p_k^{j_k}}\right\rfloor\notag\\
&\ll x\sum_{\substack{p_1<\ldots<p_k
\\ p_1^2p_2\cdots p_k\le x}}(p_1\cdots p_k)^{\ell}
\sum_{\substack{j_1,\ldots,j_k=1\\
j_1\cdots j_k>1}}^{\infty}\frac{1}{p_1^{j_1}\cdots p_k^{j_k}}\notag\\
&\ll x\sum_{\substack{p_1<\ldots<p_k
\\ p_1^2p_2\cdots p_k\le x}}(p_1\cdots p_k)^{\ell}
\lt \frac{1}{(p_1-1)\cdots(p_k-1)} - \frac{1}{p_1\cdots p_k}\rt\notag\\
&\ll x\sum_{\substack{p_1<\ldots<p_k
\\ p_1^2p_2\cdots p_k\le x}}
\frac{(p_1\cdots p_k)^{\ell}}{p_1^2p_2\cdots p_k}\notag\\
&\ll x\sum_{p\le x^{\frac{1}{k+1}}}\lt
p^{\ell-2}\sum_{\substack{n\le x/p
\\ \omega(n)=\Omega(n)=k-1}}n^{\ell-1}\rt\notag\\
&\ll x\sum_{p\le x^{\frac{1}{k+1}}}\lt
p^{\ell-2}\frac{(x/p)^{\ell}(\log{\log{(x/p)}})^{k-2}}{\log{(x/p)}}\rt\notag\\
&=x^{\ell+1}\sum_{p\le x^{\frac{1}{k+1}}}
\frac{(\log{\log{(x/p)}})^{k-2}}{p^2\log{(x/p)}}\notag\\
&\ll \frac{x^{\ell+1}(\log{\log{x}})^{k-2}}{\log{x}}.\label{e:sa:13}
\end{align}

Let us now bound the terms of Theorem 5 corresponding to $r<k$. We
require the following power series identity:
\begin{equation*}
\sum_{n=j}^{\infty}\binom{n-1}{j-1}x^n=\lt\frac{x}{1-x}\rt^j,
\end{equation*}
which holds for $|x|<1$. We have
\begin{align}
\sum_{r=1}^{k-1}\sum_{p_1<\ldots<p_r}\sum_{\substack{i_1+\cdots +i_r=k\\
i_1,\ldots,i_r>0}} &(p_1^{i_1}\cdots p_r^{i_r})^{\ell}%\times\notag\\
\lt
\sum_{j_1,\ldots,j_r=1}^{\infty}\binom{j_1-1}{i_1-1}\cdots\binom{j_r-1}{i_r-1}\left\lfloor
\frac{x}{p_1^{j_1}\cdots p_r^{j_r}}\right\rfloor \rt\notag\\
&\ll x\sum_{r=1}^{k-1}\sum_{\substack{i_1+\cdots +i_r=k\\
i_1,\ldots,i_r>0}}\sum_{\substack{p_1<\ldots<p_r\\ p_1^{i_1}\cdots
p_r^{i_r}\le x }} (p_1^{i_1}\cdots
p_r^{i_r})^{\ell}\prod_{m=1}^r\sum_{j_m=i_m}^{\infty}\binom{j_m-1}{i_m-1}\frac{1}{p^{j_m}}\notag\\
&= x\sum_{r=1}^{k-1}\sum_{\substack{i_1+\cdots +i_r=k\\
i_1,\ldots,i_r>0}}\sum_{\substack{p_1<\ldots<p_r\\ p_1^{i_1}\cdots
p_r^{i_r}\le x }} (p_1^{i_1}\cdots
p_r^{i_r})^{\ell}\prod_{m=1}^r\lt\frac{1}{p_m-1}\rt^{i_m}\notag\\
&\ll x\sum_{r=1}^{k-1}\sum_{\substack{i_1+\cdots +i_r=k\\
i_1,\ldots,i_r>0}}\sum_{\substack{p_1<\ldots<p_r\\ p_1^{i_1}\cdots
p_r^{i_r}\le x }}
(p_1^{i_1}\cdots p_r^{i_r})^{\ell-1}\notag\\
&= x\sum_{\substack{n\le x\\ \omega(n)<k=\Omega(n)}}n^{\ell-1}\notag\\
&= x\int_{2^-}^{x}t^{\ell-1}\,d(\pi_k^*(t)-\pi_k(t)).\label{e:sa:9}
\end{align}
Applying integration by parts to~\eqref{e:sa:9}, and using the
bound~\eqref{e:sa:10}, we have that
\begin{align}
\sum_{r=1}^{k-1}\sum_{p_1<\ldots<p_r}\sum_{\substack{i_1+\cdots +i_r=k\\
i_1,\ldots,i_r>0}} (p_1^{i_1}\cdots p_r^{i_r})^{\ell}%\times\notag\\
&\times \lt
\sum_{j_1,\ldots,j_r=1}^{\infty}\binom{j_1-1}{i_1-1}
\cdots\binom{j_r-1}{i_r-1}\left\lfloor
\frac{x}{p_1^{j_1}\cdots p_r^{j_r}}\right\rfloor \rt\notag\\
&\ll \frac{x^{\ell+1}(\log{\log{x}})^{k-2}}{\log{x}}.\label{e:sa:11}
\end{align}

Combining Theorem 5 with~\eqref{e:sa:12}, ~\eqref{e:sa:13},
and~\eqref{e:sa:11}, we have proved the following theorem:

\ni
{\bf Theorem 7}\label{t:sa:2} Let $k\ge2$, and let $\ell\ge1$. Then
\begin{equation*}
\sum_{n\le x}s_{k,\ell}(n) =
\frac{\zeta(\ell+1)x^{\ell+1}(\log{\log{x}})^{k-1}}{(\ell+1)(k-1)!\log{x}}
+ O\lt\frac{x^{\ell+1}(\log{\log{x}})^{k-2}}{\log{x}}\rt.
\end{equation*}


\newpage

\noindent{\bf Acknowledgements} I would like to thank my supervisor,
Dr. Izabella Laba, for her guidance, as well as Dr. Greg Martin, for
invaluable knowledge and insights.


\vskip 30pt





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%%%%\section*{\normalsize Acknowledgments}




\end{document}