Asymptotic formula for the multiplicative function $\frac{d(n)}{k^{\omega(n)}}$

Meselem Karras Faculty of Science and Technology Djilali Bounaama Khemis Miliana University, 44225, Algeria karras.m@hotmail.fr

Abstract

For a fixed integer $k$ , we define the multiplicative function

D_{k,\omega}(n):=\frac{d(n)}{k^{\omega(n)}},

where $d(n)$ is the divisor function and $\omega(n)$ is the number of distinct prime divisors of $n$ . The main purpose of this paper is the study of the mean value of the function $D_{k,\omega}(n)$ by using elementary methods.

keywords:

Divisor function, number of distinct prime divisors, mean value.

\msc

11N37, 11A25, 11N36 \VOLUME31 \YEAR2023 \NUMBER1 \DOIhttps://doi.org/10.46298/cm.10104 {paper}

1 Introduction

Let $k\geq 2$ be a fixed integer. We recall that $d(n):=\sum_{d\mid n}1$ is the number of divisors of $n$ , and $\omega(n):=\sum_{p\mid n}1$ is the number of distinct prime divisors of $n$ . We define the function $D_{k,\omega}(n)$ by

D_{k,\omega}(n):=\frac{d(n)}{k^{\omega(n)}}.

(1)

Notice that for every fixed integer $k\geq 2$ , the function $D_{k,\omega}(n)$ is multiplicative and for every prime number $p$ and every integer $m$ the relation

D_{k,\omega}(p^{m})=\frac{m+1}{k},

(2)

holds. By using (2), we get

D_{k,\omega}(n)=\prod\limits_{p^{m}\|n}\frac{m+1}{k}

where $p^{m}\|n$ means $p^{m}\mid$ $n$ and $p^{m+1}\nmid n$ . In the particular case $k=2$ , the function $D_{2,\omega}(n)$ is exactly $D(n)=\dfrac{d(n)}{d^{\ast}(n)},(\text{see~{}\cite[cite]{[\@@bibref{}{D.K}{}{}]}})$ . For $k\geq 3$ , we can easily check that

\sum_{n\leq x}D_{k},_{\omega}(n)\ll_{k}x(\log x)^{2/k-1}.

(3)

Indeed, for any integer $n$ , we have $D_{k},_{\omega}(n)\leq d(n)\ll_{\varepsilon}n^{\varepsilon}$ . Furthermore, the hypotheses of Shiu’s theorem are satisfied; see Theorem $1$ in [Shui] and [Nar-Ten, p.1]. One gets

\sum_{n\leq x}D_{k},_{\omega}(n)\ll_{k}\frac{x}{\log x}\exp\Bigl{(}\sum_{p\leq x}\frac{2}{kp}\Bigr{)}.

Now, by using Lemma $4.63$ in [O.BORD], it follows that

\sum_{n\leq x}D_{k},_{\omega}(n)\ll_{k}\frac{x}{\log x}\exp\Bigl{(}\frac{2}{k}\log(2e^{\gamma}\log x)\Bigr{)}\ll_{k}x(\log x)^{2/k-1}.

2 Main result

In this section, we establish two results concerning the mean value of the function $D_{k,\omega}(n)$ . We begin by giving a weaker result.

Theorem 2.1.

Let $k\geq 2$ be a fixed integer. For all $x\geq 1$ large enough, we have

\sum_{n\leq x}D_{k,\omega}(n)=\frac{x(\log x)^{2/k-1}}{\Gamma(2/k)}\prod\limits_{p}\Bigl{(}1-\frac{1}{p}\Bigr{)}^{2/k}\Bigl{(}1+\frac{2p-1}{kp(p-1)^{2}}\Bigr{)}+O(x(\log x)^{-1})(\log\log x)^{4/k}.

The proof of this result is based on Tulyaganov’s theorem; this theorem is summarized as follows:

Theorem 2.2.

Let $f$ be a complex valued multiplicative function. Suppose there exists $z\in\mathbb{C}$ , independent of $p$ , with $|z|\leq c_{1}$ and

a)

$\sum_{p\leq x}f(p)\log p=zx+O(xe^{-c_{2}\sqrt{\log x}})$
b)

$\sum_{p\leq x}|f(p)|\log p\ll x$
c)

$\sum_{p\leq x}\sum_{\alpha=2}^{\infty}\frac{|f(p^{\alpha})|\log p^{\alpha}}{p^{\alpha}}\ll(\log\log x)^{2}$
d)

$\sum_{p}\frac{|f(p)|^{2}\log p}{p^{2}}<c_{3}$

for some real numbers $c_{1}$ , $c_{2}$ and $c_{3}$ . Then, for all $x\geq 1$ sufficiently large, we have

	$\displaystyle\sum_{n\leq x}f(n)={}$	$\displaystyle{}\frac{x(\log x)^{z-1}}{\Gamma(z)}\prod\limits_{p}\Bigl{(}1-\frac{1}{p}\Bigr{)}^{z}\Bigl{(}1+\sum_{\alpha=1}^{\infty}\frac{f(p^{\alpha})}{p^{\alpha}}\Bigr{)}\Bigl{\{}1+O(\frac{(\log\log x)^{2}}{\log x})\Bigr{\}}$
		$\displaystyle{}+O(x(\log x)^{\max(0,\mathrm{Re}\,z-1)-1})(\log\log x)^{2(A-\max(0,\mathrm{Re}\,z-1))},$

where $A>0$ satisfies

\sum_{u<p\leq v}|f(p)|p^{-1}\leq A\log(\log v/\log u)+O(1).

Proof 2.3.

This theorem is a consequence of Theorem $4$ in [Tulyag], where we take $g=f$ .

To complete the demonstration of the main result we have the following lemmas.

Lemma 2.4.

For any fixed integer $k\geq 2$ , we have the estimate

\sum_{p\leq x}|D_{k,\omega}(p)|\log p\ll x.

Proof 2.5.

By Chebyshev’s estimates [Disar], we have

\sum_{p\leq x}|D_{k,\omega}(p)|\log p=\frac{2}{k}\sum_{p\leq x}\log p<\frac{2}{k}(1.000081x)\ll x.

Lemma 2.6.

For any fixed integer $k\geq 2$ , there is a constant $c>0$ , such that

\sum_{p\leq x}D_{k,\omega}(p)\log p=\frac{{\footnotesize 2}}{k}x+O(xe^{-c\sqrt{\log x}}).

Proof 2.7.

We have

\sum_{p\leq x}D_{k,\omega}(p)\log p=\frac{{\footnotesize 2}}{k}\sum_{p\leq x}\log p=\frac{{\footnotesize 2}}{k}\theta(x),

and by Theorem $6.9$ in [H.L.MandR.C.V], there is a constant $c>0$ such that

\theta(x)=x+O(xe^{-c\sqrt{\log x}}),

which implies the desired result.

Lemma 2.8.

For any fixed integer $k\geq 2$ , we have

\sum_{p}\frac{|D_{k,\omega}(p)|^{2}}{p^{2}}\log p<\infty.

Proof 2.9.

We first check the inequality $\sum\limits_{m=2}^{\infty}\dfrac{\log m}{m(m-1)}\leq\log 4$ , and using the following

\sum_{p}\frac{\log p}{p^{2}}<\sum_{m=2}^{\infty}\frac{\log m}{m^{2}}\leq\sum_{m=2}^{\infty}\frac{\log m}{m(m-1)},

then we have

\sum_{p}\frac{|D_{k,\omega}(p)|^{2}}{p^{2}}\log p=\frac{4}{k^{2}}\sum_{p}\frac{\log p}{p^{2}}<\frac{4\log 4}{k^{2}}.

Lemma 2.10.

For any fixed integer $k\geq 2$ , we have

\sum_{p\leq x}\sum_{\alpha=2}^{\infty}\frac{|D_{k,\omega}(p^{\alpha})|\log(p^{\alpha})}{p^{\alpha}}\leq\frac{28}{k}.

Proof 2.11.

For every integer $k\geq 3$ , we write

	$\displaystyle\sum_{p\leq x}\sum_{\alpha=2}^{\infty}\frac{\|D_{k,\omega}(p^{\alpha})\|\log(p^{\alpha})}{p^{\alpha}}$	$\displaystyle=\frac{1}{k}\sum_{p\leq x}\log p\sum_{\alpha=2}^{\infty}\frac{\alpha(\alpha+1)}{p^{\alpha}}$
		$\displaystyle=\frac{1}{k}\sum_{p\leq x}\frac{\log p}{p}\sum_{\alpha=2}^{\infty}\frac{\alpha(\alpha+1)}{p^{\alpha-1}},$

and the infinite series $\sum\limits_{\alpha=2}^{\infty}\dfrac{\alpha(\alpha+1)}{p^{\alpha-1}}$ converges to $\dfrac{2}{(1-1/p)^{3}}-2$ , since

	$\displaystyle\sum_{p\leq x}\sum_{\alpha=2}^{\infty}\frac{\|D_{k,\omega}(p^{\alpha})\|\log(p^{\alpha})}{p^{\alpha}}$	$\displaystyle=\frac{2}{k}\sum_{p\leq x}\frac{3p^{2}-3p+1}{p(p-1)^{3}}\log p$
		$\displaystyle\leq\frac{28}{k}\sum_{p\leq x}\frac{\log p}{p^{2}}.$

By Lemma $70.1$ in [HALL-TENENBAUM], we have $\sum\limits_{p}\dfrac{\log p}{p^{\alpha}}<\dfrac{1}{\alpha-1}$ for all $\alpha>1$ , consequently

\sum_{p\leq x}\sum_{\alpha=2}^{\infty}\frac{|D_{k,\omega}(p^{\alpha})|\log(p^{\alpha})}{p^{\alpha}}<\frac{28}{k}.

Finally, by Lemma 2.4, 2.6, 2.8 and 2.10 we have shown that the function $D_{k,\omega}(n)$ satisfies the conditions of Theorem 2.2. As we have

\sum_{u<p\leq v}\frac{|D_{k,\omega}(p)|}{p}=\frac{2}{k}\sum_{u<p\leq v}\frac{1}{p}\leq\frac{2}{k}\log\frac{\log v}{\log u}+O(1),

then the constant $A$ in Theorem $2$ is $\frac{2}{k}$ .

The next result is improved over the previous one.

Theorem 2.12.

Let $k\geq 2$ be a fixed integer. For all $x\geq 1$ large enough, we have

\sum_{n\leq x}D_{k,\omega}(n)=\frac{x(\log x)^{2/k-1}}{\Gamma(2/k)}\prod\limits_{p}\Bigl{(}1-\frac{1}{p}\Bigr{)}^{2/k}\Bigl{(}1+\frac{2p-1}{k(p-1)^{2}}\Bigr{)}+O_{k}(x(\log x)^{2/k-2}).

The demonstration is based on the following lemmas:

Lemma 2.13.

Let $k\geq 2$ be a fixed integer. For every $s:=\sigma+it\in\mathbb{C}$ such that $\sigma>1$ and $L(s,D_{k,\omega}(n)):=\sum\limits_{n=1}^{\infty}\dfrac{D_{k,\omega}(n)}{n^{s}}$ , we have

L(s,D_{k,\omega}(n))=\zeta(s)^{2/k}L(s,g_{k}),

or $L(s,g_{k})$ is a series of Dirichlet absolutely convergent in the half-plane $\sigma>\frac{1}{2}$ .

Proof 2.14.

If $\sigma>1$ , then

	$\displaystyle L(s,D_{k,\omega}(n))$	$\displaystyle=\prod\limits_{p}\Bigl{(}1+\sum\limits_{\alpha=1}^{\infty}\dfrac{D_{k,\omega}(p^{\alpha})}{p^{\alpha s}}\Bigr{)}$
		$\displaystyle=\prod\limits_{p}\Bigl{(}1+\sum\limits_{\alpha=1}^{\infty}\dfrac{\alpha+1}{kp^{\alpha s}}\Bigr{)}$
		$\displaystyle=\prod\limits_{p}\Bigl{(}1+\dfrac{2p^{s}-1}{k(p^{s}-1)^{2}}\Bigr{)},$

on the other hand we have

\Bigl{(}1+\frac{2p^{s}-1}{k(p^{s}-1)^{2}}\Bigr{)}=((1-p^{-s})^{-2/k})\Bigl{(}1+\frac{h(s)}{k(p^{s}-1)^{2}}\Bigr{)},

such that

h(s)=(1-p^{-s})^{2/k}(kp^{2s}-2(k-1)p^{s}+k-1)-k(p^{s}-1)^{2}.

Since

(1-p^{-s})^{2/k}=1-\frac{2}{kp^{s}}-\frac{k-2}{k^{2}p^{2s}}-O\Bigl{(}\frac{k}{p^{3\sigma}}\Bigr{)},

he comes

	$\displaystyle h(s)$	$\displaystyle=\Bigl{(}1-\frac{2}{kp^{s}}-\frac{k-2}{k^{2}p^{2s}}-O\bigl{(}\frac{k}{p^{3\sigma}}\bigr{)}\Bigr{)}(kp^{2s}-2(k-1)p^{s}+k-1)-k(p^{s}-1)^{2}$
		$\displaystyle=2\bigl{(}1-\frac{1}{k}\bigr{)}+O(p^{-\sigma}),$

which implies the announced result.

Lemma 2.15 ([Selb]).

Let $A>0$ . Uniformly for $x\geq 2$ and $z\in\mathbb{C}$ such that $|z|\leq A$ , we have

\sum_{n\leq x}\tau_{z}(n)=\frac{x(\log x)^{z-1}}{\Gamma(z)}+O_{A}(x(\log x)^{\mathrm{Re}\,z-2}).

$\tau_{z}(n)$ is the multiplicative function defined by $\tau_{z}(p^{\alpha})=\binom{z+\alpha-1}{\alpha}$ .

Proof 2.16 (Proof of Theorem 3).

According to the Lemma 2.13, we have $D_{k,\omega}=\tau_{2/k}\ast g_{k}$ . Then, by Lemma 2.15

	$\displaystyle\sum_{n\leq x}D_{k,\omega}(n)$	$\displaystyle=\sum_{d\leq x}g_{k}(d)\sum_{m\leq\frac{x}{d}}\tau_{2/k}(m)$
		$\displaystyle=\sum_{d\leq x}g_{k}(d)\Bigl{(}\frac{x(\log\frac{x}{d})^{2/k-1}}{d\Gamma(2/k)}+O_{k}\bigl{(}\frac{x}{d}(\log\frac{x}{d})^{2/k-2}\bigr{)}\Bigr{)}$
		$\displaystyle=\sum_{d\leq x}g_{k}(d)\Bigl{(}\frac{x(\log x)^{2/k-1}}{d\Gamma(2/k)}+O_{k}((\log x)^{2/k-2}\log d)$
		$\displaystyle\qquad+O_{k}\bigl{(}\frac{x}{d}(\log\frac{x}{d})^{2/k-2}\bigr{)}\Bigr{)}$
		$\displaystyle=\frac{x(\log x)^{2/k-1}}{\Gamma(2/k)}\sum_{d\leq x}\frac{g_{k}(d)}{d}+O_{k}\Bigl{(}x(\log x)^{2/k-2}\sum_{d\leq x}\frac{\|g_{k}(d)\|(1+\log d)}{d}\Bigr{)}.$

The series $L(s,g_{k})$ is absolutely convergent on the half-plane $\sigma>\frac{1}{2}$ , then for all $\varepsilon>0$

\sum_{d\leq x}|g_{k}(d)|\ll_{k,\varepsilon}x^{1/2+\varepsilon},

hence by partial summation

\sum_{d\leq x}\frac{|g_{k}(d)|(1+\log d)}{d}\ll_{k,\varepsilon}x^{-1/2+\varepsilon}

and therefore

\sum_{n\leq x}D_{k,\omega}(n)=L(1,g_{k})\frac{x(\log x)^{2/k-1}}{\Gamma(2/k)}+O_{k}(x(\log x)^{2/k-2})+O_{k,\omega}(x^{1/2+\varepsilon}).

Which completes the demonstration.

Acknowledgments

The author would like to sincerely thank Professor Olivier Bordellès for his help and interest in this work and Professor Karl Dilcher for his generosity in reviewing this paper.

References

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29 September, 201928 January, 2020Karl Dilcher

Asymptotic formula for the multiplicative function d​(n)kω​(n)\frac{d(n)}{k^{\omega(n)}}