A generalization on the average ratio of the smallest and largest prime divisor of $n$

TL;DR

This paper extends Erd"os and van Lint's 1982 work by estimating the average of the positive integer powers of the ratio between the smallest and largest prime divisors of integers, using C.H. Jia's method.

Contribution

It generalizes previous results by providing estimates for higher powers of the ratio, offering a broader understanding of prime divisor distributions.

Findings

Derived estimates for the average of powers of the ratio of prime divisors

Applied C.H. Jia's method to extend previous bounds

Enhanced understanding of prime divisor ratios in integers

Abstract

In 1982, Erd\"os and van Lint showed an estimate for the average of the ratio of the smallest and largest prime divisor of $n$. In this note, we apply C.H. Jia's method to give an estimate for the average of positive integer power of the ratio.

Equations

$$S(x)=\sum _{n⩽x}\frac{p(n)}{P(n)}.$$

$$S(x)=\frac{x}{\log x}+\frac{3x}{{\log }^{2}x}+o\left(\frac{x}{(\log x{)}^2}\right).$$

$$S(x)=\frac{x}{\log x}+\frac{3x}{{\log }^{2}x}+\frac{15x}{{\log }^{3}x}+o\left(\frac{x}{(\log x{)}^3}\right).$$

$$S_{\lambda ,\alpha }(x)=\sum _{n⩽x}\lambda (\omega (n)){\left(\frac{p(n)}{P(n)}\right)}^{\alpha }.$$

$$S_{\lambda ,\alpha }(x)=(\lambda (1)+O(1))\pi (x)$$

$$\begin{array}{rl}{S}_{\lambda ,\alpha }(x)& =\lambda (1)\frac{x}{\log x}+\left(\frac{2}{\alpha }\lambda (2)+\lambda (1)\right)\frac{x}{{\log }^{2}x}\\ & +\left(\frac{9}{{\alpha }^2}\lambda (3)+\frac{4}{\alpha }\lambda (2)+2\lambda (1)\right)\frac{x}{{\log }^{3}x}+O\left(\frac{x}{(\log x{)}^4}\right).\end{array}$$

$$\pi (x)=\sum _{p⩽x}1={\int }_2^x\frac{dt}{\log t}+O\left(\frac{x}{(\log x{)}^A}\right).$$

$$\sum _{y<p⩽x}f(p)={\int }_y^x\frac{f(t)}{\log t}dt+O\left(\frac{x^{c+1}}{(\log x{)}^A}\right),c>-1,$$

$$\sum _{y<p⩽x}f(p)={\int }_y^x\frac{f(t)}{\log t}dt+O\left(\frac{y^{c+1}}{(\log y{)}^A}\right),c<-1.$$

$$y=\exp \left(\frac{\log x}{\log \log x}\right),$$

$$\Psi (x,y)=O\left(\frac{x}{(\log x{)}^A}\right).$$

$$S_{\lambda ,\alpha }(x)=\sum _{i=1}^{\infty }\lambda (i){\Sigma }^{(i)}+O\left(\frac{x}{(\log x{)}^A}\right).$$

$$\sum _{n⩽x,\mu (n)=0}{\left(\frac{p(n)}{P(n)}\right)}^{\alpha }$$

$$\displaystyle=O\left(\frac{x}{(\log x)^{A}}\right)+O\Big{(}\sum_{\begin{smallmatrix}n\leqslant x,\mu(n)=0\\ P(n)<p(n)(\log x)^{\frac{A}{\alpha}}\end{smallmatrix}}1\,\,\Big{)}$$

$$=O\left(\frac{x}{(\log x{)}^A}\right)+O\left(\frac{x}{(\log x{)}^{\frac{A}{\alpha }}}\right)=O\left(\frac{x}{(\log x{)}^A}\right).$$

$$S_{\lambda ,\alpha }(x)=\sum _{2⩽n⩽x,\mu (n)\ne 0}\lambda (\omega (n)){\left(\frac{p(n)}{P(n)}\right)}^{\alpha }+O\left(\frac{x}{(\log x{)}^A}\right)$$

$${\Sigma }^{(1)}=\pi (x)=\frac{x}{\log x}+\frac{x}{{\log }^{2}x}+\frac{2x}{{\log }^{3}x}+O\left(\frac{x}{{\log }^{4}x}\right).$$

$${\Sigma }^{(2)}=\sum _{\begin{array}{c}{p}_1{p}_2⩽x\\ p_1<p_2\end{array}}\frac{p_1^{\alpha }}{p_2^{\alpha }}$$

$$=I_1+I_2+O\left(\frac{x}{{\log }^{4}x}\right)$$

$$\sum _{p_1<p_2}{p}_1^{\alpha }={\int }_2^{p_2}\frac{t^{\alpha }}{\log t}dt+O\left(\frac{p_2^{\alpha +1}}{(\log p_2{)}^3}\right).$$

$$I_1$$

$$=\frac{2}{\alpha +1}\frac{x}{{\log }^{2}x}+\frac{4\alpha +8}{(\alpha +1{)}^2}\frac{x}{{\log }^{3}x}+O\left(\frac{x}{{\log }^{4}x}\right)$$

$$I_2$$

$$\sum _{\sqrt{x}<p_2⩽\sqrt{x}(\log x{)}^{\frac{4}{\alpha }}}\frac{1}{p_2^{2\alpha +1}\log \frac{x}{p_2}}$$

$$=\frac{1}{x^{\alpha }}{\int }_1^{(\log x{)}^{\frac{4}{\alpha }}}\frac{dt}{t^{2\alpha +1}\log \frac{\sqrt{x}}{t}\log (\sqrt{x}t)}+O\left(\frac{1}{x^{\alpha }(\log x{)}^4}\right).$$

$${\int }_1^{(\log x{)}^{\frac{4}{\alpha }}}\frac{dt}{t^{2\alpha +1}\log \frac{\sqrt{x}}{t}\log (\sqrt{x}t)}$$

$$={\left[-\frac{1}{2\alpha }\frac{1}{t^{2\alpha }\log \frac{\sqrt{x}}{t}\log (\sqrt{x}t)}\right]}_1^{(\log x{)}^{\frac{4}{\alpha }}}+\frac{1}{\alpha }{\int }_1^{(\log x{)}^{\frac{4}{\alpha }}}\frac{\log t}{t^{2\alpha +1}{\log }^2\frac{\sqrt{x}}{t}{\log }^2(\sqrt{x}t)}dt$$

$$=\frac{2}{\alpha }\frac{1}{{\log }^{2}x}+O\left(\frac{1}{(\log x{)}^4}\right).$$

$$\sum _{\sqrt{x}<p_2⩽\sqrt{x}(\log x{)}^{\frac{4}{\alpha }}}\frac{1}{p_2^{2\alpha +1}\log \frac{x}{p_2}}=\frac{2}{\alpha }\frac{1}{x^{\alpha }{\log }^{2}x}+O\left(\frac{1}{x^{\alpha }(\log x{)}^4}\right).$$

$$\sum _{\sqrt{x}<p_2⩽\sqrt{x}(\log x{)}^{\frac{4}{\alpha }}}\frac{1}{p_2^{2\alpha +1}(\log \frac{x}{p_2}{)}^2}$$

$$I_2=\frac{2}{\alpha (\alpha +1)}\frac{x}{{\log }^{2}x}+\frac{4}{\alpha (\alpha +1{)}^2}\frac{x}{{\log }^{3}x}+O\left(\frac{x}{(\log x{)}^4}\right)$$

$${\Sigma }^{(2)}=\frac{2}{\alpha }\frac{x}{{\log }^{2}x}+\frac{4}{\alpha }\frac{x}{{\log }^{3}x}+O\left(\frac{x}{{\log }^{4}x}\right).$$

$${\Sigma }^{(3)}$$

$$\sum _{x^{\frac{1}{3}}<p_3⩽x^{\frac{1}{3}}(\log x{)}^{\frac{4}{\alpha }}}\frac{1}{p_3^{\alpha }}\sum _{\sqrt{\frac{x}{p_3}}<p_2<p_3}\sum _{p_1⩽\frac{x}{p_2{p}_3}}{p}_1^{\alpha }+O\left(\frac{x}{{\log }^{4}x}\right)$$

Full text

1 Introduction

Let $n>1$ be an integer. Denote by $p(n)$ the smallest prime divisor of $n$ and $P(n)$ the largest prime divisor of $n$. Let $S(x)$ be the average of the ratio of the smallest and largest prime divisor of $n$:

[TABLE]

In 1982, Erd$\ddot{\text{o}}$s and van Lint [erdos82] proved that

[TABLE]

In 1987, C.H. Jia [jia87] proved that

[TABLE]

In this note, we consider a generalization of the average $S(x)$ and find an estimate by applying C.H. Jia’s method in [jia87]. Let $\omega(n)$ be the number of distinct prime divisors of $n$. Suppose $\lambda:\mathbb{N}\rightarrow\mathbb{C}$ is a bounded arithmetic function. For a positive real number $\alpha>0$, let $S_{\lambda,\alpha}(x)$ be the weighted average of the $\alpha$-th power of the ratio of the smallest and largest prime divisor of $n$ with respect to $\lambda$ as follows

[TABLE]

Note that when $\lambda\equiv 1$ and $\alpha=1$, $S_{\lambda,\alpha}(x)$ turns to be $S(x)=\sum\limits_{n\leqslant x}\frac{p(n)}{P(n)}.$

Theorem 1.1**.**

Let $S_{\lambda,\alpha}(x)$ be defined as above and $\pi(x)=\sum_{p\leqslant x}1$ be the prime counting function. Then for $\alpha>\frac{4}{5}$ we have

[TABLE]

and

[TABLE]

Remark 1.2*.*

Using the method of the proof, in principle one can find out the coefficient of the term $\frac{x}{\log^{4}x}$ and so on. But the computation is very complicated.

2 Some Lemmas

Before going to the proof of Theorem 1.1, we cite/improve some lemmas below.

Lemma 2.1** ([bom62]).**

For any constant $A>0$, we have

[TABLE]

Lemma 2.2** (Lemma 3, [jia87]).**

Suppose $c$ is a constant, $f(x)$ is a function satisfying $f(x)=O\left(x^{c}\right),f^{\prime}(x)=O\left(x^{c-1}\right)$, $x\geqslant 1$. Then for any constant $A>0$ and $\frac{3}{2}<y\leqslant x$, we have

[TABLE]

[TABLE]

Lemma 2.3** (Lemma 4, [jia87]).**

Let $\Psi(x,y)$ be the number of positive integers in $[1,x]$ whose prime factors are $\leqslant y$. If

[TABLE]

then for any $A>0$, we have

[TABLE]

Lemma 2.4**.**

For integer $i\geqslant 1$, let $\displaystyle{\Sigma^{(i)}=\sum_{n\leqslant x,\mu(n)\neq 0,\omega(n)=i}\left(\frac{p(n)}{P(n)}\right)^{\alpha}},$ where $\mu(n)$ is the M$\ddot{\text{o}}$bius function. Then for any constant $A>0$, we have

[TABLE]

Proof.

Since $\lambda$ is bounded, we have $\sum\limits_{n\leqslant x,\mu(n)=0}\lambda(\omega(n))\left(\frac{p(n)}{P(n)}\right)^{\alpha}=O\left(\sum\limits_{n\leqslant x,\mu(n)=0}\left(\frac{p(n)}{P(n)}\right)^{\alpha}\right)$. If $\alpha\geqslant 1$, then $\left(\frac{p(n)}{P(n)}\right)^{\alpha}\leqslant\frac{p(n)}{P(n)}$. By the proof of Lemma 5 in [jia87], we have $\sum\limits_{n\leqslant x,\mu(n)=0}\frac{p(n)}{P(n)}=O\left(\frac{x}{(\log x)^{A}}\right)$. So $\sum\limits_{n\leqslant x,\mu(n)=0}\left(\frac{p(n)}{P(n)}\right)^{\alpha}=O\left(\frac{x}{(\log x)^{A}}\right)$.

If $0<\alpha<1$, then $\left(\frac{p(n)}{P(n)}\right)^{\alpha}\leqslant 1$, we have

[TABLE]

Here the upper bound for the second $O$-term comes from the proof of Lemma 5 in [jia87].

Therefore, for any constant $A>0$,

[TABLE]

and the lemma follows immediately. ∎

By Lemma 2.4, to estimate $S_{\lambda,\alpha}(x)$, it suffices to estimate $\Sigma^{(i)}$ for each $i$. We shall use Lemma 2.2 repeatedly to get the estimates.

3 Proof of Theorem 1.1

3.1 Computation of $\Sigma^{(1)}$

Clearly, $\Sigma^{(1)}=\pi(x)$. By Lemma 2.1, we get that

[TABLE]

3.2 Computation of $\Sigma^{(2)}$

[TABLE]

By Lemma 2.2, we get

[TABLE]

It follows that

[TABLE]

For $I_{2}$, notice that $\log\frac{x}{p_{2}}\asymp\log x$ for $\sqrt{x}<p_{2}\leqslant\sqrt{x}(\log x)^{\frac{4}{\alpha}}$. So by Lemma 2.2 again,

[TABLE]

By Lemma 2.2 and using substitution of variables, we can get

[TABLE]

For the integral, we use integration by parts to get

[TABLE]

So

[TABLE]

Similarly, we have

[TABLE]

Plugging equations (10) and (11) into equation (9), we get

[TABLE]

Therefore, adding estimates for $I_{1}$ and $I_{2}$, we get

[TABLE]

3.3 Computation of $\Sigma^{(3)}$

For $\Sigma^{(3)}$, similar to the $\Sigma^{(3)}$ in [jia87], we have

[TABLE]

For $I_{3}$, similar to the computation of $I_{1}$, we have

[TABLE]

For $I_{4}$, similar to the computation of $I_{2}$, we have

[TABLE]

For $I_{5}$, first we have

[TABLE]

Then by Lemma 2.2,

[TABLE]

Notice that $\log{\frac{x}{p_{3}}}\asymp\log x$ for $x^{\frac{1}{3}}<p_{3}\leqslant x^{\frac{1}{3}}(\log x)^{\frac{4}{\alpha}}$. Plugging (18) into (17) we get

[TABLE]

Now, similar to equation (10), we have

[TABLE]

and

[TABLE]

Plugging them into equation (3.3), we can get

[TABLE]

Therefore,

[TABLE]

3.4 Computation of the remainder

Similar to the proof for the estimates of $\Sigma^{(4)}$ in [jia87], one can show that

[TABLE]

Notice that for $P(n)<p(n)\log^{\frac{4}{\alpha}}x$, we have

[TABLE]

By Lemma 2.1 and Lemma 2.3, we have

[TABLE]

If $\alpha>\frac{4}{5}$, then

[TABLE]

Proof of Theorem1.1. Theorem 1.1 follows by combining the results in sections 3.1-3.4 and taking $A=4$ in Lemma 2.4.

4 An application

Let $k\geqslant 1$ be an integer. Taking $\lambda\equiv 1$ and $\alpha=k$, we can get an estimation of $S_{k}(x)=\sum\limits_{n\leqslant x}\left(\frac{p(n)}{P(n)}\right)^{k}$ by Theorem 1.1. Suppose $f$ is a real analytic function with convergence radius greater than 1 and $f(0)=0$, then $f(x)=\sum\limits_{i=1}^{\infty}a_{i}x^{i}$. Let $S_{f}(x)=\sum\limits_{n\leqslant x}f\left(\frac{p(n)}{P(n)}\right)$. Then $S_{f}(x)=\sum\limits_{i=1}^{\infty}a_{i}S_{i}(x)$. Thus, by Theorem 1.1 we can get another generalization of $S(x)$.

Theorem 4.1**.**

Suppose $f$ is a real analytic function with convergence radius greater than 1 and $f(0)=0$. Let $S_{f}(x)=\sum\limits_{n\leqslant x}f\left(\frac{p(n)}{P(n)}\right)$, then

[TABLE]

Acknowledgments. The author would like to thank his advisor Prof. Xiaoqing Li for her useful suggestions and would also like to thank Jiseong Kim for reading this note.