Table of Contents
Journal of Numbers
Volume 2014, Article ID 537606, 6 pages
http://dx.doi.org/10.1155/2014/537606
Research Article

On the Distribution of -Tuples of -Free Numbers

Department of Mathematics, Northwest University, Xi’an, Shaanxi 710069, China

Received 24 January 2014; Accepted 27 March 2014; Published 13 April 2014

Academic Editor: Andrej Dujella

Copyright © 2014 Ting Zhang and Huaning Liu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

For k, being a fixed integer ≥2, a positive integer n is called k-free number if n is not divisible by the kth power of any integer >1. In this paper, we studied the distribution of r-tuples of k-free numbers and derived an asymptotic formula.

1. Introduction

A positive integer is called square-free number if it is not divisible by a perfect square except . Let be the characteristic function of the sequence of square-free numbers. That is,

From [1] we know that

Mirsky [2] studied the frequency of pairs of square-free numbers with a given difference and proved the asymptotic formula:

Heath-Brown [3] investigated the number of consecutive square-free numbers not more than and obtained the following result:

Pillai [4] gave an asymptotic formula for

Tsang [5] proved the following.

Proposition 1. Let be distinct integers with and
For we have where is the number of distinct residue classes moduli represented by the numbers .

For , being a fixed integer ≥2, a positive integer is called -free number if is not divisible by the th power of any integer >1. Let be the characteristic function of the sequence of -free integers. Gegenbauer [6] proved that

Mirsky [7] showed that and in [2] Mirsky improved the error term to

Meng [8] further improved this result as follows:

Moreover, some recent results on pairs of -free numbers are given in [9, 10].

In this paper, we will study the distribution of -tuples of -free numbers by using the Buchstab-Rosser sieve and the methods in [5]. Our main result is the following.

Theorem 2. Let be distinct integers with and
For we have where is the number of distinct residue classes moduli represented by the numbers .

Remark 3. Taking in Theorem 2, we immediately get Proposition 1.

The main tool in our argument is the Buchstab-Rosser sieve. Let be a sequence of positive numbers which lie between and and let be a real number. Define . For any square-free number with , , let , where is the Möbius function and when the following set of inequalities is satisfied. Otherwise, .

Similarly, let , where when the following set of inequalities is satisfied. Otherwise, .

From [5] we know that for any positive integer and for any positive integer whose smallest prime factor does not exceed .

In our case we take . Throughout this paper is an unspecified absolute constant.

2. Proof of Theorem 2

To prove the theorem we need the following lemma.

Lemma 4. For any and any positive integers , , there exist absolute constants and such that

Proof. This lemma can be proved by using the methods of the lemma in [5] with a slight modification. For completeness we give a proof.
From the result of Ramanujan [11] and the observation we can obtain (19).
For any , the inequalities and imply where is an absolute constant. Now we define
If , we have . While if , we have .
Without loss of generality, we assume that . Then
This ends the proof of (20).

New we prove Theorem 2. Define , , for . Then

Let ; then we have

Define ; then any    that is not -free has a divisor ≤. By (18) we have where

It is easy to show that the congruence has solutions modulo . Therefore the congruence has solutions modulo , by the Chinese remainder theorem. Then we have

Therefore

Let

By (16) we have where

It is easy to show that

On the other hand, the inequality implies

Then from (20) we have

Now combining (32)–(36) we get

Furthermore, by (16), we have

From (13) of [5] we know that

And by (19) we get

Therefore

Now from (30), (37), and (41) we immediately get

Taking we have

since

Choosing from (42) and (44), we immediately deduce that

This proves (13).

Using the similar methods we have where

By (15) and (20) we get

Moreover, by using the argument leading to (41), we can have

Therefore

Taking we have

Therefore

This completes the proof of (14).

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This work is supported by the National Natural Science Foundation of China under Grant no. 11201370, the Natural Science Foundation of Shaanxi Province of China under Grant nos. 2013JM1017 and 2011JQ1010, and the Natural Science Foundation of the Education Department of Shaanxi Province of China under Grant no. 2013JK0558.

References

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