JN Journal of Numbers 2314-842X Hindawi Publishing Corporation 537606 10.1155/2014/537606 537606 Research Article On the Distribution of r -Tuples of k -Free Numbers Zhang Ting http://orcid.org/0000-0002-6480-8071 Liu Huaning Dujella Andrej Department of Mathematics Northwest University Xi’an, Shaanxi 710069 China nwu.ac.za 2014 1242014 2014 24 01 2014 27 03 2014 13 4 2014 2014 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.

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 n is called square-free number if it is not divisible by a perfect square except 1 . Let q 2 be the characteristic function of the sequence of square-free numbers. That is, (1) q 2 ( n ) = { 1 , if n is a square-free number , 0 , otherwise .

From  we know that (2) n x q 2 ( n ) = 6 π 2 x + O ( x 1 / 2 ) .

Mirsky  studied the frequency of pairs of square-free numbers with a given difference and proved the asymptotic formula: (3) n x q 2 ( n ) q 2 ( n + r ) = x p ( 1 - 2 p 2 ) p 2 r ( 1 + 1 p 2 - 2 ) n x q 2 ( n ) q 2 ( n + r ) = + O r ( x 2 / 3 ( log x ) 4 / 3 ) .

Heath-Brown  investigated the number of consecutive square-free numbers not more than x and obtained the following result: (4) n x q 2 ( n ) q 2 ( n + 1 ) = x p ( 1 - 2 p 2 ) + O ( x 7 / 11 ( log x ) 7 ) .

Pillai  gave an asymptotic formula for (5) n x q 2 ( n + l 1 ) q 2 ( n + l r ) .

Tsang  proved the following.

Proposition 1.

Let l 1 , l 2 , , l r be distinct integers with | l i | L x and (6) r 1 25 ( log x log log x ) .

For x 3 we have (7) n x q 2 ( n + l 1 ) q 2 ( n + l r ) x p ( 1 - u ( p ) p 2 ) - O ( r log x L x + x 3 / 5 r 12 / 5 ( log x ) 8 / 5 ) , n x q 2 ( n + l 1 ) q 2 ( n + l r ) x p ( 1 - u ( p ) p 2 ) + O ( x 2 / 3 ( r log x ) 4 / 3 ) , where u ( p ) is the number of distinct residue classes moduli p 2 represented by the numbers l 1 , l 2 , , l r .

For k , being a fixed integer ≥2, a positive integer n is called k -free number if n is not divisible by the k th power of any integer >1. Let q k ( n ) be the characteristic function of the sequence of k -free integers. Gegenbauer  proved that (8) n x q k ( n ) = x ζ ( k ) + O ( x 1 / k ) , k 2 .

Mirsky  showed that (9) n x q k ( n ) q k ( n + r ) = x p ( 1 - 2 p k ) p k r ( p k - 1 p k - 2 ) n x q k ( n ) q k ( n + r ) = + O r , k ( x ( 2 / ( k + 1 ) ) + ϵ ) , and in  Mirsky improved the error term to (10) n x q k ( n ) q k ( n + r ) = x p ( 1 - 2 p k ) p k r ( p k - 1 p k - 2 ) n x q k ( n ) q k ( n + r ) = + O r , k ( x 2 / ( k + 1 ) ( log x ) ( k + 2 ) / ( k + 1 ) ) .

Meng  further improved this result as follows: (11) n x q k ( n ) q k ( n + r ) = x p ( 1 - 2 p k ) p k r ( p k - 1 p k - 2 ) n x q k ( n ) q k ( n + r ) = + O r , k ( x 2 / ( k + 1 ) ) .

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

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

Theorem 2.

Let l 1 , l 2 , , l r be distinct integers with | l i | L x and (12) r 1 5 ( 2 k + 1 ) · log x log log x .

For x 3 we have (13) n x q k ( n + l 1 ) q k ( n + l r ) x p ( 1 - v ( p ) p k ) - O ( r log x ( L x ) 1 / k + x 3 / ( 2 k + 1 ) r 6 k / ( 2 k + 1 ) ( log x ) ( 2 k + 4 ) / ( 2 k + 1 ) ) , (14) n x q k ( n + l 1 ) q k ( n + l r ) x p ( 1 - v ( p ) p k ) n x q k ( n + l 1 ) q k ( n + l r ) + O ( x 2 / ( k + 1 ) ( r log x ) 2 k / ( k + 1 ) ) , where v ( p ) is the number of distinct residue classes moduli p k represented by the numbers l 1 , l 2 , , l r .

Remark 3.

Taking k = 2 in Theorem 2, we immediately get Proposition 1.

The main tool in our argument is the Buchstab-Rosser sieve. Let Z 0 , Z 1 , Z 2 , be a sequence of positive numbers which lie between x 1 / k and x and let β > 0 be a real number. Define λ 1 = λ 1 = 1 . For any square-free number d = p 1 p 2 p s with p 1 < p 2 < < p s , s 1 , let λ d = θ d μ ( d ) , where μ ( d ) is the Möbius function and θ d = 1 when the following set of inequalities (15) p s β + 1 Z 1 p s - 2 β + 1 p s - 1 p s Z 3 p s - 2 t β + 1 p s - 2 t + 1 p s Z 2 t + 1 , t 1 2 ( s - 1 ) is satisfied. Otherwise, θ d = 0 .

Similarly, let λ d = θ d μ ( d ) , where θ d = 1 when the following set of inequalities (16) p s Z 0 p s - 1 β + 1 p s Z 2 p s - 3 β + 1 p s - 2 p s - 1 p s Z 4 p s - 2 t + 1 β + 1 p s Z 2 t , t s 2 is satisfied. Otherwise, θ d = 0 .

From  we know that (17) d n λ d d n μ ( d ) for any positive integer n and (18) d n λ d d n μ ( d ) for any positive integer n whose smallest prime factor does not exceed Z 0 .

In our case we take β = 1 . Throughout this paper c is an unspecified absolute constant.

2. Proof of Theorem <xref ref-type="statement" rid="thm1.1">2</xref>

To prove the theorem we need the following lemma.

Lemma 4.

For any Z > 2 and any positive integers h , k 2 , there exist absolute constants B and A such that (19) p 1 < p 2 < < p h p 1 p 2 p h Z v ( p 1 ) v ( p 2 ) v ( p h ) B r h ( log log Z + A ) h - 1 Z ( h - 1 ) ! log Z , (20) p 1 < p 2 < < p h p 1 p 2 p h > Z v ( p 1 ) v ( p 2 ) v ( p h ) ( p 1 p h ) - k B k r h ( log log Z + A ) h - 1 ( h - 1 ) ! ( k - 1 ) Z k - 1 log Z .

Proof.

This lemma can be proved by using the methods of the lemma in  with a slight modification. For completeness we give a proof.

From the result of Ramanujan  (21) p 1 < p 2 < < p h p 1 p 2 p h Z 1 B ( log log Z + A ) h - 1 Z ( h - 1 ) ! log Z and the observation v ( p ) r we can obtain (19).

For any t > 2 , the inequalities p 1 p 2 p h t and p 1 < p 2 < < p h imply (22) h c log t log log t , where c is an absolute constant. Now we define (23) ψ ( t ) = p 1 < p 2 < < p h p 1 p 2 p h t 1 .

If h c log t / log log t , we have ψ ( t ) ( B ( log log t + A ) h - 1 t ) / ( ( h - 1 ) ! log t ) . While if h > c log t / log log t , we have ψ ( t ) = 0 .

Without loss of generality, we assume that h c log t / log log t . Then (24) p 1 < p 2 < < p h p 1 p 2 p h > Z ( p 1 p 2 p h ) - k = Z t - k d ψ ( t ) = - Z - k ψ ( Z ) + k Z ψ ( t ) t - k - 1 d t B k ( h - 1 ) ! Z ( log log t + A ) h - 1 t k log t d t B k ( h - 1 ) ! ( k - 1 ) ( log log Z + A ) h - 1 Z k - 1 log Z .

This ends the proof of (20).

New we prove Theorem 2. Define ϱ ( 0 ) = 1 , ϱ ( n ) = p k n p , for n 0 . Then (25) d ϱ ( n ) μ ( d ) = { 1 , if n is k -free , 0 , if n is not k -free .

Let η ( n ) = ϱ ( n + l 1 ) ϱ ( n + l r ) ; then we have (26) S ( x ) = n x q k ( n + l 1 ) q k ( n + l r ) = n x d η ( n ) μ ( d ) .

Define Z 0 = ( L + 1 ) x k ; then any n + l i    ( 1 i r ) that is not k -free has a divisor ≤ Z 0 . By (18) we have (27) S ( x ) = n x d η ( n ) μ ( d ) n x d η ( n ) λ ( d ) = d λ d N d ( x ) , where (28) N d ( x ) = # { n x : η ( n ) 0 ( mod d ) } .

It is easy to show that the congruence η ( n ) 0 ( mod p ) has v ( p ) solutions modulo p k . Therefore the congruence η ( n ) 0 ( mod d ) has v ( d ) = p d v ( p ) solutions modulo d k , by the Chinese remainder theorem. Then we have (29) N d ( x ) = x d k v ( d ) + E ( d ) , | E ( d ) | v ( d ) .

Therefore (30) S ( x ) x d λ d v ( d ) d k + d λ d E ( d ) S ( x ) = x d μ ( d ) v ( d ) d k + d λ d E ( d ) S ( x ) . = - x d ( 1 - θ d ) μ ( d ) v ( d ) d k S ( x ) x d μ ( d ) v ( d ) d k S ( x ) . . = + d λ d E ( d ) - x V .

Let (31) f ( d ) = v ( d ) d k , G ( y ) = p < y ( 1 - f ( p ) ) .

By (16) we have (32) V = j 1 H j , where (33) H 1 = - p > Z 0 f ( p ) G ( p ) , H 2 = p 1 < p 2 Z 0 p 1 2 p 2 > Z 2 f ( p 1 p 2 ) G ( p 1 ) , H j + 1 = p 1 < p 2 < < p 2 j Z 0 p 1 2 p 2 p 2 j > Z 2 j f ( p 1 p 2 j ) G ( p 1 ) , for j 1 .

It is easy to show that (34) H 1 < 0 , p 1 < p 2 p 1 2 p 2 > Z 2 f ( p 1 p 2 ) G ( p 1 ) r 2 p 1 < p 2 p 1 2 p 2 > Z 2 1 p 1 k p 2 k = r 2 ( p 1 Z 2 1 / 3 1 p 1 k p 2 > Z 2 / p 1 2 1 p 2 k + p 1 > Z 2 1 / 3 1 p 1 k p 2 > p 1 1 p 2 k ) c r 2 Z 2 ( 2 - 2 k ) / 3 ( log Z 2 ) - 2 .

On the other hand, the inequality p 1 2 p 2 p 2 j > Z 2 j implies (35) p 1 p 2 p 2 j > Z 2 j 2 j / ( 2 j + 1 ) .

Then from (20) we have (36) 0 H j + 1 c k r 2 j ( log log Z 2 j + A ) 2 j - 1 ( 2 j - 1 ) ! ( k - 1 ) log Z 2 j · Z 2 j - 2 j ( k - 1 ) / ( 2 j + 1 ) c r ( r ( log log x + A ) ) 2 j - 1 ( 2 j - 1 ) ! log x · Z 2 j - 2 j ( k - 1 ) / ( 2 j + 1 ) , for j 2 .

Now combining (32)–(36) we get (37) V c ( r log x ) 2 Z 2 ( 2 - 2 k ) / 3 V = + j 2 c r ( r ( log log x + A ) ) 2 j - 1 ( 2 j - 1 ) ! log x · Z 2 j - 2 j ( k - 1 ) / ( 2 j + 1 ) .

Furthermore, by (16), we have (38) d λ d E ( d ) p Z 0 v ( p ) d λ d E ( d ) = + p 1 < p 2 < p 3 Z 0 p 2 2 p 3 Z 2 v ( p 1 p 2 p 3 ) d λ d E ( d ) = + j 2 p 1 < p 2 < < p 2 j + 1 p 1 p 2 p 2 j + 1 Z 2 j v ( p 1 p 2 j + 1 ) .

From (13) of  we know that (39) p 1 < p 2 < p 3 Z 0 p 2 2 p 3 Z 2 1 Z 2 log Z 2 .

And by (19) we get (40) j 2 p 1 < p 2 < < p 2 j + 1 p 1 p 2 p 2 j + 1 Z 2 j v ( p 1 p 2 j + 1 ) j 2 c r 2 j + 1 ( 2 j ) ! log Z 2 j ( log log Z 2 j + A ) 2 j Z 2 j .

Therefore (41) d λ d E ( d ) c r Z 0 log Z 0 + c r 3 Z 2 log x d λ d E ( d ) = + j 2 c r ( 2 j ) ! log x ( r ( log log x + A ) ) 2 j Z 2 j .

Now from (30), (37), and (41) we immediately get (42) S ( x ) x p ( 1 - v ( p ) p k ) - c r Z 0 log Z 0 - c r 2 log 2 x = × ( x Z 2 ( 2 - 2 k ) / 3 + r Z 2 log x ) = - j 2 c r ( r ( log log x + A ) ) 2 j - 1 ( 2 j - 1 ) ! log x = 1111 × ( x Z 2 j - 2 j ( k - 1 ) / ( 2 j + 1 ) + Z 2 j r log log x ) .

Taking (43) Z 2 j = ( x r log    log x ) ( 2 j + 1 ) / ( 2 j k + 1 ) , for j = 2,3 , , we have (44) j 2 c r ( r ( log log x + A ) ) 2 j - 1 ( 2 j - 1 ) ! log x × ( x Z 2 j - 2 j ( k - 1 ) / ( 2 j + 1 ) + Z 2 j r log log x ) j 2 r ( r ( log log x + A ) ) 2 j - 1 ( 2 j - 1 ) ! log x · x ( 2 j + 1 ) / ( 2 j k + 1 ) ( r log log x ) 2 j ( k - 1 ) / ( 2 j k + 1 ) r log x · x 5 / ( 4 k + 1 ) · ( r log log x ) 4 ( k - 1 ) / ( 4 k + 1 ) × j 2 ( r ( log log x + A ) ) 2 j - 1 ( 2 j - 1 ) ! x 5 / ( 4 k + 1 ) log x · e r ( log log x + A ) x ( 3 / ( 2 k + 1 ) ) - ϵ ,

since (45) r 1 5 ( 2 k + 1 ) · log x log log x .

Choosing (46) Z 2 = ( x r log x ) 3 / ( 2 k + 1 ) , from (42) and (44), we immediately deduce that (47) S ( x ) x p ( 1 - v ( p ) p k ) - c r log x ( L x ) 1 / k S ( x ) = - c x 3 / ( 2 k + 1 ) r 6 k / ( 2 k + 1 ) ( log x ) ( - 2 k + 4 ) / ( 2 k + 1 ) .

This proves (13).

Using the similar methods we have (48) S ( x ) x d μ ( d ) v ( d ) d k + d λ d E ( d ) + x V , where (49) V = d ( 1 - θ d ) μ ( d ) v ( d ) d k .

By (15) and (20) we get (50) 0 V = j 1 p 1 < p 2 < < p 2 j - 1 p 1 2 p 2 p 2 j - 1 > Z 2 j - 1 f ( p 1 p 2 j - 1 ) G ( p 1 ) 0 V j 1 p 1 < p 2 < < p 2 j - 1 p 1 p 2 p 2 j - 1 > Z 2 j - 1 ( 2 j - 1 ) / 2 j v ( p 1 ) v ( p 2 ) v ( p 2 j - 1 ) 0111111 11111 11111111111 = × ( p 1 p 2 p 2 j - 1 ) - k 0 V r Z 1 ( k - 1 ) / 2 log x + j 2 r ( 2 j - 2 ) ! log x ( r ( log log x + A ) ) 2 j - 2 0 0001 r Z 1 ( k - 1 ) / 2 log x + j 2 · Z 2 j - 1 - ( 2 j - 1 ) ( k - 1 ) / 2 j .

Moreover, by using the argument leading to (41), we can have (51) d λ d E ( d ) ( r log x ) 2 Z 1 + j 2 r ( 2 j - 1 ) ! log x ( r ( log log x + A ) ) 2 j - 1 Z 2 j - 1 .

Therefore (52) S ( x ) x p ( 1 - v ( p ) p k ) + c r log x ( x Z 1 ( k - 1 ) / 2 + r Z 1 log x ) = . + c r log x j 2 ( r ( log log x + A ) ) 2 j - 2 ( 2 j - 2 ) ! = . + c r log x 111 × ( x Z 2 j - 1 - ( 2 j - 1 ) ( k - 1 ) / 2 j + r Z 2 j - 1 log log x ) .

Taking (53) Z 1 = ( x log x r ) 2 / ( k + 1 ) , Z 2 j - 1 = ( x r log log x ) 2 j / ( 2 j k - k + 1 ) , we have (54) c r log x ( x Z 1 ( k - 1 ) / 2 + r Z 1 log x ) x 2 / ( k + 1 ) ( r log x ) 2 k / ( k + 1 ) , c r log x j 2 ( r ( log log x + A ) ) 2 j - 2 ( 2 j - 2 ) ! c r log x j 2 × ( x Z 2 j - 1 - ( 2 j - 1 ) ( k - 1 ) / 2 j + r Z 2 j - 1 log log x ) r log x j 2 ( r ( log log x + A ) ) 2 j - 2 ( 2 j - 2 ) ! r log x j 2 · x 2 j / ( 2 j k - k + 1 ) ( r log log x ) ( 2 j - 1 ) ( k - 1 ) / ( 2 j k - k + 1 ) r log x · x 4 / ( 3 k + 1 ) ( r log log x ) 3 ( k - 1 ) / ( 3 k + 1 ) × j 2 ( r ( log log x + A ) ) 2 j - 2 ( 2 j - 2 ) ! r x 4 / ( 3 k + 1 ) e r ( log log x + A ) r x ( 43 k + 21 ) / ( 5 ( 3 k + 1 ) ( 2 k + 1 ) ) .

Therefore (55) S ( x ) x p ( 1 - v ( p ) p k ) + c x 2 / ( k + 1 ) ( r log x ) 2 k / ( k + 1 ) .

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.

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