Some new results on k-free numbers

Journal of Number Theory 121 (2006) 45–66 www.elsevier.com/locate/jnt

Some new results on k-free numbers Zaizhao Meng School of Mathematical Sciences, Peking University, Beijing 100871, PR China Received 1 June 2004; revised 23 October 2005 Available online 3 March 2006 Communicated by David Goss

Abstract In this paper we obtain an improved asymptotic formula on the frequency of k-free numbers with a given difference. We also give a new upper bound of Barban–Davenport–Halberstam type for the k-free numbers in arithmetic progressions. © 2006 Elsevier Inc. All rights reserved. MSC: 11N13

1. Introduction In this paper, we give some new results on the average behaviour of the remainder terms of the k-free numbers in arithmetic progressions. Such result is analogous to the Barban–Davenport– Halberstam theorem for the primes in arithmetic progressions. Let μk (n) be the characteristic function of the k-free numbers, then we have  μ(d), (1.1) μk (n) = d k |n

where μ(n) is the Möbius function. For fixed positive integer r and positive real number ε > 0, in [10] Mirsky obtained       2  2  pk − 1 1− k x + Or,k x k+1 +ε , μk (n)μk (n + r) = k p p −2 k p nx

p |r

E-mail address: [email protected]. 0022-314X/$ – see front matter © 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.jnt.2006.01.010

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Z. Meng / Journal of Number Theory 121 (2006) 45–66

and in [11], Mirsky improved this to 

μk (n)μk (n + r) =

 p

nx

2 1− k p

  p k |r

  2 k+2  pk − 1 x + Or,k x k+1 (log x) k+1 . pk − 2

We improve this result as follows: Theorem 1. For fixed positive integer r  1, k  2, we have 

μk (n)μk (n + r) =

      2  2  pk − 1 1− k x + Ok (x + r) log log 3r k+1 . k p p −2 k p p |r

nx

Especially, we have 

μk (n)μk (n + r) =



1−

p

nx

2 pk

  p k |r

  2  pk − 1 x + Or,k x k+1 . k p −2

For k = 2, r = 1, Heath-Brown [6] obtained O(x 7/11 (log x)7 ) in place of O(x 2/3 (log x)4/3 ) in Mirsky’s theorem. Write M(x; q, a) =



 Φa (m) =

μk (n),

nx n≡a (mod q)

m, 0,

if m | a, otherwise,

(1.2)

μ(d)(d k , q) , dk

(1.3)

and F (x; q, a) =

  ∞  Φa ((p k , q)) x −1 1− = xq q p pk

d=1 (d k ,q)|a

where the product is over prime numbers. This formula can be obtained as Hooley had done in [8,9]. When a and q are positive integers, define E(x; q, a) by the relation M(x; q, a) = F (x; q, a) + E(x; q, a). Also, define S(x, Q) =





qQ 1aq

E 2 (x; q, a).

(1.4)

Z. Meng / Journal of Number Theory 121 (2006) 45–66

47

When k = 2, Warlimont [15] obtained the following bounds for S(x, Q): ⎧ 1+ε ⎨ x Q, S(x, Q)  xQ, ⎩ xQ,

for 1  Q  x, for 1  Q  x 1/3 , for x 1/3 log10/3 x  Q  x.

He also proved that   2(2α+1) S x, x α  x 3 +ε ,

for

1  α  1. 2

In [16], Warlimont improved the above results for x 3/4  Q  x, namely S(x, Q)  x 1/2 Q3/2 + x 5/3+ε . Later, Croft [4] improved these results for x 5/8  Q  x 2/3 , that is S(x, Q)  x 1/2 Q3/2 + x 3/2 log7/2 x. When k > 2, Brüdern and others [2,3] obtained 2

2

S(x, Q)  x k +ε Q2− k ,

for 1  Q  x.

(1.5)

In this paper, we obtain the following: Theorem 2. Suppose that Q and x are positive real numbers greater than 1, and that k is an integer with k > 2. Then 1

k+4

1

2k+6

S(x, Q)  x k Q2− k + x k+2 (log x) k+2 ,

for 1 < Q  x,

(1.6)

where the —constant will depend at most on k. k 2 +2k−4

This theorem improves (1.5), for x 2(k−1)(k+2) < Q < x. Theorem 2 also improves Theorem 1.2 of Vaughan [13] in the case an = μk (n). It should be noticed that we cannot use (1.6) to the case k = 2 directly. When k > 2, we cannot obtain asymptotic formula of Montgomery–Hooley type as in the prime numbers in arithmetic progressions [5,7,14]. We insert the definition of E(x; q, a) in (1.4) and square out. Thus    2 M(x; q, a) − F (x; q, a) = J1 − 2J2 + J3 , (1.7) S(x, Q) = qQ 1aq

where J1 =





M(x; q, a)2 ,

qQ 1aq

J3 =





J2 =





M(x; q, a)F (x; q, a),

qQ 1aq

F (x; q, a)2 .

qQ 1aq

We shall prove Theorem 1 in Section 2, and Theorem 2 in Section 5.

(1.8)

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Z. Meng / Journal of Number Theory 121 (2006) 45–66

Notation. Let k > 1 denote a positive integer. Throughout, the implicit constants in Vinogradov’s notation , and in Landau’s O-notation, will depend at most on k unless it is pointed out depend upon the corresponding parameters. n ≡ a (mod q) may be written as n ≡ a(q). The greatest common divisor and the least common multiple of integers a, b are denoted by (a, b) and [a, b], respectively; μ(n) denotes the Möbius function and τ (n) denotes the divisor function; ω(n) denotes the number of distinct prime factors of n; [e] denotes the integer part of e; ψ(v) = [v] − v + 12 . The letter p denotes a prime number, and write p t  n when p t | n but p t+1 † n. Let x denote a sufficiently large real number and Q be a positive real number with Q  x. 2. Lemmas for J1 and the proof of Theorem 1 By (1.2), we have J1 =







qQ 1aq

2



μk (n)

= J11 + 2J12 ,

(2.1)

nx n≡a(q)

where J11 =





qQ 1aq

Now, J11 =



 qQ

nx



J12 =

μk (n)2 ,





qQ 1aq

nx n≡a(q)



μk (m)μk (n).

(2.2)

m
μk (n). Using (2) in [12], we have 

 1 μk (n) = ζ −1 (k)x + O x k ,

nx

therefore  1 J11 = ζ −1 (k)x[Q] + O Qx k .

(2.3)

Write Tq =







μk (n)μk (n − h) =



lxq −1 lq
1hx h
and for a fixed positive integer r Wr (x) =



μk (n)μk (n − r),

r
then J12 =

 qQ

Tq .

μk (n)μk (n − lq),

(2.4)

Z. Meng / Journal of Number Theory 121 (2006) 45–66

49

For fixed integers u and v, we define the function N (u, v, x, r) by 

N (u, v, x, r) =

1,

uc−vd=r ucx

where c and d are positive integers. Lemma 2.1. For positive integers u and v, we have  x−r (u, v) + O(1), N (u, v, x, r) = uv 0,

if (u, v) | r, otherwise.

Proof. If (u, v) † r, the lemma is obvious. We suppose that (u, v) | r. Let c0 , d0 be any fixed solution of the equation uc − vd = r, then all the solutions of this equation may be represented v u t, d = d0 + (u,v) t, where t is any integer, and uc0 − vd0 = r. We need to count as c = c0 + (u,v) the number of t that satisfy the following conditions: uc0 +

uv t  x, (u, v)

c0 +

v t  1, (u, v)

d0 +

u t  1. (u, v)

Then, we have   v − vd0 x − uc0 u − uc0 (u, v), (u, v)  t  (u, v), max uv uv uv the number of such t is

x−r uv (u, v) + O(1),

and the lemma follows.

2

Lemma 2.2. For positive integer r and positive real number y, we have Wr (x) = (x − r)

 μ(a)μ(b)(a, b)k  μ(a)μ(b)(a, b)k − (x − r) k k a b a k bk k k

(a,b) |r





+

μ(a)μ(b) + O

(a,b) |r ab>y



2

2

μ (a)μ (b) .

(a,b)k |r

a k c−bk d=r a k cx, ab>y

aby

Proof. By (1.1), we have Wr (x) =

  r
μ(a)

a k |n



μ(b) =

bk |n−r



μ(a)μ(b) =

a k c−bk d=r a k cx

 1

+



where  1

=

 a k c−bk d=r a k cx, aby

μ(a)μ(b),

 2

=

 a k c−bk d=r a k cx, ab>y

μ(a)μ(b).

2

,

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Z. Meng / Journal of Number Theory 121 (2006) 45–66

By Lemma 2.1, we have  1



=

(a,b)k |r aby

    k k  (a, b)k μ(a)μ(b)N a , b , x, r = μ(a)μ(b) (x − r) + O(1) a k bk k (a,b) |r aby

= (x − r)

 μ(a)μ(b)(a, b)k  μ(a)μ(b)(a, b)k − (x − r) a k bk a k bk k k

(a,b) |r

+O

(a,b) |r ab>y



2

2

μ (a)μ (b) ,

(a,b)k |r aby

and the lemma follows.

2

Write Kk (n) =



pm .

p m n mk

Lemma 2.3. For y  2, we have  μ(a)μ(b)(a, b)k k  y − 2 2ω(Kk (r)) log y. k k a b k

(a,b) |r ab>y

If k  3 then  μ(a)μ(b)(a, b)k k  y − 2 2ω(Kk (r)) . k bk a k

(a,b) |r ab>y

Proof. Write I=

 μ(a)μ(b)(a, b)k , a k bk k

(a,b) |r ab>y

then I=

  μ(a)μ(b)(a, b)k   2 −k  μ (t)t a −k a k bk k k t |r (a,b)=t ab>y

t |r

a1

 b>yt −2 a −1

b−k .

Z. Meng / Journal of Number Theory 121 (2006) 45–66

51

We need to estimate the sums of a and b more carefully. We have 



a −k

b>yt −2 a −1

a1



b−k 



a −k +

a>yt −2

 1−k a −k ya −1 t −2 ,

ayt −2

hence I



μ2 (t)t −k

  a>yt −2

t k |r



a −k + y 1−k t 2k−2

 a −1  I1 + I2 + I3 ,

ayt −2

where I1 =



μ2 (t)t −k

I3 = y 1−k

a −k ,



I2 =

a>yt −2

t k |r

t 2 y



μ2 (t)t −k

t k |r



a −k ,

a1

t 2 >y

 t k |r

μ2 (t)t k−2



a −1 .

ayt −2

t 2 y

We estimate I1 , I2 and I3 separately. We have I1 



  1−k k−2  μ2 (t)t −k yt −2  y 1−k μ2 (t)t k−2  y 1−k y 2 μ2 (t),

t k |r t 2 y

t k |r

t k |r t 2 y

therefore k

I1  y − 2 2ω(Kk (r)) .

(2.5)

Also I2 



k

μ2 (t)t −k  y − 2



k

μ2 (t)  y − 2 2ω(Kk (r)) .

(2.6)

t k |r

t k |r t 2 >y

Similarly, we have I3  y 1−k log y



k

μ2 (t)t k−2  y − 2 log y

t k |r t 2 y

By (2.5)–(2.7), we obtain the first part of the lemma.

 t k |r

k

μ2 (t)  y − 2 log y2ω(Kk (r)) .

(2.7)

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Z. Meng / Journal of Number Theory 121 (2006) 45–66

If k  3, then



I3 = y 1−k

μ2 (t)t k−2

y

1−k

a −1 +

μ (t)t

k−2

t k |r

y log 2 + t

9t 2 y

and the lemma follows.



a −1

ayt −2

y<9t 2 9y 2



μ2 (t)t k−2

t k |r

9t 2 y





ayt −2

t k |r





2

μ (t)t

k−2

k

 y − 2 2ω(Kk (r)) ,

t k |r y<9t 2 9y

2

Lemma 2.4. For y  2, we have 

μ2 (a)μ2 (b)  y log y.

(a,b)k |r aby

Proof. This is a trivial result, we give a proof for completeness. We have 

μ2 (a)μ2 (b) 

ay bya −1

(a,b)k |r aby

and the lemma follows.

 

1y



a −1  y log y,

ay

2

Write f (r) =

 μ(a)μ(b)(a, b)k . a k bk k

(2.8)

(a,b) |r

Lemma 2.5. For r  x and y  2, we have 

Wr (x) = (x − r)f (r) +

  k μ(a)μ(b) + O (x − r)y − 2 2ω(Kk (r)) log y + y log y .

a k c−bk d=r a k cx, ab>y

If k  3, then Wr (x) = (x − r)f (r) +

 a k c−bk d=r a k cx, ab>y

  k μ(a)μ(b) + O (x − r)y − 2 2ω(Kk (r)) + y log y .

Z. Meng / Journal of Number Theory 121 (2006) 45–66

53

Proof. By Lemmas 2.2–2.4, we have 

  k Wr (x) = (x − r)f (r) + O (x − r)y − 2 2ω(Kk (r)) log y +

μ(a)μ(b)

a k c−bk d=r a k cx, ab>y

+ O(y log y), and the first part of the lemma follows. We may obtain the second part similarly.

2

Write ∞ 

g(t) = μ2 (t)t −k

μ(n)τ (n)n−k .

(2.9)

n=1 (n,t)=1

Lemma 2.6. For positive integer r, we have f (r) =



g(t) =

 p

t k |r

1−

2 pk

  p k |r

 pk − 1 . pk − 2

Proof. By the definition of f (r), we have f (r) =

  μ(a)μ(b)(a, b)k   μ(ab) 2 −k = μ (t)t a k bk a k bk k k t |r (a,b)=t

=



t |r

 μ(n)τ (n)  = g(t), nk k

μ2 (t)t −k

t k |r

(ab,t)=1

t |r

(n,t)=1

this is the first part of the formula. We continue in this fashion obtaining f (r) =



μ2 (t)t −k

t k |r

=

 p

     2 −1 1 − 2p −k = 1 − 2p −k 1 − 2p −k μ (t)t −k p

p†t

1−

2 pk

and the lemma follows.

  p k |r

t k |r

 pk − 1 , pk − 2

p|t

2

Combining Lemma 2.5 and the method of Atkinson and Cherwell [1] gives  nx

μk (n)μk (n + r) =

 p

2 1− k p

  p k |r

  2  pk − 1 x + Or,k x k+1 log x . pk − 2

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Z. Meng / Journal of Number Theory 121 (2006) 45–66

Write D1 (n) =





μ(v)

v k |n

uk |n+r

vz

D3 (n) =





D2 (n) = μk (n)

μ(u),

μ(u),

uk |n+r u>z

uz

μ(v)μk (n + r),

D4 (n) = −

v k |n v>z





μ(v)

v k |n v>z

μ(u).

uk |n+r u>z

Lemma 2.7. For fixed positive integer r and positive real number z, we have μk (n)μk (n + r) = D1 (n) + D2 (n) + D3 (n) + D4 (n). Proof. By (1.1),



μk (n)μk (n + r) = μk (n)

μ(u) +

uk |n+r

 uk |n+r

u>z



μ(u) = D2 (n) + μk (n)

μ(u),

uk |n+r

uz

uz

and 

μk (n)

μ(u) = D1 (n) +

uk |n+r

μ(v)

v k |n v>z



μ(u) =

uk |n+r uz

 v k |n v>z

and the lemma follows.



uz

μ(v)

v k |n v>z

μ(u).

uk |n+r

v>z

μ(v)μk (n + r) −



μ(v)

v k |n

uz

We have 





μ(u) = D3 (n) + D4 (n),

uk |n+r u>z

2

Now, we have 

μk (n)μk (n + r) =

4  

Dj (n) =

j =1 nx

nx

E1 =

 u,vz

  = xE11 + O z2 ,   t −k E11 = t k |r



μ(u)μ(v)

4 

1=

nx n+r≡0(uk ), n≡0(v k )

m,szt −1 (m,s)=1

Ej ,

j =1



say,  μ(u)μ(v)

u,vz (u,v)k |r

say, −k

μ(mt)μ(st)(ms)

  = f (r) + O z1−k log log 3r .

= f (r) + O

 t k |r

t

 x + O(1) [u, v]k

−k

 s>zt −1

s

−k



Z. Meng / Journal of Number Theory 121 (2006) 45–66

55

Hence   E1 = f (r)x + O xz1−k log log 3r + z2 . For j = 2, 3, 

Ej 



1  (x + r)z1−k .

uk x+r nx+r u>z uk |n

For j = 4,

E4 





nx+r

 (x + r)



2





1



dk

d=1

1  (x + r)

u>z v>z [uk ,v k ]|n nx+r

uk |n u>z ∞ 





[u, v]−k  (x + r)

u,v>z

(uv)−k  (x + r)

∞ 

d −k

u,v>z

t −k

2

t>zd −1

d=1

(u,v)=d

 

 (u, v)k uk v k u,v>z

  2     (x + r) d −k t −k + d −k . t>zd −1

dz

d>z

Hence       1−k 2 E4  (x + r) d −k zd −1 + z1−k  (x + r) z2−2k d k−2 + z1−k dz

dz

 (x + r)z1−k . Thus 

  μk (n)μk (n + r) = f (r)x + O (x + r)z1−k log log 3r + z2 ,

(2.10)

nx 1

so that by choosing z = ((x + r) log log 3r) k+1 in (2.10), this completes the proof of Theorem 1. 3. The formula for Tq 2

From now on we make the assumption: 2  y  x k , k > 2. By Lemma 2.5 and (2.4), we have

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Z. Meng / Journal of Number Theory 121 (2006) 45–66

Tq =

 lxq −1

=



Wlq (x)



(x − lq)f (lq) +

lxq −1



μ(a)μ(b) + O (x − lq)y

− k2 ω(Kk (lq))

2

+ y log y



a k c−bk d=lq

  = Uq + Vq + O Zq∗ ,

a k cx, ab>y

where Uq =



(x − lq)f (lq),

lxq −1

Zq∗ =



Vq =



μ(a)μ(b),

lxq −1 a k c−bk d=lq a k cx, ab>y

    k xq −1 − l qy − 2 2ω(Kk (lq)) + y log y . lxq −1

Lemma 3.1. We have k

Zq∗  x 2 q −1 y − 2 2ω(Kk (q)) log x + xq −1 y log x. Proof. By noting that ω(Kk (lq))  ω(Kk (l)) + ω(Kk (q)), we have    xq −1 − l τ (l) + xq −1 y log x.

k

Zq∗  2ω(Kk (q)) qy − 2

lxq −1

By Dirichlet’s formula for divisor function, we have −1

xq     2   −1 xq − l τ (l)  τ (l) dt  xq −1 log x, lxq −1

and the lemma follows.

1

lt

2

Write k

Zq = x 2 q −1 y − 2 2ω(Kk (q)) log x + xq −1 y log x, then Tq = Uq + Vq + O(Zq ), and J12 =

 qQ

Tq =

  Uq + Vq + O(Zq ) = A + B + O(C), qQ

(3.1)

Z. Meng / Journal of Number Theory 121 (2006) 45–66

57

where 

A=

Uq ,

B=

qQ



Vq ,



C=

qQ

Zq .

qQ

Lemma 3.2. We have k

C  x 2 y − 2 log3 x + xy log2 x. Proof. By the definition of C, we have k

C  x 2 y − 2 log x



q −1 τ (q) + xy log x

qQ



q −1 ,

qQ

again, by Dirichlet’s formula for the divisor function and noting that have 

q −1 τ (q)  log2 x,



qQ q

−1

 log Q, we

k

C  x 2 y − 2 log2 x log x + xy log x log x,

qQ

and the lemma follows.

2

4. The formula for J1 In this section we will give the formula for J1 . Lemma 4.1. Suppose that f (t) and g(t) are defined by (2.8) and (2.9), respectively, that u > 0 and that v = u(t k , q)t −k . We have 

f (mq) =

∞  t=1

mu

   ∞   1 1 + g(t) v − g(t)ψ(v) + O (uq) k −1 . 2 t=1

Proof. By Lemma 2.6, we have 

f (mq) =

  mu t k |mq

mu

=

 1

t(uq) k

=

 1

t(uq) k

g(t) =



g(t) 1

t(uq) k

 mu t k |mq

1=



    g(t) u t k , q t −k 1

t(uq) k

    k  −k   k  −k 1 g(t) u t , q t − + ψ u t , q t 2    1 + g(t) v − g(t)ψ(v). 2 1 t(uq) k

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Z. Meng / Journal of Number Theory 121 (2006) 45–66

By (2.9), we have g(t)  t −k , and 

g(t)(v + 1)  1

   1 t −k 1 + ut −k q, t k  (uq) k −1 , 1

t>(uq) k

and the lemma follows.

 t>(uq) k

2

Write

H (x, q) = 2q

∞ 

 −1 g(t)t k q, t k

k )q −1 t −k x(q,t

t=1

ψ(v) dv. 0

Lemma 4.2. We have ∞    1 1 1  1  −1  A = − ζ −1 (k)x[Q] + H (x, q) + x 2 q g(t)t −k q, t k + O Qx k . 2 2 2 qQ

qQ

t=1

Proof. By Lemma 4.1, we have −1 xq  

Uq = q 0

=q

 mu

−1 xq   ∞

0

t=1

−1

xq       ∞ ∞   1 1 −1 k du f (mq) du = q g(t) v − g(t)ψ(v) + O (uq) + 2

0

t=1

t=1

−1

xq

    ∞ 1 1 + g(t) v − g(t)ψ(v) du + O q (uq) k −1 du . 2

t=1

0

The error term is 1 1 1  q k xq −1 k  x k . Therefore,

Uq = q

∞ 

−1 xq  

g(t)

t=1

0

−1

xq   ∞   1 1 v− du + q g(t) ψ(v) du + O x k . 2 t=1

0

Also −1 xq  

v− 0

   1 1 1 du = x 2 t −k q, t k q −2 − xq −1 , 2 2 2

Z. Meng / Journal of Number Theory 121 (2006) 45–66 −1 xq 

59

t −k (q,t k ) xq −1

 −1 ψ(v) du = t q, t k k

ψ(v) dv,

0

0 ∞ 

g(t) = ζ −1 (k).

t=1

Thus ∞

∞

t=1

t=1

  1   1  1 1 g(t)t −k q, t k − x g(t) + H (x, q) + O x k , Uq = x 2 q −1 2 2 2 and the lemma follows.

(4.1)

2 2

Lemma 4.3. Suppose that k > 2 and that 2  y  x k . Then 3k

k

B  x 2 y 1−k log5/2 x + x 3/2 y 1− 2 log7/2 x + x 2 y 1/2− 4 log3/2 x. Proof. We refine the argumentation in [4].We have B  J (x), where J (x) =







1

qQ lxq −1 a k c−bk d=lq a k cx, ab>y





τ (m)

1.

a k c−bk d=m a k cx, ab>y

mx

We write K = K(x; a, b) =





τ (m)

mx

1,

a k c−bk d=m cxa −k

then J (x) 



(4.2)

K(x; a, b).

a k x, bk x ab>y

By the Cauchy inequality and Dirichlet’s formula for the divisor function K



1/2 τ (m) L1/2  x 1/2 log3/2 xL1/2 , 2

mx

where L = L(x; a, b) =

 a k c−bk d=a k γ −bk δ c,γ xa −k

1.

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Z. Meng / Journal of Number Theory 121 (2006) 45–66

Writing c − γ = u and d − δ = v, we have L

x2 a k bk



  (a, b)k x2 x3 x2 x + 1  2k 2k (a, b)k + k k . k k k k a b a b a b a b

1

a k u=bk v |u|xa −k , |v|xb−k

Hence k x 3/2 x (a, b) 2 + k k , k k a b a2b2

L1/2  and

k

K  x 2 log3/2 x

(a, b) 2 1 + x 3/2 log3/2 x k k . a k bk a2b2

(4.3)

By (4.2) and (4.3), J (x)  x log 2

3/2



x

k  (a, b) 2 1 3/2 3/2 + x log x . k k a k bk 1 a2b2

1

a,bx k ab>y

a,bx k ab>y

Thus J (x)  x 2 log3/2 xL1 + x 3/2 log3/2 xL2 ,

(4.4)

where L1 =

 1 a,bx k



k

(a, b) 2 , a k bk

L2 =

1 a,bx k

ab>y

1 k 2

k

.

a b2

ab>y

We have L1 

 1

dx k



k

d 2 −2k

 2

yd −2 nx k d −2

 τ (n)  −3k d 2 + nk √ y

1ny

τ (n)  nk

n d

 2

1nx k

 2

τ (n) nk √



d

−3k 2

1 y k −1/2 n dx n

τ (n) . nk

y
Again, by Dirichlet’s formula for the divisor function, we have 1

3k

L1  y 2 − 4 + y 1−k log x.

(4.5)

Z. Meng / Journal of Number Theory 121 (2006) 45–66

61

We see at once that 

L2 

1 y

1 a,bx k

k

k 2 −1

ab

 y 1− 2 log2 x,

(4.6)

ab>y

and the lemma then follows from (4.4)–(4.6).

2

Lemma 4.4. For k > 2, we have J1 =



H (x, q) + x 2

qQ



q −1

∞ 

  g(t)t −k q, t k

t=1

qQ

 1 k 1 3k + O Qx k + x 2 y 1−k log5/2 x + x 3/2 y 1− 2 log7/2 x + x 2 y 2 − 4 log3/2 x  k + x 2 y − 2 log3 x + xy log2 x . Proof. By (2.1)–(2.3), we have  1 J1 = ζ −1 (k)x[Q] + O Qx k + 2J12 . By (3.1), (4.1), Lemmas 3.2, 4.2 and 4.3, we have ∞    1 1 1  1  −1  H (x, q) + x 2 q g(t)t −k q, t k + O Qx k J12 = − ζ −1 (k)x[Q] + 2 2 2 qQ

qQ

t=1

 k + O x 2 y 1−k log5/2 x + x 3/2 y 1− 2 log7/2 x  1 3k k + x 2 y 2 − 4 log3/2 x + x 2 y − 2 log3 x + xy log2 x , and the lemma follows.

2

5. The formulas for J2 , J3 and the proof of Theorem 2 In this section we give the complete proof of Theorem 2. For k > 2, there are not simple properties for the function M(x; q, a) as in the case k = 2 [4,16], our method is suitable for all k  2. Lemma 5.1. For k  2, we have J2 = x 2

 qQ

Proof. By (1.1)–(1.3) and (1.8),

q −1

∞  t=1

    1 g(t)t −k q, t k + O x 1+ k log x .

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Z. Meng / Journal of Number Theory 121 (2006) 45–66

J2 =





 

qQ 1aq



=x

q −1

nx t k |n n≡a(q)

d=1 (d k ,q)|a

∞  μ(d)(q, d k ) 

dk

d=1

qQ

μ(t)xq

∞ 

−1

μ(t)

t k x

μ(d)(d k , q) dk

q 



1.

a=1 t k |nx (q,d k )|a n≡a(q)

Write q 

G=



1.

a=1 t k |nx (q,d k )|a n≡a(q)

Writing a = (q, d k )j and n = t k m, we have G=



k −1 q(q,d )



j =1

mxt −k t k m≡(q,d k )j (q)

1=

 x(q, d k , t k ) x(q, d k , t k ) = + O(1). k k (q, d )t (q, d k )t k

Hence J2 = J21 + O(J22 ),

(5.1)

where J21 = x 2



q −1

qQ

J22 = x



q −1

∞  μ(d)  d=1

  μ(t) q, d k , t k t −k ,

1

tx k

∞  μ2 (d)(q, d k ) 

dk

d=1

qQ

dk

μ2 (t).

(5.2)

1 tx k

By (2.9), proceeding as in the proof of Lemma 2.6, we get J21 = x 2



q −1

∞ 

   g(t)t −k q, t k − x 2 q −1 Δ21 ,

t=1

qQ

qQ

where, Δ21 =

∞  μ(d)  d=1

We have

dk

1

t>x k

  μ(t) q, d k , t k t −k .

(5.3)

Z. Meng / Journal of Number Theory 121 (2006) 45–66

63

∞ ∞       μ(d)  μ(d)  k −k q, m Δ21 = μ(t)t = μ(m) q, mk m−k k dk d 1 m|d

d=1

=

∞ 

∞ 

  μ(m) q, mk m−k

m=1

d=1 m|d

m|d

d=1

t>x k (t,d)=m



μ(t)t −k

1 t>x k m−1

(t,d)=1



μ(d) dk

μ(t)t −k = Δ211 + Δ212 ,

1

t>x k m−1 (t,d)=1

where 

Δ211 =

1 mx k

Δ212 =



∞  −k   μ(d) k μ(m) q, m m dk d=1 m|d

  μ(m) q, mk m−k

1

∞  d=1 m|d

m>x k



μ(t)t −k ,

1 t>x k m−1

(t,d)=1

μ(d) dk



μ(t)t −k .

1

t>x k m−1 (t,d)=1

We have Δ211 

   q, mk m−2k 1

mx k 1

 x k −1



    1 1−k q, mk m−2k x k m−1

t −k 

1

1

t>x k m−1

mx k

  q, d k d −k−1 , 1

dx k

hence        1 1  x2 q, d k d −k−1  x 1+ k q −1 Δ211  x 2 q −1 x k −1 d −k−1 q −1 q, d k qQ

1

qQ 1

 x 1+ k



1

dx k

d −k−1



m|d k

1

dx k



m

dx k 1

q −1  x 1+ k log x

(q,d k )=m qQ

qQ



  τ d k d −k−1 .

1

dx k

Thus x2



1

q −1 Δ211  x 1+ k log x.

(5.4)

qQ

Since Δ212  x2



 qQ

1

m>x k

(q, mk )m−2k , we have

q −1 Δ212  x 2

 qQ

q −1

    q, mk m−2k  x 2 q −1 m−k 1

m>x k

 1 1−k  x 2 log x x k ,

qQ

1

m>x k

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Z. Meng / Journal of Number Theory 121 (2006) 45–66

therefore, x2



1

q −1 Δ212  x 1+ k log x.

(5.5)

qQ

By (5.2), we have 1

J22  x 1+ k



q −1

∞  μ2 (d)(q, d k )

dk

d=1

qQ

1

 x 1+ k

∞ 



d −k

d=1

  q −1 q, d k ,

qQ

as in the estimation of (5.4), we have 1

J22  x 1+ k log x,

(5.6)

2

and the lemma follows from (5.1), (5.3)–(5.6). Lemma 5.2. For k  2, we have J3 = x 2



q −1

qQ

∞ 

  g(t)t −k q, t k .

t=1

Proof. By the definition of J3 , we have J3 =

 

xq

−1

d=1 (d k ,q)|a

qQ 1aq

= x2



q −1

qQ

∞ 

∞ 

μ(d)(d k , q) dk

2 =x

2

 qQ

q

−1

∞   μ(d1 )μ(d2 )  q, d1k , d2k k k d1 d2 d ,d =1 1

2

  g(t)t −k q, t k ,

t=1

and the lemma follows.

2

By (1.7), Lemmas 4.4, 5.1 and 5.2, for k > 2, S(x, Q) =



  1 k 1 3k H (x, q) + O Qx k + x 2 y 1−k log5/2 x + x 3/2 y 1− 2 log7/2 x + x 2 y 2 − 4 log3/2 x

qQ

  k 1 + O x 2 y − 2 log3 x + xy log2 x + x 1+ k log x . A good choice for y is 2

y = (x log x) k+2 , then S(x, Q) =

 qQ

 k+4 2k+6  H (x, q) + O x k+2 (log x) k+2 .

(5.7)

Z. Meng / Journal of Number Theory 121 (2006) 45–66

65

Lemma 5.3. For k  2, we have 

1

1

H (x, q)  x k Q2− k .

qQ

Proof. We have g(t)  t

−k

,

    

   ψ(v) dv   1, 

k )q −1 t −k x(q,t

  ψ(v)  1.

0

Hence  H (x, q)  q



 k −1 q, t + 1

xq

−1 −k

t



 1   1−k  q xq −1 k + qxq −1 xq −1 k

1

t<(xq −1 ) k 1

 t(xq −1 ) k

1

 x k q 1− k , and 

1

H (x, q)  x k

qQ

and the lemma follows.



1

1

1

q 1− k  x k Q2− k ,

qQ

2

Finally, by Lemma 5.3 and (5.7), we have 1

1

k+4

2k+6

S(x, Q)  x k Q2− k + x k+2 (log x) k+2 , and Theorem 2 follows. Acknowledgment I am grateful to the referee for his/her many helpful comments and constructive suggestions, which enabled me to strengthen the results of the earlier version of this paper. References [1] F.V. Atkinson, Lord Cherwell, The mean-values of arithmetical functions, Quart. J. Math. Oxford 20 (1949) 65–79. [2] J. Brüdern, A. Granville, A. Perelli, R.C. Vaughan, T.D. Wooley, On the exponential sum over k-free numbers, Philos. Trans. R. Soc. Lond. Ser. A 356 (1998) 739–761. [3] J. Brüdern, A. Perelli, T.D. Wooley, Twins of k-free numbers and their exponential sum, Michigan Math. J. 47 (2000) 173–190. [4] M.J. Croft, Square-free numbers in arithmetic progressions, Proc. London Math. Soc. 30 (1975) 143–159. [5] D.A. Goldston, R.C. Vaughan, On the Montgomery–Hooley asymptotic formula, in: G.R.H. Greaves, G. Harman, M.N. Huxley (Eds.), Sieve Methods, Exponential Sums, and Their Applications in Number Theory, Cardiff, 1995, Cambridge Univ. Press, Cambridge, 1996, pp. 117–142. [6] D.R. Heath-Brown, The square sieve and consecutive square-free numbers, Math. Ann. 266 (1984) 251–259. [7] C. Hooley, On the Barban–Davenport–Halberstam theorem: I, J. Reine Angew. Math. 274/275 (1975) 206–223.

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[8] C. Hooley, On the Barban–Davenport–Halberstam theorem: XII, in: K. Györy, H. Iwaniec, J. Urbanowicz (Eds.), Number Theory in Progress, vol. II, Zakopane-Ko´scielisko, Poland, 1997, de Gruyter, Berlin, 1999, pp. 893–910. [9] C. Hooley, On the Barban–Davenport–Halberstam theorem: XIII, Acta Arith. 94 (1) (2000) 53–86. [10] L. Mirsky, Note on an asymptotic formula connected with r-free integers, Quart. J. Math. Oxford 18 (1947) 178– 182. [11] L. Mirsky, On the frequency of pairs of square-free numbers with a given difference, Bull. Amer. Math. Soc. 55 (1949) 936–939. [12] H.L. Montgomery, R.C. Vaughan, The distribution of squarefree numbers, in: Recent Progress in Analytic Number Theory, Symposium Durham, 1979, vol. 1, Academic, London, 1981, pp. 247–256. [13] R.C. Vaughan, On a variance associated with the distribution of general sequences in arithmetic progressions. II, Philos. Trans. R. Soc. Lond. Ser. A 356 (1998) 793–809. [14] R.C. Vaughan, On a variance associated with the distribution of primes in arithmetic progressions, Proc. London Math. Soc. (3) 82 (2001) 533–553. [15] R. Warlimont, On squarefree numbers in arithmetic progressions, Monatsh. Math. 73 (1969) 433–448. [16] R. Warlimont, Über die kleinsten quadratfreien Zahlen in arithmetischen Progressionen mit primen Differenzen, J. Reine Angew. Math. 253 (1972) 19–23.