# On the Distribution ofr-Tuples of k-Free Numbers

Authors: Ting Zhang; Huaning Liu
Journal: Journal of Numbers
 (2014)

---

## Abstract

Fork, 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.


---

## Body

## 1. Introduction

A positive integern 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 [1] we know that
(2)
∑
n
≤
x
q
2
(
n
)
=
6
π
2
x
+
O
(
x
1
/
2
)
.Mirsky [2] 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 [3] 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 [4] gave an asymptotic formula for
(5)
∑
n
≤
x
q
2
(
n
+
l
1
)
⋯
q
2
(
n
+
l
r
)
.Tsang [5] proved the following.Proposition 1.
Letl
1
,
l
2
,
…
,
l
r be distinct integers with |
l
i
|
≤
L
x and
(6)
r
≤
1
25
(
log
⁡
x
log
⁡
⁡
log
⁡
x
)
.
Forx
≥
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.Fork, 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. Let q
k
(
n
) be the characteristic function of the sequence of k-free integers. Gegenbauer [6] proved that
(8)
∑
n
≤
x
q
k
(
n
)
=
x
ζ
(
k
)
+
O
(
x
1
/
k
)
,
k
≥
2
.Mirsky [7] 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 [2] 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 [8] 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 ofk-free numbers are given in [9, 10].In this paper, we will study the distribution ofr-tuples of k-free numbers by using the Buchstab-Rosser sieve and the methods in [5]. Our main result is the following.Theorem 2.
Letl
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
.
Forx
≥
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.
Takingk
=
2 in Theorem 2, we immediately get Proposition 1.The main tool in our argument is the Buchstab-Rosser sieve. LetZ
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, letλ
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 [5] 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 takeβ
=
1. Throughout this paper c is an unspecified absolute constant.


## 2. Proof of Theorem2

To prove the theorem we need the following lemma.Lemma 4.
For anyZ
>
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 [5] with a slight modification. For completeness we give a proof.
From the result of Ramanujan [11]
(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 anyt
>
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
.
Ifh
≤
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 thath
≤
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 Theorem2. 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
.Letη
(
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
)
.DefineZ
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 congruenceη
(
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 inequalityp
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 [5] 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).



---

*Source: 537606-2014-04-13.xml*