Introduction. Let P (x) denote the greatest prime factor of Y

LXV.4 (1993)

The greatest prime factor of the integers in an interval

by

Hong-Quan Liu (Harbin)

Introduction. Let P (x) denote the greatest prime factor of Y

x≤n≤x+x^1/2

n ,

where x is a large positive number. To estimate the lower bound of P (x) is one of the “greatest prime factor” problems to which the Chebyshev–

Hooley’s machinery can be applied. Chronologically Ramachandra [11], [12], Graham [4], Baker [1], and Jia [7] have contributed to the topic. In partic- ular, Baker [1] proved

P (x) > x

^0.7

.

Jia [7] improved the above result. But as [7] contained a substantial mistake concerning a multiple exponential sum (which was already picked out by the present author in 1987), the announced estimate P (x) > x

^0.71

was actually not attainable there.

We prove the following better estimate.

Theorem.

P (x) > x

^0.723

. We begin with

x

^1/2

lnx + O(x

^1/2

) = X

p≤P (x)

N (p) lnp, p primes , where

N (p) = X

x≤pn≤x+x^1/2

1 . Thus to prove the Theorem we need to show

(1) X

:= X

p≤x^0.723

N (p) lnp < (1 − ε)x

^1/2

lnx , ε > 0 .

For a suitable number σ, 1/2 < σ < 0.723, we write X

1

= X

p≤x^σ

N (p) lnp , X

2

= X

x^σ≤p≤x^0.723

N (p) lnp . We need to establish an asymptotic formula for P

1

with σ as large as possible, and to estimate P

2

via the sieve.

The proof, as usual, depends mainly on treating some exponential sums, in particular we can prove an asymptotic formula for P

1

with σ = 0.6 − ε. We also make an innovation in the sieve part. To save space we do not analyze the method term by term; the interested reader can find the advantages by himself.

Notations. e(ξ) = exp(2πiξ). [ξ] is the integer part of ξ, and {ξ} = ξ − [ξ], kξk = min({ξ}, 1 − {ξ}).

X

(x1,x2,...)∈D

means summation over all lattice points (x

1

, x

2

, . . .) inside a domain D.

f (x, y) e ∆g(x, y) means that

f

_xⁱ_y^j

(x, y) = g

_xⁱ_y^j

(x, y) + O(∆g

_xⁱ_y^j

(x, y))

whenever both sides make sense. x ∼ X means X ≤ x < 2X, x ∼ = X means C

1

X ≤ x ≤ C

2

X for some constants C

1

and C

2

. The meaning of x = O(X) or x  X is as usual. ε, of course, is a sufficiently small number.

1. An asymptotic formula. In this section, we prove Proposition 1. For σ = 0.6 − ε, we have

(2) X

1

= (0.6 − ε)x

^1/2

lnx + O(x

^1/2

) . As

N (n) = x

^1/2

n + ψ  x + x

^1/2

n



− ψ  x n



, ψ(ξ) = 1

2 − {ξ} ,

in view of the Prime Number Theorem, to prove (2) it remains to show that (with Λ(n) being the von Mangoldt function)

X

x^1/2≤n≤x^0.6−ε

Λ(n)



ψ  x + x

^1/2

n



− ψ  x n



 x

^1/2

.

We split the range x

^1/2

≤ n ≤ x

^0.6−ε

into ranges of the form v ≤ n ≤ v

⁰

, v

⁰

≤ 2v. Then we need to show that

(∗) X

v≤n≤v⁰

Λ(n)f (n)  x

^1/2

(lnx)

⁻¹

, f (n) = ψ  x + x

^1/2

n



− ψ  x n



.

In view of [1], (∗) holds for x

^1/2

 v  x

^13/22−ε

, thus we only need to consider the range x

^0.59

≤ v ≤ x

^0.6−ε

. We appeal to the following decom- position.

Lemma 1.1. Let 3 ≤ u < w < z ≤ 2v, suppose that z − 1/2 is an integer , and z ≥ 4u

²

, v ≥ 32z

²

u, w

³

≥ 64v. Set

U = max

B

∞

X

m=1

d

3

(m)   

X

z<n<B v≤mn≤v⁰

f (mn)    ,

W = sup

g

∞

X

m=1

d

4

(m)   

X

u≤n≤w v≤mn≤v⁰

g(n)f (mn)    ,

where the supremum is taken over all arithmetic functions g(n) such that

|g(n)| ≤ d

₃

(n). Then

(3) U  x

^1/2

(lnx)

⁻¹⁰

and W  x

^1/2

(lnx)

⁻¹⁰

imply the estimate

X

v≤n≤v⁰

Λ(n)f (n)  x

^1/2

(lnx)

⁻¹

. P r o o f. See [5].

Choose

u = vx

^−1/2+15η

, w = x

²

v

⁻³

x

^−10η

, z = [x

^1/4−10η

] +

¹₂

in Lemma 1.1, where η = ε

²

, x

^0.59

 v  x

^0.6−ε

. Then it is easy to verify that Lemma 1.1 is applicable. As in [1], by a reduction using the Fourier expansion of the function ψ(ξ), in order to show (3) it suffices to demonstrate, with H = vx

^−1/2+η

, the following estimates:

(4) X

0<h≤H

   

X

m∼M

a(m) X

n∼N

e  hx

⁰

mn

   

 vx

^−η

whenever v ∼ = M N , N  z, |a(m)| ≤ x

^η

, x ∼ = x

⁰

; and

(5) X

0<h≤H

   

X

m∼M

a(m) X

n∼N

b(n)e  hx

⁰

mn

   

 vx

^−η

whenever v ∼ = M N , u  N  w, |a(m)|, |b(n)| ≤ x

^η

, x ∼ = x

⁰

.

Lemma 1.2. (5) holds for v in the (larger ) range v ≤ x

^5/8−ε

. P r o o f. This is Lemma 9 of [1].

Now it remains to show

Lemma 1. (4) holds.

We cite three lemmata of basic importance.

Lemma 1.3. Let I be a subinterval of [X, 2X], Q be a positive number , and z

n

(X ≤ n ≤ 2X) be complex numbers. Then

  

X

n∈I

z

n

  

2

≤ (1 + XQ

⁻¹

) X

|q|≤Q

(1 − |q|Q

⁻¹

) X

n,n+q∈I

z

n

z

n+q

.

P r o o f. This is Weyl’s inequality. Cf. [1].

Lemma 1.4. Let f (x) and g(x) be algebraic functions in the interval [a, b], and

|f

⁰⁰

(x)| ∼ = R

⁻¹

, |f

⁰⁰⁰

(x)|  (RU )

⁻¹

,

|g(x)| ≤ H , |g

⁰

(x)|  HU

₁⁻¹

, U, U

1

≥ 1 . Then

X

a≤n≤b

g(n)e(f (n)) = X

α≤u≤β

b

u

g(n(u))

pf

⁰⁰

(n(u)) e(f (n(u)) − un(u) + 1/8) + O(H ln(β − α + 2) + H(b − a + R)(U

⁻¹

+ U

₁⁻¹

)) + O(H min(R

^1/2

, max(1/hαi, 1/hβi))) ,

where [α, β] is the image of [a, b] under the mapping y = f

⁰

(x), n(u) is determined by the equation f

⁰

(n(u)) = u, b

u

= 1/2 or 1 according as u is one of α and β or not , and hxi is defined as follows:

hxi =  kxk if x is not an integer , β − α if x is an integer.

Moreover , √

f

⁰⁰

> 0 if f

⁰⁰

> 0, √

f

⁰⁰

= ip|f

⁰⁰

| if f

⁰⁰

< 0.

P r o o f. This is Theorem 2.2 of S. H. Min [10].

Lemma 1.5. Let f (x, y) be an algebraic function in the rectangle D

0

= {(x, y) | x ∼ X, y ∼ Y }, f (x, y) e ∆Ax

^α

y

^β

hold throughout D

0

, and D be a subdomain of D

0

bounded by O(1) algebraic curves. Suppose that X ≥ Y , N = XY , A > 0, F = AX

^α

Y

^β

, αβ(α + β − 1)(α + β − 2) 6= 0, 0 < ∆ < ε

0

, where ε

0

is a small number depending at most on α and β. Then

X

(x,y)∈D

e(f (x, y)) 

ε,α,β

(

⁶

√

F

²

N

³

+ N

^5/6

+

¹⁰

√

∆

⁴

Y

⁴

F

²

N

⁵

+

⁸

√

F

⁻¹

X

⁻¹

N

⁸

+ N F

^−1/4

+

⁴

√

∆X

⁻¹

N

⁴

+ N Y

^−1/2

)(N F )

^η

.

P r o o f. This is Lemma 1 of [9].

We also need the following auxiliary lemma.

Lemma 1.6. Let M ≤ N < N

1

≤ M

₁

, and a

n

(M ≤ n ≤ M

1

) be complex numbers. Then

  

X

N <n≤N1

a

n

   ≤

∞

R

−∞

K(θ)   

X

M <m≤M1

a

m

e(θm)    dθ , with K(θ) = min(M

1

− M + 1, (π|θ|)

⁻¹

, (πθ)

⁻²

), and

∞

R

−∞

K(θ) dθ ≤ 3 ln(2 + M

1

− M ) . P r o o f. This is Lemma 2.2 of [2].

P r o o f o f L e m m a 1. For notational simplicity, we let x

⁰

= x in (4).

For 1 ≤ h ≤ H, let

S(h) = X

m∼M

a(m) X

n∼N

e  hx mn

 . By Lemma 1.4, after a partial summation, we obtain (6) x

^−η

S(h)   vN

²

hx



1/2

X

m∼M

  

X

u∈I(m)

e(F (u, m))    +  v

³

hx



1/2

+ v

²

hx +M , where I(m) is a subinterval of [C

1

U, C

2

U ], U = hxv

⁻¹

N

⁻¹

, F (u, m) = C

3

(hxum

⁻¹

)

^1/2

(for i ≥ 1, C

i

denotes a constant). By Lemma 1.6 we get

x

^−η

X

m∼M

  

X

u∈I(m)

e(F (u, m))  

  X

m∼M

  

X

u∼=U

e(G(u, m))  

 = S

⁰

(h) , say , where G(u, m) = F (u, m) + θu, θ is independent of u and m, and 0 ≤ θ < 1.

Let

Q = min(

⁸

√

hxM

⁵

N

⁻⁵

, hxM

⁻¹

N

⁻²

)x

^−η

. If Q ≤ 100, then we get, trivially,

 vN

²

hx



1/2

S

⁰

(h)  (hxM N

⁻¹

Q

⁻¹

)

^1/2

 (M N

^1/2

+

¹⁶

p

(hx)

⁷

M

³

N

⁻³

)x

^η

 (vx

^−1/8

+ (v

⁵

x)

^1/8

)x

^10η

 x

^1/2−10η

.

If Q > 100, by Cauchy’s inequality, Lemma 1.3, and partial summation, we get, with some Q

1

, 1 ≤ Q

1

≤ Q, the inequality

(7) |S

⁰

(h)|

²

 (M U )

²

Q

⁻¹

+M

^3/2

U Q

⁻¹

  

X

(u,q)∈D

X

m∼M

m

^−1/2

e(f (u, q, m))





 ,

where D = {(u, q) | q ∼ Q

1

, u ∼ = U, u + q ∼ = U } and f (u, q, m) =

C

3

(hxm

⁻¹

)

^1/2

(u

^1/2

− (u + q)

^1/2

) − qθ. We apply Lemma 1.4 to transform

the sum over m to a sum over w, where w ∼ = W = N Q

1

M

⁻¹

. Then we change the order of summation, and estimate the sum over w trivially, to get, with some w, the estimate

(8) M

^3/2

U Q

⁻¹

  

X

(u,q)∈D

X

m∼M

m

^−1/2

e(f (u, q, m))   

 hxN

^−3/2

Q

^−1/2

  

X

(u,q)∈D1

e(g(u, q))   

+ (hx)

²

Q

^−1/2

N

^−9/2

+ (hx)

²

Q

⁻¹

N

⁻⁵

+ (hx)

²

M

⁻¹

N

⁻⁴

, where

D

1

= D ∩ {(u, q) | 1 ≤ |C

3

(2w)

⁻¹

(hx)

^1/2

M

^−3/2

(u

^1/2

− (u + q)

^1/2

)| ≤ √

³

4} , g(u, q) = C

4

(hxw(u

^1/2

− (u + q)

^1/2

)

²

)

^1/3

− qθ .

It is easy to verify that

g(u, q) e ∆C

5

(hxw)

^1/3

u

^−1/3

q

^2/3

, ∆ = Q

1

U

⁻¹

.

Choosing X = U , Y = Q

1

, F ∼ = N Q

1

, ∆ = Q

1

U

⁻¹

in Lemma 1.5, we find that

(9) hxN

^−3/2

Q

^−1/2

  

X

(u,q)∈D1

e(g(u, q))   

 ( p

⁶

(hx)

⁹

Q

²

M

⁻³

N

⁻¹³

+ p

⁶

(hx)

¹¹

Q

²

M

⁻⁵

N

⁻¹⁹

+

¹⁰

p

(hx)

¹¹

Q

¹⁰

M

⁻¹

N

⁻¹⁵

+ p

⁸

(hx)

¹⁵

Q

³

M

⁻⁷

N

⁻²⁷

+ p

⁴

(hx)

⁸

QM

⁻⁴

N

⁻¹⁵

+ p

⁴

(hx)

⁶

Q

³

M

⁻²

N

⁻¹⁰

+ p

(hx)

⁴

M

⁻²

N

⁻⁷

)x

^2η

. From (6)–(9), we deduce that

 vN

²

hx



1/2

S

⁰

(h)  (

¹²

p

(hx)

³

M

³

N

⁵

Q

²

+

¹²

p

(hx)

⁵

M N

⁻¹

Q

²

(10)

+

²⁰

p

(hx)M

⁹

N

¹⁵

Q

¹⁰

+

¹⁶

p

(hx)

⁷

M N

⁻³

Q

³

+ p

⁸

(hx)

⁴

N

⁻³

Q + p

⁸

(hx)

²

v

²

Q

³

+ p

⁴

(hx)

²

N

⁻¹

+ p

⁴

(hx)

²

M

²

N

⁻³

Q

⁻¹

+ p

hxM N

⁻¹

Q

⁻¹

)x

^10η

. As we have already seen,

p hxM N

⁻¹

Q

⁻¹

 x

^1/2−20η

,

thus

p

4

(hx)

²

M

²

N

⁻³

Q

⁻¹

 p

⁴

(hx)M N

⁻²

x 

⁸

√

v

⁴

x

³

N

⁻⁶

(11)

 (v

⁸

x

³

)

^1/16

x

^10η

 x

^1/2−20η

. From (9)–(11), we get

 vN

²

hx



1/2

S

⁰

(h)  (

⁴⁸

p

(hx)

¹³

M

¹⁷

N

¹⁵

+

⁴⁸

p

(hx)

²¹

M

⁹

N

⁻⁹

+

⁸⁰

p

(hx)

⁹

M

⁶¹

N

³⁵

+

¹²⁸

p

(hx)

⁵⁹

M

²³

N

⁻³⁹

+

⁶⁴

p

(hx)

³³

M

⁵

N

⁻²⁹

+

⁶⁴

p

(hx)

¹⁹

M

³¹

N + p

⁴

(hx)

²

N

⁻¹

)x

^10η

+ x

^1/2−10η

 (

⁹⁶

√

v

⁶⁰

x

¹³

N

⁻⁴

+

⁹⁶

√

v

⁶⁰

x

²¹

N

⁻³⁶

+

¹⁶⁰

√

v

¹⁴⁰

x

⁹

N

⁻⁵²

+

²⁵⁶

√

v

¹⁶⁴

x

⁵⁹

N

⁻¹²⁴

+

¹²⁸

√

v

⁷⁶

x

³³

N

⁻⁶⁸

+

¹²⁸

√

v

¹⁰⁰

x

¹⁹

N

⁻⁶⁰

+

⁴

√

v

²

xN

⁻¹

)x

^20η

+ x

^1/2−10η

 (

⁹⁶

√

v

⁶⁰

x

¹²

+

¹⁶⁰

√

v

¹⁴⁰

x

⁻⁴

+

¹²⁸

√

v

⁸²

x

¹⁴

+

⁶⁴

√ v

³⁸

x

⁸

+

⁶⁴

√

v

⁵⁰

x

²

+

¹⁶

√

v

⁸

x

³

)x

^30η

+ x

^1/2−10η

 x

^1/2−10η

. From (6), (7) and the above estimate, we get

S(h)   v

³

hx



1/2

+ v

²

hx + vx

^−1/4



x

^20η

+ x

^1/2−8η

, hence

X

1≤h≤H

|S(h)|  (v

²

x

^−3/4

)x

^30η

+ vx

^−η

 vx

^−η

. The proof of Lemma 1 is finished.

R e m a r k. The proof of Lemma 6 of [7] is false. On p. 191 we find the equality E

1

= E

2

, where

E

1

=

Q

X

q=1

X

(h,m)∈Pq

X

(h⁰,m⁰)∈Pq

ε

h

ε

h⁰

X

n∼N

e  x n

 h m − h

⁰

m

⁰



,

E

2

= X

h∼J

X

h⁰∼J

ε

h

ε

h⁰

X

0≤k≤M hh⁰(J Q)⁻¹

X

m∼M,m⁰∼M m⁰h−mh⁰=k

X

n∼N

e

 kx mm

⁰

n

 ,

and

P

_q

=



(h, m) | h ∼ J, m ∼ M, M (q − 1)

J Q ≤ m

h ≤ M q J Q

 . But we observe that

0 ≤ k ≤ M hh

⁰

(J Q)

⁻¹

and m

⁰

h − mh

⁰

= k

cannot imply that there is a q, 1 ≤ q ≤ Q, such that both (h, m) ∈ P

q

and (h

⁰

, m

⁰

) ∈ P

q

; this means that there is no one-to-one correspondence between the summation variables of E

1

and those of E

2

, thus in general E

1

6= E

₂

.

2. Multiple exponential sums. Here we give several results on expo- nential sums, thus preparing an application of the sieve.

Lemma 2. For v ∼ = M N , H = vx

^−1/2+η

, x

^0.6

 vx

^ε

 x

^2/3

,

|a(n)|, |b(m)| ≤ x

^η

, and vx

^−1/2+30η

 N  (vx

^1/2

)

^1/7

x

^−30η

, we have X

1≤h≤H

X

n∼N

X

m∼M

a(n)b(m)e  hx mn



 vx

^−η

. To prove Lemma 2, we need two lemmata.

Lemma 2.1. Let H

1

≥ H

₁⁰

≥ 1, H

₂

≥ H

₂⁰

≥ 1, n

₁

and n

2

be positive integers with (n

1

, n

2

) = 1. Then

ω(n

1

, n

2

; r) := X

^∗

h1n1−h2n2=r

1 =

1

R

0

ω(n b

1

, n

2

; θ)e(θr) dθ and

1

R

0

| ω(n b

1

, n

2

; θ)| dθ  (1 + H

1

H

2

n

⁻¹₁

n

⁻¹₂

)

^1/2

(ln(2H

1

H

2

))

²

, where ∗ means the condition H

₁⁰

≤ h

₁

≤ H

₁

, H

₂⁰

≤ h

₂

≤ H

₂

.

P r o o f. This is Lemma 8 of [3].

Lemma 2.2. Suppose α = a/q + θ/q

²

, (a, q) = 1, q ≥ 1, |θ| ≤ 1. Then for any β, U > 0, P ≥ 1, we have

P

X

x=1

min



U, 1

kαx + βk



≤ 6  P q + 1



(U + q lnq) . P r o o f. This is Lemma 6 of Chapter 5 of [8].

P r o o f o f L e m m a 2. We assume that x is irrational. Pick an integer j such that M ∼ M

1

= 2

^j

. We denote the triple exponential sum of Lemma 2 by S(M, N ). By Cauchy’s inequality, we have (m ' M

1

means M

1

≤ m

≤ 4M

₁

)

x

^−2η

|S(M, N )|

²

 M

^3/2

X

m'M1

m

^−1/2

   

H

X

h=1

X

n∼N

a(n)e  hx mn

   

2

= M

^3/2

X

h,h1,n,n1

a(n

1

)a(n) X

m'M1

m

^−1/2

e  h

₁

n

1

− h n

 x m



.

The terms with h

1

n = hn

1

contribute at most O(M

²

HN x

^η

). We classify the remaining terms according to the value of (n, n

1

). After a familiar argument, we get

(2.1) x

^−3η

|S(M, N )|

²

 M

^3/2

X

d∼D

X

n,n1∼N1

(n,n1)=1

   

X

r∼R

ω(n, n

1

; r) X

m'M1

e

 −rx dmnn

1

 m

^−1/2

   

+ M

²

HN ,

for some D, 1 ≤ D ≤ N , N

1

= N/d, and some R, 1 ≤ R ≤ HN/D. By Lemma 1.4, the innermost sum is equal to

(2.2) X

α≤u≤β

u

^−1/2

e

 C

6

 urx dnn

1



1/2

 C

7

+ O



M

^−1/2

min  M

³

N

²

RDx



1/2

, max

 1 kαk , 1

kβk



+ O  M

^3/2

N

²

RDx x

^η



+ O(M

^−1/2

x

^η

) , where

α = rx

(4M

1

)

²

dnn

1

, β = rx

M

₁²

dnn

1

. We consider the sum

S

^∗

(D, R) = M X

d∼D

X

n,n1∼N1

(n,n1)=1

X

r∼R

ω(n, n

1

; r) min  M

³

N

²

RDx



1/2

, 1 kβk



;

a similar sum with β being replaced by α can be treated analogously.

We shall prove the following estimate for S

^∗

(D, R):

(2.3) S

^∗

(D, R)  v

²

x

^−10η

if N ≤ x

^1/4−20η

. If D ≥ (M N )

²

x

^−1−η

, then we trivially get

S

^∗

(D, R)  M  M

³

N

²

RDx



1/2

D X

|nh1−n1h|≤2R h,h1≤H,n,n1≤2N/D

1

 (M N )

^5/2

H

^3/2

x

^−1/2

D

⁻¹

x

^η

 x

^1/2

(M N )

^1/2

H

^3/2

x

^2η

 v

²

x

^−1/4+20η

. If D ≤ (M N )

²

x

^−1−η

, then we see that

β r = 1

q + θ

q

²

, with q = [dnn

1

M

₁²

x

⁻¹

] ≥ 1 and |θ| ≤ 1 ,

hence by Lemmata 2.1 and 2.2, we have the estimate S

^∗

(D, R)  M D N

²

D

²



1 + DH N



d,n,n

max

1

X

r∼R

min  M

³

N

²

RDx



1/2

, 1 kβk



 M N

²

 RDx

M

²

N

²

+ 1  N

²

M

³

RDx



1/2

+ N

²

M

²

Dx

 x

^η

 (vN x

^1/4

+ v

²

x

^−1/2

N

²

+ v

^5/2

N

^1/2

x

^−1/2

+ v

³

N x

⁻¹

)x

^10η

 v

²

x

^−10η

provided only N ≤ x

^1/4−20η

. Thus anyhow (2.3) holds. By inserting (2.2) in (2.1), and taking into account (2.3), we get, after changing the order of summation, the following estimate (with U = RDxv

⁻²

):

x

^−3η

|S(M, N )|

²

 N M

^5/2

(RDx)

^1/2

X

d∼D

X

n,n1∼N1

X

u∼=U

   

X

r∈I

ω(n, n

1

; r)e

 C

6

 urx dnn

1



1/2

    + v

²

x

^−10η

,

where I is some subinterval of [R, 2R], which may depend on the variables outside the absolute value symbol. By Lemma 1.6 we get

(2.4) x

^−4η

|S(M, N )|

²

 N M

^5/2

(RDx)

^1/2

X

d∼D

X

n,n1∼N1

X

u∼=U

   

X

r∼=R

ω(n, n

1

; r)e

 C

6

 urx dnn

1



1/2

+ θr

    + v

²

x

^−10η

,

with some θ, which is independent of the variables r, u, n, n

1

and d. Now we have arrived at (6.1) of [3], p. 325. The argument in what follows is exactly the same as in [3], and we get (cf. p. 329 of [3]), by the assumption of Lemma 2,

x

^−10η

|S(M, N )|

⁴

 v

²

 Hx v



1/2

H

²

N

⁴

+  Hx v



H

^3/2

N

^7/2

+  Hx v



1/2

N

³

H

²

M

^1/2

+  Hx

v



H

^4/3

M

^1/3

N

^8/3



+ (v

²

x

^−10η

)

²

 v

⁴

x

^−20η

.

Lemma 2 is proven.

Lemma 3. For v ∼ = M N , H = vx

^−1/2+η

, x

^0.6

 vx

^ε

 x

^3/4

,

|a(n)|, |b(m)| ≤ x

^η

, (k, λ) an exponent pair , and

vx

^−1/2+10η

≤ N ≤ min((v

^1−λ+k

x

^{−k+λ/2−1/4}

)

^1/(1−k+λ)

, x

^1/4

)x

^−20η

, we have

X

1≤h≤H

X

m∼M

X

n∼N

a(n)b(m)e  hx mn



 vx

^−η

. P r o o f. Note that (2.4) holds provided that

(2.5) v ≤ x

^3/4−ε

, N ≤ x

^1/4−20η

. In view of Lemma 2.1, (2.4) and (2.5), we get (2.6) x

^−4η

|S(M, N )|

²

 N

³

M

^3/2

(RD)

^1/2

  

X

r∼R

e(F (r, d, n, n

1

, u))  

 + v

²

x

^−10η

, where F (r, d, n, n

1

, u) = C

6

(urx/(dnn

1

))

^1/2

+ θr, for certain d, n, n

1

, u and θ with |θ| ≤ 1. It is easy to see that F

_r⁰

(r, d, n, n

1

, u) ∼ = xD/(vN )  1, thus

(2.7) X

r∼R

e(F (r, d, n, n

1

, u))  R

^λ

 Dx vN



k

. Lemma 3 follows from (2.6), (2.7) and the fact that R  HN .

The last result of the section is

Lemma 4. For v ∼ = M N , H = vx

^−1/2+η

, x

^0.6

 vx

^ε

 x

^3/4

, |a(n)| ≤ x

^η

and N ≤ x

^3/8−ε

, we have

X

1≤h≤H

X

m∼M

X

n∼N

a(n)e  hx mn



 vx

^−η

. To prove Lemma 4 we need again two lemmata.

Lemma 2.3. Let α, β, γ be real constants such that (α − 1)βγ 6= 0. Let M, R, S, x ≥ 1 and let φ

m

and ψ

rs

be complex numbers with modulus not exceeding 1. Then

X

m∼M

X

r∼R

X

s∼S

φ

m

ψ

rs

e  xm

^α

r

^β

s

^γ

M

^α

R

^β

S

^γ



 (x

^1/4

M

^1/2

(RS)

^3/4

+ M

^7/10

RS + M (RS)

^3/4

+ x

^−1/4

M

^11/10

RS)(ln(10M RS))

⁵

.

P r o o f. This is Theorem 3 of [3].

Lemma 2.4. Let X ≥ 100, Y ≥ 100, A > 0, f (x, y) = Ax

^α

y

^β

, F =

AX

^α

Y

^β

, α and β being rational numbers (not positive integers). Suppose

F

⁻²

X

⁴

≤ Y

³

N

^−η

, and for a

x

and b

y

being complex numbers with modulus not exceeding 1, define

S(X, Y ) := X

(x,y)∈D

a

x

b

y

e(f (x, y)) ,

where D is some region contained in the rectangle {(x, y) | x ∼ X, y ∼ Y } such that for a fixed e x, x ∼ X, the intersection D ∩ {( e x, y) | y ∼ Y } has at e most O(1) segments. Then, for W = X

⁵

+ Y

⁵

,

S(X, Y )  (

⁴⁰

√

F

⁸

X

²⁴

Y

²⁷

W +

⁴⁰

√

F

⁻⁴

X

²⁸

Y

³⁹

W +

⁴

√

F

⁻¹

X

³

Y

⁵

(2.8)

+

²⁰

√

F X

¹³

Y

¹⁹

+

⁴

√

F X

²

Y

³

+

²⁰

√

F

³

X

¹⁴

Y

⁷

+

²⁰

√

F

⁻³

X

¹⁶

Y

¹³

+

¹⁰

√

F

⁻³

X

⁶

Y

¹³

)(F XY )

^η

=: E . P r o o f. Put δ = η

²

. By Lemma 1.6, we have (with N = F XY )

N

^−δ

|S(X, Y )|  X

x∼X

  

X

y∼Y

C(y)e(f (x, y))    ,

where C(y) = b

y

e(θy) with some real number θ (which is independent of x and y). We choose

Q = (F

⁻¹

X

²

Y )

^2/5

≤ Y N

^−δ

(by assumption).

If Q ≤ N

^η

, then we trivially get

|S(X, Y )|  N

^η/2

XY Q

^−1/2

 N

^η/2

(

⁵

√

F X

³

Y

⁴

)  E . Assume that Q ≥ N

^η

. By Cauchy’s inequality and Lemma 1.3, we get (2.9) N

^−3δ

|S(X, Y )|

²

 (XY )

²

Q + XY

Q

X

(y,q)∈D1

C(y)C(y + q) X

x∼X

e(Ax

^α

t(y, q)) , where D

1

= {(y, q) | y, y + q ∼ Y, q ∼ Q

1

} for some Q

1

, 1 ≤ Q

1

≤ Q, and t(y, q) = (y + q)

^β

− y

^β

. Applying Lemma 1.4 to the innermost sum, we get (2.10) X

x∼X

e(Ax

^α

t)

= X

u∈I

C

7

|(At)

^γ

u

^−1/2−γ

|e(C

8

(At)

^2γ

u

^1−2γ

)

+ O



min  X

²

Y Q

1

F



1/2

, 1

kg

1

(y, q)k + 1 kg

2

(y, q)k



+ XY

Q

1

F + lnN + R(y, q)

 ,

where I = (C

9

AX

^α−1

|t|, C

₁₀

AX

^α−1

|t|), γ = 1/(2(1 − α)), g

₁

= αAX

^α−1

t,

g

2

= αA(2X)

^α−1

t, and R(y, q) = 0 or O((X

²

Y Q

⁻¹₁

F

⁻¹

)

^1/2

) according as

none of the end points of I is an integer or otherwise. For each fixed q, q ∼ Q

1

, we consider the sum

Φ = X

y∼Y

min((X

²

Y Q

⁻¹₁

F

⁻¹

)

^1/2

, 1/kg(y)k) ,

where g(y) is either g

1

(y, q) or g

2

(y, q). As both g(y) and g

⁰

(y) are mono- tonic, and g

⁰

(y) ∼ = F Q

1

X

⁻¹

Y

⁻²

, we can classify y according to the integer part of g(y), and it is easy to get

Φ  (1 + F Q

1

X

⁻¹

Y

⁻¹

)((X

²

Y Q

⁻¹₁

F

⁻¹

)

^1/2

+ F

⁻¹

Q

⁻¹₁

XY

²

lnN ) , which contributes to the RHS of (2.9) at most



¹⁰

√

F

⁻³

X

¹⁶

Y

¹³

+

⁵

√

F

⁻³

X

⁶

Y

¹³

+

¹⁰

√

F

³

X

¹⁴

Y

⁷

+

√ XY

²

(2.11)

 E

²

N

^−2η

,

and similarly for the contribution of R(y, q). From (2.9)–(2.11), after chang- ing the order of summation, and estimating the sum over u trivially, we get, with some u, |u| ∼ = F Q

1

X

⁻¹

Y

⁻¹

, the estimate

(2.12) N

^−4δ

|S(X, Y )|

²

 XY Q



1 + F Q XY



A

^γ

 XY F Q

1



γ+1/2

|S

₁

| + N

^−η

E

²

, S

1

= S

1

(u) = X

(y,q)∈D2

C(y)C(y + q)t

^γ

e(C

8

(At)

^2γ

u

^1−2γ

) , D

2

= D

1

∩ {(y, q) | (C

₁₀

AX

^α−1

)|t| ≥ u ≥ (C

9

AX

^α−1

)|t|} . If Q

1

≤ N

^η

, then we trivially have

XY Q



1 + F Q XY



A

^γ

 XY F Q

1



γ+1/2

|S

₁

|

 XY Q



1 + F Q XY



XY

^3/2

F

^−1/2

N

^1/2



¹⁰

√

F

⁻¹

X

¹²

Y

²¹

+

√

F X

²

Y

³

 √

F

⁻¹

X

³

Y

⁵

+

¹⁰

√

F X

¹³

Y

¹⁹

+

√

F X

²

Y

³

 N

^−η

E

²

. Assume that Q

1

≥ N

^η

. By Lemma 1.6 we get

(2.13) N

^−δ

|S

₁

|  X

y∼Y

  

X

q∼Q1

C(y + q)e(ϕq)t

^γ

e(C

8

(At)

^2γ

u

^1−2γ

)





 ,

with some real number ϕ, independent of y and q. Assume that t(y, q) > 0

(otherwise we consider −t). Applying Cauchy’s inequality and Lemma 1.3

to (2.13), we get, with Q

2

= Q

^1/2₁

N

^−δ

, G = (Q

1

Y

^β−1

)

^γ

, (2.14) N

^−3δ

|S

₁

|

²

 (Y Q

1

G)

²

Q

2

+ Y Q

1

Q

2

X

1≤|q1|≤Q2

|S

₂

(q

1

)| ,

S

2

(q

1

) = X

(y,q)∈D3

C(y + q)C(y + q + q

1

)t

1

(y, q, q

1

)e(t

2

(y, q, q

1

)) , D

3

= {(y, q) | y ∼ Y, q, q + q

1

∼ Q

₁

} ,

t

1

(y, q, q

1

) = (t(y, q)t(y, q + q

1

))

^γ

, t

2

(y, q, q

1

) = C

8

u  A

u



2γ

(t(y, q + q

1

)

^2γ

− t(y, q)

^2γ

) . By writing y + q = z we get

S

2

(q

1

) = X

(z,q)∈D4

C(z)C(z + q

1

)T

1

(z, q, q

1

)e(T

2

(z, q, q

1

)) , D

4

= {(z, q) | q, q + q

1

∼ Q

₁

, z − q ∼ Y } ,

T

1

(z, q, q

1

) = (t(z − q, q)t(z − q, q + q

1

))

^γ

, T

2

(z, q, q

1

) = C

8

u  A

u



2γ

(t(z − q, q + q

1

)

^2γ

− t(z − q, q)

^2γ

) . Again by Lemma 1.6, we get, with I

1

= [0.5Y, 2.5Y ],

N

^−δ

|S

2

(q

1

)|  X

z∈I1

  

X

q∼Q1

T

1

(z, q, q

1

)e(ξq)e(T

2

(z, q, q

1

))    ,

with some real number ξ, independent of z and q. Applying Cauchy’s in- equality and Weyl’s lemma, we obtain, with Q

3

= Q

1

N

^−δ

,

(2.15) N

^−3δ

|S

₂

(q

1

)|

²

 (Y Q

₁

G

²

)

²

Q

⁻¹₃

+ Y Q

1

Q

⁻¹₃

X

1≤|q2|≤Q3

X

q∼Q1

|S

₃

(q, q

1

, q

2

)| ,

S

3

(q, q

1

, q

2

) = X

z∈I1

T

3

(z, q, q

1

, q

2

)e(T

4

(z, q, q

1

, q

2

)) , T

3

(z, q, q

1

, q

2

) = T

1

(z, q, q

1

)T

1

(z, q + q

2

, q

1

) , T

4

(z, q, q

1

, q

2

) = T

2

(z, q + q

2

, q

1

) − T

2

(z, q, q

1

) . Let

U (z) = T

3

(z, q, q

1

, q

2

) , V (z) = T

4

(z, q, q

1

, q

2

) . It is an easy exercise to verify that, for z ∈ I

1

,

U (z) ∼ = G

⁴

, U

⁰

(z) ∼ = G

⁴

Y

⁻¹

, V (z) ∼ = |F q

1

q

2

|Y

⁻¹

Q

⁻¹₁

, V

⁰

(z) ∼ = F q

1

q

2

Y Q

1

Y .

As V (z) has nice properties with respect to the variable z, by partial sum- mation and the exponent pair (1/2, 1/2), we obtain

(2.16) S

3

(q, q

1

, q

2

)  G

⁴

((|F q

1

q

2

|Y

⁻²

Q

⁻¹₁

)

^1/2

Y

^1/2

+ Y

²

Q

1

(|F q

1

q

2

|)

⁻¹

) . In view of (2.14) and (2.15), the first term in (2.16) contributes to (2.12) at most

 N

^η/2

XY Q



1 + F Q XY



A

^γ

 XY F Q

1



1/2+γ

G

¹⁶

q

Y

¹⁰

Q

¹⁷₁

F

²





1 + F Q XY



16

p

F

⁻⁶

X

³²

Y

³⁴

Q

⁻⁷

N

^η/2

 (

²⁰

√

F

⁸

X

²⁴

Y

²⁷

W +

²⁰

√

F

⁻⁴

X

²⁸

Y

³⁹

W )N

^η/2

 E

²

N

^−η

;

and the second term in (2.16) contributes to (2.12) at most

 N

^η/2

XY Q



1 + F Q XY



A

^γ

 XY F Q

1



1/2+γ

G(Q

⁷₁

Y

¹⁰

F

⁻²

)

^1/8

 p

⁸

Q

⁻⁵

X

¹⁶

Y

²²

F

⁻⁶



1 + F Q XY

 N

^η/2

 ( √

F

⁻¹

X

³

Y

⁵

+

¹⁰

√

F X

¹³

Y

¹⁹

)N

^η/2

 E

²

N

^−η

.

Finally, the term (Y Q

1

G

²

)

²

Q

⁻¹₃

together with the term (Y Q

1

G)

²

Q

⁻¹₂

con- tributes to (2.12) at most

N

^η/2

XY Q



1 + F Q XY



A

^γ

 XY F Q

1



1/2+γ

Y GQ

^3/4₁

 X

²

Y

^5/2

F

^−1/2

Q

^−3/4



1 + F Q XY



N

^η/2

 E

²

N

^−η

. The estimate (2.8) follows from the above observations.

P r o o f o f L e m m a 4. By Lemma 1.4, we obtain (2.17) X

m∼M

e



− hx mn



= X

u∈I

C

11

(nu

⁻³

h

⁻¹

)

^1/4

e

 C

12

 hux n



1/2



+ O  M

³

N hx



1/2



+ O  M

²

N hx



+ O(lnx) , where I = (α, β), α = hx/(4M

²

n), β = hx/(M

²

n). We divide the sum in question into subsums of the form

(2.18) X

h∼H1

   

X

n∼N

a(n) X

m∼M

e



− hx mn

   

= S(H

1

) , say ,

where H

1

≤ H. By substituting (2.17) in (2.18), and exchanging the order of summation, we get, with an application of Lemma 1.6,

S(H

1

)   M

³

N H

1

x



1/2

X

w∼W

   

X

n∼N

b(n)e(θn)e

 C

12

 wx n



1/2

   

x

^η

+ vx

^−10η

, where W = H

₁²

xv

⁻¹

M

⁻¹

and θ is some number (independent of the vari- ables).

If H

₁⁸

≤ x

^η

(N x

⁻⁵

v

⁸

), then by Lemma 2.3 with (r, s, m) = (1, w, n), we get the estimate

x

^−3η

|S(H

1

)|



⁴

q

H

₁⁵

x

²

N v

⁻¹

+

¹⁰

q

H

₁¹⁵

x

⁵

N

⁷

v

⁻⁵

+

⁴

q

H

₁⁴

xN

³

+

²⁰

q

H

₁²⁵

x

⁵

N

²²

v

⁻⁵

 v(

⁸⁰

√

N

⁷¹

x

⁻³⁵

+

⁸

√

N

⁷

x

⁻³

+

¹⁶⁰

√

N

²⁰¹

x

⁻⁸⁵

+

³²

√

N

¹³

x

⁻⁹

) + vx

^−ε

 vx

^−ε

. If H

₁⁸

> x

^η

(N x

⁻⁵

v

⁸

), then we can apply Lemma 2.4, with (x, y) = (n, w), to get

x

^−3η

|S(H

₁

)|   M

³

N H

1

x



1/2

(

⁴⁰

q

H

₁⁶²

x

³⁵

v

⁻⁶²

N

⁵⁶

+

⁴⁰

q

H

₁⁷²

x

⁴⁰

v

⁻⁷²

N

⁵⁶

+

⁴⁰

q

H

₁⁷⁴

x

³⁵

v

⁻⁷⁴

N

⁷²

+

⁴⁰

q

H

₁⁸⁴

x

⁴⁰

v

⁻⁸⁴

N

⁷²

+

⁴

q

H

₁⁹

x

⁴

v

⁻⁹

N

⁸

+

²⁰

q

H

₁³⁹

x

²⁰

v

⁻³⁹

N

³²

+

⁴

q

H

₁⁷

x

⁴

v

⁻⁷

N

⁵

+

²⁰

q

H

₁¹⁷

x

¹⁰

v

⁻¹⁷

N

²¹

+

²⁰

q

H

₁²³

x

¹⁰

v

⁻²³

N

²⁹

+

¹⁰

q

H

₁²³

x

¹⁰

v

⁻²³

N

¹⁹

)



⁴⁰

q

H

₁⁴²

x

¹⁵

v

⁻²

N

¹⁶

+

⁴⁰

q

H

₁⁵²

x

²⁰

v

⁻¹²

N

¹⁶

+

⁴⁰

q

H

₁⁵⁴

x

¹⁵

v

⁻¹⁴

N

³²

+

⁴⁰

q

H

₁⁶⁴

x

²⁰

v

⁻²⁴

N

³²

+

⁴

q

H

₁⁷

x

²

v

⁻³

N

⁴

+

²⁰

q

H

₁²⁹

x

¹⁰

v

⁻⁹

N

¹²

+

⁴

q

H

₁⁵

x

²

v

⁻¹

N +

²⁰

q

H

₁⁷

v

¹³

N +

²⁰

q

H

₁¹³

v

⁷

N

⁹

+

¹⁰

q

H

₁¹⁸

x

⁵

v

⁻⁸

N

⁹

 vx

^−10η

.

The proof of Lemma 4 is finished.

3. Sieve methods

3.1. Outline of setting. Let v be a number such that

x

^0.6−ε

 v  x

^0.723

.

The sequence A is defined as

A = {n | v ≤ n ≤ ev and there exists an m with x ≤ mn ≤ x + x

^1/2

} , where e = 2.71828 . . . is the base of the natural logarithms. As usual, S(A, z) denotes the number of elements in A having no prime factors less than z.

For r a positive integer, let

|A

r

| = X

n∈A,n≡0 (mod r)

1 . It is easy to see that

|A

_r

| = x

^1/2

/r + R(A, r) , R(A, r) = X

v≤rs≤ev



ψ  x + x

^1/2

rs



− ψ  x rs



+ O  x

^1/2

v

 . Let V (z) = Q

p<z

(1 − 1/p). With the above property of A, we have Lemma 3.1.1. We have

(3.1.1) S(A, z) ≤ x

^1/2

V (z)



F  lnD lnz



+ O(ε)



+ R

⁺

if 2 ≤ z ≤ D , where

R

⁺

= X

(D)

X

r<D^η r|P (D^η)

C

(D)

(r, η) X

Di≤pi≤min(z,D_i^1+η) 1≤i≤t

R(A, rp

1

. . . p

t

) ,

t  1, P

(D)

is summation over all sequences {D

i

}

^t_i=1

with each D

i

of the form

D

^η2(1+η2)n

, n = 0, 1, 2, . . . , such that

D

1

≥ . . . ≥ D

_t

≥ D

^η2

, D

1

. . . D

2s

D

_2s+1³

≤ D for all 0 ≤ s ≤ (t − 1)/2 , and

P (D

^η

) = Y

p<D^η

p , |C

_(D)

(r, η)| ≤ 1 . Also, for p ∼

¹₂

M , 2 ≤ M ≤ D

^1/2

, we have

(3.1.2) S(A

p

, M ) ≥ x

^1/2

p V (M )



f  lnD lnM



+ O(ε)



− X

d<D d|P (M )

λ

d

R(A, pd) ,

λ

d

being some numbers with |λ

d

| ≤ D

^η

. The functions F and f are the well-known functions in the linear sieve.

P r o o f. Both (3.1.1) and (3.1.2) come from [6].

We choose P = vx

^−1/2+50η

, and x

^10ε

D =

 (x

¹⁰

v

⁻¹⁵

)

^1/2

for x

^0.6

 vx

^ε

≤ x

^11/18

, x

^3/8

for x

^11/18

< vx

^ε

≤ x

^0.723+ε

,

x

^50η

Q =



 



 



v

⁻³

x

²

for x

^0.6

 vx

^ε

≤ x

^27/44

, (vx

^1/2

)

^1/7

for x

^27/44

< vx

^ε

≤ x

^67/104

, (v

²⁰

x

⁻⁷

)

^1/36

for x

^67/104

< vx

^ε

≤ x

^0.665

, (v

⁵⁰

x

⁻²¹

)

^1/70

for x

^0.665

< vx

^ε

≤ x

^0.7

. Lemma 5. We have

S(A, P ) ≤ x

^1/2

V (P )



F  lnD lnP



+ O(ε)

 . P r o o f. By (3.1.1) it suffices to show that

X

p1∼A1

. . . X

pt∼At

X

v⁰≤p1...ptn≤ev⁰

 ψ

 x + x

^1/2

rp

1

. . . p

t

n



− ψ

 x

rp

1

. . . p

t

n



 x

^1/2−0.75η

, where

A

1

 A

₂

 . . .  A

_t

,

A

1

. . . A

2s

A

³_2s+1

≤ D

^1+η

for 0 ≤ s ≤ (t − 1)/2 , v

⁰

= vr

⁻¹

, r < D

^η

.

Then, by a standard reduction, it is enough to establish

(3.1.3) X

1≤h≤H

X

p1∼A1

. . . X

pt∼At

X

v≤rp1...ptn≤ev

e

 hx

⁰

p

1

. . . p

t

rn



 vx

^−η

, where H = vx

^−1/2+η

, x ∼ = x

⁰

. Our aim is to arrange {r, p

1

, . . . , p

t

, n} into two subsets, so that we can produce from (3.1.3) an exponential sum of the type of Section 2, and then (3.1.3) will follow from the estimate given there.

We claim that either

(3.1.4) there exists a subset S of {1, 2, . . . , t} with P ≤ Y

i∈S

A

i

≤ Qx

^−2η

:= Q

1

, or

(3.1.5) A

1

. . . A

t

≤ x

^3/8−22η

:= A

0

.

We assume the contrary, that is, neither (3.1.4) nor (3.1.5) is true, and deduce a contradiction. For v ≥ x

^11/18−ε

, (3.1.5) is obvious. For v <

x

^11/18−ε

we reason as follows. From A

³₁

≤ D

^1+η

< Q

³₁

, we have A

1

< P . As

A

0

< A

1

. . . A

t

 P

²

A

3

. . . A

t

, we have A

3

. . . A

t

> Q

1

. If A

3

A

4

< P , there

must be a least j such that A

3

A

4

. . . A

j

> P , hence A

3

A

4

. . . A

j

> Q

1

; but then

Q

1

< A

3

A

4

. . . A

j

 (A

3

. . . A

j−1

)(A

3

A

4

)

^1/2

< P

^3/2

,

a contradiction. If A

3

A

4

≥ P , then A

₃

A

4

> Q

1

, thus A

3

> Q

^1/2₁

. But now Q

^5/2₁

 A

₁

A

2

A

³₃

≤ D

^1+η

,

also a contradiction. The proof of Lemma 5 is finished.

Lemma 6. We have X

P ≤p<Q⁰

S(A

p

, p) ≥ X

P ≤p<Q⁰⁰

x

^1/2

V (p) p



f  ln(v/p) lnp



+ O(ε)

 , where

2Q

⁰⁰

= Q

⁰

=

 v

^10/37

for x

^0.6

 vx

^ε

≤ x

^11/18

, Q for x

^11/18

< vx

^ε

≤ x

^0.7

.

P r o o f. Let J be the integer such that 1 ≤ Q

⁰

/(2

^J

P ) < 2. Then

(3.1.6) X

P ≤p<Q⁰

S(A

p

, p) ≥ X

1≤j≤J

X

p∼2^j−1P

S(A

p

, 2

^j

P ) .

By (3.1.2), we have, with D

j

= (v/(2

^j

P ))x

^−ε

, P

j

= 2

^j

P , (3.1.7) S(A

p

, 2

^j

P ) ≥ x

^1/2

p V (2

^j

P )



f  lnD

_j

lnP

j

 + O(ε)



− X

d<Dj

λ

d

R(A, pd) . We now show that for each fixed j, 1 ≤ j ≤ J , we have

(3.1.8) X

p∼Pj/2,d<Dj

λ

d

R(A, pd)  x

^1/2−η/4

.

Again, by the standard argument using the Fourier expansion of ψ(ξ), to prove (3.1.8) it suffices to prove

X

1≤h≤H

X

p∼Pj/2

X

d≤Dj

X

v≤pds≤ev

e  hx

⁰

pds



 vx

^−η

,

where H = vx

^−1/2+η

, x ∼ = x

⁰

. At this stage the condition v ≤ pds ≤ ev can be removed by a familiar lemma (cf. [1]). Thus it is enough to get

(3.1.9) X

1≤h≤H

X

n∼N

X

m∼M

a(n)b(m)e  hξ mn



 vx

^−η

,

where N = P

j

/2, M N ∼ = v, ξ ∼ = x, |a(n)| ≤ x

^η

, |b(m)| ≤ x

^η

. The above

exponential sum is just of the form we considered in Section 2. As P 

N = P

j

/2  Q for x

^0.6

 vx

^ε

≤ x

^0.7

, we find that

(3.1.9) follows from



 

 

 

 

 

 

Lemma 1.2 when x

^0.6

 vx

^ε

≤ x

^27/44

; Lemma 2 when x

^27/44

< vx

^ε

≤ x

^67/104

; Lemma 3 with (k, λ) = (2/7, 4/7) when

x

^67/104

< vx

^ε

≤ x

^0.665

;

Lemma 3 with (k, λ) = (11/30, 16/30) when x

^0.665

< vx

^ε

≤ x

^0.7

.

Hence (3.1.8) is true. Using the asymptotic formula for V (·) and the well-known property of f (·), we have, for p ∼ P

j

/2 = 2

^j−1

P ,

(3.1.10)

V (2

^j

P ) = V (p)

 1 + O

 1 lnx



, f  lnD

j

lnP

j



= f  ln(v/p) lnp



+ O(ε) . Lemma 6 follows from (3.1.6)–(3.1.8) and (3.1.10).

3.2. The contribution of the range (0.6 − ε, 11/18 − ε). We prove Proposition 2. We have

X

21

:= X

x^0.6−ε≤p≤x^11/18−ε

N (p) lnp < 0.02278x

^1/2

lnx .

Unless otherwise specified, all symbols have the same meaning as in Subsection 3.1. Now, v satisfies

e

⁻¹

x

^0.6−ε

≤ v ≤ x

^11/18−ε

. Lemma 3.2.1.

X

p1p2∈A Q⁰≤p1≤min(p2,Q)

1 = x

^1/2

lnv



ln  5.4 − 8.1s 4s − 2



+ O(ε)

 ,

where s = (lnv)/(lnx).

P r o o f. It is clear that X

p1p2∈A Q⁰≤p1≤min(p2,Q)

1 = X

Q⁰≤p1≤Q

X

v≤p1p2≤ev

X

x≤p1p2m≤x+x^1/2

1 = U + V ,

where

U = X

Q⁰≤p1≤Q

X

v≤p1p2≤ev

x

^1/2

p

1

p

2

,

V = X

Q⁰≤p1≤Q

X

v≤p1p2≤ev



ψ  x + x

^1/2

p

1

p

2



− ψ

 x p

1

p

2



.

By the Prime Number Theorem, we easily deduce that U = x

^1/2

 X

Q⁰≤p1≤Q

1

p

1

ln(v/p

1

) + O((lnx)

⁻²

)



= x

^1/2

lnv



ln  5.4 − 8.1s 4s − 2



+ O(ε)



; and the argument in the proof of Lemma 6 gives

V = O(x

^1/2

(lnx)

⁻²

) . This proves Lemma 3.2.1.

P r o o f o f P r o p o s i t i o n 2. Let A = 0.6 − ε, B = 11/18 − ε, and L = lnx. Then

X

21

≤ X

AL−1≤k≤BL

(k + 1) X

e^k≤p≤e^k+1

N (p) . For v = e

^k

, by Buchstab’s identity it is easy to verify that

X

v≤p≤ev

N (p) ≤ S(A, P ) − X

P ≤p<Q⁰

S(A

p

, p) − X

p1p2∈A Q⁰≤p1≤min(p2,Q)

1 ,

which, in conjunction with Lemmata 5, 6, and 3.2.1, leads to the estimate X

21

≤ x

^1/2

L(I

1

− I

₂

− I

₃

+ O(ε)) , where (γ being the Euler constant)

e

^γ

I

1

=

B

R

A

s

s − 1/2 F  10 − 15s 2s − 1

 ds ,

e

^γ

I

2

=

B

R

A g(s)

R

2.7

f (t) dt ds , g(s) = 1/(2s − 1) ,

I

3

=

B

R

A

ln  5.4 − 8.1s 4s − 2

 ds . The following formulae are well known:

(3.2.1) F (u) = 2e

^γ

u (1 + B(u)) , B(u) =



 

 

0 for 0 < u ≤ 3,

u−1

R

2

ln(t − 1)

t dt for 3 ≤ u ≤ 5;

(3.2.2) f (u) = 2e

^γ

ln(u − 1)

u for 2 ≤ u ≤ 4 ;