Sequences with bounded l.c.m. of each pair of terms by

LXXXIV.1 (1998)

Sequences with bounded l.c.m. of each pair of terms

by

Yong-Gao Chen (Nanjing)

0. Introduction. Let A

x

be a set of positive integers with the least common multiple of each pair of terms not exceeding x and |A

_x

| being the largest. In 1951, P. Erd˝os [3] proposed the following problem: what is the value of |A

_x

|? It is known that

q

9

8

x + O(1) ≤ |A

_x

| ≤ √

4x + O(1).

For a proof one may see Erd˝os [4]. The problem is problem E2 and a part of problem B26 in the well known problem book [5] of Guy. Choi [1] improved the upper bound to 1.638 √

x, and later [2] to 1.43 √ x.

In number theory, it is rare to give an asymptotic formula for such a problem. In this paper an asymptotic formula for |A

_x

| is given. Further, let B

_x

be the union of the set of positive integers not exceeding p

x/2 and the set of even integers between p

x/2 and √

2x. It is clear that the least common multiple of each pair of terms of B

x

does not exceed x. We will show that A

_x

is almost the same as B

_x

. That is,

Theorem. We have

|A

_x

\ B

_x

| = o( √ x).

In particular ,

|A

x

| = |B

x

| + o( √ x) =

q

9

8

x + o( √ x).

Note. From the proof of the Theorem we will see that o( √

x) can be given explicitly. By the Theorem we have

|A

x

∩ B

x

| = |A

x

| − |A

x

\ B

x

| = q

9

8

x + o( √ x)

1991 Mathematics Subject Classification: Primary 11B83.

Supported by Fok Ying Tung Education Foundation and the National Natural Science Foundation of China.

[71]

and

|B

x

\ A

x

| = |B

x

| − |A

x

∩ B

x

| = o( √ x).

1. Preliminary lemmas

Lemma 1. Let M be an integer with M ≥ 3, and let c

₀

, c

₁

and c

₂

be real numbers with c

₁

≥ c

₀

> 0. Then there exists an x

₀

= x

₀

(M, c

₀

) such that if x ≥ x

0

and a

i

, b

i

(1 ≤ i ≤ t ≤ M/2) are integers with (a

i

, b

i

) = 1 (1 ≤ i ≤ t) and with each prime factor of Q

_t

i=1

(a

_i

n + b

_i

) exceeding M for any integer n, then there exists an integer k ∈ (c

₁

x

^1/2

+ c

₂

, c

₁

(x

^1/2

+ x

^1/4

) + c

₂

) such that each prime factor of Q

_t

i=1

(a

i

k + b

i

) exceeds 1

6 log M log x.

P r o o f. We employ the standard Eratosthenes–Legendre sieve. One may refer to [6], p. 31, Theorem 1.1. We take

A = n Y

^t

i=1

(a

i

k + b

i

) : k ∈ (c

1

x

^1/2

+ c

2

, c

1

(x

^1/2

+ x

^1/4

) + c

2

) o

, P = P

1

, z = 1

6 log M log x, X = c

1

x

^1/4

, A

0

=

¹₂

M, ω(p) being the number of solutions of

Y

t i=1

(a

_i

n + b

_i

) ≡ 0 (mod p).

Noting that P = ∅ we have |R

_d

| = |r

_d

| ≤ ω(d) if µ(d) 6= 0. By Theorem 1.1 of [6], p. 31, we have

S(A; P, z) = XW (z) + θ(1 + A

₀

)

^z

= c

₁

x

^1/4

Y

p≤z



1 − ω(p) p

 + θ

 1 + 1

2 M



_z

≥ c

0

x

^1/4

Y

M <p≤z

 1 − M

2p



− M

^z

 x

^1/4

(log log x)

^M/2

, where |θ| ≤ 1 and  depends only on M and c

0

. From this we obtain the assertion of Lemma 1.

Note. a

i

and b

i

may depend on x, c

0

, c

1

and c

2

. x

0

(M, c

0

) can be effec- tively computed. For a stronger result one should use Brun’s sieve. Here the conclusion is sufficient for the present paper.

Lemma 2. Let c

i

(3 ≤ i ≤ 6) be nonnegative real numbers with c

4

> c

3

.

Let D and M be integers with |D| ≤ c

₅

x

^c6

and with each prime factor of D

exceeding

1

6 log M log x.

Then the number of a with (a, D) > 1, a ∈ [c

3

x

^1/2

, c

4

x

^1/2

] is O( √

x/log log x), where O depends only on M and c

_i

(3 ≤ i ≤ 6).

P r o o f. If D = 0, then x ≤ M

¹²

and the conclusion is trivial. Now we assume that D 6= 0. Let |D| = p

^l₁¹

p

^l₂²

. . . p

^l_r^r

be the standard factorization of

|D|. Then r log

 1

6 log M log x



≤ X

r i=1

log p

_i

≤ log |D|  log x.

Thus

r  log x log log x . Hence

X

a∈[c3

√x,c4

√x]

(a,D)>1

1 ≤ X

r i=1

X

a∈[c3

√x,c4

√x]

pi|a

1 ≤ X

r i=1

 (c

₄

− c

₃

) √ x p

i

+ 1



≤ (c

₄

− c

₃

)r √ x

log x 6 log M + r 

√ x log log x . This completes the proof of Lemma 2.

2. General lemmas. For an interval I = (a, b], let

|I √

x ∩ A

_x

| = α(I)|I| √ x,

|I √

x ∩ A

_x

∩ (2Z)| = α

⁽⁰⁾

(I)|I| √ x,

|I √

x ∩ A

_x

∩ (2Z + 1)| = α

⁽¹⁾

(I)|I| √ x,

where |X| denotes the number of elements of X or the length of an interval X, and 2Z and 2Z+1 denote the sets of all even integers and all odd integers respectively. Let I = {I

1

, . . . , I

l

} be a set of pairwise disjoint intervals with I

_i

= (a

_i

, b

_i

] and 0 < a

₀

< a

₁

< . . . < a

_l

. Let

α

i

= α(I

i

), α

⁽⁰⁾_i

= α

⁽⁰⁾

(I

i

), α

⁽¹⁾_i

= α

⁽¹⁾

(I

i

), M = 4(1 + [a

²_l

]), where [a

²_l

] denotes the integral part of a

²_l

. It is clear that α

i

= α

⁽⁰⁾_i

+ α

⁽¹⁾_i

.

Lemma 3. Let r

_ij

(j = 1, . . . , k

_i

; i = 1, . . . , l) be distinct integers with

|r

_ij

− r

_uv

| ≤ g(r

_ij

, r

_uv

)a

_i

a

_u

,

where g(a, b) = 1 +

¹₄

(1 − (−1)

^a

)(1 − (−1)

^b

). Let k

_i⁽⁰⁾

= |{r

ij

: 2 | r

ij

, j = 1, . . . , k

i

}|, k

_i⁽¹⁾

= k

_i

− k

⁽⁰⁾_i

, i = 1, . . . , l.

Then

X

l i=1

(k

⁽⁰⁾_i

α

_i⁽⁰⁾

+ k

⁽¹⁾_i

α

⁽¹⁾_i

) ≤ 1

2 + O((log log x)

^−1/2

), where O depends only on I.

P r o o f. Let K = P

_l

i=1

k

_i

. If K = 0 or 1, then by the definitions of α

⁽⁰⁾_i

and α

⁽¹⁾_i

the assertion of Lemma 3 is true. In the following we assume that K ≥ 2. Let δ be a small positive number which will be determined later, and let

I

_i

(t) = (a

_i

+ tδ, a

_i

+ (t + 1)δ].

For the (index) set

{t

ij

: 0 ≤ t

ij

≤ |I

i

|/δ − 1, t

ij

∈ Z, j = 1, . . . , k

i

; i = 1, . . . , l}

we first show that  

 [

i,j

(I

i

(t

ij

) √

x ∩ A

x

∩ (2Z + r

ij

))    ≤ 1

2 δ √ x + O

 √

x log log x

 , where O depends only on I. To do this we consider the set

∆(a) = [

i,j

{M !l

_ij

+ r

_ij

+ 2a},

where l

ij

are integers which will be determined later such that (1) (a

_i

+ t

_ij

δ) √

x ≤ M !l

_ij

+ r

_ij

≤ (a

_i

+ t

_ij

δ)( √

x + x

^1/4

) hold for j = 1, . . . , k

_i

; i = 1, . . . , l. For convenience we rewrite ∆(0) as

∆(0) = {M !l

1

+ r

1

, M !l

2

+ r

2

, . . . , M !l

K

+ r

K

}.

Since a

_i

a

_u

< M/4 (i, u = 1, . . . , l), by the conditions of Lemma 3 we have

|r

_i

− r

_j

| < M/2, i, j = 1, . . . , K,

whence K ≤ M/2. Now we take l

1

satisfying (1). Suppose that we have chosen l

₁

, . . . , l

_u

(u < K). By Lemma 1 for x ≥ x

₀

(M, a

₀

/M !) there exists a l

_u+1

satisfying (1) such that each prime factor of

Y

u i=1

 M ! r

u+1

− r

i

l

_u+1

− M ! r

u+1

− r

i

l

_i

+ 1



exceeds

1

6 log M log x.

Thus by induction we have determined all l

u

(1 ≤ u ≤ K). Let

D = Y

1≤v<u≤K

 M !

r

_u

− r

_v

l

u

− M !

r

_u

− r

_v

l

v

+ 1

 . Then each prime factor of D exceeds

1

6 log M log x and by (1),

|D| ≤ Y

1≤v<u≤K

|M !l

_u

+ r

_u

− M !l

_v

− r

_v

|

≤ (2b

l

√ x)

^K(K−1)/2

≤ (2b

l

)

^{M (M −1)}

x

^{M (M −1)}

. By Lemma 2, the number of a such that (a, D) > 1 and a ∈ (0, b

l

√ x] is O( √

x/log log x), where O depends only on I. Let B =

n a : a ∈

[

l i=1

(I

i

√ x ∩ Z), (a, D) = 1 o

. If a ∈ (0, δ √

x/2] and

M !l

_u

+ r

_u

+ 2a ∈ B, M !l

_v

+ r

_v

+ 2a ∈ B, then for u 6= v we have

(2) (M !l

_u

+ r

_u

+ 2a, M !l

_v

+ r

_v

+ 2a)

= (M !l

u

+ r

u

+ 2a, M !(l

v

− l

u

) + r

v

− r

u

)

= (M !l

_u

+ r

_u

+ 2a, r

_v

− r

_u

) ≤ g(r

_u

, r

_v

)

⁻¹

|r

_u

− r

_v

|.

Thus for a ∈ (0, δ √

x/2] with M !l

_ij

+ r

_ij

+ 2a ∈ B,

M !l

uv

+ r

uv

+ 2a ∈ B, (i − u)

²

+ (j − v)

²

6= 0, by (1), (2) and the conditions of the lemma we have

l.c.m.{M !l

_ij

+ r

_ij

+ 2a, M !l

_uv

+ r

_uv

+ 2a}

= (M !l

ij

+ r

ij

+ 2a)(M !l

uv

+ r

uv

+ 2a) (M !l

_ij

+ r

_ij

+ 2a, M !l

_uv

+ r

_uv

+ 2a)

> (a

i

+ t

ij

δ)(a

u

+ t

uv

δ)x

|r

ij

− r

uv

| g(r

ij

, r

uv

)

≥ (a

_i

+ t

_ij

δ)(a

_u

+ t

_uv

δ)x a

i

a

u

≥ x.

So |∆(a) ∩ B ∩ A

x

| ≤ 1. Since (see (1)) I

_i

(t

_ij

) √

x ∩ (2Z + r

_ij

)

⊆ ((M !l

_ij

+ r

_ij

, M !l

_ij

+ r

_ij

+ δ √ x]

∪ ((a

_i

+ t

_ij

δ) √

x, (a

_i

+ t

_ij

δ)( √

x + x

^1/4

)]) ∩ (2Z + r

_ij

)

⊆

 [

0<a≤δ√ x/2

{M !l

ij

+ r

ij

+ 2a}



∪ (((a

i

+ t

ij

δ) √

x, (a

i

+ t

ij

δ)( √

x + x

^1/4

)] ∩ Z), we have

[

i,j

(I

_i

(t

_ij

) √

x ∩ (2Z + r

_ij

))

⊆

 [

0<a≤δ√ x/2

[

i,j

{M !l

_ij

+ r

_ij

+ 2a}



∪  [

i,j

(((a

i

+ t

ij

δ) √

x, (a

i

+ t

ij

δ)( √

x + x

^1/4

)] ∩ Z)



⊆

 [

0<a≤δ√ x/2

∆(a)



∪  [

i,j

(((a

_i

+ t

_ij

δ) √

x, (a

_i

+ t

_ij

δ)( √

x + x

^1/4

)] ∩ Z)

 . Hence

   [

i,j

(I

_i

(t

_ij

) √

x ∩ A

_x

∩ B ∩ (2Z + r

_ij

))   

≤ 1 2 δ √

x + X

i,j

((a

_i

+ t

_ij

δ)x

^1/4

+ 1) ≤ 1 2 δ √

x + X

i,j

((a

_i

+ |I

_i

|)x

^1/4

+ 1)

≤ 1 2 δ √

x + X

i,j

(b

i

x

^1/4

+ 1) ≤ 1 2 δ √

x + K max

i

b

i

x

^1/4

+ K

≤ 1 2 δ √

x + O(x

^1/4

),

where O depends only on I (note that K ≤ M ), whence  

 [

i,j

(I

_i

(t

_ij

) √

x ∩ A

_x

∩ (2Z + r

_ij

))   

≤ 1 2 δ √

x + O(x

^1/4

) + O

 √

x log log x



≤ 1 2 δ √

x + O

 √

x log log x



,

where O depends only on I. Since I

₁

, . . . , I

_l

are pairwise disjoint, we have

   [

i≤l−1

[

j

(I

i

(t

ij

) √

x ∩ A

x

∩ (2Z + r

ij

))    +

  

k_l

[

j=1 2|rlj

(I

l

(t

lj

) √

x ∩ A

x

∩ (2Z))   

+   

kl

[

j=1 2

-

r_lj

(I

_l

(t

_lj

) √

x ∩ A

_x

∩ (2Z + 1))   

≤ 1 2 δ √

x + O

 √

x log log x

 . Hence, if k

⁽¹⁾_l

≥ 1 and u ≥ 0, then

   [

i≤l−1

[

j

(I

_i

(t

_ij

) √

x ∩ A

_x

∩ (2Z + r

_ij

))    +

  

k_l

[

j=1 2|r_lj

(I

_l

(t

_lj

) √

x ∩ A

_x

∩ (2Z))   

+   

k⁽¹⁾_l

[

−1 r=0 k⁽¹⁾_l u+r≤|Il|/δ−1

(I

_l

(k

_l⁽¹⁾

u + r) √

x ∩ A

_x

∩ (2Z + 1))   

≤ 1 2 δ √

x + O

 √

x log log x

 . Thus 

|I

_l

| k

_l⁽¹⁾

δ

 + 1

   [

i≤l−1

[

j

(I

_i

(t

_ij

) √

x ∩ A

_x

∩ (2Z + r

_ij

))   

+

 |I

_l

| k

⁽¹⁾_l

δ

 + 1

  

kl

[

j=1 2|rlj

(I

_l

(t

_lj

) √

x ∩ A

_x

∩ (2Z))   

+ X

0≤u≤[|Il|/(k⁽¹⁾_l δ)]

  

k⁽¹⁾_l

[

−1 r=0 k⁽¹⁾_l u+r≤|Il|/δ−1

(I

l

(k

_l⁽¹⁾

u + r) √

x ∩ A

x

∩ (2Z + 1))   

≤ 1 2 δ √

x

 |I

l

| k

⁽¹⁾_l

δ

 + 1

 + O

 |I

l

| k

⁽¹⁾_l

δ

 + 1

 √

x log log x

 . Hence

|I

l

| k

_l⁽¹⁾

δ

   [

i≤l−1

[

j

(I

_i

(t

_ij

) √

x ∩ A

_x

∩ (2Z + r

_ij

))   

+ |I

_l

| k

_l⁽¹⁾

δ

  

k_l

[

j=1 2|r_lj

(I

_l

(t

_lj

) √

x ∩ A

_x

∩ (2Z))

 



+  

 [

0≤t≤|Il|/δ−1

(I

_l

(t) √

x ∩ A

_x

∩ (2Z + 1))   

≤ 1 2 δ √

x

 |I

_l

| k

⁽¹⁾_l

δ + 1

 + O

 |I

_l

| k

_l⁽¹⁾

δ + 1

 √

x log log x

 . So 

  [

i≤l−1

[

j

(I

i

(t

ij

) √

x ∩ A

x

∩ (2Z + r

ij

))   

+   

k_l

[

j=1,2|rlj

(I

_l

(t

_lj

) √

x ∩ A

_x

∩ (2Z))   

+ k

_l⁽¹⁾

δ

|I

_l

|  

 [

0≤t≤|Il|/δ−1

(I

_l

(t) √

x ∩ A

_x

∩ (2Z + 1))   

≤ 1 2 δ √

x + 1

2 · k

_l⁽¹⁾

δ

²

|I

_l

|

√ x + O



1 + k

⁽¹⁾_l

δ

|I

_l

|

 √

x log log x



≤ 1 2 δ √

x + 1 2 · Kδ

²

|I

_l

|

√ x + O



1 + Kδ

|I

_l

|

 √

x log log x



≤ 1 2 δ √

x + O

 δ

²

√

x +

√ x log log x

 . Noting that

 

 [

0≤t≤|I_l|/δ−1

(I

l

(t) √

x ∩ A

x

∩ (2Z + 1))   

= |I

_l

√

x ∩ A

_x

∩ (2Z + 1)| − θ

_l⁽¹⁾

δ √

x (0 ≤ θ

⁽¹⁾_l

≤ 1)

= α

⁽¹⁾_l

|I

_l

| √

x − θ

⁽¹⁾_l

δ √ x, we have

(3)    [

i≤l−1

[

j

(I

i

(t

ij

) √

x ∩ A

x

∩ (2Z + r

ij

))   

+   

kl

[

j=1,2|rlj

(I

_l

(t

_lj

) √

x ∩ A

_x

∩ (2Z))  

 + k

⁽¹⁾_l

α

_l⁽¹⁾

δ √ x

≤ 1 2 δ √

x + θ

⁽¹⁾_l

k

⁽¹⁾_l

δ

²

|I

l

|

√ x + O

 δ

²

√

x +

√ x log log x



≤ 1 2 δ √

x + O

 δ

²

√

x +

√ x log log x



.

It is clear that if k

_l⁽¹⁾

= 0, (3) also holds. Similarly, we have  

 [

i≤l−1

[

j

(I

_i

(t

_ij

) √

x ∩ A

_x

∩ (2Z + r

_ij

))  

 + k

⁽⁰⁾_l

α

⁽⁰⁾_l

δ √

x + k

_l⁽¹⁾

α

⁽¹⁾_l

δ √ x

≤ 1 2 δ √

x + O

 δ

²

√

x +

√ x log log x

 . Continuing this procedure we have

X

l i=1

(k

_i⁽⁰⁾

α

⁽⁰⁾_i

+ k

⁽¹⁾_i

α

⁽¹⁾_i

)δ √ x ≤ 1

2 δ √ x + O

 δ

²

√

x +

√ x log log x

 , where O depends only on I. Taking δ = (log log x)

^−1/2

, we have

X

l i=1

(k

⁽⁰⁾_i

α

_i⁽⁰⁾

+ k

⁽¹⁾_i

α

⁽¹⁾_i

) ≤ 1

2 + O((log log x)

^−1/2

).

This completes the proof of Lemma 3.

Corollary. Let the conditions be as in Lemma 3 and m

⁽⁰⁾₁

, . . . , m

⁽⁰⁾_l

, m

⁽¹⁾₁

, . . . , m

⁽¹⁾_l

be nonnegative integers with

(4)

X

t i=1

m

^(v)_i

≤ X

t i=1

k

^(v)_i

, t = 1, . . . , l; v = 0, 1.

Then

X

l i=1

(m

⁽⁰⁾_i

α

_i⁽⁰⁾

+ m

⁽¹⁾_i

α

_i⁽¹⁾

) ≤ 1

2 + O((log log x)

^−1/2

), where O depends only on I.

P r o o f. Let m

i

= m

⁽⁰⁾_i

+ m

⁽¹⁾_i

. By (4) we may rearrange {r

ij

} as {w

_ij

: i = 1, . . . , l; j = 1, . . . , m

_i

} ∪ A

such that w

_ij

= r

_uv

implies that u ≤ i, and

m

⁽⁰⁾_i

= |{w

_ij

: 2 | w

_ij

, j = 1, . . . , m

_i

}|.

Thus

|w

ij

− w

i⁰j⁰

| = |r

uv

− r

u⁰v⁰

| ≤ g(r

uv

, r

u⁰v⁰

)a

u

a

u⁰

≤ g(w

ij

, w

i⁰j⁰

)a

i

a

i⁰

. Then the Corollary follows from Lemma 3.

Lemma 4. Let m, n

1

, . . . , n

r

be nonnegative integers with m ≤ n

1

+ . . . . . . + n

_r

. Then there exist nonnegative integers m

₁

, . . . , m

_r

such that

m = m

1

+ . . . + m

r

and m

i

≤ n

i

, i = 1, . . . , r.

The proof is clear.

Lemma 5. Let the conditions be as in Lemma 3. Let β

₁^(v)

, . . . , β

^(v)_l

(v = 0, 1) be nonnegative real numbers with

(5)

X

t i=1

β

_i^(v)

≤ X

t i=1

k

_i^(v)

, t = 1, . . . , l; v = 0, 1.

Then

X

l i=1

(β

⁽⁰⁾_i

α

_i⁽⁰⁾

+ β

_i⁽¹⁾

α

⁽¹⁾_i

) ≤ 1

2 + O((log log x)

^−1/2

), where O depends only on I.

P r o o f. Let n and n

^(v)_i

(1 ≤ i ≤ l; v = 0, 1) be nonnegative integers with

n

^(v)_i

n ≤ β

_i^(v)

< n

^(v)_i

+ 1

n , i = 1, . . . , l; v = 0, 1.

Then (5) implies that (6)

X

t i=1

n

^(v)_i

≤ n X

t

i=1

k

_i^(v)

, t = 1, . . . , l; v = 0, 1.

Now we use induction on t to prove the following proposition P (t): There exist nonnegative integers n

^(v)_ij

(1 ≤ i ≤ t; 1 ≤ j ≤ n; v = 0, 1) such that

n

^(v)_i

= X

n j=1

n

^(v)_ij

, v = 0, 1; i = 1, . . . , t,

and X

s

i=1

n

^(v)_ij

≤ X

s i=1

k

_i^(v)

, s = 1, . . . , t; j = 1, . . . , n; v = 0, 1.

By (6) and Lemma 4, P (1) is true. Suppose that P (t) (1 ≤ t < l) is true.

Now by (6) and the induction hypothesis we have n

^(v)_t+1

≤ n

X

t+1 i=1

k

^(v)_i

− X

t

i=1

n

^(v)_i

≤ X

n j=1

 X

^t+1

i=1

k

_i^(v)

− X

t i=1

n

^(v)_ij



and

X

t+1 i=1

k

_i^(v)

− X

t i=1

n

^(v)_ij

≥ 0.

By Lemma 4 there exist nonnegative integers n

^(v)_(t+1)j

(1 ≤ j ≤ n; v = 0, 1) such that

n

^(v)_(t+1)j

≤ X

t+1 i=1

k

_i^(v)

− X

t i=1

n

^(v)_ij

, j = 1, . . . , n,

and

n

^(v)_t+1

= X

n j=1

n

^(v)_(t+1)j

, v = 0, 1.

So P (t + 1) is true. Hence P (t) is true for all t, 1 ≤ t ≤ l. In particular, P (l) is true, that is, there exist nonnegative integers n

^(v)_ij

(1 ≤ i ≤ l; 1 ≤ j ≤ n;

v = 0, 1) such that n

^(v)_i

=

X

n j=1

n

^(v)_ij

, v = 0, 1; i = 1, . . . , l, and

X

t i=1

n

^(v)_ij

≤ X

t i=1

k

^(v)_i

, t = 1, . . . , l; j = 1, . . . , n; v = 0, 1.

By the Corollary of Lemma 3 we have X

l

i=1

(n

⁽⁰⁾_ij

α

⁽⁰⁾_i

+ n

⁽¹⁾_ij

α

_i⁽¹⁾

) ≤ 1

2 + O((log log x)

^−1/2

), j = 1, . . . , n.

Hence

X

l i=1

(n

⁽⁰⁾_i

α

⁽⁰⁾_i

+ n

⁽¹⁾_i

α

⁽¹⁾_i

) ≤ 1

2 n + O(n(log log x)

^−1/2

), that is,

X

l i=1

 n

⁽⁰⁾_i

n α

⁽⁰⁾_i

+ n

⁽¹⁾_i

n α

⁽¹⁾_i



≤ 1

2 + O((log log x)

^−1/2

),

where O depends only on I. Letting n → ∞ we obtain the statement of Lemma 5.

Lemma 6. Let X

l i=1

(k

_ij⁽⁰⁾

α

⁽⁰⁾_i

+ k

⁽¹⁾_ij

α

⁽¹⁾_i

) ≤ 1

2 + O((log log x)

^−1/2

)

(j = 1, . . . , r) be r relations obtained by using Lemma 3 (not necessarily from the same {r

_ij

}). Let β

₁^(v)

, . . . , β

_l^(v)

, δ

₁

, . . . , δ

_r

(v = 0, 1) be nonnegative real numbers with

(7)

X

t i=1

β

_i^(v)

≤ X

t

i=1

X

r j=1

δ

_j

k

^(v)_ij

, t = 1, . . . , l; v = 0, 1.

Then X

l i=1

(β

_i⁽⁰⁾

α

⁽⁰⁾_i

+ β

_i⁽¹⁾

α

⁽¹⁾_i

) ≤ 1 2

X

r j=1

δ

_j

+ O

 X

^r

j=1

δ

_j

(log log x)

^−1/2



,

where O depends only on I.

P r o o f. As in Lemma 4, if u and v

j

(1 ≤ j ≤ r) are nonnegative real numbers with u ≤ P

_r

j=1

δ

_j

v

_j

, then there exist nonnegative real numbers u

₁

, . . . , u

_r

such that

u = X

r j=1

δ

j

u

j

, u

j

≤ v

j

, j = 1, . . . , r.

Using this fact and (7) we infer, as in the proof of Lemma 5, that there exist nonnegative real numbers β

_ij^(v)

(i = 1, . . . , l; j = 1, . . . , r; v = 0, 1) such that

β

_i^(v)

= X

r j=1

δ

_j

β

_ij^(v)

, i = 1, . . . , l; v = 0, 1, and

X

t i=1

β

_ij^(v)

≤ X

t i=1

k

^(v)_ij

, t = 1, . . . , l; j = 1, . . . , r; v = 0, 1.

By Lemma 5 we have X

l

i=1

(β

⁽⁰⁾_ij

α

_i⁽⁰⁾

+ β

_ij⁽¹⁾

α

⁽¹⁾_i

) ≤ 1

2 + O((log log x)

^−1/2

).

Hence X

l i=1

 X

^r

j=1

δ

_j

β

_ij⁽⁰⁾

α

⁽⁰⁾_i

+ X

r j=1

δ

_j

β

_ij⁽¹⁾

α

⁽¹⁾_i



≤ 1 2

X

r j=1

δ

_j

+ O

 X

^r

j=1

δ

_j

(log log x)

^−1/2

 . That is,

X

l i=1

(β

_i⁽⁰⁾

α

⁽⁰⁾_i

+ β

_i⁽¹⁾

α

⁽¹⁾_i

) ≤ 1 2

X

r j=1

δ

_j

+ O

 X

^r

j=1

δ

_j

(log log x)

^−1/2

 , where O depends only on I. This completes the proof of Lemma 6.

Lemma 7. Let r

_ij

(j = 1, . . . , k

_i

; i = 1, . . . , l) be distinct integers with

|r

_ij

− r

_uv

| ≤ a

_i

a

_u

. Then

X

l i=1

k

_i

α

_i

≤ 1 + O((log log x)

^−1/2

), where O depends only on I.

P r o o f. By Lemma 3 we have (8)

X

l i=1

(k

⁽⁰⁾_i

α

_i⁽⁰⁾

+ k

⁽¹⁾_i

α

⁽¹⁾_i

) ≤ 1

2 + O((log log x)

^−1/2

).

Let

w

_ij

= r

_ij

+ 1,

n

_i

= k

_i

, j = 1, . . . , n

_i

; i = 1, . . . , l.

Then

|w

ij

− w

uv

| ≤ a

i

a

u

, n

⁽⁰⁾_i

= k

⁽¹⁾_i

, n

⁽¹⁾_i

= k

⁽⁰⁾_i

. By Lemma 3 we have

X

l i=1

(n

⁽⁰⁾_i

α

_i⁽⁰⁾

+ n

⁽¹⁾_i

α

⁽¹⁾_i

) ≤ 1

2 + O((log log x)

^−1/2

).

That is, (9)

X

l i=1

(k

⁽¹⁾_i

α

_i⁽⁰⁾

+ k

⁽⁰⁾_i

α

⁽¹⁾_i

) ≤ 1

2 + O((log log x)

^−1/2

).

By (8), (9), k

_i⁽⁰⁾

+ k

⁽¹⁾_i

= k

i

and α

⁽⁰⁾_i

+ α

⁽¹⁾_i

= α

i

, we have X

l

i=1

k

i

α

i

≤ 1 + O((log log x)

^−1/2

).

This completes the proof of Lemma 7.

Lemma 8. Let X

l i=1

k

ij

α

i

≤ 1 + O((log log x)

^−1/2

), j = 1, . . . , r,

be r relations obtained by using Lemma 7 (not necessarily from the same {r

_ij

}). Let β

₁

, . . . , β

_l

, δ

₁

, . . . , δ

_r

be nonnegative real numbers with

(10)

X

t i=1

β

_i

≤ X

t i=1

X

r j=1

δ

_j

k

_ij

, t = 1, . . . , l.

Then

X

l i=1

β

_i

α

_i

≤ X

r j=1

δ

_j

+ O

 X

^r

j=1

δ

_j

(log log x)

^−1/2

 , where O depends only on I.

P r o o f. By (10) and k

ij

= k

_ij⁽⁰⁾

+ k

⁽¹⁾_ij

we have X

t

i=1

β

i

≤ X

t i=1

X

r j=1

δ

j

k

_ij⁽⁰⁾

+ X

t i=1

X

r j=1

δ

j

k

⁽¹⁾_ij

, t = 1, . . . , l.

By the argument used in the proof of Lemma 5 there exist nonnegative real numbers β

_i^(v)

(1 ≤ i ≤ l; v = 0, 1) such that

X

t i=1

β

_i^(v)

≤ X

t i=1

X

r j=1

δ

j

k

_ij^(v)

, t = 1, . . . , l; v = 0, 1.

By the argument used in the proof of Lemma 7 we have for j = 1, . . . , r, X

l

i=1

(k

⁽⁰⁾_i

α

_i⁽⁰⁾

+ k

⁽¹⁾_i

α

⁽¹⁾_i

) ≤ 1

2 + O((log log x)

^−1/2

), X

l

i=1

(k

⁽¹⁾_i

α

_i⁽⁰⁾

+ k

⁽⁰⁾_i

α

⁽¹⁾_i

) ≤ 1

2 + O((log log x)

^−1/2

).

By Lemma 6 we have X

l

i=1

(β

_i⁽⁰⁾

α

⁽⁰⁾_i

+ β

_i⁽¹⁾

α

⁽¹⁾_i

) ≤ 1 2

X

r j=1

δ

_j

+ O

 X

^r

j=1

δ

_j

(log log x)

^−1/2

 , X

l

i=1

(β

_i⁽¹⁾

α

⁽⁰⁾_i

+ β

_i⁽⁰⁾

α

⁽¹⁾_i

) ≤ 1 2

X

r j=1

δ

_j

+ O

 X

^r

j=1

δ

_j

(log log x)

^−1/2

 . Since α

i

= α

⁽⁰⁾_i

+ α

⁽¹⁾_i

and β

i

= β

_i⁽⁰⁾

+ β

_i⁽¹⁾

, we have

X

l i=1

β

_i

α

_i

≤ X

r j=1

δ

_j

+ O

 X

^r

j=1

δ

_j

(log log x)

^−1/2

 . This completes the proof of Lemma 8.

3. The asymptotic formula for |A

x

|. Let L and S be suitable large integers and

q = 2

^1/(2L)

, I

_i

= (q

ⁱ

, q

ⁱ⁺¹

], T = 2LS − 1.

For positive real numbers α, β, let

B(α, β) = {a : a ∈ Z, 1 ≤ a ≤ αβ}

∪ {a : a ∈ Z, − min{αβ, α

⁻¹

β − 1} ≤ a ≤ 0}, A

_ij

=

 B(q

^j

, q

ⁱ

) if i ≥ j,

∅ if i < j.

In the following we make the convention that P

a∈∅

h(a) = 0 for any function h(t).

Lemma 9. Let 0 < α ≤ min{β, γ}. If a ∈ B(α, β) and b ∈ B(α, γ), then

|a − b| ≤ βγ.

P r o o f. If ab ≥ 0, then |a − b| ≤ max{|a|, |b|} ≤ max{αβ, αγ} ≤ βγ.

Now we assume that ab < 0. Without loss of generality, we may assume that a > 0 and b < 0. In this case we have αβ ≥ 1 and αγ ≥ 1. Thus

|a − b| = a − b ≤ αβ + min{αγ, α

⁻¹

γ − 1}

≤ αβ + α

⁻¹

γ − 1 = βγ + (β − α

⁻¹

)(α − γ) ≤ βγ.

This completes the proof of Lemma 9.

To use Lemma 8, let α = (10 − 7 √

2)/32,

k

ij

= |A

ij

\ A

_(i−1)j

|, −T ≤ i ≤ T, −T ≤ j ≤ L − 1, k

_iL

= 0 (−T ≤ i ≤ T, i 6= 0), k

_0L

= 1,

β

i

= q

ⁱ

(q − 1), −T ≤ i ≤ L − 1, β

_i

= (1 + α)q

ⁱ

(q − 1), L ≤ i ≤ T, δ

j

= q

^j

(q − 1), −T ≤ j ≤ −1,

δ

_j

=

¹₂

(q − 1)(q

^j

− q

^−j−1

), 0 ≤ j ≤ L − 1, δ

L

= 1 − q

⁻¹

.

Lemma 10. For −T ≤ j ≤ L, we have X

−T ≤i≤T

k

_ij

α

_i

≤ 1 + O((log log x)

^−1/2

), where O depends only on L and S.

P r o o f. The inequality X

T i=−T

k

_iL

α

_i

≤ 1 + O((log log x)

^−1/2

) can be deduced from Lemma 7 by taking

{r

ij

: j = 1, . . . , k

iL

; i = −T, . . . , T } = {r

01

= 1}.

Now we assume that −T ≤ j ≤ L − 1. If i < j, then

|A

ij

\ A

_(i−1)j

| = ∅.

If j ≤ i ≤ i

⁰

≤ T and

a ∈ A

_ij

\ A

_(i−1)j

, b ∈ A

_i0j

\ A

_(i0−1)j

,

then by Lemma 9 we have |a − b| ≤ q

ⁱ⁺ⁱ⁰

. Then Lemma 10 follows from Lemma 7.

Lemma 11. There exists a L

0

such that if L ≥ L

0

, then X

t

i=−T

β

i

≤ X

t i=−T

X

L j=−T

δ

j

k

ij

, t = −T, −T + 1, . . . , T.

P r o o f. For convenience let f (t) =

X

t i=−T

X

L j=−T

δ

_j

k

_ij

. That is,

f (t) = X

−T ≤i≤t

 X

−T ≤j≤−1

|A

_ij

\ A

_(i−1)j

|q

^j

(q − 1)

+ 1 2

X

0≤j≤L−1

|A

_ij

\ A

_(i−1)j

|(q − 1)(q

^j

− q

^−j−1

)



+ ε

_t

(1 − q

⁻¹

), where ε

_t

= 0 if t ≤ −1, and ε

_t

= 1 if t ≥ 0. Since

A

_ij

⊇ A

_(i−1)j

, A

_{(−T −1)j}

= ∅, j ≥ −T, we have

X

t i=−T

|A

_ij

\ A

_(i−1)j

| = |A

_tj

|, j ≥ −T.

Hence

f (t) = X

−T ≤j≤−1

X

−T ≤i≤t

|A

_ij

\ A

_(i−1)j

|q

^j

(q − 1) (11)

+ 1 2

X

0≤j≤L−1

X

−T ≤i≤t

|A

_ij

\ A

_(i−1)j

|(q − 1)(q

^j

− q

^−j−1

) + ε

_t

(1 − q

⁻¹

)

= X

−T ≤j≤−1

|A

_tj

|q

^j

(q − 1)

+ 1 2

X

0≤j≤L−1

|A

tj

|(q − 1)(q

^j

− q

^−j−1

) + ε

t

(1 − q

⁻¹

).

Case 1: −T ≤ t ≤ −1. Then by (11) we have f (t) ≥ X

−T ≤j≤t

|A

_tj

|q

^j

(q − 1) ≥ X

−T ≤j≤t

q

^j

(q − 1) ≥ X

−T ≤i≤t

β

_i

. Case 2: 0 ≤ t ≤ L − 1. Then by (11) we have

f (t) ≥ X

−T ≤j≤−t−1

|A

tj

|q

^j

(q − 1) + X

−t≤j≤−1

|A

tj

|q

^j

(q − 1)

+ 1 2

X

0≤j≤t

|A

_tj

|(q − 1)(q

^j

− q

^−j−1

) + 1 − q

⁻¹

≥ X

−T ≤j≤−t−1

q

^j

(q − 1) + 2 X

−t≤j≤−1

q

^j

(q − 1)

+ X

0≤j≤t

(q − 1)(q

^j

− q

^−j−1

) + 1 − q

⁻¹

≥ q

^t+1

− q

^−T

+ (1 − q

⁻¹

)(1 − q

^−t

)

≥ q

^t+1

− q

^−T

≥ X

−T ≤i≤t

β

_i

.

Case 3: L ≤ t ≤ 2L − 1. Then by (11) we have

f (t) ≥ X

−T ≤j≤−t−1

|A

tj

|q

^j

(q − 1) + X

−t≤j≤t−2L

|A

tj

|q

^j

(q − 1)

+ X

t−2L<j≤−1

|A

_tj

|q

^j

(q − 1) + 1 2

X

0≤j<2L−t

|A

_tj

|(q − 1)(q

^j

− q

^−j−1

)

+ 1 2

X

2L−t≤j≤L−1

|A

tj

|(q − 1)(q

^j

− q

^−j−1

) + 1 − q

⁻¹

≥ X

−T ≤j≤−t−1

q

^j

(q − 1) + 3 X

−t≤j≤t−2L

q

^j

(q − 1) + 2 X

t−2L<j≤−1

q

^j

(q − 1)

+ X

0≤j<2L−t

(q − 1)(q

^j

− q

^−j−1

) + 3 2

X

2L−t≤j≤L−1

(q − 1)(q

^j

− q

^−j−1

) + 1 − q

⁻¹

≥ q

^t+1

− q

^−T

+ α(q

^t+1

− q

^L

) + 9 4

√ 2 + √

2α + 1 − q

⁻¹

+ q

^t



− 1 2 q − 1

4 − αq



− 3q

^−t

≥ q

^t+1

− q

^−T

+ α(q

^t+1

− q

^L

) + 9 4

√ 2 + √

2α + 1 − q

⁻¹

+ min

 q

^L



− 1 2 q − 1

4 − αq



− 3q

^−L

, q

^2L−1



− 1 2 q − 1

4 − αq



− 3q

^−2L+1



≥ q

^t+1

− q

^−T

+ α(q

^t+1

− q

^L

)

≥ X

−T ≤i≤t

β

_i

.

The last inequality but one holds for all sufficiently large L (note that q = 2

^1/(2L)

).

Case 4: T ≥ t ≥ 2L. Then by (11) we have f (t) ≥ X

−T ≤j≤−1

|A

tj

|q

^j

(q − 1) + 1 2

X

0≤j≤L−1

|A

tj

|(q − 1)(q

^j

− q

^−j−1

)

(12)

= X

−T ≤j≤−1

q

^j

(q − 1)

 X

1≤a≤q^j+t

1 + X

− min{q^j+t,q^t−j−1}≤a≤0

1



+ 1 2

X

0≤j≤L−1

(q − 1)(q

^j

− q

^−j−1

)

 X

1≤a≤q^j+t

1 + X

1−q^t−j≤a≤0

1

 .

Now we estimate each part in (12):

X

−T ≤j≤−1

q

^j

(q − 1) X

1≤a≤q^j+t

1 = X

1≤a≤q^t−1

X

−T ≤j≤−1 j≥2L log₂a−t

q

^j

(q − 1)

≥ X

1≤a≤q^t−1

(1 − q

^{2L log2a−t+1}

)

≥ X

1≤a≤q^t

(1 − q

^{2L log2a−t+1}

)

≥ X

1≤a≤q^t

(1 − aq

^−t+1

)

≥ [q

^t

] − 1

2 q

^−t+1

[q

^t

]([q

^t

] + 1);

X

−T ≤j≤−1

q

^j

(q − 1) X

− min{q^t+j,q^t−j−1}≤a≤0

1

= X

−T ≤j≤−1

q

^j

(q − 1) X

1≤a≤min{q^t+j+1,q^t−j}

1

≥ X

1≤a≤1/2+q^t

X

−T ≤j≤−1 min{q^t+j+1,q^t−j}≥a

q

^j

(q − 1)

=: X

2≤a≤q^t

X

−T ≤j≤−1 min{q^t+j+1,q^t−j}≥a

q

^j

(q − 1) + δ(t) + 1 − q

^−T

= X

2≤a≤q^t

X

2L log₂(a−1)−t≤j≤−1

q

^j

(q − 1) + δ(t) + 1 − q

^−T

≥ X

2≤a≤q^t

(1 − q

^{2L log2(a−1)−t+1}

) + δ(t) + 1 − q

^−T

≥ X

2≤a≤q^t

(1 − (a − 1)q

^−t+1

) + δ(t) + 1 − q

^−T

≥ [q

^t

] − 1

2 q

^−t+1

[q

^t

]([q

^t

] − 1) − q

^−T

+ δ(t);