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arXiv:1101.1653v2 [math.NT] 13 Jan 2011

NOTE ON THE ADDITIVE COMPLEMENTS OF PRIMES

LI-XIA DAI AND HAO PAN

1. Introduction

For a set A ⊆ N = {0,1,2,3,...}, we say a set B ⊆ N is an additive complement of

A, provided that for every n ∈ N, there exist a ∈ A and b ∈ B such that n = a + b.

Furthermore, for the sets A, B ⊆ N, if the sumset

A + B = {a + b : a ∈ A, b ∈ B}

has lower density 1, i.e., almost all positive integers n can be represented as n = a + b

with a ∈ A and n ∈ B, then we say B is an almost additive complement of A.

Let P denote the set of all primes. In [2], Erd˝ os proved that there exists an additive

complement A of P satisfying A(x) = O((logx)2), where A(x) = |A ∩ [1,x]|. Subse-

quently, Ruzsa [7] improved the results of Wolke [9] and Kolountzakis[6], and showed

that for every function w(x) with limx→+∞w(x) = +∞, there exists an almost additive

complement A of P with A(x) = O(w(x)logx). In 2001, Vu extended the result of Erd˝ os,

and proved that P has an additive complement A of order 2 with A(x) = O(logx), i.e.,

every sufficiently large integer can be represented as a1+ a2+ p where a1,a2∈ A and

p ∈ P.

The well-known Vinogradov’s three primes theorem asserts that every sufficiently large

odd integer can be represented as the sum of three primes. And for the binary Goldbach

problem, we know that almost all even positive integers can be represented as the sum

of two primes.

In this note, we shall extend the results of Ruzsa and Vu.

Theorem 1.1. There exists a set A ⊆ P with A(x) = O(logx) such that every suffi-

ciently large odd integer can be represented as a1+ a2+ p where a1,a2∈ A and p ∈ P.

Theorem 1.2. For every function w(x) with limx→+∞w(x) = +∞, there exists a set

B ⊆ P with B(x) = O(w(x)logx) such that almost all even positive integers can be

represented as a + p where a ∈ B and p ∈ P.

The proofs of Theorems 1.1 and 1.2 will be given in the next sections.

2010 Mathematics Subject Classification. Primary 11P32; Secondary 05D40, 11B13, 11N36.

The first author is supported by National Natural Science Foundation of China (Grant No.

10801075) and the Natural Science Foundation of Jiangsu Higher Education Institutions of China (Grant

No.08KJB11007). The second author is supported by National Natural Science Foundation of China

(Grant No. 10901078).

1

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2LI-XIA DAI AND HAO PAN

2. Proof of the Theorem 1.1

The key of our proof is the following lemma.

Lemma 2.1. There exists a positive constant c0 such that if x is sufficiently large,

0 < M < x1−c0and 0 ≤ y ≤ x − M, then for all even integers n with x ≤ n ≤ x + M,

except for O(M(logx)−2) exceptional values, we always have

?

n=p1+p2

y≤p1≤y+M

x−y−M≤p2≤x−y+M

1 ≫ C(n)

M

(logx)2

(2.1)

where

C(n) =

?

p∤n

?

1 −

1

(p − 1)2

??

p|n

?

1 +

1

p − 1

?

(2.2)

and the implied constant in (2.1) only depends on c0.

Proof. This lemma can be proved by the method of Jia in [5], although he only discussed

the case y = x/2. In fact, Jia proved that Lemma 2.1 holds whenever c0> 7/12.

?

For each x ∈ P, we choose x to be in A with the probability

̺x=clogx

x

,

where c > 0 is a constant to be chosen later. Then almost surely, A(n) = O(logn) for

every n, since

?

p≤n

p is prime

Let tx be the binary random variable representing the choice of x, i.e., tx = 1 with

probability ̺xand 0 with probability 1 − ̺x. Consider

Yn=

?

p<n

p is prime

i,j are prime

logp

p

≪ logn.

?

i+j=n−p

titj.

We need to prove that P(Yn> 0) ≥ 1/2 for every sufficiently large n. Let 0 < ǫ < c0/2

and let M = n1−2ǫ, where c0is the one appearing in Lemma 2.1. Let

?

p≤n−n1−ǫ

p is prime

Y∗

n=

?

i,j≥M

i,j are prime

i+j=n−p

titj.

Clearly P(Yn> 0) ≥ P(Y∗

Lemma 2.2. Let z1,··· ,zmbe independent indicator random variables and Y = ΣαIα

where each Iαis the product of few z′

Iαand Iβ. Let ∆ =?

P(Y ≤ (1 − ε)E(Y )) ≤ exp

n> 0). The following lemma is due to Janson.

js. Define α ∼ β if there is some zjwhich is in both

α∼βE(IαIβ). For any Y and any positive number ε we have

(εE(Y ))2

2(E(Y ) + ∆)

?

−

?

.

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NOTE ON THE ADDITIVE COMPLEMENTS OF PRIMES3

Let

X = {(i,j) : i,j ≥ M,i + j ≥ n1−ǫand i,j,n − i − j are all prime}.

Clearly

Y∗

n=

?

(i,j)∈X

titj=

?

α∈X

Iα,

where Iα= titjfor α = (i,j). In view of Lemma 2.2, we only need to estimate

E(Y∗

n) = E

??

α∈X

Iα

?

and ∆ = E

??

α,β∈X

α∼β

IαIβ

?

.

By Lemma 2.1, we have

E

??

α∈X

Iα

?

=E

?

?

p≤n−n1−ǫ

p is prime

?

i,j≥M

i,j are prime

i+j=n−p

titj

?

≥E

?

?

0≤t≤n1−ǫ/M

tM≤n−n1−ǫ−p<(t+1)M

p is prime

?

1≤s≤(n−p)/M−2

i+j=n−p

sM≤i<(s+1)M

i,j are prime

titj

?

≥E

?

?

0≤t≤n1−ǫ/M

1≤s≤n1−ǫ/M+t−2

?

tM≤n−n1−ǫ−p<(t+1)M

p is prime

?

i+j=n−p

sM≤i<(s+1)M

i,j are prime

titj

?

≫

?

0≤t≤n1−ǫ/M

1≤s≤n1−ǫ/M+t−2

M

logM·

M

(logM)2·

c2(logM)2

sM · (n1−ǫ+ (t + 1)M − sM).

Clearly,

?

0≤t≤n1−ǫ/M

1≤s≤n1−ǫ/M+t−2

?n1−ǫ/M

1

?n1−ǫ/M

1

?n1−ǫ/M

1

1

sM · (n1−ǫ+ (t + 1)M − sM)

≫

??n1−ǫ/M+t−2

1

1

n1−ǫ+ tM

1

sM(n1−ǫ+ tM − sM)ds

??n1−ǫ/M+t−2

1

log(n1−ǫ+ tM)

M(n1−ǫ+ tM)dt ≫(log(2n1−ǫ))2

?

dt

≫

?

1

sM+

1

n1−ǫ+ tM − sM

≫(logn)2

M2

?

ds

?

dt

≫

M2

.

Thus we get

E(Y∗

n) ≫ c2logn.

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4 LI-XIA DAI AND HAO PAN

Now we turn to ∆.

E

??

α,β∈X

α∼β

IαIβ

?

=E

?

?

i≥M

i is prime

?

p1+j1=p2+j2=n−i

p1,p2≥n−n1−ǫ

j1,j2≥M

p1,p2,j1,j2are prime

clogi

i

p1+j1=p2+j2=n−i

p1,p2≥n−n1−ǫ

(titj1)(titj2)

?

≤

?

i≥M

i is prime

?

j1,j2≥M

p1,p2,j1,j2are prime

clogj1

j1

·clogj2

j2

≤

?

M≤i≤n−n1−ǫ

i is prime

clogn

i

?

?

j≥M

p+j=n−i

p,j are prime

clogn

j

?2

.

For a set U of positive integers, by the partial summation, we have

?

M≤j≤x

j∈U

1

j≪U(x)

x

−U(M)

M

+

?

M<y≤x

U(y)

y2.

Note that by the sieve method, we have

|M ≤ j ≤ y : both j and m − j are primes| ≪C(m)(y − M)

(log(y − M))2≪C(m)y

(logy)2

since y/(logy)2is increasing for y > e. Hence

?

j≥M

p+j=n−i

p,j are prime

1

j≪

1

(n − i)·C(n − i) · (n − i)

(log(n − i))2

+

1

M+

?

M+e<y≤n−i

C(n − i)

y(logy)2

≪C(n − i)

logM

≪C(n − i)

logn

.

Thus

E

??

α∼β

IαIβ

?

≪ c3logn

?

M≤i≤n−n1−ǫ

i is prime

C(n − i)2

i

.

Note that

C(n − i)2≪

?

p|n−i

?

1 +

1

p − 1

?2

≪

?

p|n−i

?

1 +1

p

?2

≪

?

p|n−i

?

1 +2

p

?

=

?

d|n−i

d is square-free

2ω(d)

d

,

where ω(d) denotes the number of the distinct prime factors of d. Hence

?

M≤i≤n−n1−ǫ

i is prime

C(n − i)2

i

≪

?

d<n−n1−ǫ−M

d is square-free

2ω(d)

d

?

M≤i≤n−n1−ǫ

i is prime

i≡n (mod d)

1

i

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NOTE ON THE ADDITIVE COMPLEMENTS OF PRIMES5

By the Brun-Titchmarsh theorem, we know

|{M ≤ i ≤ y : i is prime and i ≡ n (mod d)}| ≪

Hence by the partial summation,

1

i≪

y − M

φ(d)log((y − M)/d).

?

M≤i≤n−n1−ǫ

i is prime

i≡n (mod d)

1

n − n1−ǫ·

n

φ(d)log(n/d)+

1

M+

?

M+ed≤y≤n−n1−ǫ

1

φ(d)ylog(y/d)

≪

1

φ(d)logn+

1

M+loglog(n/d) − loglog(M/d)

φ(d)

.

If d ≤√M, then

loglog(n/d) − loglog(M/d) ≤ loglogn − loglog

If d >√M, then

√M ≪ 1.

loglog(n/d) − loglog(M/d)

φ(d)

≪loglogn

d1−ǫ

≪

1

M1/2−ǫ.

Thus we always have

?

M≤i≤n−n1−ǫ

i is prime

i≡n (mod d)

1

i≪

1

M1/2−ǫ+

1

φ(d)≪

1

M1/2−ǫ+

1

d1−ǫ.

Note that

?

d<n−n1−ǫ−M

d is square-free

2ω(d)

d

·

1

d1−ǫ≪

?

d

1

d2−2ǫ< +∞,

and

?

d<n−n1−ǫ−M

d is square-free

2ω(d)

d

·

1

M1/2−ǫ≪

1

M1/2−ǫ

?

d<n−n1−ǫ−M

d is square-free

1

d1−ǫ≪

nǫ

M1/2−ǫ≪ 1.

It follows that

?

M≤i≤n−n1−ǫ

i is prime

C(n − i)2

i

≪ 1,

i.e.,

∆ ≪ c3logn.

Now

E(Y∗

E(Y∗

n)2

n) + ∆=

E(Y∗

1 +

n)

∆

E(Y∗

n)

≫

c2logn

1 +c3logn

c2logn

=

c2

1 + clogn.

So we may choose sufficiently large c such that

E(Y∗

E(Y∗

n)2

n) + ∆≥ 100logn.