On a generalization of a theorem of Sárközy and Sós

European Journal of Combinatorics 54 (2016) 201–206

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European Journal of Combinatorics journal homepage: www.elsevier.com/locate/ejc

On a generalization of a theorem of Sárközy and Sós Yong-Gao Chen a , Min Tang b a

School of Mathematical Sciences and Institute of Mathematics, Nanjing Normal University, Nanjing 210023, PR China b

School of Mathematics and Computer Science, Anhui Normal University, Wuhu 241003, PR China

article

abstract

info

Article history: Received 21 July 2015 Accepted 30 December 2015 Available online 28 January 2016

Let N0 be the set of all nonnegative integers and ℓ ≥ 2 be a fixed integer. For A ⊆ N0 and n ∈ N0 , let rℓ′ (A, n) denote the number of solutions of a1 + · · · + aℓ = n with a1 , . . . , aℓ ∈ A and a1 ≤ · · · ≤ aℓ . Let k be a fixed positive integer. In this paper, we prove that, for any given distinct positive integers ui (1 ≤ i ≤ k) and positive rational numbers αi (1 ≤ i ≤ k) with α1 +· · ·+αk = 1, there are infinitely many sets A ⊆ N0 such that rℓ′ (A, n) ≥ 1 for all n ≥ 0 and the set of n with rℓ′ (A, n) = ui has density αi for all 1 ≤ i ≤ k. © 2016 Elsevier Ltd. All rights reserved.

1. Introduction Let N be the set of all positive integers and N0 be the set of all nonnegative integers. Let ℓ ≥ 2 be a fixed integer. For A ⊆ N0 , n ∈ N0 , and N ∈ N, let rℓ (A, n) = ♯{(a1 , a2 , . . . , aℓ ) ∈ Aℓ : a1 + a2 + · · · + aℓ = n},

rℓ′ (A, n) = ♯{(a1 , a2 , . . . , aℓ ) ∈ Aℓ : a1 + a2 + · · · + aℓ = n, a1 ≤ a2 ≤ · · · ≤ aℓ },

Su(ℓ) (A) = {n ∈ N : rℓ′ (A, n) = u},

Su(ℓ) (A, N ) = ♯{n ≤ N : rℓ′ (A, n) = u}. The subset A of N0 is called a basis of order ℓ if rℓ′ (A, n) ≥ 1 for all n ≥ 0. The well-known Erdős–Turán conjecture [3] asserts that if A is a basis of order 2, then r2 (A, n) is unbounded. It is also well known by now that the counterpart of the Erdős–Turán conjecture does

E-mail addresses: [email protected] (Y.-G. Chen), [email protected] (M. Tang). http://dx.doi.org/10.1016/j.ejc.2015.12.016 0195-6698/© 2016 Elsevier Ltd. All rights reserved.

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not hold in many families of semigroups. Unfortunately, this conjecture itself is still a major unsolved problem in additive number theory. Several mathematicians improved the known lower bound of lim supn→∞ r2 (A, n) for all bases A. In 2003, Grekos et al. [4] proved that if A is a basis of order 2, then lim supn→∞ r2 (A, n) ≥ 6. In 2005, Borwein et al. [1] improved 6 to 8. In 2013, Konstantoulas [5] proved that, if the upper density of the set of numbers not represented as sums of two elements of A is less than 1/10, then lim supn→∞ r2 (A, n) ≥ 6. In 2012, the first author of this paper [2] proved that there exists a basis A of order 2 of N such that the set of n with r2 (A, n) = 2 has density one. In 2013, Yang [8] generalized Chen’s method to prove that for any integer k ≥ 2, there exists a basis A of order k such that the set of n with rk (A, n) = k! has density one. The second author of this paper [7] developed Chen and Yang’s method of proof to establish the following more general result: For any fixed integers k ≥ 2 and u ≥ 1, there exists a basis A of order k such that rk (A, n) ≥ 1 for all n ≥ 0 and the set of n with rk (A, n) = k!u has density one. In 1997, Sárközy and Sós [6] considered a similar problem and they showed that for every finite set U ∈ N there is a set A such that, apart from a ‘‘thin’’ set of integers n, r2′ (A, n) assumes only the prescribed values u ∈ U with about the same frequency. In detail, they proved the following result. Theorem A. Let k ∈ N and let u1 < u2 < · · · < uk be positive integers. Then there is an infinite set A ⊂ N0 such that writing B = N \ (∪ki=1 Su(2i ) (A)) we have

Su(2i ) (A, N ) =

N k

+ O(N α )

and B(N ) = O(N α ) where α = log 3/ log 4 and B(N ) = |B ∩ [1, N ]|.

k

Let ri ∈ Q, 1 ≤ i ≤ k with i=1 ri = 1. Sárközy and Sós (See [6, Remark 4.1]) remarked that using the same idea as in the proof of Theorem A, they can prove the existence of an infinite set A ⊂ N0 for which

Su(2i ) (A, N ) = ri N + O(N α ),

1≤i≤k

with some 0 < α < 1. In this paper, we extend Sárközy and Sós’s result to ℓ ≥ 2. We find that it is difficult to handle the cases ℓ ≥ 3 by using Sárközy and Sós’s method. The method used here is different from Sárközy and Sós’s method. Theorem 1. Let k, ℓ ∈ N with ℓ ≥ 2 and let u1 < u2 < · · · < uk be positive integers. Let αi (1 ≤ i ≤ k) be positive rational numbers with α1 + · · · + αk = 1. Then there are infinitely many bases A of order ℓ such that

Su(ℓ) (A, N ) = αi N + O(N α ), i

1 ≤ i ≤ k,

where α = α(A) with 0 < α < 1. (ℓ)

Let B = N \ (∪ki=1 Sui (A)). If (1) holds, then B(N ) = O(N α ). 2. Proofs For Xi ⊆ Z (1 ≤ i ≤ t ), let X1 + · · · + Xt = {x1 + · · · + xt : xi ∈ Xi (1 ≤ i ≤ t )}. For X ⊆ Z and n ∈ N, let n × X = {nx : x ∈ X }.

(1)

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203

Proof of Theorem 1. Fixed a positive integer g ≥ 2. Let

 At =

∞ 

 εi g

ℓi+t

: εi ∈ {0, 1, . . . , g − 1} ,

t = 0, 1, . . . , ℓ − 1,

i=0

where in each sum there are only finitely many εi ̸= 0. Since αi (1 ≤ i ≤ k) are positive rational numbers with α1 + · · · + αk = 1, there exist positive integers h, h1 , . . . , hk such that αi = hi /h and h = h1 + · · · + hk . Put I1 = [1, h1 ] ∩ N,

Ii =

 i−1 

ht + 1,

t =1

i 

 ht

∩ N,

2 ≤ i ≤ k.

t =1

For any 1 ≤ j ≤ h, there exists an 1 ≤ i ≤ k such that j ∈ Ii . Write

Hj = {0, 1, 2, . . . , ui − 1}, Lj = h × (A0 + Hj ) + {j},

j = 1, 2, . . . , h.

Let

B0 =

h 

Lj ,

Bµ = h × Aµ ,

µ = 1, 2, . . . , ℓ − 1.

j =1

Let A(g ) =

ℓ−1 

Bt .

t =0

Since g ∈ A1 , it follows that hg ∈ B1 . So hg ∈ A(g ) . We shall prove that set A(g ) satisfies the desired properties. First we prove that A(g ) is a basis of order ℓ. Let n ∈ N. Then there exist two integers m, j with m ≥ 0 and 1 ≤ j ≤ h such that n = hm + j. Since

A0 + · · · + Aℓ−1 = N0 , it follows that m = a0 + · · · + aℓ−1 ,

at ∈ At (0 ≤ t ≤ ℓ − 1).

Noting that 0 ∈ Hj , we have ha0 + j ∈ Lj and then ha0 + j ∈ B0 . It is clear that haµ ∈ Bµ for all 1 ≤ µ ≤ ℓ − 1. Thus n = hm + j = (ha0 + j) + ha1 + · · · + haℓ−1 ∈ B0 + · · · + Bℓ−1 . Hence rℓ′ (A(g ) , n) ≥ 1 for all n ≥ 1. It is clear that rℓ′ (A(g ) , 0) = 1 since 0 ∈ A(g ) . That is, A(g ) is a basis of order ℓ. Let U be the set of all positive integers n such that if n = a0 + · · · + aℓ−1 with at ∈ A(g ) (0 ≤ t ≤ ℓ − 1), then at least two of a0 , . . . , aℓ−1 are in the same Bt . Let V = N \ U. To complete the proof, it is enough to prove that (i) |U ∩ [1, N ]| = O(N α ) for some 0 < α < 1; (ii) For n ∈ V and n ≡ j (mod h) with j ∈ Ii and n being large enough (depending on ui ), we have rℓ′ (A, n) = ui . Thus,

Su(ℓ) (A(g ) , N ) = |V ∩ {n ≤ N : n ≡ j (mod h), j ∈ Ii }| + O(|U ∩ [1, N ]|) i

| Ii |

N + O(N α ) h = αi N + O(N α ).

=

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Now we prove that |U ∩ [1, N ]| = O(N α ) for some 0 < α < 1. By the definition of U, we have U =

ℓ−1 

 

Bt + Bt +

t =0 0≤j1 ≤···≤jℓ−2 ≤ℓ−1

ℓ−2 

 B js

.

s=1

Thus,

|U ∩ [1, N ]| ≤

ℓ−1 



t =0 0≤j1 ≤···≤jℓ−2

≤

ℓ−1 

   ℓ−2      B js ∩ [ 1 , N ]  Bt + Bt +   s =1 ≤ℓ−1



|(Bt + Bt ) ∩ [1, N ]|

t =0 0≤j1 ≤···≤jℓ−2 ≤ℓ−1

ℓ−2    Bj ∩ [1, N ] . s

(2)

s=1

Now we estimate |(Bt + Bt ) ∩ [1, N ]| and |Bt ∩ [1, N ]|. For 1 ≤ t ≤ ℓ − 1, we have

|Bt ∩ [1, N ]| ≤ |At ∩ [1, N ]| ≤ g log N /(ℓ log g )+1 ≤ gN 1/ℓ

(3)

and

|(Bt + Bt ) ∩ [1, N ]| ≤ |(At + At ) ∩ [1, N ]| ≤ (2g − 1)log N /(ℓ log g )+1 ≤ (2g − 1)N log(2g −1)/(ℓ log g ) . For t = 0, we have

   h      |B0 ∩ [1, N ]| ≤  Lj ∩ [1, N ]  j =1  ≤

h    Lj ∩ [1, N ] j =1

≤

h 

|A0 ∩ [1, N ]|

j =1

≤ hg log N /(ℓ log g )+1 ≤ hgN 1/ℓ and

   h h       |(B0 + B0 ) ∩ [1, N ]| ≤  Li + Lj ∩ [1, N ]  i=1  j =1 ≤

h      Li + Lj ∩ [1, N ] i,j

≤

h 

|(A0 + A0 ) ∩ [1, N ]|

i,j

≤ h2 (2g − 1)log N /(ℓ log g )+1 ≤ h2 (2g − 1)N log(2g −1)/(ℓ log g ) . It follows from (2) that

 ℓ−2 |U ∩ [1, N ]| ≤ ℓℓ−1 h2 (2g − 1)N log(2g −1)/(ℓ log g ) hgN 1/ℓ = O (N α ) ,

(4)

Y.-G. Chen, M. Tang / European Journal of Combinatorics 54 (2016) 201–206

205

where

α=

log(2g − 1)

ℓ log g

+

  ℓ−2 1 log(2g − 1) =1− 2− . ℓ ℓ log g

It is clear that 0 < α < 1. Now we prove that, for n ∈ V and n ≡ j (mod h) with j ∈ Ii and n being large enough (depending on ui ), we have rℓ′ (A(g ) , n) = ui . Let n = hm + j with j ∈ Ii . If n = b0 + · · · + bℓ−1 with bt ∈ A(g ) (0 ≤ t ≤ ℓ − 1), then, by n ̸∈ U, no two of bt are in the same Bx . Without loss of generality, we may assume that bt ∈ Bt for all t. Let b0 = h(a0 + s) + j (0 ≤ s ≤ ui − 1) and bt = hat (1 ≤ t ≤ ℓ − 1). Then m = (a0 + s) + a1 + · · · + aℓ−1 . For any given 0 ≤ s ≤ ui − 1, the equation m − s = a0 + a1 + · · · + aℓ−1 with at ∈ At (0 ≤ t ≤ ℓ − 1) has unique solution. Hence, rℓ′ (A(g ) , n) = ui for all n = hm + j ≥ h max{ui : 1 ≤ i ≤ k}. Now we have proved that, for any integer g ≥ 2, the set A(g ) satisfies the desired properties. (gi ) Finally we prove that there is a strictly increasing sequence {gi }∞ are distinct. i=1 such that sets A (gi ) Let g1 = 2. Suppose that we have {gi }m with g < · · · < g such that sets A ( 1 ≤ i ≤ m) are 1 m i =1 distinct. By (3) and (4), we have

|A(gi ) ∩ [1, N ]| =

ℓ−1 

|Bt ∩ [1, N ]| ≤ (ℓ + h − 1)gi N 1/ℓ ,

i = 1, . . . , m.

t =0

Thus,

   m      (gi ) A ∩ [1, N ] ≤ (ℓ + h − 1)(g1 + · · · + gm )N 1/ℓ .   i=1  Since

     hl : gm < l ≤ N  > N − gm − 1 > (ℓ + h − 1)(g1 + · · · + gm )N 1/ℓ  h  h for sufficiently large N, there is an integer gm+1 > gm such that hgm+1 ̸∈

m 

A(gi ) .

i=1

Noting that hgm+1 ∈ A(gm+1 ) , we have A(gm+1 ) ̸= A(gi ) ,

1 ≤ i ≤ m.

(gi ) By induction, there is a strictly increasing sequence {gi }∞ are distinct. i=1 such that sets A This completes the proof of Theorem 1. 

Acknowledgments We sincerely thank the referee for his/her valuable comments. This work was supported by the National Natural Science Foundation of China (Grant Nos.11371195 and 11471017) and a project funded by the Priority Academic Program Development of Jiangsu Higher Education Institutions (PAPD). References [1] P. Borwein, S. Choi, F. Chu, An old conjecure of Erdős–Turán on additive bases, Math. Comp. 75 (2005) 475–484. [2] Y.G. Chen, On the Erdős–Turán conjecture, C. R. Acad. Sci., Paris I 350 (2012) 933–935.

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[3] P. Erdős, P. Turán, On a problem of Sidon in additive number theory, and on some related problems, J. Lond. Math. Soc. 16 (1941) 212–215. [4] G. Grekos, L. Haddad, C. Helou, J. Pihko, On the Erdős–Turán conjecture, J. Number Theory 102 (2003) 339–352. [5] I. Konstantoulas, Lower bounds for a conjecture of Erdős and Turán, Acta Arith. 159 (2013) 301–313. [6] A. Sárközy, V.T. Sós, On additive representation functions, in: The Mathematics of Paul Erdős I, in: Algorithms Combin., vol. 13, Springer, Berlin, 1997, pp. 129–150. [7] M. Tang, On the Erdős–Turán conjecture, J. Number Theory 150 (2015) 74–80. [8] Q.H. Yang, A generalization of Chen’s theorem on the Erdős–Turán conjecture, Int. J. Number Theory 9 (2013) 1683–1686.