Unique Difference Bases of Z

Transcription

1 Journal of Integer Sequences, Vol. 14 (011), Article Unique Difference Bases of Z Chi-Wu Tang, Min Tang, 1 and Lei Wu Department of Mathematics Anhui Normal University Wuhu P. R. China tmzzz000@163.com Abstract For n Z, A Z, let δ A (n) denote the number of representations of n in the form n = a a, where a, a A. A set A Z is called a unique difference basis of Z if δ A (n) = 1 for all n 0 in Z. In this paper, we prove that there exists a unique difference basis of Z whose growth is logarithmic. These results show that the analogue of the Erdős-Turán conjecture fails to hold in (Z, ). 1 Introduction For sets A and B of integers and for any integer c, we define the set and the translations The counting function for the set A is For n Z, we write A B = {a b : a A,b B}, A c = {a c : a A}, c A = {c a : a A}. A(y,x) = card{a A : y a x}. δ A (n) = card{(a,a ) A A : a a = n}, 1 Corresponding author. The second author was supported by the National Natural Science Foundation of China, Grant No and the SF of the Education Department of Anhui Province, Grant No. KJ010A16. 1

2 σ A (n) = card{(a,a ) A A : a + a = n}. We call A Z a difference basis of Z if δ A (n) 1 for all n Z, and a unique difference basis of Z if δ A (n) = 1 for all n 0 in Z. We call A a subset of N an additive asymptotic basis of N if there is n 0 = n 0 (A) such that σ A (n) 1 for all n n 0. The celebrated Erdős- Turán conjecture [1] states that if A N is an additive asymptotic basis of N, then the representation function σ A (n) must be unbounded. In 1990, Ruzsa [5] constructed a basis of A N for which σ A (n) is bounded in the square mean. Pŭs [4] first established that the analogue of the Erdős-Turán conjecture fails to hold in some abelian groups. Nathanson [3] constructed a family of arbitrarily sparse unique additive representation bases for Z. In 004, Haddad and Helou [] showed that the analogue of the Erdős-Turán conjecture does not hold in a variety of additive groups derived from those of certain fields. Let K be a finite field of characteristic and G the additive group of K K. Recently, Chi-Wu Tang and Min Tang [6] proved there exists a set B G such that 1 δ B (g) 14 for all g 0. It is natural to consider the analogue of the Erdős-Turán conjecture in (Z, ). In this paper, we obtain the following results. Theorem 1. There exists a family of unique difference bases of Z. Theorem. There exists a unique difference basis A of Z such that log(3x + 3) log 5 < A(0,x) log(x + 3) for all x 1. Proof of Theorem 1. We shall construct an ascending sequence A 1 A of finite sets of nonnegative integers such that A k = k, for all k 1, We shall prove that the infinite set δ Ak (n) 1, for all n 0. A = k=1 is a unique difference basis of Z. We construct the sets A k by induction. Let A 1 = {0, 1}. We assume that for some k 1 we have constructed sets A 1 A k such that A i = i and δ Ai (n) 1 for all 1 i k and all integers n 0. We define the integers A k d k = max{a : a A k }, b k = min{ b : b A k A k }. To construct the set A k+1, we choose an integer c k such that c k > d k. Let A k+1 = A k {c k,b k + c k }.

3 Then A k+1 = k + = (k + 1) for all k 1, and A k [0,d k ], A k A k [ d k,d k ]. Note that A k+1 A k+1 = (A k A k ) (A k (b k + c k )) ((b k + c k ) A k ) (A k c k ) (c k A k ) {b k, b k }. (1) We shall show that A k+1 A k+1 is the disjoint union of the above six sets. If u A k A k, then d k u d k. () If v 1 A k (b k + c k ) and v (b k + c k ) A k, then there exist a,a A k such that v 1 = a (b k + c k ) and v = (b k + c k ) a. Since 0 a,a d k, we have If w 1 A k c k and w c k A k, similarly, we have b k c k v 1 b k c k + d k, (3) b k + c k d k v b k + c k. (4) c k w 1 c k + d k, (5) c k d k w c k. (6) For any n Z, if n A k A k, then n A k A k, thus by the definition of b k, we have b k A k A k and b k A k A k. (7) Assume (c k A k ) ((b k +c k ) A k ), then there exist a,a A k such that b k +c k a = c k a, b k = a a A k A k which contradicts with the fact b k A k A k. Similarly, we have (A k (b k + c k )) (A k c k ) =. Moreover, we have d k A k A k, hence b k d k. If b k < d k, it is easy to see that the set { b k,b k } is disjoint with the other five sets. If b k > d k, since d k A k A k and by the definition of b k, we have b k = d k + 1. Then c k d k (d k + 1) d k = d k + > b k and c k + d k (d k + 1) + d k = d k < b k, thus the set { b k,b k } is disjoint with the other five sets. By Eq. (1) (6) and the above discussion, we know that the sets A k A k, A k (b k + c k ), (b k +c k ) A k, A k c k, c k A k, {b k, b k } are pairwise disjoint. That is, δ Ak+1 (n) 1 for all integers n 0. Let A = A k. Then for all k 1, by (7) and the definition of b k, we have k=1 { b k + 1, b k +,, 1, 1,,b k,b k 1} A k A k A A, and the sequence {b k } k 1 is strictly increasing, since A k A k A k+1 A k+1 and ±b k A k+1 A k+1 but ±b k A k A k. Thus A is a difference basis of Z. If δ A (n) for some n, then by construction, δ Ak (n) for some k, which is impossible. Therefore, A is a unique difference basis of Z. It completes the proof of Theorem 1. 3

4 3 Proof of Theorem. We apply the method of Theorem 1 with c k = d k + 1 for all k 1. This is essentially a greedy algorithm construction, since at each iteration we choose the smallest possible value of c k. It is instructive to compute the first few sets A k. Since A 1 = {0, 1}, A 1 A 1 = { 1, 0, 1}, we have b 1 =,d 1 = 1, and c 1 = d =. Then A = {0, 1, 4, 6}, A A = { 6, 5, 4, 3,, 1, 0, 1,, 3, 4, 5, 6}, hence b = 7,d = 6, c = d + 1 = 7. The next iteration of the algorithm produces the sets A 3 = {0, 1, 4, 6, 14, 1}, A 3 A 3 = { 1, 0, 17, 15, 14, 13, 10, 8, so we obtain b 3 = 9,d 3 = 1,c 3 =, and 7, 7, 8, 10, 13, 14, 15, 17, 0, 1} (A A ), A 4 = {0, 1, 4, 6, 14, 1, 44, 53}. We shall compute upper and lower bounds for the counting function A(0, x). We observe that if x d 1 and k is the unique integer such that d k x < d k+1, by the construction of A, we know A k = k and A k+1 = A k {c k, c k + b k }, then { k, if d k x < c k, A(0,x) = A k+1 (0,x) = k + 1, if c k x < c k + b k = d k+1. For k 1, we have 1 < b k d k + 1 = c k and c k+1 = d k = c k + b k + 1, hence c k + < c k+1 3c k + 1. Since c 1 = d =, it follows by induction on k that and so k+1 c k 5 3k 1 1, log 6 ( ck ) k log c k + for all k 1. We obtain an upper bound for A(0,x) as follows. If d k x < c k, then c k x + 1, and log c k + A(0,x) = A k+1 (0,x) = k 4 = log(c k + ) log(x + 3).

5 If c k x < d k+1, then c k x, and A(0,x) = A k+1 (0,x) = k + 1 log c k log ( x 4 + 1) + 1 = log(x + 4) 3. Therefore, A(0,x) log(x + 3) for all x 1. Similarly, we obtain a lower bound for A(0,x). If d k x < c k, then log 6 ( ck + 1 A(0,x) = k 5 ) > (x + 1) 5 = log(3x + 3) If c k x < d k+1, then d k+1 = b k + c k 3c k. So c k 1 3 d k+1 > 1 3 x and log 5. A(0,x) = k + 1 log 6 ( ck ) + 1 > log 6 (x ) + 1 = log(x + 3) + 1 log 5. Therefore, A(0,x) > log(3x + 3) This completes the proof of Theorem. log 5 for all x 1. 4 Acknowledgements We would like to thank the referee for his/her many helpful suggestions. References [1] P. Erdős and P. Turán, On a problem of Sidon in additive number theory, and on some related problems, J. London Math. Soc. 16 (1941), [] L. Haddad and C. Helou, Bases in some additive groups and the Erdős-Turán conjecture, J. Combin. Theory Series A 108 (004), [3] M. B. Nathanson, Unique representation bases for integers, Acta Arith. 108 (003), 1 8. [4] V. Pǔs, On multiplicative bases in abelian groups, Czech. Math. J. 41 (1991), [5] I. Z. Ruzsa, A just basis, Monatsh. Math 109 (1990), [6] Chi-Wu Tang and Min Tang, Note on a result of Haddad and Helou, Integers 10 (010), 9 3. Available electronically at 5

6 000 Mathematics Subject Classification: Primary 11B13; Secondary 11B34. Keywords: Erdős-Turán conjecture; difference bases; counting function. Received October 1 010; revised version received January Published in Journal of Integer Sequences, February Return to Journal of Integer Sequences home page. 6