A problem on unique representation bases

In this paper we construct a unique representation basis whose growth is more than x1/2−ε for infinitely many positive integers x , which solves a problem posed by Nathanson in [M.B. Nathanson, Unique representation bases for the integers, Acta Arith. 108 (2003) 1–8]. c © 2005 Elsevier Ltd. All rights reserved. MSC: 11B13; 11B34; 11B05 Let A be a set of integers, and let rA(n) = #{(a, b) : a, b ∈ A, a ≤ b, a + b = n}. A set A of integers is called an additive basis for the integers if rA(n) ≥ 1 for all n ∈ Z, and a unique representation basis if rA(n) = 1 for all n ∈ Z. A set B of integers is called a Sidon set if rB(n) ≤ 1 for all n ∈ Z. Thus a unique representation basis is a Sidon set that is a basis for the integers. Recently, Nathanson [2] proved that a unique representation basis for the integers can be arbitrarily sparse. An interesting problem is to find a dense unique representation basis. Nathanson [2] constructed a unique representation basis A whose growth is logarithmic in the sense that the number of elements a ∈ A with |a| ≤ x is bounded above and below by constant multiples of log x . Nathanson [3] studied related problems with given representation functions. Let A(y, x) = #{a ∈ A : y ≤ a ≤ x}. Nathanson [2] asked the following problem: I Supported by the National Natural Science Foundation of China, Grant No 10471064 and the Teaching and Research Award Program for Outstanding Young Teachers in Nanjing Normal University. E-mail address: [email protected]. 0195-6698/$ see front matter c © 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.ejc.2005.10.004 34 Y.-G. Chen / European Journal of Combinatorics 28 (2007) 33–35 Does there exist a number θ < 1/2 such that A(−x, x) ≤ x for every unique representation basis A and for all sufficiently large x? In this note, we show that the answer to the problem is negative. Theorem. For any ε > 0, there exists a unique representation basis A for the integers such that for infinitely many positive integers x, we have A(−x, x) ≥ x1/2−ε. For a set A and any integer c, define A + c = {a + c : a ∈ A}. Lemma 1. Let A be a nonempty finite set of integers with rA(n) ≤ 1 for all n ∈ Z and 0 6∈ A. If m is an integer with rA(m) = 0, then there exists a finite set B of integers such that A ⊆ B, rB(n) ≤ 1 for all n ∈ Z, rB(m) = 1 and 0 6∈ B. Proof. Let b = max{|a| : a ∈ A}. Take c = 4b + |m| and B = A ∪ {−c, c + m}. It is easy to verify that the four sets 2A, A − c, A + c + m, {m,−2c, 2c + 2m} are disjoint each other. Hence rB(n) ≤ 1 for all n ∈ Z and 0 6∈ B by c 6= 0 and c + m 6= 0. This completes the proof of Lemma 1. Lemma 2. Let A be a nonempty finite set of integers with rA(n) ≤ 1 for all n ∈ Z and 0 6∈ A. Then, for any ε > 0 and M > 0, there exists an integer x > M and a finite set B of integers with 0 6∈ B, A ⊆ B, rB(n) ≤ 1 for all n ∈ Z and B(−x, x) ≥ x1/2−ε. Proof. It is well known that for any positive integer m there exists a Sidon set S ⊆ [1,m] with |S| ≥ √ m + o( √ m) (see [1,4]). Thus there exists an integer x > M + (25T )1/(2ε) and a set D of positive integers with rD(n) ≤ 1 for all positive integers n and D(1, x/(5T )) ≥ 1 2 √ x/(5T ), where T = max{|a| : a ∈ A}. Let B = A ∪ {5Tb : b ∈ D}. Then 0 6∈ B and B(−x, x) ≥ D(1, x/(5T )) ≥ 1 2 √ x/(5T ) ≥ x1/2−ε. It is easy to verify that rB(n) ≤ 1 for all n ∈ Z. This completes the proof of Lemma 2. Proof of the Theorem. We shall use induction to construct an ascending sequence A1 ⊆ A2 ⊆ · · · of finite sets of integers and a sequence {xi } i=1 of positive integers with xi+1 > xi for all i such that for any positive integer k, we have (i) rAk (n) ≤ 1 for all n ∈ Z; (ii) rA2k (n) = 1 for all n ∈ Z with |n| ≤ k; (iii) A2k−1(−xk, xk) ≥ x 1/2−ε k ; (iv) 0 6∈ Ak . Let A1 = {−1, 1} and x1 = 1. Suppose that we have A1, A2, . . . , A2l−1 and positive integers x1 < x2 < · · · < xl . Let m be an integer with minimum absolute value and rA2l−1(m) = 0. If l = 1, then |m| = 1 = l. If l > 1, then by A2l−2 ⊆ A2l−1 we have rA2l−2(m) = 0. By the Y.-G. Chen / European Journal of Combinatorics 28 (2007) 33–35 35 inductive hypothesis and (ii) we have m ≥ l. By Lemma 1 there exists a finite set B of integers such that A2l−1 ⊆ B, rB(n) ≤ 1 for all n ∈ Z, rB(m) = 1 and 0 6∈ B. If rB(−m) = 0, then there exists a finite set B ′ of integers such that B ⊆ B , rB′(n) ≤ 1 for all n ∈ Z, rB′(m) = 1 and 0 6∈ B . Now, let A2l = B if rB(−m) 6= 0, and let A2l = B ′ if rB(−m) = 0. Then A2l satisfies (i), (ii), (iv) and A2l−1 ⊆ A2l . By Lemma 2 there exists a finite set A2l+1 of integers and an integer xl+1 > xl such that (i), (iii) and (iv) hold, and A2l ⊆ A2l+1. Let A = ∪∞k=1 Ak . By (ii), we have that rA(n) = 1 for all n ∈ Z. So A is a unique representation basis for the integers. By (iii), we have A(−xk, xk) ≥ x 1/2−ε k . This completes the proof. Finally, we pose the following open problems: (1) Does there exist a real number c > 0 and a unique representation basis A such that A(−x, x) ≥ c √ x for infinitely many positive integers x? (2) Does there exist a real number c > 0 and a unique representation basis A such that A(−x, x) ≥ c √ x for all real numbers x ≥ 1? (3) Does there exist a real number θ < 2 such that for any unique representation basis A there are infinitely many positive integers x with A(−x, x) < x?