Discrete Mathematics 312 (2012) 2069–2075
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Discrete Mathematics
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Thin bases in additive number theory
Melvyn B. Nathanson ∗
Department of Mathematics, Lehman College (CUNY), Bronx, NY 10468, United States CUNY Graduate Center, New York, NY 10016, United States article info a b s t r a c t
Article history: The set A of nonnegative integers is called a basis of order h if every nonnegative integer can Received 1 July 2010 be represented as the sum of exactly h not necessarily distinct elements of A. An additive Received in revised form 7 March 2012 basis A of order h is called thin if there exists c > 0 such that the number of elements of Accepted 19 March 2012 A not exceeding x is less than cx1/h for all x ≥ 1. This paper describes a construction of Available online 6 April 2012 Shatrovski˘i of thin bases of order h. © 2012 Elsevier B.V. All rights reserved. Keywords: Additive number theory Thin bases Bases of finite order Shatrovski˘i bases
1. Additive bases of finite order
The central theme in additive number theory is the representation of a nonnegative integer as the sum of a bounded number of elements chosen from a fixed set of integers, such as the squares, the cubes, or the primes. Let N0 and Z denote the sets of nonnegative integers and integers, respectively. For real numbers a and b with a ≤ b, let [a, b] denote the interval of integers between a and b, that is, [a, b] = {n ∈ Z : a ≤ n ≤ b}.
Let h ≥ 2, and let (A1,..., Ah) be an h-tuple of subsets of N0. We define the sumset
A1 + · · · + Ah = {a1 + · · · + ah : ai ∈ Ai for i = 1,..., h}. The h-fold sumset of a set A of nonnegative integers is
hA = A + · · · + A . h terms
The h-tuple (A1,..., Ah) is called an additive system of order h if A1 + · · · + Ah = N0. The set A is an additive basis, or, simply, a basis, of order h if hA = N0. Note that every basis must contain 0 and 1. If (A1,..., Ah) is an additive system of = h order h, then A i=1 Ai is a basis of order h. The set A is an asymptotic basis of order h if the sumset hA contains all sufficiently large integers. For example, Lagrange’s theorem states that the squares are a basis of order 4, and Wieferich’s theorem states that the nonnegative cubes are a basis of order 9. Let r1,..., rh be relatively prime positive integers, and let
Ai = ri ∗ N0 = {rivi : vi ∈ N0}
∗ Correspondence to: Department of Mathematics, Lehman College (CUNY), Bronx, NY 10468, United States. E-mail address: [email protected].
0012-365X/$ – see front matter © 2012 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2012.03.026 2070 M.B. Nathanson / Discrete Mathematics 312 (2012) 2069–2075 for i = 1,..., h. There exists an integer C such that every integer n ≥ C can be represented in the form n = r1v1 +· · ·+rhvh with v1, . . . , vh ∈ N0 [8, Section 1.6], and so the set h A = [0, C − 1] ∪ Ai i=1 is an additive basis of order h. Let g ≥ 2. We can use the g-adic representation of the nonnegative integers to construct another class of additive bases of order h. Let K1,..., Kh be pairwise disjoint subsets of N0 such that
N0 = K1 ∪ · · · ∪ Kh.
For i = 1,..., h, let F (Ki) denote the set of all finite subsets of Ki, and let k Ai = εkg : F ∈ F (Ki) and εk ∈ {0, 1, 2,..., g − 1} . (1) k∈F h = { } = h Then (A1,..., Ah) is an additive system of order h with i=1 Ai 0 , and A i=1 Ai is a basis of order h. The counting function of a set A of integers is A(x) = 1. a∈A 1≤a≤x
If A is a basis of order h, then for all x ∈ N0 we have A(x) + h (A(x) + h)h x + 1 ≤ ≤ h h! and so A(x) lim inf ≥ (h!)1/h. x→∞ x1/h The basis A of order h is called a thin basis if A(x) lim sup < ∞. 1/h x→∞ x One of the earliest general problems in additive number theory was that of the existence and construction of thin bases of finite order [11]. In 1937, Raikov [10] and Stöhr [14] independently solved this problem. Using the 2-adic representation and = h = { ∈ : ≡ − } = = h the partition N0 i=1 Ki, where Ki k N0 k i 1(mod h) for i 1,..., h, they proved that A i=1 Ai is a thin basis of order h, where Ai is the set defined by (1) with g = 2. Many thin bases have been constructed using variations of the g-adic construction; see, for example, Cassels [2], Grekos [4], Nathanson [9] and Jia and Nathanson [6]. Recently, Blomer [1], Hofmeister [5], and Schmitt [12] constructed other examples of thin bases. In 1940 Shatrovski˘i [13] described a beautiful class of thin bases of order h that depends on the linear diophantine 1 equation n = a1x1 + · · · + ahxh and does not use g-adic representations. This is an almost forgotten paper. It has been mentioned in a few surveys of solved and unsolved problems in additive number theory (e.g. Erdős and Nathanson [3], Nathanson [7] and Stöhr [15]), but MathSciNet lists no publication that cites Shatrovski˘i’s paper. This paper contains an exposition of Shatrovski˘i’s construction of thin bases.
2. The construction
Fix an integer h ≥ 2, and define
1 k0 = k0(h) = + 1. (2) 21/h − 1 Then 1 1 + < 21/h k for all k ≥ k0.
1 Shatrovski˘i published his paper in Russian, with a summary in French. The transliteration of the author’s name in the French summary is Chatrovsky, and this is the spelling that appears in Mathematical Reviews. I prefer the AMS and Library of Congress system for transliteration of Cyrillic into English. M.B. Nathanson / Discrete Mathematics 312 (2012) 2069–2075 2071
Let r1,..., rh be a strictly increasing sequence of pairwise relatively prime positive integers, that is, ri < rj and (ri, rj) = 1 for 1 ≤ i < j ≤ h. Let P be a positive integer such that
(i) P ≥ rh − r1, and (ii) if 1 ≤ i < j ≤ h and if the prime p divides rj − ri, then p divides P. For every positive integer k and for i = 1,..., h, we define the integers
si,k = ri + kP. (3)
Then s1,k < s2,k < ··· < sh,k, and s1,k ≥ s1,1 = r1 + P ≥ 2. Let 1 ≤ i < j ≤ h. If a prime number p divides both si,k and sj,k, then p divides sj,k − si,k = rj − ri and so p divides P. Therefore, p divides both ri and rj, which contradicts (ri, rj) = 1. It follows that s1,k, s2,k,..., sh,k is a strictly increasing sequence of pairwise relatively prime positive integers. Moreover, if 1 ≤ i < j ≤ h, then
sj,k rj + kP 1 < = si,k ri + kP
rj − ri = 1 + r1 + kP rh − r1 ≤ 1 + r1 + kP P ≤ 1 + r1 + kP 1 < 1 + . k 1/h If k ≥ k0, then si,k < sj,k < 2 si,k. For every positive integer k, we define the integer
h h Sk = si,k = (ri + kP). (4) i=1 i=1 h h h h ≥ Then s1,k < Sk < sh,k, and sh,k < 2s1,k < 2Sk for k k0. For all k ≥ 1 we have h Sk+1 ri + (k + 1)P 1 < = S r + kP k i=1 i h P = 1 + r + kP i=1 i 1 h < 1 + . k
If k ≥ k0, then
Sk < Sk+1 < 2Sk. For every positive integer k and for i = 1,..., h, we define the integers
h Sk ai,k = = sj,k. (5) si,k j=1 j̸=i
Then a1,k > a2,k > ··· > ah,k. For k ≥ k0, we have
S k = 1/h 1/h sh,k < 2 Sk . (6) ah,k
Suppose that there is a prime number p that divides ai,k for all i = 1,..., h. Because p divides a1,k, it follows from (5) that p divides sℓ,k for some ℓ ∈ {2, 3,..., h}. Because p also divides aℓ,k, it follows from (5) that p divides sm,k for some m ∈ {1, 2, 3,..., h}\{ℓ}. This is impossible because (sℓ,k, sm,k) = 1. Therefore, a1,k, a2,k,..., ah,k is a strictly decreasing sequence of relatively prime positive integers. 2072 M.B. Nathanson / Discrete Mathematics 312 (2012) 2069–2075
Lemma 1. Let h ≥ 2, and let r1,..., rh be a strictly increasing sequence of pairwise relatively prime positive integers. For every positive integer k and i = 1,..., h, let si,k,Sk, and ai,k be the positive integers defined by (3)–(5), respectively. Let k1 ≥ k0(h). For every positive integer t, there is a unique nonnegative integer ℓ(t) defined by
t S + ≤ S < S + + . (7) k1 ℓ(t) k1 k1 ℓ(t) 1 = { }∞ Then ℓ(1) 0 and the sequence ℓ(t) t=1 is strictly increasing. Moreover, 1 2 < t . (8) S + S k1 ℓ(t) k1
{ }∞ Proof. Because the sequence Sk k=1 is strictly increasing, for every positive integer t there is a unique nonnegative integer 1 ℓ(t) that satisfies inequality (7). In particular, S ≤ S < S + and so ℓ(1) = 0. If k ≥ k ≥ k , then S + < 2S and so k1 k1 k1 1 1 0 k 1 k
t t t+1 S < S + + < 2S + ≤ 2S < S < S + + + . k1 k1 ℓ(t) 1 k1 ℓ(t) k1 k1 k1 ℓ(t 1) 1 It follows that ℓ(t) < ℓ(t + 1). Inequality (8) follows from
t S < S + + < 2S + . k1 k1 ℓ(t) 1 k1 ℓ(t) This completes the proof.
Lemma 2. Let h ≥ 2, and let r1,..., rh be a strictly increasing sequence of pairwise relatively prime positive integers. For every positive integer k and i = 1,..., h, let si,k,Sk, and ai,k be the positive integers defined by (3)–(5), respectively. For every integer n such that (h − 1)Sk < n ≤ L, there exist positive integers v1, v2, . . . , vh such that
a1,kv1 + a2,kv2 + · · · + ah,kvh = n and 1 ≤ vi ≤ si,k for i = 1,..., h − 1 and 1 ≤ vh < L/ah,k.
Proof. By Lemma 1, the integers a1,k, a2,k,..., ah,k are relatively prime, and so there exist integers u1, u2,..., uh, not necessarily positive, such that
a1,ku1 + a2,ku2 + · · · + ah,kuh = n.
For each i = 1, 2,..., h − 1 there is a unique integer vi ∈ {1, 2,..., si,k} such that
vi ≡ ui(mod si,k). Then h n = ai,kui i=1 − − h 1 h 1 = ai,kvi + ai,k(ui − vi) + ah,kuh i=1 i=1 h−1 h−1 ui − vi = ai,kvi + ai,ksi,k + ah,kuh s i=1 i=1 i,k h−1 h−1 ui − vi = ai,kvi + Sk + ah,kuh s i=1 i=1 i,k h−1 h−1 ui − vi = ai,kvi + ah,ksh,k + ah,kuh s i=1 i=1 i,k
h = ai,kvi i=1 where h−1 ui − vi vh = sh,k + uh ∈ Z. s i=1 i,k M.B. Nathanson / Discrete Mathematics 312 (2012) 2069–2075 2073
The inequality − − h 1 h 1 ai,kvi ≤ ai,ksi,k = (h − 1)Sk i=1 i=1 implies that if n > (h − 1)Sk, then − h 1 ah,kvh = n − ai,kvi ≥ n − (h − 1)Sk > 0 i=1 and so vh ≥ 1. Moreover, ah,kvh < n ≤ L and so vh < L/ah,k. This completes the proof.
Theorem 1. Let h ≥ 2, and let r1,..., rh be a strictly increasing sequence of pairwise relatively prime positive integers. For every positive integer k and i = 1,..., h, let si,k,Sk, and ai,k be the positive integers defined by (3)–(5), respectively. Let k1 ≥ k0(h). For every positive integer t, let ℓ(t) be the integer defined by (7), and let
(h − 1)S + + = : ∈ ∈ [ − ] ∈ k1 ℓ(t 1) Vt ai,k1+ℓ(t)vi vi 1, si,k1+ℓ(t) for i 1, h 1 and vh 1, . ah,k1+ℓ(t) Then: (i) − 2(h 1)Sk1+ℓ(t+1) |Vt | < . ah,k1+ℓ(t) (ii) ⊇ − + − hVt (h 1)Sk1+ℓ(t) 1,(h 1)Sk1+ℓ(t+1) .
(iii) For every positive integer t1, the set ∞ = At1 Vt t=t1 is an asymptotic basis of order h. (iv) The set ∞ = [ ] ∪ A 0, Sk1 Vt t=1 is a basis of order h. Proof. For i = 1,..., h − 1 we have
S + S + + = k1 ℓ(t) k1 ℓ(t 1) si,k1+ℓ(t) < ai,k1+ℓ(t) ah,k1+ℓ(t) and so h−1 − (h 1)Sk1+ℓ(t+1) |Vt | ≤ si,k +ℓ(t) + 1 a i=1 h,k1+ℓ(t) 2(h − 1)S + + < k1 ℓ(t 1) . ah,k1+ℓ(t) This proves (i). ≥ Lemma 2 implies statement (ii), and statement (ii) implies that the h-fold sumset hAt1 contains every integer n − + ≥ − + (h 1)Sk1+ℓ(t1) 1. Thus, At1 is an asymptotic basis of order h. In particular, hA1 contains every integer n (h 1)Sk1 1. Because [ − ] ⊆ [ ] = [ ] 0,(h 1)Sk1 0, hSk1 h 0, Sk1 = [ ] ∪ it follows that A 0, Sk1 A1 is a basis of order h. This completes the proof.
Theorem 2. The set ∞ = [ ] ∪ A 0, Sk1 Vt t=1 constructed in Theorem 1 is a thin basis of order h. 2074 M.B. Nathanson / Discrete Mathematics 312 (2012) 2069–2075
≥ ∗ Proof. Let A(x) be the counting function of the set A. For all x Sk1 , there is a smallest positive integer t such that ∗ ≤ Sk1+ℓ(t ) x. Then ∗ t +1 S + ∗ ≤ x < S + ∗+ ≤ S k1 ℓ(t ) k1 ℓ(t 1) k1 and 1 1 ∗+ < . (9) St 1 x k1
Because Vt (x) ≤ |Vt | for all x, we have ∗ ∞ t ∞ ≤ + ≤ + | | + A(x) Sk1 Vt (x) Sk1 Vt Vt (x). t=1 t=1 t=t∗+1 We shall estimate the two sums separately. From Theorem 1 (i) and inequality (6), we obtain ∗ ∗ t t − 2(h 1)Sk1+ℓ(t+1) |Vt | ≤ a t=1 t=1 h,k1+ℓ(t) ∗ t t+1 Sk ≤ 2(h − 1) 1 a t=1 h,k1+ℓ(t) ∗ t St k1 = 2(h − 1)Sk 1 a t=1 h,k1+ℓ(t) ∗ t Sk1+ℓ(t)+1 < 2(h − 1)Sk 1 a t=1 h,k1+ℓ(t) ∗ t Sk1+ℓ(t) ≤ 4(h − 1)Sk 1 a t=1 h,k1+ℓ(t) ∗ t 2+1/h 1/h < 2 (h − 1)Sk S 1 k1+ℓ(t) t=1 ∗ t + t/h ≤ 22 1/h(h − 1)S S k1 k1 t=1 1+1/h 1+1/h − ∗ 2 h(h 1)Sk t /h 1 S < 1/h k S − 1 1 k1 + 1+1/h 22 1/h(h − 1)S k1 1/h S ∗ < 1/h k +ℓ(t )+1 S − 1 1 k1 + 1+1/h 22 2/h(h − 1)S k1 1/h S ∗ < 1/h k +ℓ(t ) S − 1 1 k1 2+2/h − 1+1/h 2 (h 1)Sk ≤ 1 x1/h 1/h . S − 1 k1 ∈ ≤ To estimate the second sum, we observe that if ai,k1+ℓ(t)vi Vt and ai,k1+ℓ(t)vi x, then x x 1 ≤ vi ≤ ≤ . ai,k1+ℓ(t) ah,k1+ℓ(t) Applying inequalities (6), (8) and (9), we obtain ∞ ∞ h x Vt (x) ≤ a + t=t∗+1 t=t∗+1 i=1 i,k1 ℓ(t) ∞ 1 < hx a + t=t∗+1 h,k1 ℓ(t) M.B. Nathanson / Discrete Mathematics 312 (2012) 2069–2075 2075
∞ 1−1/h 1 < 21/hhx S + t=t∗+1 k1 ℓ(t) ∞ 1−1/h 2 < 21/hhx S + + t=t∗+1 k1 ℓ(t) 1 ∞ 1 < 2hx (1−1/h)t = ∗+ S t t 1 k1 1−1/h 2hS x = k1 1−1/h (1−1/h)(t∗+1) S − 1 S k1 k1 1−1/h 2hS k1 x < 1−1/h − S − 1 x1 1/h k1 1−1/h 2hSk = 1 x1/h 1−1/h . S − 1 k1
This completes the proof.
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