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All sums of h distinct terms of a sequence
European Journal of Combinatorics 41 (2014) 289–297
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All sums of h distinct terms of a sequence✩ Yong-Gao Chen a , Jin-Hui Fang b a
School of Mathematical Sciences and Institute of Mathematics, Nanjing Normal University, Nanjing 210023, PR China b Department of Mathematics, Nanjing University of Information Science & Technology, Nanjing 210044, PR China
article
abstract
info
Article history: Received 10 September 2013 Accepted 13 May 2014
For A ⊆ Z, we study the gaps in the sequence of all sums of h pairwise distinct elements of A. For example, the following result is proved: for any integer h ≥ 3, there exists A ⊆ Z such that every integer can be uniquely (neglecting the order) represented as a sum of h not necessarily distinct elements of A, and for any integer ℓ ≥ 1, in the sequence of all sums of ℓ pairwise distinct elements of A, the gaps can be arbitrarily large. Several questions are posed in this paper. © 2014 Elsevier Ltd. All rights reserved.
1. Introduction For a sequence of integers A = {a1 < a2 < · · ·}, define
∆(A) = lim sup(ai+1 − ai ), i→+∞
let h × A be the set of all sums of h pairwise distinct elements of A and let hA be the set of all sums of h not necessarily distinct elements of A. Let N be the set of all positive integers. Let B ∼ N denote that a set of integers B contains all but finitely many positive integers. In 1996, Erdős and Burr [1] asked the following question: is it true that if h(A ∪ {0}) ∼ N, then
∆(A ∪ 2 × A ∪ · · · ∪ h × A) < +∞?
✩ This work was supported by the National Natural Science Foundation of China, Grant Nos. 11371195 and 11201237.
E-mail addresses: [email protected], [email protected] (Y.-G. Chen), [email protected] (J.-H. Fang). http://dx.doi.org/10.1016/j.ejc.2014.05.002 0195-6698/© 2014 Elsevier Ltd. All rights reserved.
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In 2007, Hegyvári, Hennecart and Plagne [5] showed that the answer to the above Erdős–Burr question is affirmative for h = 2 and negative for h ≥ 3. That is, Theorem A (Hegyvári–Hennecart–Plagne). If (A ∪ 2A) ∼ N, then
∆(A ∪ 2 × A) ≤ 2. If 2A ∼ N, then ∆(2 × A) ≤ 2. Let h ≥ 3. There exists a set A such that h(A ∪ {0}) ∼ N and
∆(A ∪ 2 × A ∪ · · · ∪ h × A) = +∞. There exists a set A such that hA ∼ N and ∆(h × A) = +∞. Hegyvári, Hennecart and Plagne [5] conjectured that, if hA ∼ N, then there exists a positive integer k such that ∆(k × A) < +∞. If this conjecture is true, then write k(h) = max min{k : k ∈ N, ∆(k × A) < +∞}. hA∼N
In [5], it is proved that k(h) ≥ 2h−2 + h − 1 for all h ≥ 2. They also conjectured that the function ∆(h × A) is non-increasing and proved that ∆(3 × A) ≤ ∆(2 × A). The restricted order of A is defined to be the least positive integer h such that A ∪ 2 × A ∪ · · · ∪ h × A ∼ N. For the other related results, one may refer to [3,6–8]. In this paper, we consider A ⊆ Z, where Z is the set of all integers. The situation is completely different when one can allow that A ⊆ Z. For example, in the proof of [5, Proposition 5], it is used that ai + (2 × A) ∩ [0, ai ) is contained in 3 × A. Now, it is not true since A contains possibly negative integers. In the sequel of this paper, we allow that the set A contains both positive and negative integers (not necessary both). Then we order A and h × A as
· · · < a− 1 < a0 < a1 < · · · , · · · < b −1 < b 0 < b 1 < · · · , respectively, where a−1 < 0 ≤ a0 and b−1 < 0 ≤ b0 . Let
∆+ (h × A) = lim sup(bi+1 − bi ),
if sup bi = +∞,
i→+∞
∆− (h × A) = lim sup(bi+1 − bi ),
if inf bi = −∞.
i→−∞
Denote −A = {−a : a ∈ A}. It is clear that −(h × A) = h × (−A). Thus
∆+ (h × A) = ∆− (h × (−A)), ∆− (h × A) = ∆+ (h × (−A)). It is clear that if A ⊆ N, then ∆+ (h × A) = ∆(h × A). Thus, Theorems 1 and 3, and Corollary 1(a) are also true for ∆(h × A) when A ⊆ N. In this paper, the following results are proved. Theorem 1. For h ≥ 2 we have
∆+ (h × A) ≤ ∆+ (2 × A),
∆− (h × A) ≤ ∆− (2 × A).
Let R(h) (A, n) be the number of solutions of equation a1 + · · · + ah = n with a1 , . . . , ah ∈ A, a1 ≤ a2 ≤ · · · ≤ ah . In [4], Hegyvári, Hennecart and Plagne confirmed a conjecture in [2]: if A ⊆ N and hA ∼ N, then h × A has a positive lower density. For A ⊆ Z and h ≥ 3, hA ∼ Z does not imply that h × A has a positive lower density. The main reason for this is that we can choose A sparsely when one can allow that A ⊆ Z.
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Theorem 2. If A is a subset of Z such that R(2) (A, n) ≥ 1 for all but finitely many n ∈ Z, then, for all integers ℓ ≥ 2, we have
∆+ (ℓ × A) ≤ 2,
∆− (ℓ × A) ≤ 2.
For any integer h ≥ 3, there exists A ⊆ Z such that R(h) (A, n) = 1 for all n ∈ Z, and A ∪ 2 × A ∪ · · · has density zero, that is d(A ∪ 2 × A ∪ · · ·) =: lim
x→+∞
|(A ∪ 2 × A ∪ · · ·) ∩ [−x, x]| = 0. 2x + 1
(1)
In particular, for all integers ℓ ≥ 1 we have
∆+ (ℓ × A) = +∞,
∆− (ℓ × A) = +∞.
(2)
For A ⊆ N, Hegyvári, Hennecart and Plagne proved that ∆((h + 1) × A) ≤ ∆(h × A) + ah − a1 (see [4, Proof of Proposition 4]). One may see that the method in [4, Proof of Proposition 4] can be used to prove that
∆+ ((h + 1) × A) ≤ ∆+ (h × A) + ah − a1 , ∆− ((h + 1) × A) ≤ ∆− (h × A) − a−h + a−1 . In this paper, the following bounds are obtained. Theorem 3. For any integer h ≥ 1, we have
∆+ ((h + 1) × A) ≤ max{∆+ (h × A), a2h−2 },
(3)
∆− ((h + 1) × A) ≤ max{∆− (h × A), −a−2h+1 }.
(4)
Corollary 1. (a) If ∆+ (h0 × A) < +∞ for some h0 , then there exists h1 ≥ h0 such that, for all h ≥ h1 , we have ∆+ (h × A) ≤ a2h−4 ; (b) If ∆− (h0 × A) < +∞ for some h0 , then there exists h1 ≥ h0 such that, for all h ≥ h1 , we have ∆− (h × A) ≤ −a−2h+3 . Motivated by Theorems 1 and 3 and the Hegyvári–Hennecart–Plagne conjecture: the function
∆(h × A) is non-increasing, we pose the following question.
Question 1. Is it true that the function ∆+ (h × A) is non-increasing? Currently, we cannot give an answer to the following weak forms of Question 1: Question 2. Is it true that for any integer h ≥ 1, there exists an integer ℓ > h such that
∆+ (ℓ × A) ≤ ∆+ (h × A)? Question 3. Is it true that for any integer h ≥ 1, there exists a constant c = c (h) such that for any integer ℓ > h we have
∆+ (ℓ × A) ≤ ∆+ (h × A) + c? Recall that all elements of h × A are · · · < b−1 < b0 < b1 < · · ·, let ∆0 (h × A) = sup(bi+1 − bi ), i
δ0 (h × A) = inf(bi+1 − bi ), i
δ+ (h × A) = lim inf(bi+1 − bi ),
if sup bi = +∞,
δ− (h × A) = lim inf(bi+1 − bi ),
if inf bi = −∞.
i→+∞
i→−∞
It is easy to see that functions δ0 (h × A), δ+ (h × A) and δ− (h × A) are non-increasing. Let A = {1, 2, 3} ∪ {n : n ≥ 5}. Then 2 × A = {n : n ∈ Z, n ≥ 3} and 3 × A = {6} ∪ {n : n ∈ Z, n ≥ 8}.
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Thus ∆0 (3 × A) = 2 > 1 = ∆0 (2 × A). It is interesting to find out the nontrivial properties of these functions. Here we pose the following questions. We do not know whether these questions are also trivial. Question 4. Given any set A ⊆ Z. Is it true that the function ∆0 (h × A) is non-increasing from some point h = h0 on? The following question is a special case of the above question. Question 5. Let A ⊆ Z such that hA = Z. Is it true that ∆0 (ℓ × A) is non-increasing from some point ℓ = ℓ0 on? 2. Proofs Proof of Theorem 1. Since ∆− (h × A) = ∆+ (h × (−A)), it is enough to prove the first inequality. Assume that ∆+ (2 × A) exists and d =: ∆+ (2 × A) < +∞. Otherwise, Theorem 1 is trivial. We shall use induction on h to prove that ∆+ (h × A) ≤ d for all integers h ≥ 2. It is clear that the conclusion is true for h = 2. Now we assume that k > 2 and the conclusion is true for 2 ≤ h < k. That is, for any 2 ≤ h < k, we have ∆+ (h × A) ≤ d. Then there exists n0 > 0 such that, for any 2 ≤ h < k and n ≥ n0 , there exists 0 ≤ i ≤ d − 1 such that n + i ∈ h × A. We shall prove that ∆+ (k × A) ≤ d. Select an ai0 +1 ≥ n0 + d and m0 = ai0 +1 + · · · + ai0 +k−2 + n0 . Let B = {m : m ≥ m0 , for any 0 ≤ i ≤ d − 1, m + i ̸∈ k × A}. We shall show that
|B| ≤ (k − 1)(k − 2)d,
(5)
i.e. B is a finite set. Indeed, assume that (5) is true. Let m′0 = max B + 1 if B ̸= ∅ and m′0 = m0 if B = ∅. Then, for any m ≥ m′0 , we have m ̸∈ B. So there exists 0 ≤ i ≤ d − 1 with m + i ∈ k × A. This implies that ∆+ (k × A) ≤ d. To prove Theorem 1, it is enough to prove that (5) holds. For m ∈ B, by m ≥ m0 = ai0 +1 + · · · + ai0 +k−2 + n0 , we have m − (ai0 +1 + · · · + ai0 +k−2 ) ≥ n0 . Hence there exists 0 ≤ i ≤ d − 1 such that m − (ai0 +1 + · · · + ai0 +k−2 ) + i ∈ 2 × A. Thus there exist two integers s < t such that m − (ai0 +1 + · · · + ai0 +k−2 ) + i = as + at . That is, m + i = ai0 +1 + · · · + ai0 +k−2 + as + at . Since m ∈ B, it follows that m + i ̸∈ k × A. So at least two of i0 + 1, . . . , i0 + k − 2, s, t are equal. Thus
{s, t } ∩ {i0 + 1, . . . , i0 + k − 2} ̸= ∅. Without loss of generality, we assume that i0 + 1 ≤ s ≤ i0 + k − 2. Since ai0 +1 + · · · + ai0 +k−2 + as − i ≥ as − d + 1 ≥ ai0 +1 − d + 1 > n0 , it follows from the inductive hypothesis that there exist 0 ≤ ji,s ≤ d − 1 and j1 < · · · < jk−1 such that ai0 +1 + · · · + ai0 +k−2 + as − i + ji,s = aj1 + · · · + ajk−1 .
(6)
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For each pair i, s with 0 ≤ i ≤ d − 1 and i0 + 1 ≤ s ≤ i0 + k − 2, according to (6), we fix ji,s , j1 , . . . , jk−1 with 0 ≤ ji,s ≤ d − 1 and j1 < · · · < jk−1 . Thus m + ji,s = aj1 + · · · + ajk−1 + at . Since m ∈ B, it follows that t ∈ {j1 , . . . , jk−1 }. Thus m = ai0 +1 + · · · + ai0 +k−2 + as + at − i can take at most d(k − 2)(k − 1) values (i can take at most d values and s can take at most k − 2 values. When i and s have been taken, the integers ji,s , j1 , . . . , jk−1 are fixed. Thus t can take at most k − 1 values). Therefore, (5) holds. This completes the proof of Theorem 1. Proof of Theorem 2. The case h = 2 is similar to the proof of [5, Theorem 1]. Since R(2) (A, n) ≥ 1 for all but finitely many n ∈ Z, it follows that all but finitely many odd numbers belong to 2 × A. This implies that
∆+ (2 × A) ≤ 2,
∆− (2 × A) ≤ 2.
(7)
By Theorem 1 and (7), for all integers ℓ ≥ 2, we have
∆+ (ℓ × A) ≤ 2,
∆− (ℓ × A) ≤ 2.
Let h ≥ 3. We construct A in the following iterate process: we shall construct a sequence 0 = i1 < i2 < · · · of integers and a sequence A1 ⊂ A2 ⊂ · · · of subsets of Z such that, for all k ≥ 1, we have (a) (b) (c) (d)
R(h) (Ak , n) = 1 for all n ∈ Z with |n| ≤ ik ; R(h) (Ak , n) ≤ 1 for all n ∈ Z; no two elements of Ak have the same absolute; di > d1 + d2 + · · · + di−1 + 4i for all 2 ≤ i ≤ tk , where d1 < · · · < dtk are all absolutes of elements in Ak . Define A=
∞
Ai .
i =1
Let i1 = 0 and A1 = {−32, 32(h − 1)}. Then (a)–(d) hold for k = 1. Suppose that we have found 0 = i1 < i2 < · · · < iℓ and A1 ⊂ A2 ⊂ · · · ⊂ Aℓ such that (a)–(d) hold for all 1 ≤ k ≤ ℓ. Let d1 , . . . , dtℓ be all absolutes of elements in Aℓ with d1 < d2 < · · · < dtℓ . Let iℓ+1 be the least positive integer for which there exists n ∈ Z with |n| = iℓ+1 and R(h) (Aℓ , n) = 0. Then, by (a), we have iℓ < iℓ+1 . Let Tℓ be the set of such integers. It is clear that |Tℓ | ∈ {1, 2}. Case 1: |Tℓ | = 1. Let n1 ∈ Z with |n1 | = iℓ+1 and R(h) (Aℓ , n1 ) = 0. We choose a′ℓ+1 > 2htℓ (dtℓ + |n1 |) + 4tℓ +2 . Now we take Aℓ+1 = Aℓ ∪ {−a′ℓ+1 , (h − 1)a′ℓ+1 + n1 }. Let dtℓ +1 = a′ℓ+1 and dtℓ +2 = (h − 1)a′ℓ+1 + n1 . By h ≥ 3 and a′ℓ+1 > 2htℓ (dtℓ + |n1 |) + 4tℓ +2 , we have dtℓ +1 > 2htℓ (dtℓ + |n1 |) + 4tℓ +2
> d1 + d2 + · · · + dtℓ + 4tℓ +1 ,
dtℓ +2 ≥ 2a′ℓ+1 − |n1 | > 2htℓ (dtℓ + |n1 |) + 4tℓ +2 + a′ℓ+1 − |n1 |
≥ d1 + d2 + · · · + dtℓ + dtℓ +1 + 4tℓ +2 .
So (c) and (d) hold for k = ℓ + 1.
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Since |Tℓ | = 1 and n1 = (h − 1)(−a′ℓ+1 ) + ((h − 1)a′ℓ+1 + n1 ), it follows that R(h) (Aℓ+1 , n) ≥ 1 for all n ∈ Z with |n| ≤ iℓ+1 . To prove (a) and (b), it is enough to prove that R(h) (Aℓ+1 , n) ≤ 1 for all n ∈ Z. Suppose that this is false. Let m be an integer with R(h) (Aℓ+1 , m) ≥ 2. Then there exist b1 , . . . , bs ∈ Aℓ with b1 ≤ b2 ≤ · · · ≤ bs , b′1 , . . . , b′s′ ∈ Aℓ with b′1 ≤ b′2 ≤ · · · ≤ b′s′ , and nonnegative integers u, v, u′ , v ′ with s + u + v = h and s′ + u′ + v ′ = h such that
(b1 , . . . , bs , u, v) ̸= (b′1 , . . . , b′s′ , u′ , v ′ ), m = b1 + · · · + bs + u(−a′ℓ+1 ) + v((h − 1)a′ℓ+1 + n1 ),
(8)
(−a′ℓ+1 )
(9)
′
′
m = b1 + · · · + bs′ + u
′
+ v ((h − ′
1)a′ℓ+1
+ n1 ).
If −u + v(h − 1) ̸= −u′ + v ′ (h − 1), then a′ℓ+1 ≤ |u(−a′ℓ+1 ) + v((h − 1)a′ℓ+1 ) − u′ (−a′ℓ+1 ) − v ′ ((h − 1)a′ℓ+1 )|
= ≤ ≤ ≤ =
|b′1 + · · · + b′s′ + v ′ n1 − (b1 + · · · + bs + v n1 )| |b′1 | + · · · + |b′s′ | + v ′ |n1 | + |b1 | + · · · + |bs | + v|n1 | s′ dtℓ + v ′ |n1 | + sdtℓ + v|n1 | hdtℓ + h|n1 | + hdtℓ + h|n1 | 2h(dtℓ + |n1 |),
a contradiction. So −u + v(h − 1) = −u′ + v ′ (h − 1). If v = v ′ , then u = u′ . Thus s = s′ , b1 + · · · + bs = b′1 + · · · + b′s′ and (b1 , . . . , bs ) ̸= (b′1 , . . . , b′s′ ). So s = s′ ≥ 2. Let m′ = b1 + · · · + bs + (h − s)b1 . Then m′ = b′1 + · · · + b′s + (h − s)b1 . This contradicts R(h) (Aℓ , m′ ) ≤ 1. Hence v ̸= v ′ . Without loss of generality, we may assume that v > v ′ . Then, by −u + v(h − 1) = −u′ + v ′ (h − 1), we have u > u′ . Since 0 ≤ u, v, v ′ , s ≤ h, it follows that u = h − v − s ≤ h − v ≤ h − 1. Hence h − 1 ≥ u − u′ = (v − v ′ )(h − 1) ≥ h − 1. This implies that u = h − 1, u′ = 0, v = 1, and v ′ = 0. So s = 0 and s′ = h. By (8) and (9), we have n1 = m = b′1 + · · · + b′s′ = b′1 + · · · + b′h , a contradiction with R(h) (Aℓ , n1 ) = 0 (n1 ∈ Tℓ ). Now we have proved that R(h) (Aℓ+1 , n) ≤ 1 for all n ∈ Z. So (a) and (b) hold. Case 2: |Tℓ | = 2. Then Tℓ = {−iℓ+1 , iℓ+1 }. We choose a′ℓ+1 > 2htℓ (dtℓ + iℓ+1 ) + 4tℓ +2 and ′ aℓ+2 > 2h(tℓ + 2)((h − 1)a′ℓ+1 + iℓ+1 + iℓ+1 ) + 4tℓ +4 . Now we take Aℓ+1 = Aℓ ∪ {−a′ℓ+1 , (h − 1)a′ℓ+1 + iℓ+1 } ∪ {−a′ℓ+2 , (h − 1)a′ℓ+2 − iℓ+1 }. Let Bℓ = Aℓ ∪ {−a′ℓ+1 , (h − 1)a′ℓ+1 + iℓ+1 }. Then Aℓ+1 = Bℓ ∪ {−a′ℓ+2 , (h − 1)a′ℓ+2 − iℓ+1 }. As in Case 1, we can prove that
(a1 ) (b1 ) (c1 ) (d1 )
R(h) (Bℓ , n) = 1 for all n ∈ Z with |n| ≤ iℓ and R(h) (Bℓ , iℓ+1 ) = 1; R(h) (Bℓ , n) ≤ 1 for all n ∈ Z; no two elements of Bℓ have the same absolute; di > d1 + d2 + · · · + di−1 + 4i for all 2 ≤ i ≤ tℓ + 2, where d1 < · · · < dtℓ +2 are all absolutes of elements in Bℓ .
Again, as in Case 1, we can prove that (a)–(d) hold for k = ℓ + 1. Now we have constructed a sequence 0 = i1 < i2 < · · · of integers and a sequence A1 ⊂ A2 ⊂ · · · of subsets of Z satisfying (a)–(d).
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Thus we have constructed A such that (a ) R(h) (A, n) = 1 for all n ∈ Z; (b′ ) no two elements of A have the same absolute; (c′ ) di > d1 + d2 + · · · + di−1 + 4i for all i ≥ 2, where d1 < d2 < · · · are all absolutes of elements in A. ′
It is clear that (b′ ) and (c′ ) hold. Now we prove that (a′ ) holds. In fact, for any n ∈ Z, there exists an integer k ≥ 1 such that |n| ≤ ik . Since Ak ⊆ A, it follows that R(h) (A, n) ≥ R(h) (Ak , n) = 1. If R(h) (A, n) ≥ 2, then n has two different expressions n = c1 + · · · + ch = c1′ + · · · + ch′ with c1 , . . . , ch ∈ A, c1 ≤ · · · ≤ ch , and c1′ , . . . , ch′ ∈ A, c1′ ≤ · · · ≤ ch′ . Since A1 ⊂ A2 ⊂ · · ·, there exists Ak′ such that c1 , . . . , ch , c1′ , . . . , ch′ ∈ Ak′ . This contradicts R(h) (Ak′ , n) ≤ 1. Now we have proved that R(h) (A, n) = 1 for all n ∈ Z. Now, we prove that (1) holds. Let ei ∈ A with |ei | = di (i = 1, 2, . . .). Let x > 4. Suppose that t ∈ (A ∪ 2 × A ∪ · · ·) ∩ [−x, x]. Then, for some positive integer ℓ, we have t = ei1 + · · · + eiℓ ,
1 ≤ i1 < i2 < · · · < iℓ .
If iℓ ≥ 2, then x ≥ |t | ≥ |eiℓ | − |eiℓ−1 | − · · · − |ei1 |
= diℓ − diℓ−1 − · · · − di1 ≥ diℓ − diℓ −1 − · · · − d1 ≥ 4iℓ . So iℓ ≤ log4 x. Hence, the number of integers t does not exceed the number of subsets of {1, 2, . . . , [log4 x]}. This implies that √ |(A ∪ 2 × A ∪ · · ·) ∩ [−x, x]| ≤ 2log4 x = x. Therefore, (1) holds. Given any integer ℓ ≥ 1. Since R(h) (A, n) = 1 for all n ∈ Z, it follows that sup A = +∞ and inf A = −∞. Thus sup(ℓ × A) = +∞,
inf(ℓ × A) = −∞.
Hence ∆+ (ℓ × A) and ∆− (ℓ × A) are well defined. If ∆+ (ℓ × A) < +∞, then all differences of two consecutive nonnegative terms of (ℓ × A) ∪ {0} are less than a positive constant c depending only on ℓ and A. So |(ℓ × A) ∩ [0, x]| ≥ (x/c ) − 1, a contradiction with d(A ∪ 2 × A ∪ · · ·) = 0. Hence ∆+ (ℓ × A) = +∞. Similarly, we have ∆− (ℓ × A) = +∞. This completes the proof of Theorem 2. Proof of Theorem 3. We will prove inequality (3). Inequality (4) can be proved similarly. By the definition of ∆+ (h × A), there exists y0 > 0 such that for any two consecutive terms y′ , y′′ of h × A with y′′ > y′ and y′′ ≥ y0 , we have y′′ − y′ ≤ ∆+ (h × A). Suppose that
∆+ ((h + 1) × A) > max{∆+ (h × A), a2h−2 }.
(10)
Then there exist infinitely many positive integers x ∈ (h + 1) × A such that for any x′ ∈ (h + 1) × A with x′ > x, we have x′ − x > max{∆+ (h × A), a2h−2 }.
(11)
Let x ∈ (h + 1) × A be such an integer with x > ah + · · · + a2h−1 + y0 . Let y be the largest integer in h × A and z be the least integer in h × A such that y ≤ x < z. Then y ≥ ah + · · · + a2h−1 . Suppose that y = ai 1 + · · · + ai h ,
i1 < · · · < ih ,
z = aj1 + · · · + ajh ,
j1 < · · · < jh .
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Since z > x ≥ y ≥ ah + · · · + a2h−1 > a0 + · · · + ah−1 , we have
{i1 , . . . , ih } ̸= {h − 1, h, . . . , 2h − 2}, {j1 , . . . , jh } ̸= {0, 1, . . . , h − 1}.
(12) (13)
By (12), there exists h − 1 ≤ j ≤ 2h − 2 such that j ̸∈ {i1 , . . . , ih }. Let i be an integer with 0 ≤ i ≤ 2h − 2 such that i ̸∈ {i1 , . . . , ih }. If y + ai > x, then we have y + ai ∈ (h + 1) × A and 0 < (y + ai ) − x ≤ (y + ai ) − y = ai ≤ a2h−2 , a contradiction with (11). Hence y + ai ≤ x. Noting that z > x > y0 and y, z ∈ h × A, if i ̸∈ {j1 , . . . , jh }, then z + ai ∈ (h + 1) × A and 0 < z − x ≤ (z + ai ) − x ≤ (z + ai ) − (y + ai ) = z − y ≤ ∆+ (h × A), a contradiction with (11). So i ∈ {j1 , . . . , jh }. If i ≥ 1 and i − 1 ∈ {i1 , . . . , ih }, then y + ai − ai−1 ∈ h × A. Since y + ai − ai−1 > y, it follows from the definition of y that y + ai − ai−1 > x. This contradicts y + ai ≤ x and ai−1 ≥ 0. So either i = 0 or i − 1 ̸∈ {i1 , . . . , ih }. Thus we have proved that, under the assumption of (10), if i is an integer with 0 ≤ i ≤ 2h − 2 such that i ̸∈ {i1 , . . . , ih }, then i ∈ {j1 , . . . , jh }
(if i ≥ 0),
i − 1 ̸∈ {i1 , . . . , ih }
(if i ≥ 1).
(14)
Let i0 be the largest integer with 0 ≤ i0 ≤ 2h − 2 such that i0 ̸∈ {i1 , . . . , ih }. By (14) we have i0 − 1, i0 − 2, . . . , 1, 0 ̸∈ {i1 , . . . , ih } and then
{0, 1, . . . , i0 } ⊆ {j1 , . . . , jh }.
(15)
By (12), we have i0 ≥ h − 1. By (13) and (15), we have i0 ≤ h − 2, a contradiction. This completes the proof of Theorem 3. Proof of Corollary 1. By Theorem 3, for ℓ > h0 , we have
∆+ (ℓ × A) ≤ ≤ = ≤ ≤
max{∆+ ((ℓ − 1) × A), a2ℓ−4 } max{max{∆+ ((ℓ − 2) × A), a2ℓ−6 }, a2ℓ−4 } max{∆+ ((ℓ − 2) × A), a2ℓ−4 }
··· max{∆+ (h0 × A), a2ℓ−4 }.
Select an integer h1 > h0 such that a2h1 −4 ≥ ∆+ (h0 × A). Then, for ℓ ≥ h1 , we have ∆+ (ℓ× A) ≤ a2ℓ−4 . This completes the proof of Corollary 1(a). Similarly, we can give a proof of Corollary 1(b). Acknowledgments We would like to thank two referees for their comments and N. Hegyvári for sending us the reprint of [5]. References [1] P. Erdős, Some of my new and almost new problems and results in combinatorial number theory, in: Number Theory (Eger 1996), de Gruyter, Berlin, 1998, pp. 169–180. [2] P. Erdős, R.L. Graham, Old and New Problems and Results in Combinatorial Number Theory, in: Monographies de L’Enseignement Mathématique, vol. 28, Genève, 1980. [3] H. Halberstam, K. Roth, Sequences, Oxford Univ. Press, 1966. [4] N. Hegyvári, F. Hennecart, A. Plagne, A proof of two Erdős’ conjectures on restricted addition and further results, J. Reine Angew. Math. 560 (2003) 199–220. [5] N. Hegyvári, F. Hennecart, A. Plagne, Answer to a question of Burr and Erdős on restricted addition and related results, Combin. Probab. Comput. 16 (5) (2007) 747–756.
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[6] F. Hennecart, On the restricted order of asymptotic bases of order two, Ramanujan J. 9 (2005) 123–130. [7] J.B. Kelly, Restricted bases, Amer. J. Math. 79 (1957) 258–264. [8] M.B. Nathanson, Additive Number Theory: Inverse Problems and the Geometry of Sumsets, in: Graduate Texts in Mathematics, vol. 165, Springer-Verlag, 1996.
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European Journal of Combinatorics journal homepage: www.elsevier.com/locate/ejc
All sums of h distinct terms of a sequence✩ Yong-Gao Chen a , Jin-Hui Fang b a
School of Mathematical Sciences and Institute of Mathematics, Nanjing Normal University, Nanjing 210023, PR China b Department of Mathematics, Nanjing University of Information Science & Technology, Nanjing 210044, PR China
article
abstract
info
Article history: Received 10 September 2013 Accepted 13 May 2014
For A ⊆ Z, we study the gaps in the sequence of all sums of h pairwise distinct elements of A. For example, the following result is proved: for any integer h ≥ 3, there exists A ⊆ Z such that every integer can be uniquely (neglecting the order) represented as a sum of h not necessarily distinct elements of A, and for any integer ℓ ≥ 1, in the sequence of all sums of ℓ pairwise distinct elements of A, the gaps can be arbitrarily large. Several questions are posed in this paper. © 2014 Elsevier Ltd. All rights reserved.
1. Introduction For a sequence of integers A = {a1 < a2 < · · ·}, define
∆(A) = lim sup(ai+1 − ai ), i→+∞
let h × A be the set of all sums of h pairwise distinct elements of A and let hA be the set of all sums of h not necessarily distinct elements of A. Let N be the set of all positive integers. Let B ∼ N denote that a set of integers B contains all but finitely many positive integers. In 1996, Erdős and Burr [1] asked the following question: is it true that if h(A ∪ {0}) ∼ N, then
∆(A ∪ 2 × A ∪ · · · ∪ h × A) < +∞?
✩ This work was supported by the National Natural Science Foundation of China, Grant Nos. 11371195 and 11201237.
E-mail addresses: [email protected], [email protected] (Y.-G. Chen), [email protected] (J.-H. Fang). http://dx.doi.org/10.1016/j.ejc.2014.05.002 0195-6698/© 2014 Elsevier Ltd. All rights reserved.
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In 2007, Hegyvári, Hennecart and Plagne [5] showed that the answer to the above Erdős–Burr question is affirmative for h = 2 and negative for h ≥ 3. That is, Theorem A (Hegyvári–Hennecart–Plagne). If (A ∪ 2A) ∼ N, then
∆(A ∪ 2 × A) ≤ 2. If 2A ∼ N, then ∆(2 × A) ≤ 2. Let h ≥ 3. There exists a set A such that h(A ∪ {0}) ∼ N and
∆(A ∪ 2 × A ∪ · · · ∪ h × A) = +∞. There exists a set A such that hA ∼ N and ∆(h × A) = +∞. Hegyvári, Hennecart and Plagne [5] conjectured that, if hA ∼ N, then there exists a positive integer k such that ∆(k × A) < +∞. If this conjecture is true, then write k(h) = max min{k : k ∈ N, ∆(k × A) < +∞}. hA∼N
In [5], it is proved that k(h) ≥ 2h−2 + h − 1 for all h ≥ 2. They also conjectured that the function ∆(h × A) is non-increasing and proved that ∆(3 × A) ≤ ∆(2 × A). The restricted order of A is defined to be the least positive integer h such that A ∪ 2 × A ∪ · · · ∪ h × A ∼ N. For the other related results, one may refer to [3,6–8]. In this paper, we consider A ⊆ Z, where Z is the set of all integers. The situation is completely different when one can allow that A ⊆ Z. For example, in the proof of [5, Proposition 5], it is used that ai + (2 × A) ∩ [0, ai ) is contained in 3 × A. Now, it is not true since A contains possibly negative integers. In the sequel of this paper, we allow that the set A contains both positive and negative integers (not necessary both). Then we order A and h × A as
· · · < a− 1 < a0 < a1 < · · · , · · · < b −1 < b 0 < b 1 < · · · , respectively, where a−1 < 0 ≤ a0 and b−1 < 0 ≤ b0 . Let
∆+ (h × A) = lim sup(bi+1 − bi ),
if sup bi = +∞,
i→+∞
∆− (h × A) = lim sup(bi+1 − bi ),
if inf bi = −∞.
i→−∞
Denote −A = {−a : a ∈ A}. It is clear that −(h × A) = h × (−A). Thus
∆+ (h × A) = ∆− (h × (−A)), ∆− (h × A) = ∆+ (h × (−A)). It is clear that if A ⊆ N, then ∆+ (h × A) = ∆(h × A). Thus, Theorems 1 and 3, and Corollary 1(a) are also true for ∆(h × A) when A ⊆ N. In this paper, the following results are proved. Theorem 1. For h ≥ 2 we have
∆+ (h × A) ≤ ∆+ (2 × A),
∆− (h × A) ≤ ∆− (2 × A).
Let R(h) (A, n) be the number of solutions of equation a1 + · · · + ah = n with a1 , . . . , ah ∈ A, a1 ≤ a2 ≤ · · · ≤ ah . In [4], Hegyvári, Hennecart and Plagne confirmed a conjecture in [2]: if A ⊆ N and hA ∼ N, then h × A has a positive lower density. For A ⊆ Z and h ≥ 3, hA ∼ Z does not imply that h × A has a positive lower density. The main reason for this is that we can choose A sparsely when one can allow that A ⊆ Z.
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Theorem 2. If A is a subset of Z such that R(2) (A, n) ≥ 1 for all but finitely many n ∈ Z, then, for all integers ℓ ≥ 2, we have
∆+ (ℓ × A) ≤ 2,
∆− (ℓ × A) ≤ 2.
For any integer h ≥ 3, there exists A ⊆ Z such that R(h) (A, n) = 1 for all n ∈ Z, and A ∪ 2 × A ∪ · · · has density zero, that is d(A ∪ 2 × A ∪ · · ·) =: lim
x→+∞
|(A ∪ 2 × A ∪ · · ·) ∩ [−x, x]| = 0. 2x + 1
(1)
In particular, for all integers ℓ ≥ 1 we have
∆+ (ℓ × A) = +∞,
∆− (ℓ × A) = +∞.
(2)
For A ⊆ N, Hegyvári, Hennecart and Plagne proved that ∆((h + 1) × A) ≤ ∆(h × A) + ah − a1 (see [4, Proof of Proposition 4]). One may see that the method in [4, Proof of Proposition 4] can be used to prove that
∆+ ((h + 1) × A) ≤ ∆+ (h × A) + ah − a1 , ∆− ((h + 1) × A) ≤ ∆− (h × A) − a−h + a−1 . In this paper, the following bounds are obtained. Theorem 3. For any integer h ≥ 1, we have
∆+ ((h + 1) × A) ≤ max{∆+ (h × A), a2h−2 },
(3)
∆− ((h + 1) × A) ≤ max{∆− (h × A), −a−2h+1 }.
(4)
Corollary 1. (a) If ∆+ (h0 × A) < +∞ for some h0 , then there exists h1 ≥ h0 such that, for all h ≥ h1 , we have ∆+ (h × A) ≤ a2h−4 ; (b) If ∆− (h0 × A) < +∞ for some h0 , then there exists h1 ≥ h0 such that, for all h ≥ h1 , we have ∆− (h × A) ≤ −a−2h+3 . Motivated by Theorems 1 and 3 and the Hegyvári–Hennecart–Plagne conjecture: the function
∆(h × A) is non-increasing, we pose the following question.
Question 1. Is it true that the function ∆+ (h × A) is non-increasing? Currently, we cannot give an answer to the following weak forms of Question 1: Question 2. Is it true that for any integer h ≥ 1, there exists an integer ℓ > h such that
∆+ (ℓ × A) ≤ ∆+ (h × A)? Question 3. Is it true that for any integer h ≥ 1, there exists a constant c = c (h) such that for any integer ℓ > h we have
∆+ (ℓ × A) ≤ ∆+ (h × A) + c? Recall that all elements of h × A are · · · < b−1 < b0 < b1 < · · ·, let ∆0 (h × A) = sup(bi+1 − bi ), i
δ0 (h × A) = inf(bi+1 − bi ), i
δ+ (h × A) = lim inf(bi+1 − bi ),
if sup bi = +∞,
δ− (h × A) = lim inf(bi+1 − bi ),
if inf bi = −∞.
i→+∞
i→−∞
It is easy to see that functions δ0 (h × A), δ+ (h × A) and δ− (h × A) are non-increasing. Let A = {1, 2, 3} ∪ {n : n ≥ 5}. Then 2 × A = {n : n ∈ Z, n ≥ 3} and 3 × A = {6} ∪ {n : n ∈ Z, n ≥ 8}.
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Thus ∆0 (3 × A) = 2 > 1 = ∆0 (2 × A). It is interesting to find out the nontrivial properties of these functions. Here we pose the following questions. We do not know whether these questions are also trivial. Question 4. Given any set A ⊆ Z. Is it true that the function ∆0 (h × A) is non-increasing from some point h = h0 on? The following question is a special case of the above question. Question 5. Let A ⊆ Z such that hA = Z. Is it true that ∆0 (ℓ × A) is non-increasing from some point ℓ = ℓ0 on? 2. Proofs Proof of Theorem 1. Since ∆− (h × A) = ∆+ (h × (−A)), it is enough to prove the first inequality. Assume that ∆+ (2 × A) exists and d =: ∆+ (2 × A) < +∞. Otherwise, Theorem 1 is trivial. We shall use induction on h to prove that ∆+ (h × A) ≤ d for all integers h ≥ 2. It is clear that the conclusion is true for h = 2. Now we assume that k > 2 and the conclusion is true for 2 ≤ h < k. That is, for any 2 ≤ h < k, we have ∆+ (h × A) ≤ d. Then there exists n0 > 0 such that, for any 2 ≤ h < k and n ≥ n0 , there exists 0 ≤ i ≤ d − 1 such that n + i ∈ h × A. We shall prove that ∆+ (k × A) ≤ d. Select an ai0 +1 ≥ n0 + d and m0 = ai0 +1 + · · · + ai0 +k−2 + n0 . Let B = {m : m ≥ m0 , for any 0 ≤ i ≤ d − 1, m + i ̸∈ k × A}. We shall show that
|B| ≤ (k − 1)(k − 2)d,
(5)
i.e. B is a finite set. Indeed, assume that (5) is true. Let m′0 = max B + 1 if B ̸= ∅ and m′0 = m0 if B = ∅. Then, for any m ≥ m′0 , we have m ̸∈ B. So there exists 0 ≤ i ≤ d − 1 with m + i ∈ k × A. This implies that ∆+ (k × A) ≤ d. To prove Theorem 1, it is enough to prove that (5) holds. For m ∈ B, by m ≥ m0 = ai0 +1 + · · · + ai0 +k−2 + n0 , we have m − (ai0 +1 + · · · + ai0 +k−2 ) ≥ n0 . Hence there exists 0 ≤ i ≤ d − 1 such that m − (ai0 +1 + · · · + ai0 +k−2 ) + i ∈ 2 × A. Thus there exist two integers s < t such that m − (ai0 +1 + · · · + ai0 +k−2 ) + i = as + at . That is, m + i = ai0 +1 + · · · + ai0 +k−2 + as + at . Since m ∈ B, it follows that m + i ̸∈ k × A. So at least two of i0 + 1, . . . , i0 + k − 2, s, t are equal. Thus
{s, t } ∩ {i0 + 1, . . . , i0 + k − 2} ̸= ∅. Without loss of generality, we assume that i0 + 1 ≤ s ≤ i0 + k − 2. Since ai0 +1 + · · · + ai0 +k−2 + as − i ≥ as − d + 1 ≥ ai0 +1 − d + 1 > n0 , it follows from the inductive hypothesis that there exist 0 ≤ ji,s ≤ d − 1 and j1 < · · · < jk−1 such that ai0 +1 + · · · + ai0 +k−2 + as − i + ji,s = aj1 + · · · + ajk−1 .
(6)
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For each pair i, s with 0 ≤ i ≤ d − 1 and i0 + 1 ≤ s ≤ i0 + k − 2, according to (6), we fix ji,s , j1 , . . . , jk−1 with 0 ≤ ji,s ≤ d − 1 and j1 < · · · < jk−1 . Thus m + ji,s = aj1 + · · · + ajk−1 + at . Since m ∈ B, it follows that t ∈ {j1 , . . . , jk−1 }. Thus m = ai0 +1 + · · · + ai0 +k−2 + as + at − i can take at most d(k − 2)(k − 1) values (i can take at most d values and s can take at most k − 2 values. When i and s have been taken, the integers ji,s , j1 , . . . , jk−1 are fixed. Thus t can take at most k − 1 values). Therefore, (5) holds. This completes the proof of Theorem 1. Proof of Theorem 2. The case h = 2 is similar to the proof of [5, Theorem 1]. Since R(2) (A, n) ≥ 1 for all but finitely many n ∈ Z, it follows that all but finitely many odd numbers belong to 2 × A. This implies that
∆+ (2 × A) ≤ 2,
∆− (2 × A) ≤ 2.
(7)
By Theorem 1 and (7), for all integers ℓ ≥ 2, we have
∆+ (ℓ × A) ≤ 2,
∆− (ℓ × A) ≤ 2.
Let h ≥ 3. We construct A in the following iterate process: we shall construct a sequence 0 = i1 < i2 < · · · of integers and a sequence A1 ⊂ A2 ⊂ · · · of subsets of Z such that, for all k ≥ 1, we have (a) (b) (c) (d)
R(h) (Ak , n) = 1 for all n ∈ Z with |n| ≤ ik ; R(h) (Ak , n) ≤ 1 for all n ∈ Z; no two elements of Ak have the same absolute; di > d1 + d2 + · · · + di−1 + 4i for all 2 ≤ i ≤ tk , where d1 < · · · < dtk are all absolutes of elements in Ak . Define A=
∞
Ai .
i =1
Let i1 = 0 and A1 = {−32, 32(h − 1)}. Then (a)–(d) hold for k = 1. Suppose that we have found 0 = i1 < i2 < · · · < iℓ and A1 ⊂ A2 ⊂ · · · ⊂ Aℓ such that (a)–(d) hold for all 1 ≤ k ≤ ℓ. Let d1 , . . . , dtℓ be all absolutes of elements in Aℓ with d1 < d2 < · · · < dtℓ . Let iℓ+1 be the least positive integer for which there exists n ∈ Z with |n| = iℓ+1 and R(h) (Aℓ , n) = 0. Then, by (a), we have iℓ < iℓ+1 . Let Tℓ be the set of such integers. It is clear that |Tℓ | ∈ {1, 2}. Case 1: |Tℓ | = 1. Let n1 ∈ Z with |n1 | = iℓ+1 and R(h) (Aℓ , n1 ) = 0. We choose a′ℓ+1 > 2htℓ (dtℓ + |n1 |) + 4tℓ +2 . Now we take Aℓ+1 = Aℓ ∪ {−a′ℓ+1 , (h − 1)a′ℓ+1 + n1 }. Let dtℓ +1 = a′ℓ+1 and dtℓ +2 = (h − 1)a′ℓ+1 + n1 . By h ≥ 3 and a′ℓ+1 > 2htℓ (dtℓ + |n1 |) + 4tℓ +2 , we have dtℓ +1 > 2htℓ (dtℓ + |n1 |) + 4tℓ +2
> d1 + d2 + · · · + dtℓ + 4tℓ +1 ,
dtℓ +2 ≥ 2a′ℓ+1 − |n1 | > 2htℓ (dtℓ + |n1 |) + 4tℓ +2 + a′ℓ+1 − |n1 |
≥ d1 + d2 + · · · + dtℓ + dtℓ +1 + 4tℓ +2 .
So (c) and (d) hold for k = ℓ + 1.
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Since |Tℓ | = 1 and n1 = (h − 1)(−a′ℓ+1 ) + ((h − 1)a′ℓ+1 + n1 ), it follows that R(h) (Aℓ+1 , n) ≥ 1 for all n ∈ Z with |n| ≤ iℓ+1 . To prove (a) and (b), it is enough to prove that R(h) (Aℓ+1 , n) ≤ 1 for all n ∈ Z. Suppose that this is false. Let m be an integer with R(h) (Aℓ+1 , m) ≥ 2. Then there exist b1 , . . . , bs ∈ Aℓ with b1 ≤ b2 ≤ · · · ≤ bs , b′1 , . . . , b′s′ ∈ Aℓ with b′1 ≤ b′2 ≤ · · · ≤ b′s′ , and nonnegative integers u, v, u′ , v ′ with s + u + v = h and s′ + u′ + v ′ = h such that
(b1 , . . . , bs , u, v) ̸= (b′1 , . . . , b′s′ , u′ , v ′ ), m = b1 + · · · + bs + u(−a′ℓ+1 ) + v((h − 1)a′ℓ+1 + n1 ),
(8)
(−a′ℓ+1 )
(9)
′
′
m = b1 + · · · + bs′ + u
′
+ v ((h − ′
1)a′ℓ+1
+ n1 ).
If −u + v(h − 1) ̸= −u′ + v ′ (h − 1), then a′ℓ+1 ≤ |u(−a′ℓ+1 ) + v((h − 1)a′ℓ+1 ) − u′ (−a′ℓ+1 ) − v ′ ((h − 1)a′ℓ+1 )|
= ≤ ≤ ≤ =
|b′1 + · · · + b′s′ + v ′ n1 − (b1 + · · · + bs + v n1 )| |b′1 | + · · · + |b′s′ | + v ′ |n1 | + |b1 | + · · · + |bs | + v|n1 | s′ dtℓ + v ′ |n1 | + sdtℓ + v|n1 | hdtℓ + h|n1 | + hdtℓ + h|n1 | 2h(dtℓ + |n1 |),
a contradiction. So −u + v(h − 1) = −u′ + v ′ (h − 1). If v = v ′ , then u = u′ . Thus s = s′ , b1 + · · · + bs = b′1 + · · · + b′s′ and (b1 , . . . , bs ) ̸= (b′1 , . . . , b′s′ ). So s = s′ ≥ 2. Let m′ = b1 + · · · + bs + (h − s)b1 . Then m′ = b′1 + · · · + b′s + (h − s)b1 . This contradicts R(h) (Aℓ , m′ ) ≤ 1. Hence v ̸= v ′ . Without loss of generality, we may assume that v > v ′ . Then, by −u + v(h − 1) = −u′ + v ′ (h − 1), we have u > u′ . Since 0 ≤ u, v, v ′ , s ≤ h, it follows that u = h − v − s ≤ h − v ≤ h − 1. Hence h − 1 ≥ u − u′ = (v − v ′ )(h − 1) ≥ h − 1. This implies that u = h − 1, u′ = 0, v = 1, and v ′ = 0. So s = 0 and s′ = h. By (8) and (9), we have n1 = m = b′1 + · · · + b′s′ = b′1 + · · · + b′h , a contradiction with R(h) (Aℓ , n1 ) = 0 (n1 ∈ Tℓ ). Now we have proved that R(h) (Aℓ+1 , n) ≤ 1 for all n ∈ Z. So (a) and (b) hold. Case 2: |Tℓ | = 2. Then Tℓ = {−iℓ+1 , iℓ+1 }. We choose a′ℓ+1 > 2htℓ (dtℓ + iℓ+1 ) + 4tℓ +2 and ′ aℓ+2 > 2h(tℓ + 2)((h − 1)a′ℓ+1 + iℓ+1 + iℓ+1 ) + 4tℓ +4 . Now we take Aℓ+1 = Aℓ ∪ {−a′ℓ+1 , (h − 1)a′ℓ+1 + iℓ+1 } ∪ {−a′ℓ+2 , (h − 1)a′ℓ+2 − iℓ+1 }. Let Bℓ = Aℓ ∪ {−a′ℓ+1 , (h − 1)a′ℓ+1 + iℓ+1 }. Then Aℓ+1 = Bℓ ∪ {−a′ℓ+2 , (h − 1)a′ℓ+2 − iℓ+1 }. As in Case 1, we can prove that
(a1 ) (b1 ) (c1 ) (d1 )
R(h) (Bℓ , n) = 1 for all n ∈ Z with |n| ≤ iℓ and R(h) (Bℓ , iℓ+1 ) = 1; R(h) (Bℓ , n) ≤ 1 for all n ∈ Z; no two elements of Bℓ have the same absolute; di > d1 + d2 + · · · + di−1 + 4i for all 2 ≤ i ≤ tℓ + 2, where d1 < · · · < dtℓ +2 are all absolutes of elements in Bℓ .
Again, as in Case 1, we can prove that (a)–(d) hold for k = ℓ + 1. Now we have constructed a sequence 0 = i1 < i2 < · · · of integers and a sequence A1 ⊂ A2 ⊂ · · · of subsets of Z satisfying (a)–(d).
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Thus we have constructed A such that (a ) R(h) (A, n) = 1 for all n ∈ Z; (b′ ) no two elements of A have the same absolute; (c′ ) di > d1 + d2 + · · · + di−1 + 4i for all i ≥ 2, where d1 < d2 < · · · are all absolutes of elements in A. ′
It is clear that (b′ ) and (c′ ) hold. Now we prove that (a′ ) holds. In fact, for any n ∈ Z, there exists an integer k ≥ 1 such that |n| ≤ ik . Since Ak ⊆ A, it follows that R(h) (A, n) ≥ R(h) (Ak , n) = 1. If R(h) (A, n) ≥ 2, then n has two different expressions n = c1 + · · · + ch = c1′ + · · · + ch′ with c1 , . . . , ch ∈ A, c1 ≤ · · · ≤ ch , and c1′ , . . . , ch′ ∈ A, c1′ ≤ · · · ≤ ch′ . Since A1 ⊂ A2 ⊂ · · ·, there exists Ak′ such that c1 , . . . , ch , c1′ , . . . , ch′ ∈ Ak′ . This contradicts R(h) (Ak′ , n) ≤ 1. Now we have proved that R(h) (A, n) = 1 for all n ∈ Z. Now, we prove that (1) holds. Let ei ∈ A with |ei | = di (i = 1, 2, . . .). Let x > 4. Suppose that t ∈ (A ∪ 2 × A ∪ · · ·) ∩ [−x, x]. Then, for some positive integer ℓ, we have t = ei1 + · · · + eiℓ ,
1 ≤ i1 < i2 < · · · < iℓ .
If iℓ ≥ 2, then x ≥ |t | ≥ |eiℓ | − |eiℓ−1 | − · · · − |ei1 |
= diℓ − diℓ−1 − · · · − di1 ≥ diℓ − diℓ −1 − · · · − d1 ≥ 4iℓ . So iℓ ≤ log4 x. Hence, the number of integers t does not exceed the number of subsets of {1, 2, . . . , [log4 x]}. This implies that √ |(A ∪ 2 × A ∪ · · ·) ∩ [−x, x]| ≤ 2log4 x = x. Therefore, (1) holds. Given any integer ℓ ≥ 1. Since R(h) (A, n) = 1 for all n ∈ Z, it follows that sup A = +∞ and inf A = −∞. Thus sup(ℓ × A) = +∞,
inf(ℓ × A) = −∞.
Hence ∆+ (ℓ × A) and ∆− (ℓ × A) are well defined. If ∆+ (ℓ × A) < +∞, then all differences of two consecutive nonnegative terms of (ℓ × A) ∪ {0} are less than a positive constant c depending only on ℓ and A. So |(ℓ × A) ∩ [0, x]| ≥ (x/c ) − 1, a contradiction with d(A ∪ 2 × A ∪ · · ·) = 0. Hence ∆+ (ℓ × A) = +∞. Similarly, we have ∆− (ℓ × A) = +∞. This completes the proof of Theorem 2. Proof of Theorem 3. We will prove inequality (3). Inequality (4) can be proved similarly. By the definition of ∆+ (h × A), there exists y0 > 0 such that for any two consecutive terms y′ , y′′ of h × A with y′′ > y′ and y′′ ≥ y0 , we have y′′ − y′ ≤ ∆+ (h × A). Suppose that
∆+ ((h + 1) × A) > max{∆+ (h × A), a2h−2 }.
(10)
Then there exist infinitely many positive integers x ∈ (h + 1) × A such that for any x′ ∈ (h + 1) × A with x′ > x, we have x′ − x > max{∆+ (h × A), a2h−2 }.
(11)
Let x ∈ (h + 1) × A be such an integer with x > ah + · · · + a2h−1 + y0 . Let y be the largest integer in h × A and z be the least integer in h × A such that y ≤ x < z. Then y ≥ ah + · · · + a2h−1 . Suppose that y = ai 1 + · · · + ai h ,
i1 < · · · < ih ,
z = aj1 + · · · + ajh ,
j1 < · · · < jh .
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Since z > x ≥ y ≥ ah + · · · + a2h−1 > a0 + · · · + ah−1 , we have
{i1 , . . . , ih } ̸= {h − 1, h, . . . , 2h − 2}, {j1 , . . . , jh } ̸= {0, 1, . . . , h − 1}.
(12) (13)
By (12), there exists h − 1 ≤ j ≤ 2h − 2 such that j ̸∈ {i1 , . . . , ih }. Let i be an integer with 0 ≤ i ≤ 2h − 2 such that i ̸∈ {i1 , . . . , ih }. If y + ai > x, then we have y + ai ∈ (h + 1) × A and 0 < (y + ai ) − x ≤ (y + ai ) − y = ai ≤ a2h−2 , a contradiction with (11). Hence y + ai ≤ x. Noting that z > x > y0 and y, z ∈ h × A, if i ̸∈ {j1 , . . . , jh }, then z + ai ∈ (h + 1) × A and 0 < z − x ≤ (z + ai ) − x ≤ (z + ai ) − (y + ai ) = z − y ≤ ∆+ (h × A), a contradiction with (11). So i ∈ {j1 , . . . , jh }. If i ≥ 1 and i − 1 ∈ {i1 , . . . , ih }, then y + ai − ai−1 ∈ h × A. Since y + ai − ai−1 > y, it follows from the definition of y that y + ai − ai−1 > x. This contradicts y + ai ≤ x and ai−1 ≥ 0. So either i = 0 or i − 1 ̸∈ {i1 , . . . , ih }. Thus we have proved that, under the assumption of (10), if i is an integer with 0 ≤ i ≤ 2h − 2 such that i ̸∈ {i1 , . . . , ih }, then i ∈ {j1 , . . . , jh }
(if i ≥ 0),
i − 1 ̸∈ {i1 , . . . , ih }
(if i ≥ 1).
(14)
Let i0 be the largest integer with 0 ≤ i0 ≤ 2h − 2 such that i0 ̸∈ {i1 , . . . , ih }. By (14) we have i0 − 1, i0 − 2, . . . , 1, 0 ̸∈ {i1 , . . . , ih } and then
{0, 1, . . . , i0 } ⊆ {j1 , . . . , jh }.
(15)
By (12), we have i0 ≥ h − 1. By (13) and (15), we have i0 ≤ h − 2, a contradiction. This completes the proof of Theorem 3. Proof of Corollary 1. By Theorem 3, for ℓ > h0 , we have
∆+ (ℓ × A) ≤ ≤ = ≤ ≤
max{∆+ ((ℓ − 1) × A), a2ℓ−4 } max{max{∆+ ((ℓ − 2) × A), a2ℓ−6 }, a2ℓ−4 } max{∆+ ((ℓ − 2) × A), a2ℓ−4 }
··· max{∆+ (h0 × A), a2ℓ−4 }.
Select an integer h1 > h0 such that a2h1 −4 ≥ ∆+ (h0 × A). Then, for ℓ ≥ h1 , we have ∆+ (ℓ× A) ≤ a2ℓ−4 . This completes the proof of Corollary 1(a). Similarly, we can give a proof of Corollary 1(b). Acknowledgments We would like to thank two referees for their comments and N. Hegyvári for sending us the reprint of [5]. References [1] P. Erdős, Some of my new and almost new problems and results in combinatorial number theory, in: Number Theory (Eger 1996), de Gruyter, Berlin, 1998, pp. 169–180. [2] P. Erdős, R.L. Graham, Old and New Problems and Results in Combinatorial Number Theory, in: Monographies de L’Enseignement Mathématique, vol. 28, Genève, 1980. [3] H. Halberstam, K. Roth, Sequences, Oxford Univ. Press, 1966. [4] N. Hegyvári, F. Hennecart, A. Plagne, A proof of two Erdős’ conjectures on restricted addition and further results, J. Reine Angew. Math. 560 (2003) 199–220. [5] N. Hegyvári, F. Hennecart, A. Plagne, Answer to a question of Burr and Erdős on restricted addition and related results, Combin. Probab. Comput. 16 (5) (2007) 747–756.
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