Generalized asymptotic Sidon basis

Let h, k ≥ 2 be integers. We say a set A of positive integers is an asymptotic basis of order k if every large enough positive integer can be represented as the sum of k terms from A. A set of positive integers A is called Bh[g] set if all positive integers can be represented as the sum of h terms from A at most g times. In this paper we prove the existence of Bh[1] sets which are asymptotic bases of order 2h + 1 by using probabilistic methods. 2010 Mathematics Subject Classification: 11B34, 11B75.


Introduction
Let N denote the set of positive integers. Let h, k ≥ 2 be integers. Let A ⊂ N be an infinite set of positive integers and let R h,A (n) denote the number of solutions of the equation a 1 + a 2 + · · · + a h = n, a 1 ∈ A, . . . , a h ∈ A, a 1 ≤ a 2 ≤ . . . ≤ a h , where n ∈ N. A set of positive integers A is called B h [g] set if for every n ∈ N, the number of representations of n as the sum of h terms in the form (1) is at most g, that is R h,A (n) ≤ g. We denote the fact that A is a B h [g] set by A ∈ B h [g]. We say a set A ⊂ N is an asymptotic basis of order k, if R k,A (n) > 0 for all large enough positive integer n, i.e., if there exists a positive integer n 0 such that R k,A (n) > 0 for n > n 0 . In [4] and [5] P. Erdős, A. Sárközy and V. T. Sós asked if there exists a Sidon set (or B 2 [1] set) which is an asymptotic basis of order 3. It is easy to see that a Sidon set cannot be an asymptotic basis of order 2. J. M. Deshouillers and A. Plagne in [3] constructed a Sidon set which is an asymptotic basis of order at most 7. In [7] it was proved the existence of Sidon sets which are asymptotic bases of order 5 by using probabilistic methods. In [1] and [9] this result was improved on by proving the existence of a Sidon set which is an asymptotic basis of order 4. It was also proved [1] that there exists a B 2 [2] set which is an asymptotic basis of order 3. In this paper we will prove a similar but more general theorem. Namely, we prove the existence of an asymptotic basis of order 2h + 1 which is a B h [1] set.
Theorem 1. For every h ≥ 2 integer there exists a B h [1] set which is an asymptotic basis of order 2h + 1.
Before we prove the above theorem, we give a short survey of the probabilistic method we are working with.

Probabilistic tools
To prove Theorem 1 we use the probabilistic method due to Erdős and Rényi. There is an excellent summary of this method in the book of Halberstam and Roth [6]. In this paper we denote the probability of an event by P, and the expectation of a random variable Y by E(Y ). Let Ω denote the set of the strictly increasing sequences of positive integers.
Then there exists a probability space (Ω, X, P) with the following two properties: (i) For every natural number n, the event E (n) = {A: A ∈ Ω, n ∈ A} is measurable, and P(E (n) ) = α n .
See Theorem 13. in [6], p. 142. We denote the characteristic function of the event E (n) by (A, n): Furthermore, for some A = {a 1 , a 2 , . . . } ∈ Ω we denote the number of solutions of Then Let R * h (n) denote the number of those representations of n in the form (1) in which there are at least two equal terms. Thus we have In the proof of Theorem 1 we use the following lemma: then with probability 1, at most a finite number of the events X j can occur.

Proof of Theorem 1
Let h be fixed and let α = 2 4h+1 . Define the sequence α n in Lemma 1 by The proof of Theorem 1 consists of three parts. In the first part we prove similarly as in [8] that with probability 1, A is an asymptotic basis of order 2h + 1. In particular, we show that R 2h+1,A (n) tends to infinity as n goes to infinity. In the second part we show that deleting finitely many elements from A we obtain a B h [1] set. Finally, we show that the above deletion does not destroy the asymptotic basis property. By (3), to prove that A is an asymptotic basis of order 2h + 1 it is enough to show r 2h+1,A (n) > 0 for every n large enough. To do this, we apply the following lemma with k = 2h + 1. The proof of Lemma 3 can be found in [8]. It is clear from (3) that where E denotes the event where c is a suitable positive constant. In the next step we prove that removing finitely elements from A we get a B h [1] set with probability 1. To do this, it is enough to show that with probability 1, R h,A (n) ≤ 1 for every n large enough. Note that in a representation of n as the sum of h terms there can be equal summands. To handle this situation we consider the terms of a representation a 1 + . . . + a h = n as a vector (a 1 , . . . , a h ) ∈ N h . We denote the set which elements are the coordinates of the vectorx as Set(x). Of course, if two or more coordinates ofx are equal, this value appears only once in Set(x). We say that two vectorsx andȳ are disjoint if Set(x) and Set(ȳ) are disjoint sets. We define r * l,A (n) as the maximum number of pairwise disjoint representations of n as sum of l elements of A, i.e., the maximum number of pairwise disjoint vectors of R l (n) with their coordinates in A. We say that A is a B * l [g] sequence if r * l,A (n) ≤ g for every n.
(ii) For every h + 1 ≤ k ≤ 2h almost always there exists a finite set A k such that r * k,A\A k (n) ≤ 4h + 1. Proof. We need the following proposition (see Lemma 3.7 in [2]). Proposition 1. For a sequence A ∈ Ω, for every k and n P(r * k,A (n) ≥ s) ≤ C k,α,s n (kα−1)s where C k,α,s depends only on k, α and s.
Since h < k ≤ 2h, we have then by the Borel-Cantelli lemma we get that almost always there exists an n k such that r * k,A (n) ≤ 4h + 1 for n ≥ n k . It follows that It follows from (4) and Lemma 4 that there exists a set A and for every 2 for n ≥ n 0 and for every 2 ≤ k ≤ h, for every h < k ≤ 2h, r * k,A (n) ≤ 4h + 1.
Set B = A \ ∪ 2h k=1 A k . In the next step we show that B is both a B h [1] set and a B 2h [g] set for some g. We apply the following proposition (see Remark 3.10 in [2]).
By using the definition of B, the fact that B * 2 [1] = B 2 [1] and (6), (7) it follows that Applying Proposition 2 with g = l = 1 we get by induction that for set. Applying Proposition 2 with g = 4h+1, l = 1 we get that B ∈ B h+1 [4h+1]. Using Proposition 2 again with g = 4h+1, . Continuing this process we obtain that for every 1 < k ≤ 2h we have B ∈ B k [g k ] for some positive integer g k . Let ∪ 2h We prove by contradiction. Assume that there exists a positive integer n with R 2h,A (n) > 2 w · max 1<k≤2h g k . Then there exist indices 1 ≤ i 1 < i 2 < . . . < i j ≤ w such that the number of representations of in the form n = d i 1 + . . . + d i j + c j+1 + . . . + c 2h , where c j+1 , . . . , c 2h ∈ B is more than max 1<k≤2h g k . It follows that which is a contradiction. Finally, we prove similarly as in [7] that B is an asymptotic basis of order 2h + 1, i.e., the deletion of the "small"elements of A does not destroy its asymptotic basis property. We prove by contradiction. Assume that there exist infinitely many positive integers M which cannot be represented as the sum of 2h + 1 numbers from B. Choose such an M large enough. In view of (5), we have R 2h+1,A (M ) > cM 1 4h+1 . It follows from our assumption that every representations of M as the sum of 2h+1 numbers from A contains at least one element from A \ B = ∪ 2h k=1 A k . Then by the pigeon hole principle there exists an y ∈ ∪ 2h k=1 A k which is in at least R 2h+1,A (M ) w representations of M . As A ∈ B 2h [G], it follows that with probability 1, which is a contradiction if M is large enough.

Acknowledgement
Supported by theÚNKP-18-4 New National Excellence Program of the Ministry of Human Capacities.