Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Cilleruelo 2000 upper bound b 2 2 sequences
theorem_1: Cilleruelo's bound that a subset of [1,N] in which every integer has at most two representations a + b with a <= b has at most sqrt(6N) + 1 elements; it bounds the counting function of every infinite such set but does not decide Problem 158.
Javier Cilleruelo, An upper bound for B_2[2] sequences. Journal of Combinatorial Theory, Series A 89 (2000), no. 1, 141-144. doi:10.1006/jcta.1999.3012.
Cilleruelo introduces a combinatorial counting method for finite B_2[2] sequences, sets in [1,N] where every integer has at most two representations as a + b with a <= b (a sum a + a counts once). Theorem 1 proves F(N,2) <= sqrt(6N) + 1, improving the density-argument bound F(N,g) <= 1.864 sqrt(gN) of Ruzsa, Trujillo and the author in the case g = 2. The proof exploits the average behavior of the difference function d(n) rather than the sum function: Lemma 2 identifies the sum of binomial(d(n),2) with 2|R_2| + |R_2'|, where R_2 and R_2' are the sets of integers with exactly two representations off and on the doubled set {2a : a in A}: the sum counts each 4-tuple a < b < c < d with a + d = b + c (one for each element of R_2) twice and each triple a < b < d with a + d = 2b (one for each element of R_2') once. Lemma 3 combines this with Lemma 1 i) to bound the sum by binomial(|A|,2), and Cauchy-Schwarz with Lemma 1 ii) then gives binomial(|A|,2) <= 3N. The paper also cites the known lower bound F(N,2) >= (3/2 + o(1)) sqrt(N) of Cilleruelo, Ruzsa and Trujillo, well below sqrt(6N). A note on p. 3 records that M. Helm proved independently, by a different method, F(N,2) <= sqrt(6N) + O(1). For problem 158 this supplies finite extremal context: an upper bound on separately optimized finite sets does not force the liminf of a single infinite B_2[2] sequence to vanish.
Version used: the labels and pages below are those of the three-page manuscript described under Source, printed pp. 1--3; the journal pagination 141--144 was not compared. Read status: claims checked for the definitions, Theorem 1 (p. 1) and Lemmas 1--3 (p. 2), read clause by clause on the page images, with the proof of Theorem 1 (pp. 2--3) followed step by step; nothing here is independently reviewed.
Source: https://doi.org/10.1006/jcta.1999.3012. The copy read for this card is a three-page author-typeset manuscript, which prints no notice; the version of record's publisher page could not be read on 2026-10-02 (DOI 10.1006/jcta.1999.3012; doi.org resolves to a linkinghub.elsevier.com redirect stub and ScienceDirect returned HTTP 403), and its Crossref record names only Elsevier's text-and-data-mining and open-archive user licenses, no Creative Commons license, none of which governs that manuscript; the term is unstated.
Bears on. #158: Theorem 1 (p. 1) applies to A ∩ [1,N] for each infinite set A of the problem, whose condition is the paper's B_2[2] (r(n) <= 2, counting a <= b), and so gives |A ∩ [1,N]| <= sqrt(6N) + 1 for every N; it bounds the ratio to N^{1/2} from above and does not decide whether its liminf is 0, which is the question. The paper does not mention the problem.
Contents.
- Theorem 1 (p. 1): F(N,2) <= sqrt(6N) + 1, where F(N,g) is the largest size of a B_2[g] subset of [1,N]; the proof gives binomial(|A|,2) <= 3N (p. 3).
- Lemma 1 (p. 2): For any finite sequence A of positive integers, binomial(|A|,2) equals both the sum of r'(n) and the sum of d(n) over n >= 1.
- Lemma 2 (p. 2): For a B_2[2] sequence A in [1,N], the sum of binomial(d(n),2) equals 2|R_2| + |R_2'|, proved by counting 4-tuples a < b < c < d with a + d = b + c (each twice) and triples a < b < d with a + d = 2b (each once).
- Lemma 3 (p. 2): For such A, the sum of binomial(d(n),2) is at most binomial(|A|,2).
- Context (p. 1): Derives the trivial bound F(N,g) <= 2 sqrt(gN) and cites from Cilleruelo, Ruzsa and Trujillo the bound F(N,g) <= 1.864 sqrt(gN) and the lower bound F(N,2) >= (3/2 + o(1)) sqrt(N).
- Note (p. 3): M. Helm's independent bound F(N,2) <= sqrt(6N) + O(1).
Results.
- Theorem 1 (p. 1): F(N,2) <= sqrt(6N) + 1.
No file of this source is held: no license on record permits its redistribution, and the card cites the edition it names above.