Wiki
Wiki

Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

Updated

An application of graph theory to additive number theory

../

problem_p203: Alon and Erdős's closing problem: a sequence is free when distinct index sets have distinct sums, Pisier's condition (6) that every finite part B has a free part of at least delta|B| terms is necessary for a finite union of free subsequences, and the authors doubt but cannot refute sufficiency.

remark_p203: Alon and Erdős's remark, stated without proof, that the first n squares contain a Sidon subsequence of c(eps) n^{2/3-eps} terms for every eps > 0, while Landau's theorem on sums of two squares bounds every such subsequence by c' n/(log n)^{1/4}.

theorem_1: Alon and Erdős's theorem that every B_2^{(k)} sequence of n terms is a union of c_2^{(k)} n^{1/3} Sidon sequences, sharp up to the constant for k >= 2, with its main ingredient (4): every such sequence contains a Sidon subsequence of at least c_4^{(k)} n^{2/3} terms.

theorem_2: Alon and Erdős's infinite analogue of their bound (4): every infinite B_2^{(k)} sequence contains a Sidon subsequence C with at least [c^{(k)} n^{2/3}] terms among its first n terms, for every n >= 1.

theorem_3: Alon and Erdős's theorem that every finite or infinite B_2^{(k)} sequence is a union of c(k) subsequences none of which contains a three-term arithmetic progression; the proof gives c(k) = 3k + 1.

theorem_4: Alon and Erdős's theorem that every B_2^{(k)} sequence of n terms is a union of c_2^{(k)} n^{1/(2k-1)} B_2^{(k-1)} subsequences, and that for k a power of 2 some B_2^{(k)} sequence of n terms is not a union of c_1^{(k)} n^{1/(2k-1)} of them.


N. Alon, P. Erdős: An application of graph theory to additive number theory, European J. Combin. 6 (1985) no. 3, 201--203, doi:10.1016/S0195-6698(85)80027-5 (MR 87d:11015; Zentralblatt 581.10029).

Theorem 1 states that every B2(k)B_2^{(k)} sequence of nn terms (one in which each integer has at most kk representations as a sum of two distinct terms) is a union of c2(k)n1/3c_2(k)n^{1/3} Sidon (B2B_2) sequences. For k≥2k\geq2 this order is sharp up to the kk-dependent constant, by the upper bound (3), Hn(k)≤Hn(2)<c n2/3H_n^{(k)}\leq H_n^{(2)}<c\,n^{2/3}, which the paper takes from a construction of Erdős it cites rather than proves (p. 201). Its main ingredient is inequality (4), Hn(k)≥c4(k)n2/3H_n^{(k)}\geq c_4(k)n^{2/3}, for the largest Sidon subset guaranteed inside such a sequence (p. 201, Theorem 1 and (4)). Repeatedly removing a subset of that size gives the asserted partition (p. 202, first paragraph).

The proof of (4) is a useful finite hypergraph template. On the indices {1,…,n}\{1,\ldots,n\}, put a 4-edge {i,j,l,m}\{i,j,l,m\} whenever ai+aj=al+ama_i+a_j=a_l+a_m. The B2(k)B_2^{(k)} hypothesis gives fewer than 12(k−1)(n2)≤14(k−1)n2\frac12(k-1)\binom n2\leq\frac14(k-1)n^2 edges. An independent vertex set indexes a Sidon subsequence. Selecting vertices independently with probability cn−1/3c n^{-1/3} gives (c+o(1))n2/3(c+o(1))n^{2/3} selected vertices spanning at most (14(k−1)+o(1))c4n2/3(\frac14(k-1)+o(1))c^4n^{2/3} edges; deleting one vertex from each edge leaves the required independent set when c=c(k)c=c(k) is small (p. 202, proof of Theorem 1). Theorem 2 adapts the selection to an infinite sequence, using probability c/i1/3c/i^{1/3} and deletion of the largest member of each bad quadruple, to obtain the prefix-by-prefix bound (5) (p. 202). Theorem 3 uses a different route: its 3-uniform hypergraph of three-term progressions has at most rkrk edges on every rr vertices, hence a vertex of degree at most 3k3k; greedy coloring and compactness then give a partition into at most 3k+13k+1 progression-free subsequences (p. 202).

For problem 772, the paper supplies the n2/3n^{2/3} lower bound (4). For problem 530, it records the Komlós--Sulyok--Szemerédi bound Hn>cn1/2H_n>cn^{1/2} for arbitrary sequences of nn integers, and Hn=(1+o(1))n1/2H_n=(1+o(1))n^{1/2} as a possible strengthening that "does not seem to be easy to prove" (p. 201, (2) and the sentence following it). On p. 203 it adds that every sequence of nn terms is likely a union of (1+o(1))n1/2(1+o(1))n^{1/2} B2B_2 subsequences, which "seems to be very difficult" and would give c=1+o(1)c=1+o(1) in (2), and that {1,…,n}\{1,\ldots,n\} is easily such a union. Theorem 4 gives the analogous B2(k)B_2^{(k)}-to-B2(k−1)B_2^{(k-1)} partition with exponent 1/(2k−1)1/(2k-1), sharp up to the constant when kk is a power of 22 (pp. 202--203). For problem 773, the closing remarks state without proof that the method easily gives a Sidon subset of {1,22,…,n2}\{1,2^2,\ldots,n^2\} of size c(ϵ)n2/3−ϵc(\epsilon)n^{2/3-\epsilon}, while Landau's theorem easily gives the upper bound c′n/(log⁡n)1/4c'n/(\log n)^{1/4}; the authors suggest that n2/3−ϵn^{2/3-\epsilon} might be replaced by n1−ϵn^{1-\epsilon} (p. 203, the paragraph before the closing problem).

Problem 774

The final paragraph on p. 203 gives the original formulation. A sequence is called free when two distinct finite sets of indices never have the same sum. For an increasing enumeration of a subset of N\mathbb N, this is exactly the property called dissociated in Problem 774. Pisier's necessary condition (6) says that there is a fixed δ>0\delta>0 such that every finite subsequence BB contains a free subsequence CC with ∣C∣≥δ∣B∣|C|\geq\delta|B|. Thus (6) is exactly proportional dissociation, and the question whether it forces a union of finitely many free subsequences is exactly E0774. The authors' judgment is explicitly negative but unresolved: they say that sufficiency "seems unlikely", while also saying that they could not find a counterexample (p. 203, final paragraph and (6)). The paper proves no implication or counterexample for this question.

There is a precise hypergraph reformulation behind the analogy. For a finite B⊂AB\subset A, make a hyperedge from the union of two distinct finite subsets of BB having the same sum (equivalently, from the support of a nonzero {−1,0,1}\{-1,0,1\} relation). Dissociated subsets are exactly independent sets in this relation hypergraph. Condition (6) gives a linear-size independent set in every finite induced subhypergraph, whereas a partition into finitely many dissociated sets asks for a uniform finite coloring of the whole relation hypergraph.

The paper's successful hypergraph arguments do not themselves bridge that gap. For the B2(k)B_2^{(k)} result, all forbidden relations are 4-uniform and the representation hypothesis gives a quadratic edge bound. For Theorem 3, every finite induced hypergraph has bounded degeneracy. In E0774 the forbidden subset-sum relations have unbounded support, and condition (6) supplies neither an edge-count bound nor bounded local degree or degeneracy. Random selection-and-deletion can recover the independent set already postulated by (6), but the paper gives no mechanism for turning that hereditary linear-independence condition into a bounded coloring. Any transfer of the method therefore needs additional structure specific to subset-sum relation hypergraphs, not merely the abstract independence-ratio hypothesis.

Source: https://users.renyi.hu/~p_erdos/1985-07.pdf. The copy read for this card is the PDF at that address, which prints "© 1985 Academic Press Inc. (London) Limited" in the footer of its first page (p. 201), every other right reserved.

Read status: claims checked for the definitions, Theorems 1 to 4, (3), (4), (5), the remarks on p. 203 and condition (6), read clause by clause on the page images; the proofs of Theorems 1 and 3 followed; the outline for Theorem 2 and the construction for Theorem 4 read for structure. The construction behind (3) and the squares bounds of p. 203 are stated in the paper without proof. Nothing here is independently reviewed. Result pages: theorem_1, theorem_2, theorem_3, theorem_4, remark_p203 and problem_p203.

Bears on. #530: the paper proves nothing about arbitrary sets; it cites the Komlós--Sulyok--Szemerédi bound (2), Hn>c n1/2H_n>c\,n^{1/2} for every sequence of nn integers, notes c≤1c\leq1 by (1), and says that Hn=(1+o(1))n1/2H_n=(1+o(1))n^{1/2} does not seem easy to prove (p. 201), which is the problem's asymptotic question for sets of integers; the p. 203 conjecture on partitions into (1+o(1))n1/2(1+o(1))n^{1/2} Sidon subsequences would give it. None of this is a result of the paper. #772: inequality (4) (p. 201) gives a Sidon subsequence of at least c4(k)n2/3c_4^{(k)}n^{2/3} terms in every B2(k)B_2^{(k)} sequence of nn terms, under the paper's count of representations as a sum of two distinct terms; this answers both of the problem's questions yes, as the problem's claim page records. #773: the remark on p. 203 states without proof the bounds c(ϵ)n2/3−ϵc(\epsilon)n^{2/3-\epsilon} and c′n/(log⁡n)1/4c'n/(\log n)^{1/4} for the largest Sidon subset of the first nn squares and suggests n1−ϵn^{1-\epsilon}; it decides nothing. #774: the closing problem (p. 203) is the problem's question, posed for free sequences, with no result either way.

Results.

  • Theorem 1 (p. 201) and inequality (4): every B2(k)B_2^{(k)} sequence of nn terms is a union of c2(k)n1/3c_2^{(k)}n^{1/3} B2B_2 sequences and contains one of at least c4(k)n2/3c_4^{(k)}n^{2/3} terms.
  • Theorem 2 (p. 202): every infinite B2(k)B_2^{(k)} sequence has a B2B_2 subsequence CC with ∣C∩{a1,…,an}∣≥[c(k)n2/3]\lvert C\cap\{a_1,\ldots,a_n\}\rvert\geq[c^{(k)}n^{2/3}] for every n≥1n\geq1.
  • Theorem 3 (p. 202): every finite or infinite B2(k)B_2^{(k)} sequence is a union of c(k)c(k) subsequences without three-term arithmetic progressions.
  • Theorem 4 (p. 202): every B2(k)B_2^{(k)} sequence of nn terms is a union of c2(k)n1/(2k−1)c_2^{(k)}n^{1/(2k-1)} B2(k−1)B_2^{(k-1)} subsequences, sharp up to the constant when k=2sk=2^s.
  • Remark (p. 203): bounds, stated without proof, for the largest Sidon subset of {1,22,…,n2}\{1,2^2,\ldots,n^2\}.
  • Closing problem (p. 203): is Pisier's necessary condition (6) sufficient for a sequence to be a finite union of free subsequences?

No file of this source is held: no license on record permits its redistribution, and the card cites the edition it names above.