Wiki
Wiki

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

Updated


Statement

For a set of positive integers 1≤a1<⋯<ak≤n1\le a_1<\cdots<a_k\le n with k≤(1+o(1))n1/2k\le(1+o(1))n^{1/2}, the paper asks "how many different sums ai+aja_i+a_j can occur below nn" and denotes the maximum by T(n)T(n) (p. 203). The sums ai+aja_i+a_j run over i≤ji\le j, equal summands allowed (the count (k+12)\binom{k+1}2 of all formal sums on pp. 196 and 205). S(n)S(n), the same maximum over Sidon sequences, satisfies S(n)≤T(n)S(n)\le T(n) (p. 203).

Proposition 1. "Given any ε>0\varepsilon>0, then for nn large enough

3/8−ε≤T(n)/n≤1/2+ε.3/8-\varepsilon\le T(n)/n\le1/2+\varepsilon.

"

Remark 2. "The proof of Proposition 1 shows that the statement remains true even if we count only those values below nn which have a unique representation as ai+aja_i+a_j."

Both as printed, Proposition 1 on p. 203 and Remark 2 on p. 204. In the notation of Problem 819, whose f(N)f(N) is the maximal ∣(A+A)∩[1,N]∣|(A+A)\cap[1,N]| over A⊆{1,…,N}A\subseteq\{1,\ldots,N\} with ∣A∣=⌊N1/2⌋|A|=\lfloor N^{1/2}\rfloor, the proposition gives (3/8−o(1))N≤f(N)≤(1/2+o(1))N(3/8-o(1))N\le f(N)\le(1/2+o(1))N: the upper bound is the count of all sums, and the lower bound transfers because adding elements of [1,N][1,N] loses no sum and removing o(N1/2)o(N^{1/2}) elements from a set of O(N1/2)O(N^{1/2}) elements loses o(N)o(N) sums (a one-line step made here; the paper allows k≤(1+o(1))n1/2k\le(1+o(1))n^{1/2} and counts the sums below nn).

Source. P. Erdős and R. Freud, On Sums of a Sidon-Sequence, J. Number Theory 38 (1991), 196--205; Proposition 1 with its proof and Remark 1 on printed p. 203 (PDF p. 8 of the publisher's open-archive scan), Remark 2 on p. 204 (PDF p. 9), read on the page images. The artifact is identified in the source digest.

Read depth. Claims checked: the definition of T(n)T(n), the statement, Remark 1 and Remark 2 were read clause by clause on the page images. The proof (one paragraph) was read in full on the page image and followed. Nothing here is independently reviewed.

Proof pointer

Page 203. The upper bound is trivial, "since the total number of sums is (1+o(1))n/2(1+o(1))n/2". For the lower bound, take a maximally dense Sidon set B={b1,b2,…}⊂[1,n/4]B=\{b_1,b_2,\ldots\}\subset[1,n/4], with (1+o(1))(n/4)1/2(1+o(1))(n/4)^{1/2} elements (see Dependencies), and adjoin its reflection 3n/4−B3n/4-B, about n1/2n^{1/2} elements in all. The only coincidence among the sums of this set is that every bi+(3n/4−bi)b_i+(3n/4-b_i) equals 3n/43n/4, and both the sums bi+bjb_i+b_j and the mixed sums bi+(3n/4−bj)b_i+(3n/4-b_j) are less than nn: about n/8n/8 values of the first kind and n/4n/4 of the second, 3n/83n/8 in all. Remark 2 follows because every sum counted, other than 3n/43n/4, has a unique representation.

Dependencies

Within the paper: none. Outside it: Sidon sequences in [1,m][1,m] of (1+o(1))m1/2(1+o(1))m^{1/2} elements, the most possible, used with m=n/4m=n/4. The paper's [1], filed as erdos_1941_problem_sidon_additive_number_theory_related, proves that a Sidon sequence in [1,n][1,n] has at most n1/2+O(n1/4)n^{1/2}+O(n^{1/4}) elements but constructs ones of only (1/2−ε)n1/2(1/\sqrt2-\varepsilon)n^{1/2} (pp. 212--214); sequences of (1−o(1))m1/2(1-o(1))m^{1/2} elements come from Singer's perfect difference sets, filed as singer_1938_theorem_finite_projective_geometry_some_applications_number_theory, with the ratio of consecutive primes tending to 1.

Bears on

  • Problem 819: the bounds (38−o(1))N≤f(N)≤(12+o(1))N(\frac38-o(1))N\le f(N)\le(\frac12+o(1))N the site attributes to the paper, with the reflected Sidon construction behind the lower bound; the site's connection to Problem 840 rests on the paper's statement (p. 204) that improving this upper bound and pushing the coefficient of the trivial quasi-Sidon bound (37) below 2\sqrt2 are equivalent problems, recorded on Definition (p. 203).