Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Source. Theorem 4, p. 5, of Greg Martin and Kevin O'Bryant,
Constructions of Generalized Sidon Sets, J. Combin. Theory Ser. A 113
(2006), no. 4, 591-607, read in the arXiv edition arXiv:math/0408081v2
(21 Feb 2005) named on the
source card.
Read depth. Claims checked: the statement and the definitions it uses were
read clause by clause on the page images; the proof (Section 3.3, p. 14) was
read for structure only. Nothing here is independently reviewed.
Statement
Setting (p. 2). S∗S(k) counts ordered pairs (s1,s2)∈S×S with
s1+s2=k, ∥S∗∥∞=maxkS∗S(k),
R(g,n)=max{∣S∣:S⊆{1,…,n},∥S∗∥∞≤g},
and σ(g)=liminfn→∞R(g,n)/⌊g/2⌋n.
Theorem 4 (p. 5). For g≥1,
σ(2g+1)≥σ(2g)≥3g2−g⌊g/3⌋+gg+2⌊g/3⌋+⌊g/6⌋.
In particular, liminfg→∞σ(g)≥11/96.
The paper adds that 11/96>1.1226, against the authors' announced upper
bound limsupg→∞σ(g)<1.8391 from a work then in preparation
(p. 5).
The right side exceeds 1 for every g≥12 (a check made here: bounding
g+2⌊g/3⌋+⌊g/6⌋≥(11g−13)/6 and
3g2−g⌊g/3⌋+g≤(8g2+5g)/3 reduces the claim to
25g2−346g+169>0, true for g≥14, and g=12,13 give 1.1055 and
1.0632). It also exceeds 1 for g=6,7,9,10 and is below 1 for
g≤5 and g=8,11, where Theorem 3 gives the stronger bounds.
Proof pointer
Section 3.3 (pp. 13--14). The first inequality is the monotonicity
R(2g+1,n)≥R(2g,n). The second applies the bound
σ(2g)≥R(g,x)/gx from the proof of Theorem 3 with
x=3g−⌊g/3⌋+1 and Theorem 2(vi). The paper records the sharper
form
R(2g,n)≥83112gn(1+O(g−1+(n/g)(α−1)/2))
as n/g and g both tend to infinity, where α<1 is any exponent such
that for all large y there is a prime between y−yα and y, for
instance α=0.525 by Baker, Harman and Pintz (p. 14); this gives the
final assertion for even g, and monotonicity gives it for odd g.
Dependencies
Theorem 2
(ii), (v) and (vi), the prime number theorem, and, for the refined form, the
Baker-Harman-Pintz theorem on primes in short intervals.
Bears on
Problem 863: with g=r,
the problem's largest B2[r] set in {1,…,N} has size R(2r,N),
and the theorem gives liminfNR(2r,N)/rN>1 for every r≥12
(and r=6,7,9,10); with
Theorem 3
for 2≤r≤11, this covers every r≥2, so if
∣A∣∼crN1/2, then cr>r. The paper does not
treat the difference sets B of that problem or the constant cr′.