problems / long-gap-dilate

Also in this folder: compute/ · figures/

A long gap in a dilate modulo a prime

Statement

Let pp be prime and A⊂Z/pZA\subset\mathbb Z/p\mathbb Z. Write g(A)g(A) for the longest run of consecutive residues missed by AA, and d⋅A={da:a∈A}d\cdot A=\{da:a\in A\}. Green #32: if ∣A∣=p|A|=\sqrt p, is there a dilate with g(d⋅A)≥100pg(d\cdot A)\ge 100\sqrt p?

Shakan (2020): for every AA with ∣A∣>1|A|>1,

sup⁡d∈Fp×g(d⋅A)  ≥  2p/∣A∣−2. \sup_{d\in\mathbb F_p^\times} g(d\cdot A)\;\ge\; 2p/|A|-2.

At ∣A∣∼p|A|\sim\sqrt p this is a gap of 2p−22\sqrt p-2. Green records this as the limit of the polynomial method.

Tonight

A certified improvement of the universal constant 2, or a documented incomplete search with replayable compute. Small-prime extremizer tables are an incomplete search unless they imply a larger universal CC.

Outcome

Incomplete search. Shakan’s 2 is still the published leading constant. Exact G(p,roundp)G(p,\mathrm{round}\sqrt p) now includes G(73,9)=24G(73,9)=24 (Cadical unsat at T=24T=24, Glucose4 witness at T=25T=25). Upper bounds G(79,9)≤26G(79,9)\le 26, G(83,9)≤27G(83,9)\le 27, G(89,9)≤30G(89,9)\le 30 have no certified floors. The search in compute/q1/ saturates the rising-factorial Alon degrees and finds no C→2C\to 2 family through p=199p=199. One new exact row is not a universal C>2C>2. Residue, not a dent.