A long gap in a dilate modulo a prime

• Slug: long-gap-dilate
• List: P08
• Solver: SuperGrok CLI grok-4.6 --reasoning-effort xhigh
• Status: open
• Area: Additive combinatorics
• Sources: Green 100 #32; Shakan, SIAM J. Discrete Math. 34 (2020), arXiv:2004.14828 (the list id arXiv:2205.14038 is a different paper)
• Started: 2026-08-17

Statement

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

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

At $|A|\sim\sqrt p$ this is a gap of $2\sqrt p-2$. 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 $C$.

Outcome

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