For the residue interval is not -sum-free
ProvedModularSchur.not_sumFree_of_ge_quotadditive-combinatoricscombinatoricsmodular-schur-numbersnumber-theorysum-free-sets
This is the second and harder upper bound behind the one-colour formula.
Let and , and suppose . Then the residue interval is not -sum-free modulo :
The obstruction comes from wrap-around: once the interval is this long, some -tuple drawn from has integer sum , which is congruent to modulo , and lies in the interval.
This is the branch that supplies the term of the one-colour formula, and it is what makes that formula a minimum of two competing quantities rather than a single expression.
Preamble
import Definitions.Def_ModularSchurBasic
import Definitions.Def_ModularSchurPartition
import Mathlib
open ModularSchur
open Finset Classical
variable {m ℓ : ℕ}
Formal statement
theorem ModularSchur.not_sumFree_of_ge_quot (hm : 2 ≤ m) (hℓ : 2 ≤ ℓ) (hlm : ℓ ≤ m) {N : ℕ}
(hN : m / ℓ + 1 ≤ N) : ¬ IsEllSumFree m ℓ (stableResidues m N) := by sorrySource
McKenna 2026, "Prime-power structure of the stable regime for modular Schur numbers", docs/paper/modular-schur.pdf in the same repository, the wrap-around upper bound in the proof of Theorem 6.1. Lean source: https://github.com/mysticflounder/modular-schur/blob/eb6098890f05eff39190e6cd8e41fdea53fa81f9/lean/ModularSchur/K1Theorem.lean#L71-L128