Counting obstruction to (Erdős 624)
ProvedErdos624.not_trackBad_countcombinatoricserdos-problems
Let be the least such that some , , satisfies for all with (Erdős Problem 624). Say holds if for every family of maps () indexed by the subsets of an -set, there are with and with for every . By an earlier result, implies for . Theorem. If
then is false: some family of maps has no common omitted value on any -set.
Proof by double counting (the first-moment method for uniformly random maps): for fixed exactly a fraction of all families omit on in all tracks, and there are pairs .
Formalization Note. and the -set is , as in Erdos624.track_amplification.
Preamble
import Mathlib import Definitions.Def_Erdos624_defs open Finset
Formal statement
namespace Erdos624
theorem not_trackBad_count {n m s : ℕ}
(h : n * n.choose m * (n - 1) ^ (2 ^ (m + s)) < n ^ (2 ^ (m + s))) :
¬ ∀ g : Finset (Fin s) → Finset (Fin n) → Fin n, ∃ Y : Finset (Fin n), #Y = m ∧
∃ u : Fin n, ∀ B : Finset (Fin s), ∀ A ∈ Y.powerset, g B A ≠ u := by sorry
end Erdos624
Source
Erdős Problem 624 (https://www.erdosproblems.com/624); definitions from Formal Conjectures 624.lean; Claim B2b of claim from GPT 6.1 Sol (second answer, Oct 23 2026), Lean-verified by Grok Bot for Ethan Cheung (Oct 23 2026) (sharpened to the exact counting hypothesis)