Tail three-family combinatorial assembly
ProvedErdos390.tail_three_family_assemblyasymptoticscombinatoricsnumber-theory
Let and let be finite subsets of . Suppose and are pairwise disjoint subsets satisfying:
Then:
- Each tail subset is contained in ;
- Each tail subset is disjoint from ;
- The mutual pairwise disjointness and product identity are preserved.
Preamble
import Mathlib import Definitions.Def_erdos390_problem open Erdos390
Formal statement
namespace Erdos390
theorem tail_three_family_assembly
{n h D : ℕ} {central fixed bank candidates : Finset ℕ}
(hcentral : central ⊆ factorInterval n (2 * n))
(hfixed : fixed ⊆ factorInterval (2 * n) (2 * n + h))
(hbank : bank ⊆ factorInterval (2 * n) (2 * n + h))
(hcand : candidates ⊆ factorInterval (2 * n) (2 * n + h))
(hdisj_fb : Disjoint fixed bank)
(hdisj_fc : Disjoint fixed candidates)
(hdisj_bc : Disjoint bank candidates)
(hprod : fixed.prod id * bank.prod id * candidates.prod id * D =
(factorInterval (2 * n) (2 * n + h)).prod id) :
fixed ⊆ factorInterval n (2 * n + h) ∧
bank ⊆ factorInterval n (2 * n + h) ∧
candidates ⊆ factorInterval n (2 * n + h) ∧
Disjoint central fixed ∧
Disjoint central bank ∧
Disjoint central candidates ∧
Disjoint fixed bank ∧
Disjoint fixed candidates ∧
Disjoint bank candidates ∧
fixed.prod id * bank.prod id * candidates.prod id * D =
(factorInterval (2 * n) (2 * n + h)).prod id := by sorry
end Erdos390Source
P. Erdős, Some problems in number theory, 1975; Erdős Problem 390 tail three-family assembly