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Tail three-family combinatorial assembly

Proved
Erdos390.tail_three_family_assembly

by doctosil · Sep 6, 2026 · Mathlib c5ea003 (Lean v4.30.0)

asymptoticscombinatoricsnumber-theory

Let n,h,D∈Nn, h, D \in \mathbb{N}n,h,D∈N and let central,fixed,bank,candidates\mathrm{central}, \mathrm{fixed}, \mathrm{bank}, \mathrm{candidates}central,fixed,bank,candidates be finite subsets of N\mathbb{N}N. Suppose central⊆(n,2n]\mathrm{central} \subseteq (n, 2n]central⊆(n,2n] and fixed,bank,candidates⊆(2n,2n+h]\mathrm{fixed}, \mathrm{bank}, \mathrm{candidates} \subseteq (2n, 2n + h]fixed,bank,candidates⊆(2n,2n+h] are pairwise disjoint subsets satisfying:

(∏a∈fixeda)⋅(∏b∈bankb)⋅(∏c∈candidatesc)⋅D=∏k∈(2n,2n+h]k.\left(\prod_{a \in \mathrm{fixed}} a\right) \cdot \left(\prod_{b \in \mathrm{bank}} b\right) \cdot \left(\prod_{c \in \mathrm{candidates}} c\right) \cdot D = \prod_{k \in (2n, 2n + h]} k.(a∈fixed∏​a)⋅(b∈bank∏​b)⋅(c∈candidates∏​c)⋅D=k∈(2n,2n+h]∏​k.

Then:

  1. Each tail subset is contained in (n,2n+h](n, 2n + h](n,2n+h];
  2. Each tail subset is disjoint from central\mathrm{central}central;
  3. 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 Erdos390
Source
P. Erdős, Some problems in number theory, 1975; Erdős Problem 390 tail three-family assembly

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