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Every index in a representation of n/2nn/2^nn/2n exceeds nnn

Proved
Erdos261.indices_gt

by Ethan Cheung Wang · Oct 23, 2026 · Mathlib 0df444a (Lean v4.33.1)

erdos-problemsnumber-theory

Write Rep⁡\operatorname{Rep}Rep for a representation n2n=∑kak2ak\frac{n}{2^n}=\sum_k \frac{a_k}{2^{a_k}}2nn​=∑k​2ak​ak​​ by pairwise distinct positive integers aka_kak​ (Erdős Problem 261). Theorem. If t≥2t\ge2t≥2 and n2n=∑k=1tak2ak\frac{n}{2^n}=\sum_{k=1}^t\frac{a_k}{2^{a_k}}2nn​=∑k=1t​2ak​ak​​ with positive aka_kak​, then ak>na_k>nak​>n for every kkk.

This is the first normalisation step for Problem 261: since x↦x/2xx\mapsto x/2^xx↦x/2x is nonincreasing on the positive integers, a single index a≤na\le na≤n already contributes at least n/2nn/2^nn/2n, leaving no room for the other (positive) terms.

Formalization Note. Injectivity of aaa is assumed (to match Erdos261Prop) but not used.

Preamble
import Mathlib
import Definitions.Def_Erdos261_defs
open Finset
Formal statement
namespace Erdos261

theorem indices_gt (n t : ℕ) (ht : 2 ≤ t) (a : Fin t → ℕ) (hinj : a.Injective) (hpos : 1 ≤ a)
    (hsum : (n : ℚ) / 2 ^ n = ∑ k, (a k : ℚ) / 2 ^ (a k)) : ∀ i, n < a i := by sorry

end Erdos261
Source
Erdős Problem 261 (https://www.erdosproblems.com/261); definitions from Formal Conjectures 261.lean; Lemma A1 of GPT 6.1 Sol; also Tengely–Ulas–Zygadło 2020 Thm 2.1 (a1≥n+1a_1\ge n+1a1​≥n+1); lemma proposed by GPT 6.1 Sol (Oct 23 2026), Lean-verified by Grok Bot for Ethan Cheung (Oct 23 2026)

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