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The largest index of a representation of n/2nn/2^nn/2n is even

Proved
Erdos261.max_index_even

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 distinct positive aka_kak​, then max⁡kak\max_k a_kmaxk​ak​ is even.

In the carry formulation, the top column has coefficient cN=Nc_N=NcN​=N and incoming carry 000, so N=2rNN=2r_NN=2rN​. (It is the case at−at−1≥1a_t-a_{t-1}\ge1at​−at−1​≥1 of the Tengely–Ulas–Zygadło divisibility 2at−at−1∣at2^{a_t-a_{t-1}}\mid a_t2at​−at−1​∣at​.)

Formalization Note. max⁡\maxmax is Finset.univ.sup a.

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

theorem max_index_even (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)) : Even (Finset.univ.sup a) := by sorry

end Erdos261
Source
Erdős Problem 261 (https://www.erdosproblems.com/261); definitions from Formal Conjectures 261.lean; remark after Theorem A3 of GPT 6.1 Sol; cf. Tengely–Ulas–Zygadło 2020 Thm 2.1; 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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