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Every Odd Number Greater Than 1 is the Sum of at Most 159 Primes

Proved
odd_sum_le_159_primes

by xuanji · Oct 4, 2026 · Mathlib 0df444a (Lean v4.33.1)

goldbachnumber-theoryschnirelmann-densitysieve-theory

Every odd natural number greater than 111 is a sum of at most 159159159 primes, with repetition allowed.

Precisely: for every n∈Nn \in \mathbb{N}n∈N with nnn odd and n>1n > 1n>1 there is a finite multiset sss of natural numbers such that

∣s∣≤159,every p∈s is prime,∑p∈sp=n.|s| \le 159, \qquad \text{every } p \in s \text{ is prime}, \qquad \sum_{p \in s} p = n.∣s∣≤159,every p∈s is prime,p∈s∑​p=n.

Here ∣s∣|s|∣s∣ counts elements with multiplicity, so the same prime may be used several times, and the order of the summands is irrelevant.

This is the campaign statement of Odd numbers as sums of primes with the value 159159159.

Formalization Note The representation is a Multiset ℕ; the bound is on Multiset.card, so repeated primes count separately.

Preamble
import Mathlib
Formal statement
theorem odd_sum_le_159_primes (n : ℕ) (hodd : Odd n) (hn : 1 < n) :
    ∃ s : Multiset ℕ, s.card ≤ 159 ∧ (∀ p ∈ s, Nat.Prime p) ∧ s.sum = n := by
  sorry
Source
AI-assisted explicit calculation (unpublished, October 2026), improving the K = 241 entry: Riesel–Vaughan small-shift second moment (Ark. Mat. 21 (1983), Lemma 8) with a fixed-shift prime-pair Selberg sieve, Mathlib's Chebyshev bound psi(x) >= x log 2 - O(log x), and the Hölder large range, giving sigma(A) >= 1/79; Mann's theorem then gives K = 159. Framework: P. Pollack, Not Always Buried Deep, Ch. 6 §6, https://www.pollack-math.net/NABDofficial.pdf
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What the Lean code literally says, in plain math · claude-opus-5-5

Theorem odd_sum_le_159_primes. Let nnn be a natural number (an element of N={0,1,2,… }\mathbb{N} = \{0, 1, 2, \dots\}N={0,1,2,…}), and assume:

  • nnn is odd, i.e. n=2k+1n = 2k + 1n=2k+1 for some natural number kkk;
  • 1<n1 < n1<n (strict inequality).

Together these hypotheses mean exactly that nnn ranges over the odd numbers n≥3n \ge 3n≥3, i.e. n∈{3,5,7,9,… }n \in \{3, 5, 7, 9, \dots\}n∈{3,5,7,9,…}. The hypotheses are satisfiable, so the statement is not vacuous.

The conclusion is that there exists a finite multiset sss of natural numbers (an unordered finite list in which repetitions are allowed and counted with multiplicity) such that all three of the following hold:

  1. the number of elements of sss, counted with multiplicity, is at most 159159159:
∣s∣≤159;|s| \le 159;∣s∣≤159;
  1. every element ppp of sss is a prime number (in the usual sense: p≥2p \ge 2p≥2 and its only positive divisors are 111 and ppp);
  2. the sum of the elements of sss, counted with multiplicity, equals nnn:
∑p∈sp=n.\sum_{p \in s} p = n.p∈s∑​p=n.

In words: every odd natural number n≥3n \ge 3n≥3 can be written as a sum of at most 159159159 primes, where the same prime may be used more than once and the order of the summands is irrelevant. The bound is non-strict (≤159\le 159≤159), and no lower bound on the number of summands is imposed: a single summand is permitted (so when nnn is itself prime, s={n}s = \{n\}s={n} suffices). The empty multiset would have sum 000, which never equals nnn under the hypotheses, so at least one prime is always used. No other conditions (distinctness of the primes, oddness of the primes, or an exact number of summands) are required.

Human review
  • Endorsed by Shuze Chen · Oct 5, 2026

    Confirmed by the moderator at approval.

  • Endorsed by xuanji · Oct 5, 2026

    Confirmed by the mission captain (proposal self-audit).

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