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Theorem 1.1 — Exact second-order asymptotic

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Erdos390.erdos390

by ShouqiaoWang · Jul 31, 2026 · Mathlib c5ea003 (Lean v4.30.0)

asymptoticscombinatoricserdos-problemsfactorialsnumber-theory

For every natural number n≥3n\ge3n≥3, let f(n)f(n)f(n) be the least possible value of the largest factor in a factorization

n!=a1a2⋯ak,n<a1<a2<⋯<ak.n!=a_1a_2\cdots a_k,\qquad n<a_1<a_2<\cdots<a_k.n!=a1​a2​⋯ak​,n<a1​<a2​<⋯<ak​.

Let

C0=402963959825970038185.C_0=\frac{4029639598}{25970038185}.C0​=259700381854029639598​.

The theorem asserts the exact second-order asymptotic

f(n)=2n+C0nlog⁡n+o ⁣(nlog⁡n)(n→∞).f(n)=2n+C_0\frac{n}{\log n} +o\!\left(\frac{n}{\log n}\right) \qquad(n\to\infty).f(n)=2n+C0​lognn​+o(lognn​)(n→∞).

Equivalently,

lim⁡n→∞(f(n)−2n)log⁡nn=C0.\lim_{n\to\infty} \frac{(f(n)-2n)\log n}{n}=C_0.n→∞lim​n(f(n)−2n)logn​=C0​.

This determines the asymptotic constant in the second-order term of Erdős Problem 390.

Formalization Note The Lean goal is literally the displayed small-ooo assertion. The equivalent normalized limit is not added as a separate conjunct. Lean sets f(n)=0f(n)=0f(n)=0 for n<3n<3n<3 only to make it total; those values do not affect the limit.

Preamble
import Definitions.Def_erdos390_problem
Formal statement
namespace Erdos390

/-- Erdős Problem 390: the exact second-order asymptotic. -/
theorem erdos390 : MainAsymptotic := by sorry

end Erdos390
Source
Shouqiao Wang, A Proposed Solution to Erdős Problem 390, p. 3, Section 1, Theorem 1.1, https://github.com/ShouqiaoW/erdos/blob/61325b10bbdc29f4fb5e0618b414b9f2189333ad/390/paper.tex#L191-L202. Closed Lean terminal: https://github.com/ShouqiaoW/erdos/blob/61325b10bbdc29f4fb5e0618b414b9f2189333ad/390/lean/Erdos390/WholePaper/BankPaperCanonicalSectionNinePostHeightSourceFirstMainAsymptoticConnector.lean#L25-L38. Literal expanded audit: https://github.com/ShouqiaoW/erdos/blob/61325b10bbdc29f4fb5e0618b414b9f2189333ad/390/lean/Erdos390/WholePaper/BankPaperCanonicalSectionNinePostHeightSourceFirstMainAsymptoticConnectorStatementAudit.lean#L55-L64.
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What the Lean code literally says, in plain math · gpt-5.6-sol

Define f(n)=0f(n)=0f(n)=0 when n=0,1,2n=0,1,2n=0,1,2, and, when n≥3n\ge3n≥3, define f(n)f(n)f(n) to be the least natural number MMM for which there exists a finite set A⊆{k∈N∣n<k≤M}A\subseteq\{k\in\mathbb N\mid n<k\le M\}A⊆{k∈N∣n<k≤M}, with distinct elements, satisfying the natural-number product equation ∏a∈Aa=n!\prod_{a\in A}a=n!∏a∈A​a=n!. The theorem asserts that the function

n⟼f(n)−(2n+402963959825970038185nlog⁡n)n\longmapsto f(n)-\left(2n+\frac{4029639598}{25970038185}\frac n{\log n}\right)n⟼f(n)−(2n+259700381854029639598​lognn​)

is little-ooo, along the natural numbers tending to infinity, of the function n↦nlog⁡nn\mapsto\frac n{\log n}n↦lognn​. Fully expanded, this says that for every real number δ>0\delta>0δ>0, there exists a natural number NNN, which may depend on δ\deltaδ, such that every n≥Nn\ge Nn≥N satisfies

∣f(n)−(2n+402963959825970038185nlog⁡n)∣≤δ∣nlog⁡n∣,\left|f(n)-\left(2n+\frac{4029639598}{25970038185}\frac n{\log n}\right)\right|\le\delta\left|\frac n{\log n}\right|,​f(n)−(2n+259700381854029639598​lognn​)​≤δ​lognn​​,

with f(n)f(n)f(n) embedded into the real numbers wherever it occurs. The real denominator in the constant is nonzero. The real logarithm and real division are totalized: in particular, log⁡0=log⁡1=0\log0=\log1=0log0=log1=0 and x/0=0x/0=0x/0=0, so the scale nlog⁡n\frac n{\log n}lognn​ is 000 at n=0,1n=0,1n=0,1 and is the usual positive quotient for n≥2n\ge2n≥2. The asymptotic quantifiers allow a different threshold NNN for each δ\deltaδ, and impose no requirement on the finite initial segment n<Nn<Nn<N.

Human review
  • Endorsed by Shuze Chen · Jul 31, 2026

  • Endorsed by ShouqiaoWang · Jul 31, 2026

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

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