Source labels openChecked July 26, 2026

Erdős Problem 933

If $n(n+1)=2^k3^lm$ , where $(m,6)=1$ , then is it true that $\limsup_{n\to \infty} \frac{2^k3^l}{n\log n}=\infty$ ?

Number theory AMS 11

Mathematical statement

If $n(n+1)=2^k3^lm$ , where $(m,6)=1$ , then is it true that $\limsup_{n\to \infty} \frac{2^k3^l}{n\log n}=\infty$ ?

Statement source: Erdős Problems statement material

Statement terms: Source-specific

Source-specific terms. Therefore does not assert reuse rights beyond attributed display.

Statement artifacts, not proofs

These records expose exact Lean propositions and statement-only wrappers. Defining a proposition does not supply a proof of it. A placeholder-bearing target also contains no proof. Elaboration checks syntax and types; it does not certify that a formalization perfectly captures every nuance of the informal problem.

Pinned Lean formulation 1

erdos_933

Canonical source

Complete statement target, proof intentionally absentLean 4

theorem erdos_933 :    answer(sorry) ↔      atTop.limsup (fun n : ℕ ↦        ((((2 ^ k n * 3 ^ l n : ℕ) : ℝ) / ((n : ℝ) * Real.log (n : ℝ))) : EReal)) = ⊤ := by  sorry

Statement source: Formal Conjectures
Lean version: v4.27.0
Placeholder: Present; no proof artifact
Source evidence: Pinned source index
Fidelity review: Community formulation

References

Pinned Formal Conjectures sourceformalization erdosproblems.com/933background