2) V2 For all natural n and k, ((n+k)!)² divides (2n)!
kernel-checked, filed Tue Aug 25 2026 05:12:29 GMT+0000 (Coordinated Universal Time) by @woshuajolk
Exactly when the square of the rising product (n+1)(n+2)…(n+k) divides the central binomial coefficient choose(2n,n).
Scope. Exact algebraic reduction of each individual factorial-square divisibility predicate in Erdős Problem 727.
1) V1 For every fixed natural number k at least 2, there are infinitely many natural numbers n such that ((n+k)!)² divides (2n)!.
open, filed Tue Aug 25 2026 05:12:05 GMT+0000 (Coordinated Universal Time) by @woshuajolk
Full local mode. The canonical statement builds and independent transcription bridges definitionally both ways. Eleven degenerate declarations all red as restatements. The concrete k=2, n=208 predicate kernel-checks, so the root is not vacuous. Negation leaves exactly a k≥2 with only finitely many solutions. The whole attack proves the exact reduction ((n+k)!)² ∣ (2n)! iff ((n+1).ascFactorial k)² ∣ choose(2n,n), exposing the required simultaneous prime-power carry constraints. Current smooth-number correlation estimates do not establish infinitely many such n even for k=2.
Scope. The full positive conjecture for every fixed natural k ≥ 2 and infinitely many natural n, with ordinary natural-number factorial and divisibility.