erdős #727
Let . Doesfor infinitely many ?
Worked, still open.
number theory · open · formalized (Lean) · 0 attempts
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vela registry pull vfr_37aec80d874a0239vela reproduce examples/erdos-problemsevidence
unverified AI candidates (2)
gpt-erdos · GPT-5.2 Pro + Deep Research · unverified
Write [ R_{n,k}:=\frac{(2n)!}{(n+k)!^2}. ] Since [ \binom{2n}{n}=\frac{(2n)!}{(n!)^2} \quad\text{and}\quad \frac{(n+k)!}{n!}=\prod_{i=1}^k(n+i), ] we have the exact identity [ R_{n,k} =\frac{(2n)!}{(n!)^2}\cdot\Big(\frac{n!}{(n+k)!}\Big)^2 =\frac{\binom{2n}{n}}{\Big(\prod_{i=1}^k (n+i)\Big)^2}. ] Therefore [ (n+k)!^2\m…
candidate solution ↗llm-hunter · gpt pro 5.2 · unverified
1 LLM attack(s) recorded (gpt pro 5.2); unverified.
candidate solution ↗formal
AMS 11 · open (literature)
theorem erdos_727 : answer(sorry) ↔ ∀ k ≥ 2,
Set.Infinite {n : ℕ | (Nat.factorial (n + k)) ^ 2 ∣ Nat.factorial (2 * n)}formal-conjectures/727.lean ↗oeis
A002503 — Numbers k such that binomial(2*k,k) is divisible by (k+1)^2.5,14,27,41,44,65,76,90,109,125,139,152,155,169,186,189,203,208,209,219,227,230,237,265,275,298,307,311,314,321,324,329,3A343507 — a(n) is the smallest nonnegative integer k such that (2*k)! / (k+n)!^2 is an integer.0,208,3475,8174,252965,3648835,72286092,159329607,2935782889A389396 — Numbers k such that binomial(2*k,k) is divisible by (k+1)^2*(k+2)^2.208,458,987,1220,1455,1597,1889,2012,2144,2330,2477,2663,2991,3353,3415,3430,3439,3475,3476,3551,3563,3568,3617,3625,372
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Catalan numbers · reference
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open