Is there some absolute constant $C > 0$ such that $p \leq n \sum 1_{p ∤ (n 2 n)} \frac{1}{p} \leq C$ for all $n$ (where the summation is restricted to primes $p \leq n$ )?

Open — best to date is a partial proof, not yet sealed.

number theory · open · formalized (Lean) · 1 attempt

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vela registry pull vfr_37aec80d874a0239

vela reproduce examples/erdos-problems

evidence

partial proof

needs verification

Erdős 377 (is ∑_{p<=n, p∤C(2n,n)} 1/p bounded?) is confirmed OPEN — not settled here. Verified contributions: (i) exact Kummer reformulation f(n)=∑ 1/p over primes p that are "p-flat" (all base-p digi

unverified AI candidates (2)

gpt-erdos · GPT-5.2 Pro + Deep Research · unverified

Let [ f(n):=\sum_{p\le n}\mathbf 1_{,p\nmid \binom{2n}{n}}\frac1p . ]

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_377 : answer(sorry) ↔
    ∃ C > (0 : ℝ), ∀ (n : ℕ), sumInvPrimesNotDvdCentralBinom n ≤ C

formal-conjectures/377.lean ↗

status

open

evidence

formal

status

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