Are there infinitely many pairs of integers $n \neq = m$ such that $(n 2 n)$ and $(m 2 m)$ have the same set of prime divisors?

unreviewedOpen. Worked here; no verified result yet.

number theory · open · formalized (Lean) · 0 attempts

machinery: Kummer-carries,base-representation-digits,central-binomial-prime-divisors,consecutive-integer-window,prime-distribution,infinite-family-construction

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Evidence

unverified AI candidates (2)

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

This is **an open problem** (still unsolved).

candidate solution ↗

llm-hunter · gpt pro 5.2 · unverified

1 LLM attack(s) recorded (gpt pro 5.2); unverified.

candidate solution ↗

Formal proof

AMS 11 · open (literature)

theorem erdos_730 : answer(sorry) ↔ S.Infinite

formal-conjectures/730.lean ↗

OEIS1

A129515 — Numbers m such that binomial(2*m, m) has the same prime factors as binomial(2*k, k) for some k > m.87,199,237,467,607,967,1127,1319,1483,1903,1943,2012,2047,2287,2348,2359,2464,2479,2495,2507,2623,2645,2719,3349,3467,35

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