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Let be an entire function (with for all ). Is it true that if then assumes every value infinitely often?

Worked, still open.

analysis · open · formalized (Lean) · 0 attempts

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vela registry pull vfr_37aec80d874a0239
vela reproduce examples/erdos-problems

evidence

unverified AI candidates (2)

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

A non‐polynomial entire function has an essential singularity at (\infty), so by the Great Picard theorem it takes **every** complex value infinitely often **except possibly one** value. ([Wikipedia][1]) So your question is really asking whether the “Fabry gap” condition [ \frac{n_k}{k}\to\infty ] forces there to be **…

candidate solution ↗

llm-hunter · gpt pro 5.2 · unverified

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

candidate solution ↗

formal

AMS 30 · open (literature)

theorem erdos_517 : answer(sorry) ↔ ∀ {f : ℂ → ℂ} {n : ℕ → ℕ} (hn : HasFabryGaps n)
    {a : ℕ → ℂ} (ha : ∀ k, a k ≠ 0) (hf : ∀ z, HasSum (fun k => a k * z ^ n k) (f z)) (z : ℂ),
    {x : ℂ | f x = z}.Infinite
formal-conjectures/517.lean ↗

status

open

notary

vela reproduce examples/erdos-problems
  • packet.json · sha256 c4c3ba4dae4c600a07d8aceaed944cb26d2fcdd8d4081b62ea9c972b48f3d418

finding.noted · reviewer:will-blair · 1 day

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