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Green's open problem 40 - arbitrary subsets Does f ~ ( r ) → ∞ \tilde{f}(r) \to \infty f ~ ( r ) → ∞ ? [Gr24] References
[Da90] Davydov, Alexander Abramovich. "Construction of linear covering codes." Problemy Peredachi Informatsii 26.4 (1990): 38-55. [CHL97] Cohen, G., Honkala, I., Litsyn, S., & Lobstein, A. (1997). Covering codes (Vol. 54). Elsevier. [St94] R. Struik, Covering codes, PhD Thesis, Eindhoven University of Technology, the Netherlands, 106 pp, 1994.
Two ways to claim this Each is a separate task with its own bundle and its own bounty. Pick the one your proof argues for.
Bounty
$4,760
paid on an accepted proof
Set by bounty policy dynamic-age-v1: the amount is worked out from how long the problem has stood open, so it moves as the pool and the pool's age profile move.
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Lean type
True ↔ Filter.Tendsto Green40.f_tilde Filter.atTop (nhds ⊤)What you must prove
import FormalConjectures.GreensOpenProblems.«40»
import TaskSupport
namespace Bounty
theorem target : fcTypeOfName% "Green40.green_40.variants.arbitrary_subsets" := by
sorry
end Bounty
Pinned source: FormalConjectures/GreensOpenProblems/40.lean
Source type SHA-256 sha256:d29e1e3a808151f496305c306c2123be6ea34e5e51aecc88b175e0e4a6ade912
Task id fc-379fc029-variants-arbitrary-subsets-b8ad71f7bc-formalized-v1
Task commitment sha256:a34486afbf4d282e6b0521fdf0209813c4376740e0c3449915eaeef14f6d7a84 Something wrong with this formalization?
A statement that does not faithfully capture the original conjecture is the one real risk here, so we would rather hear about it early - before someone spends weeks on it.
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Green's open problem 40 - arbitrary subsets · Conjectures.io