Combinatorics
Date added
$5,226 bounty1 piece from 1 personlast one 2 months ago
Nobody has attempted this. A complete proof or refutation takes the whole bounty.
A lemma, a definition, a special case. There is no race: when someone closes the problem, up to half of its bounty can go to the recognised pieces that led there.
contrib new erdos-9Command line, then a pull request on GitHub.
1 piece · 43 lemmas
1 Sept 2026
Crocker-type infinite family: infinitely many odd numbers not of the form p + 2^a + 2^b
5FqLp5…FfZZiK · 43 lemmas
Formal statement
Proof target
import Mathlib
import FormalConjectures.ErdosProblems.«9»
import TaskSupport
namespace Bounty
theorem target : fcTypeOfName% "Erdos9.erdos_9" := by
sorry
end Bounty
Original conjecture (Lean type)
True ↔ 0 < Erdos9.Erdos9A.upperDensityOriginal conjecture source: FormalConjectures/ErdosProblems/9.lean
References
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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