Combinatorics
Green15.green_15
Does there exist a Lipschitz function whose graph is free of 3-term progressions?References
- [BJP14] T. Brown, V. Jungić and A. Poelstra, "On double 3-term arithmetic progressions", Integers 14 (2014), Paper No. A43.
- [CCS14] J. Cassaigne, J. D. Currie, L. Schaeffer and J. Shallit, "Avoidance of additive cubes and related results", Adv. in Appl. Math. 56 (2014), 25–66.
No one has attempted this yet.
Formal statement
Lean type
True ↔ ∃ K f, LipschitzWith K f ∧ {x | ∃ n, (↑n, f n) = x}.IsAPOfLengthFree 3What you must prove
import FormalConjectures.GreensOpenProblems.«15»
import TaskSupport
namespace Bounty
theorem target : fcTypeOfName% "Green15.green_15" := by
sorry
end Bounty
Pinned source: FormalConjectures/GreensOpenProblems/15.lean
- Source type SHA-256
- sha256:e8499f839608d37b026c975a89730544b39d77ba20964d5b93c766d914eaf019
- Task id
- fc-379fc029-green15-green-15-3c2ad99b82-formalized-v1
- Task commitment
- sha256:dc20224737542a57d1a483bbe43fa077bdcb026435529b13f0eaa854983e6fba
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.