Verstraëte’s conjecture on consecutive even cycle lengths

For every integer k1k\ge 1 and every nn-vertex graph GG, if e(G)>(2k+1)(n1)2e(G)>\frac{(2k+1)(n-1)}{2}, then there exists an integer r2r\ge 2 such that GG contains cycles of every length in the set {2r,2r+2,,2r+2(k1)}\{2r,2r+2,\ldots,2r+2(k-1)\}. Equivalently, the sharp extremal bound for graphs containing no kk consecutive even cycle lengths is conjectured to be (2k+1)(n1)2\frac{(2k+1)(n-1)}{2}.

References

Primary source

arXiv

Progress summary

Refreshed
Claimed progress

A new unrefereed paper settles the conjecture only for sufficiently long runs of even cycle lengths; all cases are not settled.

Verstraëte posed the conjecture in 2016. It predicts the sharp edge threshold forcing kk consecutive even cycle lengths in every nn-vertex graph.

Known results

  • k=1k=1 is trivial.
  • k=2k=2: Gao, Li, Ma, and Xie proved the sharp threshold e(G)5(n1)2e(G)\ge \frac{5(n-1)}{2}, with characterized K5K_5-block exceptions (2025).
  • Verstraëte proved that average degree at least 8k8k forces kk consecutive even cycle lengths, a weaker type of bound.
  • A 2026 paper proves the conjectured bound when 2k+2n4k+12k+2\le n\le 4k+1, with additional partial results.

August 2026 sufficiently-large-tt claim

On August 27, 2026, The Erdos--Gallai bound for consecutive even cycle lengths claimed the conjectured sharp bound, including its equality structure, for every sufficiently large number tt of consecutive even lengths. This advances the large-tt range but does not cover arbitrary tt, and the preprint has not been independently verified.

Current status (as of August 2026): The conjecture is claimed for sufficiently large tt, but the general cases remain open and that claim is unverified.

Sources

Solutions 0

No solutions have been posted yet.