Even and composite parameter values

Determine whether the cell bound remains tight for every even q in the complete-graph incidence family with n=2q, and establish whether the perfect one-factorization certificate can be eliminated from the construction for even q or composite odd q.

Background

The even-order construction relies on a perfect one-factorization of K_{2q}, together with a transferred recursive-line certificate. This is unavailable in the unresolved ranges where q is even or odd composite, although explicit configurations settle q=4 and q=6. The paper asks whether the numerical bound remains exact without this structural input.

References

Does the cell bound remain tight there? The first two members of this branch, n=8 (q=4) and n=12 (q=6), are settled in the affirmative by the explicit configurations of Appendices~\ref{app:288} and~\ref{app:6612}, obtained by restricting a certified configuration to the edges of K_8 respectively K_{12} and re-pairing the cells whose partners are thereby lost; as at n=9, the resulting configurations contain degenerate two-edges and therefore do not come from the one-factorization scheme. Does the cell bound remain tight for every even q, and can the perfect one-factorization certificate required in the construction of Section~\ref{sec:even} be dispensed with altogether?

— Exact Second-Order Zarankiewicz Numbers for Complete-Graph Incidence Families  (2609.25974 - Chen et al., 22 Sep 2026) in Section 5, Open problems, item 3 (Open Problem~\ref{op:evenq})

What happens for m<\binom n2? The extremal skeleton is no longer unique, and the cell bound may not be tight.

— Exact Second-Order Zarankiewicz Numbers for Complete-Graph Incidence Families  (2609.25974 - Chen et al., 22 Sep 2026) in Section 5, Open problems, item 5