Ma’s saturation-number conjecture

Let C[4,r]={C4,C5,,Cr}\mathcal{C}_{[4,r]}=\{C_4,C_5,\ldots,C_r\}. For every integer r5r\ge 5, there exists an integer n0(r)n_0(r) such that, for every nn0(r)n\ge n_0(r), sat(n,C[4,r])=5n432\operatorname{sat}(n,\mathcal{C}_{[4,r]})=\left\lceil\frac{5n}{4}-\frac{3}{2}\right\rceil, where sat(n,C[4,r])\operatorname{sat}(n,\mathcal{C}_{[4,r]}) is the minimum number of edges in an nn-vertex graph containing no cycle CiC_i for 4ir4\le i\le r, but such that adding any missing edge creates a copy of at least one such cycle.

Progress summary

Partially solved

A new paper disproves the conjecture once the forbidden cycle range includes length six, proves the length-six case exactly, and leaves longer ranges unresolved.

Ma’s conjecture predicts that, for every r5r \ge 5, the saturation number for the interval of cycles from C4C_4 through CrC_r eventually equals 5n432\left\lceil \frac{5n}{4}-\frac{3}{2}\right\rceil.

Known results

  • The exact formula sat(n,C{4,5})=5n432\operatorname{sat}(n,\mathcal{C}_{\{4,5\}})=\left\lceil \frac{5n}{4}-\frac{3}{2}\right\rceil was known for every positive integer nn.

19 August 2026 disproof for r6r \ge 6

The new paper proves a stronger upper bound with an rr-dependent subtraction, establishes sat(n,C[4,6])\operatorname{sat}(n,\mathcal{C}_{[4,6]}) exactly, and thereby disproves the conjectured eventual formula for every r6r \ge 6. It does not determine the exact saturation number for every r7r \ge 7.

Current status (as of August 2026): The conjecture is false for every r6r \ge 6, and the case r=6r=6 is settled exactly; the exact values for all r7r \ge 7 remain open.

Sources
Sources & referencesView supporting material

Primary source

arXiv

Solutions 0

No solutions have been posted yet.