Second-order Zarankiewicz-number conjecture for complete-graph incidence families

For every integer n≥6n\ge 6, let m=(n2)m=\binom{n}{2} and consider the incidence family of the complete graph KnK_n, with one column for each vertex and one row for each edge. If z2(m,n)z_2(m,n), zSL(m,n)z_{SL}(m,n), and zRL(m,n)z_{RL}(m,n) denote its second-order, signed, and recursive-line Zarankiewicz numbers, then

z2(m,n)=zSL(m,n)=zRL(m,n)=Z(n),z_2(m,n)=z_{SL}(m,n)=z_{RL}(m,n)=Z(n),

where

Z(n)=⌊n(n−1)(n+2)4⌋={n(n−1)(n+2)4,n≡0,1(mod4),n(n−1)(n+2)−24,n≡3(mod4).Z(n)=\left\lfloor\frac{n(n-1)(n+2)}{4}\right\rfloor=\begin{cases}\dfrac{n(n-1)(n+2)}{4},&n\equiv0,1\pmod 4,\\[6pt]\dfrac{n(n-1)(n+2)-2}{4},&n\equiv3\pmod 4.\end{cases}
References

Primary source

arXiv

Additional references

Progress summary

Refreshed
Claimed progress

A new paper claims substantial progress by settling many cases, but the conjecture remains open for infinitely many orders.

The conjecture concerns exact second-order Zarankiewicz values for incidence families arising from complete graphs. Its origin and proposer are not identified in the retrieved material.

September 2026 exact-value claims

On September 22, 2026, Yannan Chen and Liqun Qi reported exact values for infinite families of even and odd orders, plus certified configurations for n=8n=8, n=9n=9, n=12n=12, and n=13n=13. The paper explicitly leaves the remaining orders, beginning with n=16n=16, as conjectural.

Current status (as of September 2026): Exact values are claimed for several infinite families and four additional orders, while the conjecture remains open from n=16n=16 onward; the claims have not been independently verified in this scan.

Sources

Solutions 0

No solutions have been posted yet.