Bukh–Griggs–Lu conjecture for the extremal number of a poset

Let PP be a finite poset and let Bn=2[n]\mathcal{B}_n=2^{[n]} be the Boolean lattice. Write La(n,P)La(n,P) for the largest size of a PP-free subfamily of Bn\mathcal{B}_n. Define e(P)e(P) to be the largest integer \ell such that, for every jj and nn,

i=1([n]i+j)\bigcup_{i=1}^{\ell}\binom{[n]}{i+j}

is PP-free. Bukh–Griggs–Lu conjecture.

La(n,P)=(1+o(1))e(P)(nn/2).La(n,P)=(1+o(1))e(P)\binom{n}{\lfloor n/2\rfloor}.

This conjecture proposes the asymptotic size of the largest PP-free family in the Boolean lattice. The paper presents it as a conjecture of Griggs and Lu and independently Bukh; no resolution is supplied in the provided text.

Sources & referencesView supporting material

Primary source

Tao Jiang, Sean Longbrake, Sam Spiro and Liana Yepremyan, “Tree Posets: Supersaturation, Enumeration, and Randomness”, arXiv:2406.11999 (2025).

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