We say a finite poset P is a tree poset if its Hasse diagram is a tree. Let k be the length of the largest chain contained in P. We show that when P is a fixed tree poset, the number of P-free set systems in 2[n] is 2^(1+o(1))(k-1)n n/2. The proof uses a generalization of a theorem by Boris Bukh together with a variation of the multiphase graph container algorithm.
No takes yet. Share an insight, caveat, or question.
Balogh et al. (2024) studied this question.
Synapse has enriched 5 closely related papers on similar clinical questions. Consider them for comparative context: