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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.
Balogh et al. (Wed,) studied this question.