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%I #12 Jan 17 2019 16:03:15
%S 2,3,10,328
%N Size of the free distributive lattice on the meet semilattice of partitions.
%e n=1: There is one partition of {1}, and the free distributive lattice on this (the unique) one-element meet-semilattice has a(1)=2 elements.
%e n=2: There are two partitions of {1,2}, and the free distributive lattice on this (the unique) two-element meet-semilattice has a(2)=3 elements.
%e n=3: There are five partitions of {1,2,3}, and the free distributive lattice on the meet semilattice {123, 12, 23, 13, top) has a(3)=10 elements.
%e n=4: There are 15 partitions of {1,2,3,4}, and the free distributive lattice on this meet-semilattice has
%e 328 = 1 +
%e C(6,0) * 2^7 +
%e C(6,1) * 2^4 +
%e C(6,2) * 2^2 +
%e C(6,3) +
%e C(6,4) +
%e C(6,5) +
%e C(6,6) +
%e 1
%e elements, where the C(n,k) are binomial coefficients.
%Y Cf. A000110, A000372.
%K nonn,more
%O 1,1
%A _David Spivak_, Nov 25 2018