OFFSET
0,2
COMMENTS
a(n) is the maximum number of antichains (including the empty antichain) among all posets of size n with a Hasse diagram corresponding to a ternary tree (each node has up to three children). Equivalently, a(n)-1 is the maximum number of subtrees containing the root among all ternary trees of size n.
a(n)^(1/n) converges, and the decimal expansion of the limit seems to start with 1.6296636...
FORMULA
a(n) = 1 + Max_{0<=i<=j<=k; i+j+k=n-1} a(i)*a(j)*a(k) for n>0, a(0) = 1.
EXAMPLE
For n = 1 we have a(1) = 1 + a(0)*a(0)*a(0) = 1 + 1*1*1 = 2.
For n = 6 we have a(6) = 1 + a(1)*a(1)*a(3) = 1 + 2*2*5 = 21.
For n = 24 we have a(24) = 1 + a(4)*a(6)*a(13) = 1+9*21*730 = 137971.
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Justin Dallant, Jul 28 2020
STATUS
approved