A336631 a(n) = 1 + Max_{0<=i<=j<=k; i+j+k=n-1} a(i)*a(j)*a(k) for n>0, with a(0) = 1.

1, 2, 3, 5, 9, 13, 21, 37, 55, 91, 163, 244, 406, 730, 1054, 1702, 2998, 4456, 7372, 13204, 19765, 32887, 59131, 85411, 137971, 243091, 361351, 597871, 1070911, 1603081, 2667421, 4796101, 6927701, 11190901, 19717301, 29309501, 48493901, 86862701, 130027601

Offset: 0

Author

Justin Dallant, Jul 28 2020

Keywords

• easy
• nonn

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...

Examples

			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

Ternary version of A091980.

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.