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a(n+1) = Sum_{k = 0..floor(2*n/3)} a(k)*a(n-k) for n >= 0 with a(0) = 1.
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%I #17 Oct 15 2021 08:38:45

%S 1,1,1,2,4,7,15,34,72,165,387,861,2039,4894,11256,27085,66021,156347,

%T 381720,940211,2261208,5578659,13846756,33654950,83539418,208608556,

%U 512069441,1278522424,3207377196,7925966000

%N a(n+1) = Sum_{k = 0..floor(2*n/3)} a(k)*a(n-k) for n >= 0 with a(0) = 1.

%H Seiichi Manyama, <a href="/A030033/b030033.txt">Table of n, a(n) for n = 0..1000</a>

%p a := proc(n) local k; option remember;

%p if n = 0 then 1;

%p else add(a(k)*a(n - 1 - k), k = 0 .. floor(2/3*n - 2/3));

%p end if;

%p end proc;

%p seq(a(n), n = 0..30); # _Petros Hadjicostas_, Nov 07 2019

%Y Cf. A000992, A030032, A030037.

%K nonn

%O 0,4

%A _N. J. A. Sloane_

%E Name edited by _Petros Hadjicostas_, Nov 07 2019