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a(0) = 1; a(n+1) = Sum_{k=0..n} a(n-k)*a(floor(k/2)).
8

%I #20 Nov 27 2023 12:21:44

%S 1,1,2,4,8,17,35,74,154,324,677,1422,2977,6246,13086,27444,57518,

%T 120600,252794,529994,1111013,2329187,4882755,10236280,21458943,

%U 44986461,94308415,197707134,414469000,868886834,1821517772,3818600772

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

%H Seiichi Manyama, <a href="/A127680/b127680.txt">Table of n, a(n) for n = 0..1000</a>

%F G.f. A(x) satisfies: A(x) = 1 / (1 - x * (1 + x) * A(x^2)). - _Ilya Gutkovskiy_, Nov 15 2021

%F a(n) ~ c * d^n, where d = 2.096382783759695271747034891835844892559952962948180418542044889824924... and c = 0.413348184087944400305975399220165744000861336139702047444087822224828... - _Vaclav Kotesovec_, Nov 16 2021

%t f[l_List] := Block[{n = Length[l] - 1},Append[l, Sum[l[[n - k + 1]]*l[[Floor[k/2] + 1]], {k, 0, n}]]];Nest[f, {1}, 33] (* _Ray Chandler_, Feb 13 2007 *)

%Y Cf. A367652, A367653, A367654.

%Y Cf. A127681.

%K easy,nonn

%O 0,3

%A _Leroy Quet_, Jan 23 2007

%E Extended by _Ray Chandler_, Feb 13 2007