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a(0)=0, a(1)=1, a(2)=1, a(3)=1, thereafter a(n) = If[Mod[ n, 2] == 0, a(Floor[n/2]), a(a(Floor[(n - 1)/2])) + a(n - a(Floor[( n - 3)/2]))].
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%I #6 Jan 24 2019 15:42:11

%S 0,1,1,1,1,2,1,2,1,2,2,3,1,4,2,3,1,4,2,3,2,4,3,5,1,4,4,5,2,5,3,6,1,4,

%T 4,5,2,5,3,6,2,4,4,5,3,5,5,5,1,4,4,5,4,5,5,7,2,5,5,7,3,3,6,4,1,8,4,5,

%U 4,9,5,7,2,5,5,7,3,3,6,4,2,8,4,9,4,9,5,11,3,5,5,5,5,5,5,7,1,6,4,5,4,7,5,7,4

%N a(0)=0, a(1)=1, a(2)=1, a(3)=1, thereafter a(n) = If[Mod[ n, 2] == 0, a(Floor[n/2]), a(a(Floor[(n - 1)/2])) + a(n - a(Floor[( n - 3)/2]))].

%H G. C. Greubel, <a href="/A178771/b178771.txt">Table of n, a(n) for n = 0..5000</a>

%t a[0] = 0; a[1] = 1; a[2] = 1; a[3] := 1; a[n_]:= a[n] = If[Mod[n, 2] == 0, a[Floor[n/2]], a[a[Floor[(n-1)/2]]] + a[n -a[Floor[(n-3)/2]]]]; Table[a[n], {n, 0, 200}]

%Y Cf. A002487, A004001, A178770, A178677

%K nonn

%O 0,6

%A _Roger L. Bagula_, Jun 11 2010