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%I #4 Mar 30 2012 18:57:23
%S 1,1,0,1,0,1,1,0,0,1,1,0,0,0,1,1,1,1,0,0,0,0,0,1,1,1,1,1,1,0,0,1,0,0,
%T 0,0,0,1,1,0,1,0,1,1,1,1,1,0,0,1,0,1,1,0,0,0,0,0,0,1,1,0,0,1,0,0,0,1,
%U 1,1,1,1,1,1,0,0,1,1,1,0,1,1,1,1,0,0,1,0,0,0,0,0,0,1,1,0,0,0,0,1,0,0,0,0,0,1,0,1,0,1,1,0,1,1,1,1,1,0,0,1,1,1,1,1,0,1,1,1,1,1,1,1,0,1,0
%N Zero-one sequence based on the sequence (5n-3): a(A016873(k))=a(k); a(A047207(k))=1-a(k), a(1)=0.
%t r=5n-3; u[n_] := Floor[n*r]; (*A016873*)
%t a[1] = 0; h = 128;
%t c = (u[#1] &) /@ Range[2h];
%t d = (Complement[Range[Max[#1]], #1] &)[c]; (*A047207*)
%t Table[a[d[[n]]] = 1 - a[n], {n, 1, h - 1}]; (*A189129*)
%t Table[a[c[[n]]] = a[n], {n, 1, h}] (*A189129*)
%t Flatten[Position[%, 0]] (*A189130*)
%t Flatten[Position[%%, 1]] (*A189131*)
%Y Cf. A188967, A189129, A047207.
%K nonn
%O 1
%A _Clark Kimberling_, Apr 17 2011
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