%I
%S 0,1,0,1,0,1,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,0,1,1,1,0,0,0,
%T 1,1,1,0,0,0,1,1,1,0,0,0,1,1,1,0,0,0,1,1,1,0,0,0,1,1,1,0,0,1,0,1,1,0,
%U 1,0,0,1,0,1,1,0,1,0,0,1,0,1,1,0,1,0,0,1,0,1,1,0,1,0,0,1,0,1,1,0,1,0,0,1,0,1,1,0,1,0,0,1,0,1,1,0,1,0,0,1,0,1,1,0,0,1,0,0,1,1,0,1,1,0,0
%N Zeroone sequence based on cubes: a(A000578(k))=a(k); a(A007412(k))=1a(k); a(1)=0.
%C Let u=A000578 and v=A007412, so that u(n)=n^3 and v=complement(u) for n>=1. Then a is a selfgenerating zeroone sequence with initial value a(1)=0 and a(u(k))=a(k); a(v(k))=1a(k).
%t u[n_] := n^3;
%t a[1] = 0; h = 128;
%t c = (u[#1] &) /@ Range[h]; (*A000578*)
%t d = (Complement[Range[Max[#1]], #1] &)[c]; (*A007412*)
%t Table[a[d[[n]]] = 1  a[n], {n, 1, h  1}];
%t Table[a[c[[n]]] = a[n], {n, 1, h}] (*A189008*)
%t Flatten[Position[%, 0]] (*A189009*)
%t Flatten[Position[%%, 1]] (*A189010*)
%Y Cf. A188967, A189009, A189010.
%K nonn
%O 1
%A _Clark Kimberling_, Apr 15 2011
