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 A189008 Zero-one sequence based on cubes:  a(A000578(k))=a(k); a(A007412(k))=1-a(k); a(1)=0. 4

%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 Zero-one sequence based on cubes: a(A000578(k))=a(k); a(A007412(k))=1-a(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 self-generating zero-one sequence with initial value a(1)=0 and a(u(k))=a(k); a(v(k))=1-a(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

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Last modified June 7 02:48 EDT 2020. Contains 334836 sequences. (Running on oeis4.)