%I #5 Mar 30 2012 18:57:23
%S 1,1,0,1,1,0,0,1,0,1,1,1,0,0,0,1,1,0,0,0,0,1,1,1,1,1,1,0,0,0,0,1,0,1,
%T 1,1,1,1,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,0,1,1,1,0,0,0,0,0,0,0,0,
%U 0,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,1,0,1,1,0,1,0
%N Zero-one sequence based on floor(sqrt(5)): a(A022839(k))=a(k); a(A108598(k))=1-a(k); a(1)=0.
%e Let u=A022839=(Beatty sequence for sqrt(5)) and v=A108598=(complement of u). Then A189091 is the sequence a given by a(1)=0 and a(u(k))=a(k); a(v(k))=1-a(k).
%t r=5^(1/2); u[n_] := Floor[n*r]; (*A022839*)
%t a[1] = 0; h = 128;
%t c = (u[#1] &) /@ Range[2h];
%t d = (Complement[Range[Max[#1]], #1] &)[c]; (*A108598*)
%t Table[a[d[[n]]] = 1 - a[n], {n, 1, h - 1}]; (*A189091*)
%t Table[a[c[[n]]] = a[n], {n, 1, h}] (*A189091*)
%t Flatten[Position[%, 0]] (*A189092*)
%t Flatten[Position[%%, 1]] (*A189093*)
%Y Cf. A188967, A189092, A189093.
%K nonn
%O 1
%A _Clark Kimberling_, Apr 16 2011