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%I #14 Oct 03 2016 09:57:26
%S 0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,
%T 0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,
%U 0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1
%N [3r]-[nr]-[3r-nr], where r=(1+sqrt(5))/2 and [.]=floor.
%C This is column 3 of the array A188294.
%C This sequence is essentially the same as A188011. - _Michel Dekking_, Oct 03 2016
%F a(n) = [3r]-[nr]-[3r-nr], where r=(1+sqrt(5))/2.
%F a(n) = 1-A188011(n) for all n>0, except for n= 3 (from [-x]=-[x]-1 for non-integer x). - _Michel Dekking_, Oct 03 2016
%t r = (1 + 5^(1/2))/2 + .0000000000001;
%t f[n_] := Floor[3r] - Floor[n*r] - Floor[3r - n*r]
%t t = Flatten[Table[f[n], {n, 1, 200}]] (* A188436 *)
%t Flatten[Position[t, 0] ] (* A188437 *)
%t Flatten[Position[t, 1] ] (* A188438 *)
%Y Cf. A188011, A188437, A188438, A188294.
%K nonn
%O 1
%A _Clark Kimberling_, Mar 31 2011
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