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[3r]-[nr]-[3r-nr], where r=(1+sqrt(5))/2 and [.]=floor.
5

%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