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%I #30 Dec 16 2017 14:02:42
%S 1,0,1,1,0,1,1,1,1,0,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,0,1,1,1,1,0,1,1,0,
%T 1,1,1,1,0,1,1,1,1,0,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,0,1,1,1,1,0,1,1,0,
%U 1,1,1,1,0,1,1,1,1,0,1,1,0,1,1,1,1,0,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,0,1,1,1,1,0,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,0,1,1
%N a(n) = [2r]-[nr]-[2r-nr], where r=(1+sqrt(5))/2 and [.]=floor.
%C This is column two of the array A188294.
%C a(n) = 1 - A123740(n-2) for n>2 (from [-x] = -[x] - 1 for non-integer x). - _Michel Dekking_, Oct 03 2016
%C Positions of 0's and 1's are given in A188434 (essentially = A101868) and A188435 (essentially = A188010). 0's are always isolated, and except for the initial term, run lengths of 1's are either 2 or 4. In that sequence of run lengths (1, 2, 4, 2, 4, 4, 2, 4, 2, ...), the 2's are always isolated and the run lengths of 4's are either 1 or 2. In that sequence of run lengths (1, 2, 1, 2, 2, 1, 2, 1, 2, 2, ...), 1's are always isolated, and the run lengths of 2's are again either 1 or 2. - _M. F. Hasler_, Oct 12 2017
%C Along the same line of Dekking's comment above, a(n) = 1 - A188009(n), for n > 2. This explains why A188435(n) = A188010(n+1) and A188434(n) = A101868(n+1) = A101866(1,n+1) for all n > 1. - _M. F. Hasler_, Dec 14 2017
%F a(n) = [2r]-[nr]-[2r-nr], where r=(1+sqrt(5))/2.
%t r = (1 + 5^(1/2))/2;
%t f[n_] := Floor[2r] - Floor[n*r] - Floor[2r - n*r]
%t A188433 = Flatten[Table[f[n], {n, 1, 200}]] (* see also A188434 and A188435 *)
%o (PARI) a(n,a=sqrt(5)/2-.5)=3-n\a-(2-n)\a \\ _M. F. Hasler_, Oct 12 2017
%Y Cf. A188294, A188434, A188435, A188009.
%K nonn
%O 1
%A _Clark Kimberling_, Mar 31 2011
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