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Let c(1) = c(2) = 0, c(3) = 1, and c(n + 3) = (c(n) - 2*c(n + 1) + c(n + 2))/n, then a(n) = ceiling (c(n)).
0

%I #28 Nov 10 2020 22:58:20

%S 0,0,1,1,0,1,1,0,0,1,1,0,1,1,0,0,1,1,0,1,1,0,0,1,0,0,1,1,0,1,1,0,0,1,

%T 0,0,1,0,0,1,1,0,1,1,0,0,1,0,0,1,1,0,1,1,0,1,1,0,0,1,0,0,1,0,0,1,1,0,

%U 1,1,0,0,1,0,0,1,0,0,1,1,0,1,1,0,1,1,0,0,1,0,0,1,0,0,1,1,0,1,1,0,1,1,0,0,1,0,0,1,0,0,1,1,0,1,1,0,1,1,0,0,1,0,0,1,0,0,1,1

%N Let c(1) = c(2) = 0, c(3) = 1, and c(n + 3) = (c(n) - 2*c(n + 1) + c(n + 2))/n, then a(n) = ceiling (c(n)).

%C Conjectured quasiperiodicity with autocorrelation function R(x) = 1/2 if x = 0, 1/4 if x > 0.

%C Some other proved or conjectured (or suspected) nonperiodic binary sequences where there are no more than two consecutive 0's or 1's include: A083035, A285305, A190843, A286059, A288213, A288551, A288473, A176405, A188321, A188398, A191162, A272170, A197879, A078588, A272532, A273129, A074937, A188297, A289128. Others?

%t c[n_]:=c[n]=(c[n-1]-2c[n-2]+c[n-3])/n;

%t c[1] = 0; c[2] = 0; c[3] = 1;

%t Table[Ceiling@c[j],{j,1,2^7}]

%Y Cf. A083035, A285305, A190843, A286059, A288213, A288551, A288473, A176405, A188321, A188398, A191162, A272170, A197879, A078588, A272532, A273129,A074937, A188297, A289128.

%K nonn

%O 0,1

%A _Andres Cicuttin_, Oct 11 2020