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s(1)={1}. s(2)={1,0}. If a(n) = 0, s(n+2) = s(n+1) U s(n) U {1}. If a(n) = 1, s(n+2) = s(n+1) U s(n+1) U {1}. (U represents concatenation of finite sequences.) {a(n)} is the limit of {s(n)} as n -> infinity.
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%I #10 Nov 13 2014 12:02:01

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

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

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

%N s(1)={1}. s(2)={1,0}. If a(n) = 0, s(n+2) = s(n+1) U s(n) U {1}. If a(n) = 1, s(n+2) = s(n+1) U s(n+1) U {1}. (U represents concatenation of finite sequences.) {a(n)} is the limit of {s(n)} as n -> infinity.

%C Number of terms in s(n) is A112361(n).

%e s(3) = {1,0,1,0,1}, s(4) = {1,0,1,0,1,1,0,1}, s(5) = {1,0,1,0,1,1,0,1,1,0,1,0,1,1,0,1,1}

%Y Cf. A114482, A062318, A112361.

%K easy,nonn

%O 1,1

%A _Leroy Quet_, Nov 30 2005

%E More terms from _Joshua Zucker_, Jul 27 2006