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 A096055 Let {s(i)}, i=0,1,2,... be a sequence of finite sequences with terms s(i)(j), j=1,2,3,... Start with s(0)={1}. Then, for k>0, let s(k)=s(k-1)Us(k-1) if s(k-1)(k)=0, s(k)=s(k-1)U{0}Us(k-1) if s(k-1)(k)=1, where s(i)(j) is the j-th element of s(i) and U denotes concatenation of the terms of the two operands. {a(n)} is the limit of s(k) as k goes to infinity. 3
 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Suggested by Leroy Quet, Jul 18 2004. Note that this is not A137843(n) mod 2. The first difference is at n=14, where a(14) = 0, while A137843(14) = 5. - Antti Karttunen, Aug 31 2017 LINKS EXAMPLE Let s(0) = {1}. Then s(1) = s(0) U {0} U s(0) = {1,0,1}, since s(0)(1) = 1, s(2) = s(2) U s(2) = {1,0,1,1,0,1}, since s(1)(2) = 0, s(3) = s(2) U {0} U s(2) ={1,0,1,1,0,1,0,1,0,1,1,0,1}, since s(2)(3) = 1. From Antti Karttunen, Aug 31 2017: (Start) And since s(3)(4) = a(4) = 1, s(4) = s(3) U {0} s(3) = {1,0,1,1,0,1,0,1,0,1,1,0,1,0,1,0,1,1,0,1,0,1,0,1,1,0,1}, Note that because each stage is a prefix of the next (starts with the same terms), we may as well write a(n) instead of s(n-1)(n). (End) CROSSREFS Cf. A112306 (the length of each stage). Cf. A137843. Sequence in context: A104974 A024711 A128174 * A125144 A115198 A005614 Adjacent sequences:  A096052 A096053 A096054 * A096056 A096057 A096058 KEYWORD nonn AUTHOR John W. Layman, Jul 20 2004 STATUS approved

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