

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(k1)Us(k1) if s(k1)(k)=0, s(k)=s(k1)U{0}Us(k1) if s(k1)(k)=1, where s(i)(j) is the jth 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
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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

Table of n, a(n) for n=1..105.


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(n1)(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



