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A157238
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0-1 sequence generated by starting with a 0, and then by using whichever of 0, 1 will result in the shortest sequence repeated at the end.
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4
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0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0
(list;
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refs;
listen;
history;
text;
internal format)
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OFFSET
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1,1
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COMMENTS
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The same as the Linus sequence (A006345): a(n) "breaks the pattern" by avoiding the longest doubled suffix, but using 0's and 1's. - Robert G. Wilson v, Dec 01 2013
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LINKS
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Table of n, a(n) for n=1..100.
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FORMULA
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a(n) = A006345(n) - 1. - Robert G. Wilson v, Dec 02 2013
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EXAMPLE
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a(6)=1 as 0,1,0,0,1,1 has a longest repeated sequence of length 1 at the end, whereas 0,1,0,0,1,0 has a longest repeated sequence of length 3 at the end. Similarly, a(7)=0 since 0,1,0,0,1,1,0 has a longest repeated sequence of length 0 at the end.
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PROG
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(Python)
.x = [0]
.while (len(x) < 1000):
..t = x[ -1]
..z = 1
..while (2*z+1 <= len(x)):
...if (x[ -z:] == x[ -(2*z+1):-(z+1)]):
....t = x[ -(z+1)]
...z += 1
..x += [1-t]
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CROSSREFS
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Cf. A006345, A283131.
Sequence in context: A084091 A080846 A082401 * A337546 A059448 A283318
Adjacent sequences: A157235 A157236 A157237 * A157239 A157240 A157241
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KEYWORD
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nonn
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AUTHOR
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Luke Pebody (luke.pebody(AT)gmail.com), Feb 25 2009
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STATUS
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approved
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