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A157238
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.
4
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
OFFSET
1,1
COMMENTS
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
FORMULA
a(n) = A006345(n) - 1. - Robert G. Wilson v, Dec 02 2013
EXAMPLE
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.
PROG
(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.append(1 - t)
print(x)
CROSSREFS
Sequence in context: A368463 A080846 A082401 * A337546 A059448 A283318
KEYWORD
nonn
AUTHOR
Luke Pebody (luke.pebody(AT)gmail.com), Feb 25 2009
STATUS
approved