

A157238


01 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
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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


LINKS

Table of n, a(n) for n=1..100.


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

Cf. A006345, A283131.
Sequence in context: A084091 A080846 A082401 * A337546 A059448 A283318
Adjacent sequences: A157235 A157236 A157237 * A157239 A157240 A157241


KEYWORD

nonn


AUTHOR

Luke Pebody (luke.pebody(AT)gmail.com), Feb 25 2009


STATUS

approved



