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A337546
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A binary sequence defined as follows: a(1)=0; thereafter choose a(n) so as to minimize the number of final consecutive repeated words, and if there is a choice, minimize the length of the repeated word.
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3
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0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1
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OFFSET
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1
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COMMENTS
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Start with a(1)=0, then select a(n)=0 or a(n)=1 so as to minimize the largest integer T such that a(0),...,a(n) has a word repeated T times at the end, and if 0 and 1 produce the same T, so as to minimize the length of the longest word repeated T times at the end.
Cousin of the so-called Linus sequence (A006345), which avoids the longest repeated suffix. Conjectures: 1. This is cubefree (the Linus sequence is not cubefree); 2. density of 0's exists and equals 1/2; 3. recurrent and mirror invariant.
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LINKS
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EXAMPLE
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Consider the first case where the sequence differs from the Linus sequence, that is at n=20. Up to n=19 we have: 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0. Then inserting 0 produces a triple at the end, while inserting 1 produces two occurrences of the word 001 at the end and no cubes. Since the number of repetitions counts more than length of the repeated suffix, a(20)=1.
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PROG
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(C) See Links section.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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