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A082691
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a(1)=1, a(2)=2, then if the first 3*2^k-1 terms are a(1), a(2), ..., a(3*2^k - 1), the first 3*2^(k+1)-1 terms are a(1), a(2), ..., a(3*2^k - 1), a(1), a(2), ..., a(3*2^k - 1), a(3*2^k-1) + 1.
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3
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1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 5, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 5, 6, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 5, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 4, 5, 6, 7, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3
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OFFSET
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1,2
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COMMENTS
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Consider the subsequence b(k) such that a(b(k))=1. Then 3k - b(k) = A063787(k+1) and b(k) = 1 + A004134(k-1).
A naive way to try and guess whether a sequence is periodic, based on its first k terms (n1, ..., nk), is to look at all sequences which have period less than k, and guess "periodic" if any of them extend (n1, ..., nk), "nonperiodic" otherwise.
a(1)=1, a(2)=2. Suppose a(1), ..., a(n) have been defined, n > 1.
1. If the above guessing method guesses that (a(1), ..., a(n)) is an initial segment of a periodic sequence, then let a(n+1) be the least nonzero number not appearing in (a(1), ..., a(n)).
2. Otherwise, let (a(n+1), ..., a(2n)) be a copy of (a(1), ..., a(n)).
This sequence thwarts the guessing attempt, tricking the guesser into changing his mind infinitely many times as n->infinity. - Sam Alexander
As n increases, the average value of the first n terms approaches 7/3 = 2.333... - Maxim Skorohodov, Dec 15 2022
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LINKS
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EXAMPLE
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To construct the sequence: start with (1, 2); concatenating those 2 terms gives (1,2,1,2). Appending 3 gives the first 5 terms: (1,2,1,2,3). Concatenating those 5 terms gives (1,2,1,2,3,1,2,1,2,3). Appending 4 gives the first 11 terms: (1,2,1,2,3,1,2,1,2,3,4), etc.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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