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2, 4, 8, 10, 14, 16, 18, 22, 24, 28, 30, 32, 36, 38, 42, 44, 48, 50, 52, 56, 58, 62, 64, 66, 70, 72, 76, 78, 82, 84, 86, 90, 92, 96, 98, 100, 104, 106, 110, 112, 114, 118, 120, 124, 126, 130, 132, 134, 138, 140, 144, 146, 148, 152, 154, 158, 160, 164, 166
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OFFSET
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1,1
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COMMENTS
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Conjectures: a(n) - a(n-1) is in {2,4} for n >= 2, and a(n)/n -> sqrt(8).
Proof of the conjectures and the formula in FORMULA:
let (b(n)):= A286907, the [0->00,1->01]-transform of (s(n)):= A080764, the Sturmian word with slope alpha = sqrt(2)/2.
Then, clearly, (watch out for the incompatible offsets) b(2n-1) = 0 for all n>=1, and b(2n) = s(n-1) for all n>=1. The positions of 1 in (s(n-1)) are given by the Beatty sequence (floor(n*beta)), with beta = 1/alpha = sqrt(2) (see "Automatic Sequences").
The conclusion is that a(n) = 2*floor(n*beta) = 2*A001951(n+1). Also,
a(n) - a(n-1) = 2*(floor(n*beta) - floor((n-1)*beta))
is in {2,4}, because (floor(n*beta) - floor((n-1)*beta)) is a Sturmian sequence.
Finally,
a(n)/n = 2*floor(n*beta)/n -> 2*beta = sqrt(8) as n->oo.
(End)
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REFERENCES
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J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 284.
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LINKS
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FORMULA
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a(n) = 2*A001951(n) for n >= 1, conjectured.
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EXAMPLE
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As a word, A286907 = 0101000101000101..., in which 1 is in positions 2,4,8,10,14,16,....
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MATHEMATICA
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s = Nest[Flatten[# /. {0 -> 1, 1 -> {1, 1, 0}}] &, {1}, 11]; (* Sturmian word A080764 *)
w = StringJoin[Map[ToString, s]];
w1 = StringReplace[w, {"0" -> "00", "1" -> "01"}]
st = ToCharacterCode[w1] - 48 ; (* A286907 *)
Flatten[Position[st, 0]]; (* A286908 *)
Flatten[Position[st, 1]]; (* A286909 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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