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1, 2, 4, 5, 6, 8, 9, 11, 12, 13, 14, 16, 17, 19, 20, 21, 23, 24, 26, 27, 28, 30, 31, 32, 34, 35, 37, 38, 39, 41, 42, 44, 45, 46, 47, 49, 50, 52, 53, 54, 56, 57, 59, 60, 61, 62, 64, 65, 67, 68, 69, 70, 72, 73, 75, 76, 77, 79, 80, 82, 83, 84, 85, 87, 88, 90
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OFFSET
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1,2
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COMMENTS
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Conjecture: lim_{n->infinity} a(n)/n = 1.36..., and if m denotes this number, then -1 < m - a(n)/n < 1 for n >= 1.
From Michel Dekking, Feb 23 2020: (Start)
Proof of the first part of this conjecture.
Let a(0):=0. We write this sequence as the sum of its first differences:
a(n) = Sum_{k=0..n-1} a(k+1)-a(k).
We know (see A288173) that A288173 can be generated as a decoration delta(t) of the fixed point t of the morphism alpha given by
alpha(A) = AB, alpha(B) = AC, alpha(C) = ABB.
Here delta is the morphism
delta(A) = 001, delta(B) = 0001, delta(C) = 00001.
Let e = A288175 be the sequence of positions of 1 in A288173. Note that if we are at the n-th 1, then we have seen e(n)-n zeros. So the position of the (e(n)-n)-th zero is e(n)-1.
Let m(n):=e(n)-n. Then
a(m(n))/m(n) = (e(n)-1)/m(n) = (e(n)-1)/n * n/(e(n)-n).
According to the comments at e = A288175, the first factor in this product converges to 3.7092753596..., and the second to 1/(3.7092753596... - 1).
It follows that as n->infinity,
a(m(n))/m(n) -> 1.36900369004... .
It is easy to see from this that the whole sequence converges, and so
a(n)/n -> 1.36900369004... . (End)
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LINKS
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Clark Kimberling, Table of n, a(n) for n = 1..10000
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MATHEMATICA
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s = {0, 0}; w[0] = StringJoin[Map[ToString, s]];
w[n_] := StringReplace[w[n - 1], {"00" -> "0010", "1" -> "001"}]
Table[w[n], {n, 0, 8}]
st = ToCharacterCode[w[11]] - 48 (* A288173 *)
Flatten[Position[st, 0]] (* A288174 *)
Flatten[Position[st, 1]] (* A288175 *)
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CROSSREFS
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Cf. A288173, A288175.
Sequence in context: A171599 A328594 A346129 * A280998 A043687 A087118
Adjacent sequences: A288171 A288172 A288173 * A288175 A288176 A288177
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KEYWORD
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nonn,easy
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AUTHOR
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Clark Kimberling, Jun 07 2017
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STATUS
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approved
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