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2, 3, 4, 7, 8, 9, 11, 12, 13, 16, 17, 18, 21, 22, 23, 25, 26, 27, 30, 31, 32, 34, 35, 36, 39, 40, 41, 44, 45, 46, 48, 49, 50, 53, 54, 55, 58, 59, 60, 62, 63, 64, 67, 68, 69, 71, 72, 73, 76, 77, 78, 81, 82, 83, 85, 86, 87, 90, 91, 92, 94, 95, 96, 99, 100, 101
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OFFSET
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1,1
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COMMENTS
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Conjecture: -2 < n*r - a(n) < 1 for n >= 1, where r = (7 + sqrt(5))/6. [Corrected by Clark Kimberling, Aug 19 2019]
Computation of a formula for a(n): we will show that (a(n)) is a union of three generalized Beatty sequences.
Let T denote the morphism {0->1, 1->000}.
The Fibonacci word xF:=A003849 is a concatenation of the two words v = 100 and w = 10 (ignoring xF(0)). It is known that the order of these words is given by the Fibonacci word itself (cf. A003622). Now note that
T(v) = T(100) = 00011, T(w) = T(10) = 0001.
We see from this that the sequence of first differences of a = A287664 for the first 0 in each triple 000 is given by 5,4,5,5,4,5,4,5,5,4,..., and that in fact these two letters occur as the Fibonacci word on the alphabet {5,4}.
Let {b^(j)} for j = 1,2,3 be the sequence of positions of the j-th 0 in the triples 000. From Lemma 8 in the paper by Allouche and Dekking it follows that these sequences are given by
b^(j)(n) = floor(n*phi) + 3*n + r^(j),
where r^(1) = -2, r^(2) = -1, and r^(3) = 0.
Note that a(3*n+j-3) = b^(j)(n) for n=1,2,3,... and j=1,2,3. (End)
Proof of Kimberling's conjecture. {} denotes fractional part.
Let r = (7+sqrt(5))/6 = phi/3+1. Using the formula above we have
(3*n+j-3)*r - a(3*n+j-3) =
(n+j/3-1)*(phi+3) - (n*phi - {n*phi} + 3n + r^(j)) =
(j/3-1)*phi + j - 3 - r^(j) + {n*phi}.
Since ({n*phi}) is equidistributed on (0,1), this implies that a(3*n+j-3) takes values in the interval (-(2/3)*phi, -(2/3)*phi+1) for j=1, in the interval (-(1/3)*phi, -(1/3)*phi+1) for j=2, and in the interval (0,1) for j=3. Combining these, we see that a(n) takes values in (-(2/3)*phi, 1), and that this is best possible. Since -(2/3)*phi = -1.078689..., this improves on the lower bound conjectured by Kimberling.
(End)
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LINKS
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FORMULA
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a(3*n+j-3) = b^(j)(n) for j=1,2,3, with b^(1)(n) = floor(n*phi) + 3*n - 2, b^(2)(n) = floor(n*phi) + 3*n - 1, b^(3)(n) = floor(n*phi) + 3*n. - Michel Dekking, Aug 22 2019
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MATHEMATICA
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s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 10] (* A003849 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"0" -> "1", "1" -> "000"}]
st = ToCharacterCode[w1] - 48 (* A287663 *)
Flatten[Position[st, 0]] (* A287664 *)
Flatten[Position[st, 1]] (* A287665 *)
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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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STATUS
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approved
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