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A286909
Positions of 1 in A286907; complement of A286908.
4
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
OFFSET
1,1
COMMENTS
Conjectures: a(n) - a(n-1) is in {2,4} for n >= 2, and a(n)/n -> sqrt(8).
Differs from A189792 first in a(29) = 82, while A189792(29) = 80. - M. F. Hasler, Apr 13 2018
From Michel Dekking, Dec 06 2018: (Start)
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)
REFERENCES
J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 284.
LINKS
FORMULA
a(n) = 2*A001951(n) for n >= 1, conjectured.
EXAMPLE
As a word, A286907 = 0101000101000101..., in which 1 is in positions 2,4,8,10,14,16,....
MATHEMATICA
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 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, May 18 2017
EXTENSIONS
Edited by Clark Kimberling, Nov 06 2018
STATUS
approved