A286909 - OEIS

A286909

Positions of 1 in A286907; complement of A286908.

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

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

Clark Kimberling, Table of n, a(n) for n = 1..10000

FORMULA

a(n) = 2*A001951(n) for n >= 1, conjectured.