A285075 Positions of 1 in A285073; complement of A285074.

2, 4, 6, 9, 11, 14, 16, 18, 21, 23, 26, 28, 30, 33, 35, 38, 40, 43, 45, 47, 50, 52, 55, 57, 59, 62, 64, 67, 69, 72, 74, 76, 79, 81, 84, 86, 88, 91, 93, 96, 98, 100, 103, 105, 108, 110, 113, 115, 117, 120, 122, 125, 127, 129, 132, 134, 137, 139, 142, 144, 146

Offset

1,1

Comments

Conjecture: -1 + sqrt(2) < n*r - a(n) < sqrt(2) for n>=1, where r = 1 + sqrt(2).

From Michel Dekking, May 29 2017: (Start)

Proof of the conjecture: the sequence d:=A285073 can be written as d=01c, where c is the homogeneous Sturmian sequence with slope sqrt(2)-1 (see comments of A285073).

In general, a homogeneous Sturmian sequence (floor((n+1)r)-floor(nr)) gives the positions of 1 in the Beatty sequence b=(floor((n+1)s)), where s=1/r.

In our case s=1+sqrt(2). It follows that for n=0,1,… one has a(n+2) = floor((n+1)(1+sqrt(2)) + 2, which directly implies the conjecture. (End)

Links

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

Example

As a word, A285073 = 01010...,..., in which 1 is in positions 2,4,6,9,11,...

Mathematica

s = Nest[Flatten[# /. {0 -> {1, 0}, 1 -> {0, 1, 0}}] &, {0}, 14]; (* A285073 *)
Flatten[Position[s, 0]];  (* A285074 *)
Flatten[Position[s, 1]];  (* A285075 *)

Crossrefs

Cf. A285073, A285074, A285076.

Sequence in context: A160813 A247515 A186220 * A186316 A265286 A186343

Adjacent sequences: A285072 A285073 A285074 * A285076 A285077 A285078

Keyword

nonn,easy

Author

Clark Kimberling, Apr 19 2017

Status

approved