A285074 Positions of 0 in A285073; complement of A285075.

1, 3, 5, 7, 8, 10, 12, 13, 15, 17, 19, 20, 22, 24, 25, 27, 29, 31, 32, 34, 36, 37, 39, 41, 42, 44, 46, 48, 49, 51, 53, 54, 56, 58, 60, 61, 63, 65, 66, 68, 70, 71, 73, 75, 77, 78, 80, 82, 83, 85, 87, 89, 90, 92, 94, 95, 97, 99, 101, 102, 104, 106, 107, 109

Offset

1,2

Comments

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

From Michel Dekking, May 30 2017: (Start)

Proof of a stronger form of the conjecture: the sequence d:=A285073 can be written as d=01c, where c is the homogeneous Sturmian sequence with slope alpha = sqrt(2)-1 (see comments of A285073). Changing from alpha to 1-alpha = 2-sqrt(2) turns c into its mirror image, so we have to find the positions of 1 in this new sequence.

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

In our case s = 1/(2-sqrt(2)) = 1+sqrt(1/2). It follows that for n=0,1,... one has a(n+2) = floor((n+1)*(1+sqrt(1/2))) + 2, which directly implies that sqrt(1/2)-1 < n*r - a(n) < sqrt(1/2), which is a strengthening of the conjecture (actually there is no strict inequality for n=1: r-a(1) = sqrt(1/2)). (End)

Links

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

Example

As a word, A285073 = 01010..., in which 0 is in positions 1,3,5,7,8,...

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, A285075, A285076.

Sequence in context: A276224 A186342 A186315 * A186219 A185050 A083034

Adjacent sequences: A285071 A285072 A285073 * A285075 A285076 A285077

Keyword

nonn,easy,changed

Author

Clark Kimberling, Apr 19 2017

Status

approved