A284626 - OEIS

A284626

Positions of 1 in A284622; complement of A284623.

2, 3, 5, 9, 11, 12, 14, 17, 19, 20, 23, 27, 29, 30, 33, 36, 38, 39, 41, 45, 47, 48, 50, 53, 55, 56, 59, 62, 64, 65, 67, 71, 73, 74, 77, 81, 83, 84, 86, 89, 91, 92, 95, 99, 101, 102, 105, 108, 110, 111, 113, 117, 119, 120, 123, 127, 129, 130, 132, 135, 137

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

a(n) - a(n-1) is in {1,2,3,4} for n >= 2.

Conjecture: a(n)/n -> 9/4.

From Michel Dekking, Aug 26 2017: (Start)

Proof of the conjecture. Let x = A010060 be the Thue Morse sequence, and let y = A284622 be the [0011->0]-transform of x. Let a = A284626 be the positions of 1 in y. There are 3 steps in the proof.

Step 1. It is easily verified that a(n)/n -> 9/4 if and only if f(1,a) = 4/9, where in general f(w,z) denotes the frequency of a word w in the infinite sequence z, if it exists.

Step 2. One has f(0011,x) = 1/12. It is well-known that the frequencies of words in any fixed point of a primitive morphism exist. This is usually proved by Perron-Frobenius theory. For a quick proof see the paper "On the Thue-Morse measure".

Step 3. Let k(n) be the number of 1's in x(1)...x(n), and m(n) the number of 0011's in x(1)...x(n). Then the number of 1's in y(1)...y(n-3m(n)) is equal to k(n)-2m(n). But we know by Step 2 that m(n)/n -> 1/12, and obviously k(n)/n -> 1/2. So f(1,y) is equal to ((1/2 - 2/12)/(1 - 3/12) = 4/9. (End)

LINKS

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

Michel Dekking,On the Thue-Morse measure, Acta Universitatis Carolinae. Mathematica et Physica 033.2 (1992), 35-40.

EXAMPLE

As a word, A284622 = 011010001011010010..., in which 1 is in positions 2,3,5,9,11,...

MATHEMATICA

s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 9] (* A010060 *)

w = StringJoin[Map[ToString, s]]

w1 = StringReplace[w, {"0011" -> "0"}]

st = ToCharacterCode[w1] - 48 (* A284622 *)

Flatten[Position[st, 0]] (* A284623 *)

Flatten[Position[st, 1]] (* A284626 *)