A284394 {101->2}-transform of the infinite Fibonacci word A003849.

0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 2, 0, 0, 2, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 2, 0, 0, 2, 0, 0, 1, 0, 0, 2, 0, 0, 2, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 2, 0, 0, 2, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 2, 0, 0, 2, 0, 0, 1, 0, 0, 2, 0, 0, 2, 0, 0, 1, 0, 0, 2, 0, 0, 1

Offset: 1

Author

Clark Kimberling, May 02 2017

Keywords

• nonn
• easy

Comments

The sequences p = A032766, q = A284395, r = A284396, of positions of 0,1,2, respectively, partition the positive integers. Let t,u,v be the slopes of p, q, r, respectively. Then t = 3/2, u = (9+3*sqrt(5))/2, v = (3+3*sqrt(5))/2, and 1/t + 1/u + 1/v = 1.

Examples

			As a word, A003849 = 01001010010010100..., and replacing each 101 by 2 gives 01002001002002001...
		

Links

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

Crossrefs

Cf. A003849, A032766, A284395, A284396.

Programs

Mathematica

s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 13]  (* A003849 *)
w = StringJoin[Map[ToString, s]]; w1 = StringReplace[w, {"101" -> "2"}]
st = ToCharacterCode[w1] - 48 (* A284394 *)
Flatten[Position[st, 0]]  (* A032766 *)
Flatten[Position[st, 1]]  (* A284395 *)
Flatten[Position[st, 2]]  (* A284396 *)