A287107 Positions of 2 in A287104.

2, 7, 11, 14, 18, 23, 26, 30, 35, 39, 44, 47, 51, 56, 60, 63, 67, 72, 76, 81, 84, 88, 93, 97, 100, 104, 109, 112, 116, 121, 125, 128, 132, 137, 141, 146, 149, 153, 158, 162, 165, 169, 174, 177, 181, 186, 190, 195, 198, 202, 207, 211, 214, 218, 223, 226, 230

Offset: 1

Author

Clark Kimberling, May 21 2017

Keywords

• nonn
• easy

Comments

From Michel Dekking, Sep 16 2019: (Start)

Let sigma be the defining morphism of A287104: 0->10, 1->12, 2->0.

Let u=201, v=2101, w=20101 be the return words of the word 2.

Under sigma u, v, and w are mapped to sigma(201) = 01012, sigma(2101) = 0121012, sigma(20101) = 010121012.

All three images have suffix 2. We can therefore move this suffix to the front of all three images, obtaining the fixed point (a(n+1)) = 20101... when iterating. This induces the morphism 3 -> 5, 4 -> 34, 5 -> 54 on the return words, coded by their lengths.

Coding the symbols according to 3<->2, 4<->1, 5<->0, this leads to the morphism 2->0, 1->21, 0->01 on the alphabet {0,1,2}. This is exactly the morphism which has A287072 as unique fixed point. So the sequence d of first differences of (a(n)) equals A287072 with the coding above. This gives the formula below.

(End)

Links

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

Crossrefs

Cf. A287104, A287105, A287106.

Programs

Mathematica

s = Nest[Flatten[# /. {0 -> {1, 0}, 1 -> {1, 2}, 2 -> 0}] &, {0}, 10] (* A287104 *)
Flatten[Position[s, 0]] (* A287105 *)
Flatten[Position[s, 1]] (* A287106 *)
Flatten[Position[s, 2]] (* A287107 *)

Formula

a(n) = 2 + Sum_{k=1..n-1} d(k), where d(k)=5 if A287072(k)=0, d(k)=4 if A287072(k)=1, and d(k)=3 if A287072(k)=2. - Michel Dekking, Sep 16 2019