

A287107


Positions of 2 in A287104.


4



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



OFFSET

1,1


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


FORMULA

a(n) = 2 + Sum_{k=1..n1} 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


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 *)


CROSSREFS

Cf. A287104, A287105, A287106.
Sequence in context: A140548 A243630 A341076 * A308550 A018308 A329868
Adjacent sequences: A287104 A287105 A287106 * A287108 A287109 A287110


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, May 21 2017


STATUS

approved



