A131989 Start with the symbol **|* and for each iteration replace * with **|*. This sequence is the number of *'s between each dash.

2, 3, 1, 2, 3, 3, 1, 2, 1, 2, 3, 1, 2, 3, 3, 1, 2, 3, 3, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 3, 1, 2, 1, 2, 3, 1, 2, 3, 3, 1, 2, 3, 3, 1, 2, 1, 2, 3, 1, 2, 3, 3, 1, 2, 3, 3, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 3, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 3, 3, 1, 2, 1, 2, 3, 1, 2, 3, 3, 1, 2, 3

Offset

1,1

Comments

If the leading a(1)=2 is dropped, at least the next 90 terms coincide with those of A026181. - R. J. Mathar, Jun 13 2008

From Michel Dekking, Oct 19 2019: (Start)

Coding * by 1 and | by 2, the procedure is the same as performing the substitution sigma: 1 -> 1121, 2 -> 2. The return words of this morphism are 12, 112 and 1112. Under sigma these words transform as

12->11212, 112-> 112111212, 1112->1121112111212.

Coding the return words by their length minus one, the corresponding derivated morphism is

1-> 21, 2-> 231, 3-> 2331.

(a(n)) is the unique fixed point of this morphism.

(End)

a(n) = x(n+1), where x is the primitive Chacon sequence A049321 on the alphabet {3,1,2} instead of {0,1,2}. This follows from the fact that

sigma: 0->0012, 1->12, 2->012 and tau: 0->2001, 1->21, 2->201 are conjugated morphisms: tau(j) = 2 sigma(j) 2^{-1} for j=0,1,2. - Michel Dekking, Oct 23 2019

Links

Table of n, a(n) for n=1..99.

Formula

Comments from N. J. A. Sloane, Oct 10 2007: (Start)

The following is a simple recursive method to generate this sequence. The sequence is lim_{ t -> oo } S_t, where S_0 = 1+, and S_{t+1} is obtained from the concatenation S_t S_t S_t by replacing the first + by the sum of the two numbers adjacent to it and deleting the second +.

Thus we have:

S_0 = 1+,

S_1 = 1+1+1+ -> 21+,

S_2 = 21+21+21+ -> 23121+,

S_3 = 23121+23121+23121+ -> 23123312123121+,

S_4 = 23123312123121+23123312123121+23123312123121+ -> 23123312123123312331212312123123312123121+, etc.

Denote the sequence by a(1), a(2), ...

Block t, that is, S_t omitting the final 1+, extends from n=1 through n=(3^t-1)/2.

Given n, to find a(n): first find t from

p = (3^(t-1)-1)/2 < n <= (3^t-1)/2.

Assume t >= 2. Then if n=(3^(t-1)+1)/2, a(n) = 3 and if n=3^(t-1), a(n) = 1.

Otherwise, a(n) = a(n'), where

n' = n-p if n<3^(t-1), otherwise n' = n-3^(t-1). (End)

Example

The symbol through a few iterations: **|*, **|***|*|**|*, **|***|*|**|***|***|*|**|*|**|***|*|**|*, etc.

Crossrefs

a(n) = length of n-th run of 1's in A133162. - N. J. A. Sloane, Oct 09 2007

Sequence in context: A328912 A356327 A130830 * A305390 A344310 A362068

Adjacent sequences: A131986 A131987 A131988 * A131990 A131991 A131992

Keyword

easy,nonn

Author

Alex H. Bishop (AlexanderBishop(AT)stmarksschool.org), Oct 07 2007

Extensions

More terms from N. J. A. Sloane, Oct 10 2007

Status

approved