

A131989


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


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, 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
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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^t1)/2.
Given n, to find a(n): first find t from
p = (3^(t1)1)/2 < n <= (3^t1)/2.
Assume t >= 2. Then if n=(3^(t1)+1)/2, a(n) = 3 and if n=3^(t1), a(n) = 1.
Otherwise, a(n) = a(n'), where
n' = np if n<3^(t1), otherwise n' = n3^(t1). (End)


EXAMPLE

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


CROSSREFS

a(n) = length of nth run of 1's in A133162.  N. J. A. Sloane, Oct 09 2007
Sequence in context: A105315 A328912 A130830 * A305390 A344310 A255890
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



