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

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Last modified November 30 20:17 EST 2021. Contains 349425 sequences. (Running on oeis4.)