

A064353


Kolakoski(1,3) sequence: the alphabet is {1,3}, and a(n) is the length of the nth run.


16



1, 3, 3, 3, 1, 1, 1, 3, 3, 3, 1, 3, 1, 3, 3, 3, 1, 1, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 1, 1, 3, 3, 3, 1, 3, 1, 3, 3, 3, 1, 1, 1, 3, 3, 3, 1, 3, 3, 3, 1, 1, 1, 3, 3, 3, 1, 3, 3, 3, 1, 1, 1, 3, 3, 3, 1, 3, 1, 3, 3, 3, 1, 1, 1, 3, 3, 3, 1, 3, 3, 3, 1, 3, 3, 3, 1, 1, 1, 3, 3, 3, 1, 3, 1, 3, 3, 3, 1, 1, 1, 3, 3
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

Historical note: the sequence (a(n)) was introduced (by me) in 1981 in a seminar in Bordeaux. It was remarked there that (a(n+1)) is a morphic sequence, i.e., a lettertoletter projection of a fixed point of a morphism. The morphism is 1>3, 2>2, 3>343, 4>212. The lettertoletter map is 1>1, 2>1, 3>3, 4>3. There it was also remarked that this allows one to compute the frequency of the letter 3, and an exact expression for this frequency involving sqrt(177) was given.  Michel Dekking, Jan 06 2018
The frequency of the number '3' is 0.6027847... See UWC link.  Jaap Spies, Dec 12 2004
Consider the Kolakoski sequence generalized to the alphabet {A,B}, where A=2p+1, B=2q+1. The fraction of symbols that are A approaches f_A, calculated as follows: x=(p+q+1)/3; y=((pq)^2)/2; lambda = x + (x^3+y+sqrt(y^2+2*x^3*y))^(1/3) + (x^3+ysqrt(y^2+2*x^3*y))^(1/3); f_A=(lambda2q1)/(2p2q). The technique is the "simple computation" mentioned by Dekking and repeated in the UWC link.  Ed Wynn, Jul 29 2019


REFERENCES

E. Angelini, "Jeux de suites", in Dossier Pour La Science, pp. 3235, Volume 59 (Jeux math'), April/June 2008, Paris.
F. M. Dekking: "What is the long range order in the Kolakoski sequence?" in: The Mathematics of LongRange Aperiodic Order, ed. R. V. Moody, Kluwer, Dordrecht (1997), pp. 115125.


LINKS

Ulrich Reitebuch, HenrietteSophie Lipschütz, and Konrad Polthier, Visualizing the Kolakoski Sequence, Bridges Conf. Proc.; Math., Art, Music, Architecture, Culture (2023) 481484.


MATHEMATICA

A = {1, 3, 3, 3}; i = 3; next = 1; While[Length[A] < 140, A = Join[A, next*Array[1&, A[[i]]]]; i++; next = 4next]; A (* JeanFrançois Alcover, Nov 12 2016, translated from MATLAB *)


PROG

(MATLAB) A = [1 3 3 3]; i = 3; next = 1; while length(A) < 140 A = [A next*ones(1, A(i))]; i = i + 1; next = 4  next; end
(Haskell)  from John Tromp's a000002.hs
a064353 n = a064353_list !! (n1)
a064353_list = 1 : 3 : drop 2
(concat . zipWith replicate a064353_list . cycle $ [1, 3])


CROSSREFS



KEYWORD

nonn,easy,nice


AUTHOR



EXTENSIONS



STATUS

approved



