

A000002


Kolakoski sequence: a(n) is length of nth run; a(1) = 1; sequence consists just of 1's and 2's.
(Formerly M0190 N0070)


270



1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2
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OFFSET

1,2


COMMENTS

Historical note: the sequence might be better called the OldenburgerKolakoski sequence, since it was discussed by Rufus Oldenburger in 1939; see links.  Clark Kimberling, Dec 06 2012. However, to avoid confusion, this sequence will be known in the OEIS as the Kolakoski sequence. It is undesirable to have some entries refer to the OldenburgerKolakoski sequence and others to the Kolakoski sequence.  N. J. A. Sloane, Nov 22 2017
It is an unsolved problem to show that the density of 1's is equal to 1/2.
A weaker problem is to construct a combinatorial bijection between the set of positions of 1's and the set of positions of 2's.  Gus Wiseman, Mar 01 2016
The sequence is cubefree and all square subwords have lengths which are one of 2, 4, 6, 18 and 54 (see A294447) [Carpi, 1994].
This is a fractal sequence: replace each run with its length and recover the original sequence.  Kerry Mitchell, Dec 08 2005
Kupin and Rowland write: We use a method of Goulden and Jackson to bound freq_1(K), the limiting frequency of 1 in the Kolakoski word K. We prove that freq_1(K)  1/2 <= 17/762, assuming the limit exists and establish the semirigorous bound freq_1(K)  1/2 <= 1/46.  Jonathan Vos Post, Sep 16 2008
freq_1(K) is conjectured to be 1/2 + O(log(K)) (see PlanetMath link).  Jon Perry, Oct 29 2014
Conjecture: Taking the sequence in word lengths of 10, for example, batch 110, 1120, etc., then there can only be 4, 5 or 6 1's in each batch.  Jon Perry, Sep 26 2012
The sequence does not contain words of the form ababa, because this would imply the impossible 111 (1 b, 1 a, 1 b) somewhere before. This demonstrates the conjecture made by Jon Perry: more than 6 1's or 6 2's in a word of 10 would necessitate something like aabaabaaba, which would imply the impossible 12121 before (word aabaababaa is also impossible because of ababa). The remark on the sextuplets below even shows that the number of 1's in any 9tuplet is always 4 or 5.
There are only 6 triples that appear in the sequence (112, 121, 122, 211, 212 and 221); and by the preceding argument, only 18 sextuplets: the 6 double triples (112112, etc.); 112122, 112212, 121122, 121221, 211212, and 211221; and those obtained by reversing the order of the triples (122112, etc.). Regarding the density of 1's in the sequence, these 12 sextuplets all have a density 1/2 of 1's, and the 6 double triples all lead to a word with this exact density after transformation by the Kolakoski rules, for example: 112112 > 12112122 (4 1's/8); this is because the second triple reverses the numbers of 1's and 2's generated by the first triple. Therefore, the sequence can be split into the double triples on one side, a part whose transformation (which is in the sequence) has a density of 1's of 1/2; and a part with the other sextuplets, which has directly the same density of 1's. (End)
If we map 1 to +1 and 2 to 1, then the mapped sequence would have a [conjectured] mean of 0, since the Kolakoski sequence is [conjectured] to have an equal density (1/2) of 1s and 2s. For the partial sums of this mapped sequence, see A088568.  Daniel Forgues, Jul 08 2015
Looking at the plot for A088568, it seems that although the asymptotic densities of 1s and 2s appear to be 1/2, there might be a bias in favor of the 2s. I.e., D(1) = 1/2  O(log(n)/n), D(2) = 1/2 + O(log(n)/n).  Daniel Forgues, Jul 11 2015
(a(n)) is the unique fixed point of the 2block substitution beta
11 > 12
12 > 122
21 > 112
22 > 1122.
A 2block substitution beta maps a word w(1)...w(2n) to the word
beta(w(1)w(2))...beta(w(2n1)w(2n)).
If the word has odd length, then the last letter is ignored.
It was noted by me in 1979 in the Bordeaux seminar on number theory that (a(n+1)) is fixed point of the 2block substitution 11 > 21, 12 > 211, 21 > 221, 22 > 2211. (End)
Named after the American artist and recreational mathematician William George Kolakoski (19441997).  Amiram Eldar, Jun 17 2021


REFERENCES

JeanPaul Allouche and Jeffrey Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 337.
Éric 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 (Waterloo, ON, 1995), 115125, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 489, Kluwer Acad. Publ., Dordrecht, 1997. Math. Rev. 98g:11022.
Michael S. Keane, Ergodic theory and subshifts of finite type, Chap. 2 of T. Bedford et al., eds., Ergodic Theory, Symbolic Dynamics and Hyperbolic Spaces, Oxford, 1991, esp. p. 50.
J. C. Lagarias, Number Theory and Dynamical Systems, pp. 3572 of S. A. Burr, ed., The Unreasonable Effectiveness of Number Theory, Proc. Sympos. Appl. Math., 46 (1992). Amer. Math. Soc.
Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence  see "List of Sequences" in Vol. 2.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Ilan Vardi, Computational Recreations in Mathematica. AddisonWesley, Redwood City, CA, 1991, p. 233.


LINKS

Jörg Endrullis, Dimitri Hendriks and Jan Willem Klop, Degrees of Streams, Integers, Vol. 11B (2011), A6.
William Kolakoski,Problem 5304, Amer. Math. Monthly, Vol. 72, No. 8 (1965), p. 674; Self Generating Runs, Solution to Problem 5304 by Necdet Üçoluk, Vol. 73, No. 6 (1966), pp. 681682.
Gheorghe Păun and Arto Salomaa, Selfreading sequences, Amer. Math. Monthly, Vol. 103, No. 2 (1996), pp. 166168.
N. J. A. Sloane, Coordination Sequences, Planing Numbers, and Other Recent Sequences (II), Experimental Mathematics Seminar, Rutgers University, Jan 31 2019, Part I, Part 2, Slides. (Mentions this sequence)


FORMULA

These two formulas define completely the sequence: a(1)=1, a(2)=2, a(a(1) + a(2) + ... + a(k)) = (3 + (1)^k)/2 and a(a(1) + a(2) + ... + a(k) + 1) = (3  (1)^k)/2.  Benoit Cloitre, Oct 06 2003
a(n+2)*a(n+1)*a(n)/2 = a(n+2) + a(n+1) + a(n)  3 (this formula doesn't define the sequence, it is just a consequence of the definition).  Benoit Cloitre, Nov 17 2003
a(n+1) = 3  a(n) + (a(n)  a(n1))*(a(b(n))  1), where b(n) is the sequence A156253.  JeanMarc Fedou and Gabriele Fici, Mar 18 2010


EXAMPLE

Start with a(1) = 1. By definition of the sequence, this says that the first run has length 1, so it must be a single 1, and a(2) = 2. Thus, the second run (which starts with this 2) must have length 2, so the third term must be also be a(3) = 2, and the fourth term can't be a 2, so must be a(4) = 1. Since a(3) = 2, the third run must have length 2, so we deduce a(5) = 1, a(6) = 2, and so on. The correction I made was to change a(4) to a(5) and a(5) to a(6).  Labos Elemer, corrected by Graeme McRae


MAPLE

M := 100; s := [ 1, 2, 2 ]; for n from 3 to M do for i from 1 to s[ n ] do s := [ op(s), 1+((n1)mod 2) ]; od: od: s; A000002 := n>s[n];
# alternative implementation based on the Cloitre formula:
local ksu, k ;
option remember;
if n = 1 then
1;
elif n <=3 then
2;
else
for k from 1 do
ksu := add(procname(i), i=1..k) ;
if n = ksu then
return (3+(1)^k)/2 ;
elif n = ksu+ 1 then
return (3(1)^k)/2 ;
end if;
end do:
end if;


MATHEMATICA

a[steps_] := Module[{a = {1, 2, 2}}, Do[a = Append[a, 1 + Mod[(n  1), 2]], {n, 3, steps}, {i, a[[n]]}]; a]
a[ n_] := If[ n < 3, Max[ 0, n], Module[ {an = {1, 2, 2}, m = 3}, While[ Length[ an] < n, an = Join[ an, Table[ Mod[m, 2, 1], { an[[ m]]} ]]; m++]; an[[n]]]] (* Michael Somos, Jul 11 2011 *)
n=8; Prepend[ Nest[ Flatten[ Partition[#, 2] /. {{2, 2} > {2, 2, 1, 1}, {2, 1} > {2, 2, 1}, {1, 2} > {2, 1, 1}, {1, 1} > {2, 1}}] &, {2, 2}, n], 1] (* Birkas Gyorgy, Jul 10 2012 *)


PROG

(PARI) my(a=[1, 2, 2]); for(n=3, 80, for(i=1, a[n], a=concat(a, 2n%2))); a
(PARI) {a(n) = local(an=[1, 2, 2], m=3); if( n<1, 0, while( #an < n, an = concat( an, vector(an[m], i, 2m%2)); m++); an[n])};
(Haskell) a = 1:2: drop 2 (concat . zipWith replicate a . cycle $ [1, 2])  John Tromp, Apr 09 2011
(Python)
# For explanation see link.
def Kolakoski():
x = y = 1
while True:
yield [2, 1][x&1]
f = y &~ (y+1)
x ^= f
y = (y+1)  (f & (x>>1))
K = Kolakoski()


CROSSREFS

Cf. A001083, A006928, A042942, A069864, A010060, A078929, A171899, A054353 (partial sums), A074286, A216345, A294447.
Kolakoskitype sequences using other seeds than (1,2):
A078880 (2,1), A064353 (1,3), A071820 (2,3), A074804 (3,2), A071907 (1,4), A071928 (2,4), A071942 (3,4), A074803 (4,2), A079729 (1,2,3), A079730 (1,2,3,4).
Cf. A088568 Partial sums of [3  2 * a(n)].


KEYWORD

nonn,core,easy,nice


AUTHOR



EXTENSIONS

Minor edits in example and PARI code by M. F. Hasler, May 07 2014


STATUS

approved



