A006928 -id:A006928 - cp's OEIS frontend

A049710 a(n)=3-k(n), where k=A006928; also, a and k have the same runlength sequence, with n-th term k(n-1) for n >= 2.

2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2

Offset: 1

Clark Kimberling

nonn

Essentially the same as A049705. [From R. J. Mathar, Oct 15 2008]

Cf. A000002 (Kolakoski sequence).

a(n) = A049705(n+1) = A088569(n) for n > 1. - Jean-Christophe Hervé, Nov 10 2014

A000002 Kolakoski sequence: a(n) is length of n-th run; a(1) = 1; sequence consists just of 1's and 2's.

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane

Historical note: the sequence might be better called the Oldenburger-Kolakoski 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 Oldenburger-Kolakoski 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 1-10, 11-20, etc., then there can only be 4, 5 or 6 1's in each batch. - Jon Perry, Sep 26 2012
From Jean-Christophe Hervé, Oct 04 2014: (Start)
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 9-tuplet 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
From Michel Dekking, Jan 31 2018: (Start)
(a(n)) is the unique fixed point of the 2-block substitution beta
    11 -> 12
    12 -> 122
    21 -> 112
    22 -> 1122.
A 2-block substitution beta maps a word w(1)...w(2n) to the word
    beta(w(1)w(2))...beta(w(2n-1)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 2-block substitution 11 -> 21, 12 -> 211, 21 -> 221,  22 -> 2211. (End)
Named after the American artist and recreational mathematician William George Kolakoski (1944-1997). - Amiram Eldar, Jun 17 2021

			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_

Jean-Paul Allouche and Jeffrey Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 337.
Éric Angelini, "Jeux de suites", in Dossier Pour La Science, pp. 32-35, 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 long-range aperiodic order (Waterloo, ON, 1995), 115-125, 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. 35-72 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. Addison-Wesley, Redwood City, CA, 1991, p. 233.

N. J. A. Sloane, Table of n, a(n) for n = 1..10502
Jean-Paul Allouche, Michael Baake, Julien Cassaigne and David Damanik, Palindrome complexity, arXiv:math/0106121 [math.CO], 2001; Theoretical Computer Science, Vol. 292 (2003), pp. 9-31.
Michael Baake and Bernd Sing, Kolakoski-(3,1) is a (deformed) model set, arXiv:math/0206098 [math.MG], 2002-2003.
Alex Bellos and Brady Haran, The Kolakoski Sequence, Numberphile video (2017).
Olivier Bordellès and Benoit Cloitre, Bounds for the Kolakoski Sequence, J. Integer Sequences, Vol. 14 (2011), Article 11.2.1.
Richard P. Brent, Fast algorithms for the Kolakoski sequence, Slides from a talk, 2016.
Éric Brier, Rémi Géraud-Stewart, David Naccache, Alessandro Pacco, and Emanuele Troiani, The Look-and-Say The Biggest Sequence Eventually Cycles, arXiv:2006.07246 [math.DS], 2020.
Arturo Carpi, On repeated factors in C^infinity-words, Information Processing Letters, Vol. 52 (1994), pp. 289-294.
Benoit Cloitre, The Kolakoski transform of words.
Benoit Cloitre, Plot of walk on the first 60000 terms (step of unit length moving right with angle Pi/2 if 2 and left with angle -Pi/2 if 1 starting at (0,0)).
F. M. Dekking, Regularity and irregularity of sequences generated by automata, Seminar on Number Theory, 1979-1980 (Talence, 1979-1980), Exp. No. 9, 10 pp., Univ. Bordeaux I, Talence, 1980.
F. M. Dekking, On the structure of self-generating sequences, Seminar on Number Theory, 1980-1981 (Talence, 1980-1981), Exp. No. 31, 6 pp., Univ. Bordeaux I, Talence, 1981. Math. Rev. 83e:10075. JSTOR
F. M. Dekking, What Is the Long Range Order in the Kolakoski Sequence?, Report 95-100, Technische Universiteit Delft, 1995.
F. M. Dekking, The Thue-Morse Sequence in Base 3/2, J. Int. Seq., Vol. 26 (2023), Article 23.2.3.
Jörg Endrullis, Dimitri Hendriks and Jan Willem Klop, Degrees of Streams, Integers, Vol. 11B (2011), A6.
David Eppstein, The Kolakoski tree and The Kolakoski sequence via bit tricks instead of recursion.
Jean-Marc Fédou and Gabriele Fici, Some remarks on differentiable sequences and recursivity, Journal of Integer Sequences, Vol. 13 (2010), Article 10.3.2.
Abdallah Hammam, Some new Formulas for the Kolakoski Sequence A000002, Turkish Journal of Analysis and Number Theory, Vol. 4, No. 3 (2016), pp. 54-59.
Mari Huova and Juhani Karhumäki, On Unavoidability of k-abelian Squares in Pure Morphic Words, Journal of Integer Sequences, Vol. 16 (2013), #13.2.9.
Clark Kimberling, Integer Sequences and Arrays, Illustration of the Kolakoski sequence.
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. 681-682.
Leonid V. Kovalev, Kolakoski sequence II.
Elizabeth J. Kupin and Eric S. Rowland, Bounds on the frequency of 1 in the Kolakoski word, arXiv:0809.2776 [math.CO], Sep 16, 2008.
Rabie A. Mahmoud, Hardware Implementation of Binary Kolakoski Sequence, Research Gate, 2015.
Kerry Mitchell, Spirolateral-Type Images from Integer Sequences, 2013.
Johan Nilsson, A Space Efficient Algorithm for the Calculation of the Digit Distribution in the Kolakoski Sequence, J. Int. Seq., Vol. 15 (2012), Article 12.6.7; arXiv preprint, arXiv:1110.4228 [math.CO], Oct 19, 2011.
J. Nilsson, Letter Frequencies in the Kolakoski Sequence, Acta Physica Polonica A, Vol. 126 (2014), pp. 549-552.
Rufus Oldenburger, Exponent trajectories in symbolic dynamics, Trans. Amer. Math. Soc., Vol. 46 (1939), pp. 453-466.
Matthew Parker, The first 50K terms (7-Zip compressed file).
Gheorghe Păun and Arto Salomaa, Self-reading sequences, Amer. Math. Monthly, Vol. 103, No. 2 (1996), pp. 166-168.
Michael Rao, Trucs et bidules sur la séquence de Kolakoski, 2012, in French.
A. Scolnicov, Kolakoski sequence, PlanetMath.org.
Bobby Shen, A uniformness conjecture of the Kolakoski sequence, graph connectivity, and correlations, arXiv:1703.00180 [math.CO], 2017.
Bernd Sing, More Kolakoski Sequences, INTEGERS, Vol. 11B (2011), A14.
N. J. A. Sloane, Handwritten notes on Self-Generating Sequences, 1970. (note that A1148 has now become A005282)
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)
Bertran Steinsky, A Recursive Formula for the Kolakoski Sequence A000002, J. Integer Sequences, Vol. 9 (2006), Article 06.3.7.
Eric Weisstein's World of Mathematics, Kolakoski Sequence
Wikipedia, Kolakoski sequence.
Gus Wiseman, Kolakoski fractal animation for n=40000.
Ed Wynn, Program to generate A000002 in Shakespeare Programming Language.
Index entries for "core" sequences

Cf. A001083, A006928, A042942, A069864, A010060, A078929, A171899, A054353 (partial sums), A074286, A216345, A294447.
Cf. A054354, bisections: A100428, A100429.
Cf. A013947, A156077, A234322 (positions, running total and percentage of 1's).
Cf. A118270.
Cf. A049705, A088569 (are either subsequences of A000002? - Jon Perry, Oct 30 2014)
Kolakoski-type 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).
Other self-describing: A001462 (Golomb sequence, see also references therein), A005041, A100144.
Cf. A088568 (partial sums of [3 - 2 * a(n)]).

a = 1:2: drop 2 (concat . zipWith replicate a . cycle $ [1,2]) -- John Tromp, Apr 09 2011

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+((n-1)mod 2) ]; od: od: s; A000002 := n->s[n];
# alternative implementation based on the Cloitre formula:
A000002 := proc(n)
    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;
end proc: # R. J. Mathar, Nov 15 2014

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 *)
KolakoskiSeq[n_Integer] := Block[{a = {1, 2, 2}}, Fold[Join[#1, ConstantArray[Mod[#2, 2, 1], #1[[#2]]]] &, a, Range[3, n]]]; KolakoskiSeq[999] (* Mikk Heidemaa, Nov 01 2024 *) (* Corrected by Giorgos Kalogeropoulos, May 09 2025 *)

my(a=[1,2,2]); for(n=3,80, for(i=1,a[n],a=concat(a,2-n%2))); a

{a(n) = local(an=[1, 2, 2], m=3); if( n<1, 0, while( #an < n, an = concat( an, vector(an[m], i, 2-m%2)); m++); an[n])};

# 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()
print([next(K) for  in range(100)]) # _David Eppstein, Oct 15 2016

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(n-1))*(a(b(n)) - 1), where b(n) is the sequence A156253. - Jean-Marc Fedou and Gabriele Fici, Mar 18 2010
a(n) = (3 + (-1)^A156253(n))/2. - Benoit Cloitre, Sep 17 2013
Conjectures from Boštjan Gec, Oct 07 2024: (Start)
a(n)*(a(n-1) + a(n-2) - 3) + a(n-1)*a(n-2) + 7 = 3*a(n-1) + 3*a(n-2).
a(n)*(a(n-1) + a(n-2) - 3) = a(n-3)*(a(n-1) + a(n-2) - 3). (End)
Comment from Kevin Ryde, Oct 07 2024: The above formulas are true: The parts identify when terms are same or different and they hold for any sequence of 1's and 2's with run lengths 1 or 2.

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

A022300 The sequence a of 1's and 2's starting with (1,1,2,1) such that a(n) is the length of the (n+2)nd run of a.

Original entry on oeis.org

1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 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, 2, 1, 2, 2, 1, 1

Offset: 1

Clark Kimberling

nonn

It appears that various properties and unsolved problems associated with the Kolakoski sequence, A000002, apply also to A022300.

			a(1) =1, so the 3rd run has length 1, so a(5) must be 2.
a(2) = 1, so the 4th run has length 1, so a(6) = 1.
a(3) = 2, so the 5th run has length 2, so a(7) = 1 and a(8) = 2.
a(4) = 1, so the 6th run has length 1, so a(9) = 1.
Globally, the runlength sequence of a is 2,1,1,1,2,1,2,1,1,2,1,1,2,...., and deleting the first two terms leaves a = A022300.

Clark Kimberling, Table of n, a(n) for n = 1..20000

Cf. A022303, A006928, A000002.

a = {1, 1, 2}; Do[a = Join[a, ConstantArray[If[Last[a] == 1, 2, 1], {a[[n]]}]], {n, 200}]; a
(* Peter J. C. Moses, Apr 01 2016 *)

Clarified and augmented by Clark Kimberling, Apr 02 2016

A022303 The sequence a of 1's and 2's starting with (1,2,1) such that a(n) is the length of the (n+2)nd run of a.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling

nonn

			a(1) =1, so the 3rd run has length 1, so a(4) must be 2.
a(2) = 2, so the 4th run has length 2, so a(5) = 2 and a(6) = 1.
a(3) = 1, so the 5th run has length 1, so a(7) = 2.
a(4) = 2, so the 6th run has length 2, so a(8) = 1 and a(9) = 1.
Globally, the runlength sequence of a is 1,1,1,2,1,2,2,1,2,2,1,1,2,1,...., and deleting the first two terms leaves a = A022303.

Clark Kimberling, Table of n, a(n) for n = 1..20000

Cf. A022300, A006928, A000002.

a = {1, 2}; Do[a = Join[a, ConstantArray[If[Last[a] == 1, 2, 1], {a[[n]]}]], {n, 100}]; a (* Peter J. C. Moses, Apr 02 2016 *)

Clarified and augmented by Clark Kimberling, Apr 02 2016

A270641 The sequence a of 1's and 2's starting with (1,1,1,1) such that a(n) is the length of the (n+1)st run of a.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 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, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2

Offset: 1

Clark Kimberling, Apr 05 2016

Guide to related sequences (with adjustments for initial terms):
1, 1, 1, 1;    a(n) = length of (n + 1)st run of a; A270641
1, 1, 1, 2;    a(n) = length of (n + 2)nd run of a; A270641
1, 1, 2, 1;    a(n) = length of (n + 3)rd run of a; A270641
1, 1, 2, 2;    a(n) = length of (n + 2)nd run of a; A270642
1, 2, 1, 1;    a(n) = length of (n + 3)rd run of a; A022300
1, 2, 1, 2;    a(n) = length of (n + 4)th run of a; A270641
1, 2, 2, 1;    a(n) = length of (n + 3)rd run of a; A270643
1, 2, 2, 2;    a(n) = length of (n + 2)nd run of a; A270644
2, 1, 1, 1;    a(n) = length of (n + 2)nd run of a; A270645
2, 1, 1, 2;    a(n) = length of (n + 3)rd run of a; A222300
2, 1, 2, 1;    a(n) = length of (n + 4)th run of a; A270641
2, 1, 2, 2;    a(n) = length of (n + 3)rd run of a; A000002 (Kolakoski)
2, 2, 1, 1;    a(n) = length of (n + 2)nd run of a; A270646
2, 2, 1, 2;    a(n) = length of (n + 3)rd run of a; A270647
2, 2, 2, 1;    a(n) = length of (n + 2)nd run of a; A270644
2, 2, 2, 2;    a(n) = length of (n + 1)st run of a; A270648

			a(1) = 1, so the 2nd run has length 1, so a(5) must be 2 and a(6) = 1.
a(2) = 1, so the 3rd run has length 1, so a(7) = 2.
a(3) = 1, so the 4th run has length 1, so a(8) = 1.
a(4) = 1, so the 5th run has length 1, so a(9) = 2.
a(5) = 2, so the 6th run has length 2, so a(10) = 2 and a(11) = 1.
Globally, the runlength sequence of a is 4,1,1,1,1,2,1,2,1,2,2,1,...., and deleting the first term leaves a = A270641.

Clark Kimberling, Table of n, a(n) for n = 1..10000

Cf. A000002, A006928, A022300,

a = {1, 1, 1, 1};
Do[a = Join[a, ConstantArray[If[Last[a] == 1, 2, 1], {a[[n]]}]], {n,   200}]; a
(* Peter J. C. Moses, Apr 01 2016 *)

A111081 Successive generations of an alternating Kolakoski rule.

Original entry on oeis.org

1, 2, 11, 21, 221, 22112, 11221211, 21221121121, 2212211212212112, 1122122112122122112112122, 12112212211212212211211221211212212211, 211212211211221211211221221121221211221221211211221221121

Offset: 1

Benoit Cloitre, Oct 11 2005

Strings are obtained using the Kolakoski substitution and the additional rule : start with 1 if previous string ends with 2, start with 2 if previous string ends with 1. The concatenation of those strings gives 1211212212211211221211...which is A006928 word. If you replace the initial 1 with 12 you get 122112122122112112212...the infinite Kolakoski word A000002.

			1-->2-->11-->21-->221-->22112-->11221211

Cf. A000002, A054349.

Conjecture : length of n-th string is asymptotic to c*(3/2)^n for some c.

A270642 The sequence a of 1's and 2's starting with (1,1,2,2) such that a(n) is the length of the (n+2)nd run of a.

Original entry on oeis.org

1, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 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

Offset: 1

Clark Kimberling, Apr 06 2016

See A270641 for a guide to related sequences.

			a(1) = 1, so the 3rd run has length 1, so a(5) must be 1.
a(2) = 1, so the 4th run has length 1, so a(6) = 2 and a(7) = 1.
a(3) = 2, so the 5th run has length 2, so a(8) = 1 and a(9) = 2.
a(4) = 2, so the 6th run has length 2, so a(10) = 2 and a(11) = 1.
Globally, the runlength sequence is 2,2,1,1,2,2,1,2,1,1,2,2,1,2,..., and deleting the first 2 terms leaves the same sequence.

Clark Kimberling, Table of n, a(n) for n = 1..10000

Cf. A000002, A006928, A022300, A270641.

a = {1, 1, 2, 2}; Do[a = Join[a, ConstantArray[If[Last[a] == 1, 2, 1], {a[[n]]}]], {n,   200}]; a  (* Peter J. C. Moses, Apr 01 2016 *)

A270643 The sequence a of 1's and 2's starting with (1,2,2,1) such that a(n) is the length of the (n+3)rd run of a.

Original entry on oeis.org

1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 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

Offset: 1

Clark Kimberling, Apr 06 2016

See A270641 for a guide to related sequences.

			a(1) = 1, so the 4th run has length 1, so a(5) must be 1 and a(6) must be 2.
a(2) = 2, so the 5th run has length 2, so a(7) = 1 and a(8) = 2.
a(3) = 2, so the 6th run has length 2, so a(9) = 2 and a(10) = 1.
Globally, the runlength sequence is 1,2,1,1,2,2,1,2,1,1,2,2,1,2,2,1,...., and deleting the first 3 terms leaves the same sequence.

Clark Kimberling, Table of n, a(n) for n = 1..10000

Cf. A000002, A006928, A022300, A270641.

a = {1, 2, 2, 1}; Do[a = Join[a, ConstantArray[If[Last[a] == 1, 2, 1], {a[[n]]}]], {n,   200}]; a  (* Peter J. C. Moses, Apr 01 2016 *)

Conjecture: a(n) = A270642(n+1). - R. J. Mathar, Jun 21 2025

A270644 The sequence a of 1's and 2's starting with (1,2,2,2) such that a(n) is the length of the (n+2)nd run of a.

Original entry on oeis.org

1, 2, 2, 2, 1, 2, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 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, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1

Offset: 1

Clark Kimberling, Apr 06 2016

See A270641 for a guide to related sequences.

			a(1) = 1, so the 3rd run has length 1, so a(5) must be 1 and a(6) = 2.
a(2) = 2, so the 4th run has length 2, so a(7) = 2 and a(8) = 1.
a(3) = 2, so the 5th run has length 2, so a(9) = 1and a(10) = 2.
Globally, the runlength sequence is 1,3,1,2,2,2,1,2,2,1,1,2,2,1,2,2,..., and deleting the first 2 terms leaves the same sequence.

Clark Kimberling, Table of n, a(n) for n = 1..10000

Cf. A000002, A006928, A022300, A270641.

a = {1, 2, 2, 2}; Do[a = Join[a, ConstantArray[If[Last[a] == 1, 2, 1], {a[[n]]}]], {n,   200}]; a  (* Peter J. C. Moses, Apr 01 2016 *)

A270645 The sequence a of 1's and 2's starting with (2,1,1,1) such that a(n) is the length of the (n+2)nd run of a.

Original entry on oeis.org

2, 1, 1, 1, 2, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 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, 1, 2, 1, 2, 2

Offset: 1

Clark Kimberling, Apr 06 2016

See A270641 for a guide to related sequences.

			a(1) = 1, so the 3rd run has length 2, so a(5) must be 2 and a(6) = 2.
a(2) = 2, so the 4th run has length 1, so a(7) = 1 and a(8) = 2.
a(3) = 1, so the 5th run has length 1, so a(9) = 1 and a(10) = 2.
Globally, the runlength sequence is 1,3,2,1,1,1,2,2,1,2,1,2,2,1,1,..., and deleting the first 2 terms leaves the same sequence.

Clark Kimberling, Table of n, a(n) for n = 1..10000

Cf. A000002, A006928, A022300, A270641.

a = {2,1,1,1}; Do[a = Join[a, ConstantArray[If[Last[a] == 1, 2, 1], {a[[n]]}]], {n,   200}]; a  (* Peter J. C. Moses, Apr 01 2016 *)

cp's OEIS Frontend

A049710 a(n)=3-k(n), where k=A006928; also, a and k have the same runlength sequence, with n-th term k(n-1) for n >= 2.

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A000002 Kolakoski sequence: a(n) is length of n-th run; a(1) = 1; sequence consists just of 1's and 2's.

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A022300 The sequence a of 1's and 2's starting with (1,1,2,1) such that a(n) is the length of the (n+2)nd run of a.

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A022303 The sequence a of 1's and 2's starting with (1,2,1) such that a(n) is the length of the (n+2)nd run of a.

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A270641 The sequence a of 1's and 2's starting with (1,1,1,1) such that a(n) is the length of the (n+1)st run of a.

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A111081 Successive generations of an alternating Kolakoski rule.

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A270642 The sequence a of 1's and 2's starting with (1,1,2,2) such that a(n) is the length of the (n+2)nd run of a.

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A270643 The sequence a of 1's and 2's starting with (1,2,2,1) such that a(n) is the length of the (n+3)rd run of a.

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