A088568 -id:A088568 - cp's OEIS frontend

A288605 Position of first appearance of each integer in A088568 (number of 1's minus number of 2's in first n terms of A000002).

0, 1, 3, 12, 32, 93, 257, 378, 471, 798, 825, 858, 1127, 1398, 1497, 1524, 1533, 6352, 6969, 7176, 7269, 7566, 7971, 20338, 20371, 21982, 22009, 25638, 25665, 25692, 27969, 39184, 39211, 42398, 43129, 43150, 48637, 48730, 48757, 49014, 49041, 49068, 49095, 49864

Offset: 1

Gus Wiseman, Jun 11 2017

nonn

For the definition of this sequence we assume A088568(0) = 0.

			A088568(12) = -2 is the first appearance of -2 in A088568, so 12 belongs to the sequence.
A088568(32) = 2 is the first appearance of 2, so 32 belongs to the sequence.

Cf. A000002, A010120, A088568, A187831.

Map[First, Values@ #] - 1 &@ PositionIndex@ Prepend[#, 0] &@ MapIndexed[3 First@ #2 - 2 #1 &, Accumulate@ 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}, 25], 1]] (* Michael De Vlieger, Jun 12 2017, after Birkas Gyorgy at A000002 *)

A294449 a(n) = index where A088568 (or equally A294448) first reaches or exceeds n in magnitude.

Original entry on oeis.org

1, 12, 93, 378, 471, 858, 1127, 1398, 1497, 1524, 1533, 20338, 20371, 21982, 22009, 25638, 25665, 25692, 27969, 39184, 39211, 42398, 43129, 43150, 48637, 48730, 48757, 49014, 49041, 49068, 49095, 49864, 50081, 50090, 50387, 50414, 50959, 51344, 52243, 52454

Offset: 1

N. J. A. Sloane, Nov 22 2017

nonn

a(n) = smallest j such that |A088568(j)| >= n.

This is closely related to the function DELTA(n) studied by Brent (2016).

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Richard P. Brent, Fast algorithms for the Kolakoski sequence, Slides from a talk, 2016.
Rémy Sigrist, C++ program

Cf. A000002, A088568, A294448.

Cf. A022327, A025503.

More terms from Rémy Sigrist, Dec 27 2022

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

A054353 Partial sums of Kolakoski sequence A000002.

Original entry on oeis.org

1, 3, 5, 6, 7, 9, 10, 12, 14, 15, 17, 19, 20, 21, 23, 24, 25, 27, 29, 30, 32, 33, 34, 36, 37, 39, 41, 42, 43, 45, 46, 47, 49, 50, 52, 54, 55, 57, 59, 60, 61, 63, 64, 66, 68, 69, 71, 72, 73, 75, 76, 77, 79, 81, 82, 84, 86, 87, 88, 90, 91, 93, 95, 96, 98, 100

Offset: 1

N. J. A. Sloane, May 07 2000

Alternate definition: n such that A000002(n) is different from A000002(n+1). - Nathaniel Johnston, May 02 2011

Nathaniel Johnston, Table of n, a(n) for n = 1..10000
O. Bordelles and B. Cloitre, Bounds for the Kolakoski Sequence, J. Integer Sequences, 14 (2011), #11.2.1.
Bertran Steinsky, A Recursive Formula for the Kolakoski Sequence A000002, J. Integer Sequences, Vol. 9 (2006), Article 06.3.7.

Cf. A000002, A074272, A074286, A074288, A078649, A156077.

Cf. A088568 (partial sums of [3 - 2*A000002(n)]).

a054353 n = a054353_list !! (n-1)
a054353_list = scanl1 (+) a000002_list
-- Reinhard Zumkeller, Aug 03 2013

a2 = {1, 2, 2}; Do[ a2 = Join[a2, {1+Mod[n-1, 2]}], {n, 3, 50}, {a2[[n]] } ]; Accumulate[a2] (* Jean-François Alcover, Jun 18 2013 *)

from itertools import accumulate
def alst(nn):
  K = Kolakoski() # using Kolakoski() in A000002
  return list(accumulate(next(K) for i in range(1, nn+1)))
print(alst(66))   # Michael S. Branicky, Jan 12 2021

A000002(a(n)) = (3+(-1)^n)/2; A000002(a(n)+1)=(3-(-1)^n)/2. - Benoit Cloitre, Oct 16 2005

a(n) = n + A074286(n) = 2*n - A156077(n) = A156077(n) + 2*A074286(n). - Jean-Christophe Hervé, Oct 05 2014

A054354 First differences of Kolakoski sequence A000002.

Original entry on oeis.org

1, 0, -1, 0, 1, -1, 1, 0, -1, 1, 0, -1, 0, 1, -1, 0, 1, 0, -1, 1, -1, 0, 1, -1, 1, 0, -1, 0, 1, -1, 0, 1, -1, 1, 0, -1, 1, 0, -1, 0, 1, -1, 1, 0, -1, 1, -1, 0, 1, -1, 0, 1, 0, -1, 1, 0, -1, 0, 1, -1, 1, 0, -1, 1, 0, -1, 0, 1, -1, 0, 1, -1, 1, 0, -1, 1, -1, 0, 1, 0, -1, 1, 0, -1, 0, 1, -1, 1, 0, -1, 1, 0, -1

Offset: 1

N. J. A. Sloane, May 07 2000

sign

The Kolakoski sequence has only 1's and 2's, and is cubefree. Thus, for all n>=1, a(n) is in {-1, 0, 1}, a(n+1) != a(n), and if a(n) = 0, a(n+1) = -a(n-1), while if a(n) != 0, either a(n+1) = 0 and a(n+2) = -a(n) or a(n+1) = -a(n). A further consequence is that the maximum gap between equal values is 4: for all n, there is an integer k, 1Jean-Christophe Hervé, Oct 05 2014
From Daniel Forgues, Jul 07 2015: (Start)
Second differences: {-1, -1, 1, 1, -2, 2, -1, -1, 2, -1, -1, 1, 1, ...}
The sequence of first differences bounces between -1 and 1 with a slope whose absolute value is either 1 or 2. We can compress the information in the second differences into {-1, 1, -2, 2, -1, 2, -1, 1, ...} since the -1 and the 1 come in pairs; which can be compressed further into {1, 1, 2, 2, 1, 2, 1, 1, ...} since the signs alternate, where we only need to know that the initial sign is negative. (End)
This appears to divide the positive integers into three sets, each with density approaching 1/3. Note there are no adjacent equal parts (as mentioned above). - Gus Wiseman, Oct 10 2024

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A000002, A054353.
Positions of 0 are A078649.
For Golomb's sequence (A001462) we have A088517.
Positions of -1 are A156242 (descents).
Positions of 1 are A156243 (ascents).
First differences (or second differences of A000002) are A376604.
The Kolakoski sequence (A000002):
- Restrictions: A074264, A100428, A100429, A156263, A156264, A288605.
- Patterns: A013947, A013948, A022292, A074262, A074263.
- Statistics: A074286, A088568, A156077, A156253.
- Transformations: A054354, A156728, A306323, A332273, A332875, A333229.
Cf. A333254.

a054354 n = a054354_list !! (n-1)
a054354_list = zipWith (-) (tail a000002_list) a000002_list
-- Reinhard Zumkeller, Aug 03 2013

a2 = {1, 2, 2}; Do[ a2 = Join[a2, {1 + Mod[n - 1, 2]}], {n, 3, 70}, {a2[[n]]}]; Differences[a2] (* Jean-François Alcover, Jun 18 2013 *)

Abs(a(n)) = (A000002(n)+A000002(n+1)) mod 2. - Benoit Cloitre, Nov 17 2003

A296658 Length of the standard Lyndon word factorization of the first n terms of A000002.

Original entry on oeis.org

1, 1, 1, 2, 3, 2, 3, 2, 2, 3, 2, 2, 3, 4, 3, 4, 5, 4, 3, 4, 3, 4, 5, 4, 5, 3, 3, 4, 5, 4, 5, 6, 5, 6, 4, 4, 5, 4, 4, 5, 6, 5, 6, 4, 4, 5, 4, 5, 6, 5, 6, 7, 6, 4, 5, 4, 4, 5, 6, 5, 6, 4, 4, 5, 4, 4, 5, 6, 5, 6, 7, 6, 7, 5, 5, 6, 5, 6, 7, 6, 5, 6, 5, 5, 6, 7, 6

Offset: 1

Gus Wiseman, Dec 18 2017

nonn

			The standard Lyndon word factorization of (12211212212211211) is (122)(112122122)(112)(1)(1), so a(17) = 5.
The standard factorizations of initial terms of A000002:
1
12
122
122,1
122,1,1
122,112
122,112,1
122,11212
122,112122
122,112122,1
122,11212212
122,112122122
122,112122122,1
122,112122122,1,1
122,112122122,112
122,112122122,112,1
122,112122122,112,1,1
122,112122122,112,112
122,112122122,1121122
122,112122122,1121122,1

Frédérique Bassino, Julien Clement, and Cyril Nicaud, The standard factorization of Lyndon words: an average point of view, Discrete Mathematics, 290-1 (2005), 1-25.

Row lengths of A329315.
The "co-" version is A329362.
Cf. A000002, A001037, A027375, A060223, A088568, A102659, A211100, A281013, A288605, A296372, A296657, A296659, A328596, A329312, A329316, A329317.

LyndonQ[q_]:=Array[OrderedQ[{q,RotateRight[q,#]}]&,Length[q]-1,1,And]&&Array[RotateRight[q,#]&,Length[q],1,UnsameQ];
qit[q_]:=If[#===Length[q],{q},Prepend[qit[Drop[q,#]],Take[q,#]]]&[Max@@Select[Range[Length[q]],LyndonQ[Take[q,#]]&]];
kolagrow[q_]:=If[Length[q]<2,Take[{1,2},Length[q]+1],Append[q,Switch[{q[[Length[Split[q]]]],Part[q,-2],Last[q]},{1,1,1},0,{1,1,2},1,{1,2,1},2,{1,2,2},0,{2,1,1},2,{2,1,2},2,{2,2,1},1,{2,2,2},1]]];
Table[Length[qit[Nest[kolagrow,1,n]]],{n,150}]

A329315 Irregular triangle read by rows where row n gives the sequence of lengths of components of the Lyndon factorization of the first n terms of A000002.

Original entry on oeis.org

1, 2, 3, 3, 1, 3, 1, 1, 3, 3, 3, 3, 1, 3, 5, 3, 6, 3, 6, 1, 3, 8, 3, 9, 3, 9, 1, 3, 9, 1, 1, 3, 9, 3, 3, 9, 3, 1, 3, 9, 3, 1, 1, 3, 9, 3, 3, 3, 9, 7, 3, 9, 7, 1, 3, 9, 9, 3, 9, 9, 1, 3, 9, 9, 1, 1, 3, 9, 9, 3, 3, 9, 9, 3, 1, 3, 9, 14, 3, 9, 15, 3, 9, 15, 1, 3

Offset: 1

Gus Wiseman, Nov 11 2019

There are no repeated rows, as row n has sum n.
We define the Lyndon product of two or more finite sequences to be the lexicographically maximal sequence obtainable by shuffling the sequences together. For example, the Lyndon product of (231) with (213) is (232131), the product of (221) with (213) is (222131), and the product of (122) with (2121) is (2122121). A Lyndon word is a finite sequence that is prime with respect to the Lyndon product. Equivalently, a Lyndon word is a finite sequence that is lexicographically strictly less than all of its cyclic rotations. Every finite sequence has a unique (orderless) factorization into Lyndon words, and if these factors are arranged in lexicographically decreasing order, their concatenation is equal to their Lyndon product. For example, (1001) has sorted Lyndon factorization (001)(1).
It appears that some numbers (such as 4) never appear in the sequence.

			Triangle begins:
   1: (1)
   2: (2)
   3: (3)
   4: (3,1)
   5: (3,1,1)
   6: (3,3)
   7: (3,3,1)
   8: (3,5)
   9: (3,6)
  10: (3,6,1)
  11: (3,8)
  12: (3,9)
  13: (3,9,1)
  14: (3,9,1,1)
  15: (3,9,3)
  16: (3,9,3,1)
  17: (3,9,3,1,1)
  18: (3,9,3,3)
  19: (3,9,7)
  20: (3,9,7,1)
For example, the first 10 terms of A000002 are (1221121221), with Lyndon factorization (122)(112122)(1), so row 10 is (3,6,1).

Row lengths are A296658.
The reversed version is A329316.
Cf. A000002, A000031, A001037, A027375, A059966, A060223, A088568, A102659, A211100, A288605, A296372, A329314, A329317, A329325.

lynQ[q_]:=Array[Union[{q,RotateRight[q,#1]}]=={q,RotateRight[q,#1]}&,Length[q]-1,1,And];
lynfac[q_]:=If[Length[q]==0,{},Function[i,Prepend[lynfac[Drop[q,i]],Take[q,i]]][Last[Select[Range[Length[q]],lynQ[Take[q,#1]]&]]]];
kolagrow[q_]:=If[Length[q]<2,Take[{1,2},Length[q]+1],Append[q,Switch[{q[[Length[Split[q]]]],q[[-2]],Last[q]},{1,1,1},0,{1,1,2},1,{1,2,1},2,{1,2,2},0,{2,1,1},2,{2,1,2},2,{2,2,1},1,{2,2,2},1]]];
kol[n_Integer]:=Nest[kolagrow,{1},n-1];
Table[Length/@lynfac[kol[n]],{n,100}]

A376604 Second differences of the Kolakoski sequence (A000002). First differences of A054354.

Original entry on oeis.org

-1, -1, 1, 1, -2, 2, -1, -1, 2, -1, -1, 1, 1, -2, 1, 1, -1, -1, 2, -2, 1, 1, -2, 2, -1, -1, 1, 1, -2, 1, 1, -2, 2, -1, -1, 2, -1, -1, 1, 1, -2, 2, -1, -1, 2, -2, 1, 1, -2, 1, 1, -1, -1, 2, -1, -1, 1, 1, -2, 2, -1, -1, 2, -1, -1, 1, 1, -2, 1, 1, -2, 2, -1, -1

Offset: 1

Gus Wiseman, Oct 02 2024

sign

Since A000002 has no runs of length 3, this sequence contains no zeros.

The densities appear to approach (1/3, 1/3, 1/6, 1/6).

			The Kolakoski sequence (A000002) is:
  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, ...
with first differences (A054354):
  1, 0, -1, 0, 1, -1, 1, 0, -1, 1, 0, -1, 0, 1, -1, 0, 1, 0, -1, 1, -1, 0, 1, -1, ...
with first differences (A376604):
  -1, -1, 1, 1, -2, 2, -1, -1, 2, -1, -1, 1, 1, -2, 1, 1, -1, -1, 2, -2, 1, 1, -2, ...

A001462 is Golomb's sequence.
A078649 appears to be zeros of the first and third differences.
A288605 gives positions of first appearances of each balance.
A306323 gives a 'broken' version.
A333254 lists run-lengths of differences between consecutive primes.
For the Kolakoski sequence (A000002):
- Restrictions: A074264, A100428, A100429, A156263, A156264.
- Patterns: A013947, A013948, A022292, A074262, A074263, A156242, A156243.
- Statistics: A054353, A074286, A088568, A156077, A156253.
- Transformations: A054354, A156728, A332273, A332875, A333229, A376604.
For second differences: A036263 (prime), A073445 (composite), A376559 (perfect-power), A376562 (non-perfect-power), A376590 (squarefree), A376593 (nonsquarefree), A376596 (prime-power), A376599 (non-prime-power).

kolagrow[q_]:=If[Length[q]<2,Take[{1,2},Length[q]+1],Append[q,Switch[{q[[Length[Split[q]]]],q[[-2]],Last[q]},{1,1,2},1,{1,2,1},2,{2,1,1},2,{2,1,2},2,{2,2,1},1,{2,2,2},1]]]
kol[n_]:=Nest[kolagrow,{1},n-1];
Differences[kol[100],2]

A329317 Length of the Lyndon factorization of the reversed first n terms of A000002.

Original entry on oeis.org

1, 2, 3, 2, 2, 3, 3, 4, 5, 4, 5, 6, 5, 3, 4, 4, 2, 3, 4, 3, 4, 3, 3, 4, 4, 5, 6, 5, 4, 5, 5, 2, 3, 3, 4, 5, 4, 5, 6, 5, 3, 4, 4, 5, 6, 5, 6, 5, 3, 4, 4, 2, 3, 4, 3, 4, 5, 4, 3, 4, 4, 5, 6, 5, 6, 7, 6, 4, 5, 5, 3, 4, 4, 5, 6, 5, 6, 5, 4, 5, 6, 5, 6, 7, 6, 5, 6

Offset: 1

Gus Wiseman, Nov 11 2019

nonn

We define the Lyndon product of two or more finite sequences to be the lexicographically maximal sequence obtainable by shuffling the sequences together. For example, the Lyndon product of (231) with (213) is (232131), the product of (221) with (213) is (222131), and the product of (122) with (2121) is (2122121). A Lyndon word is a finite sequence that is prime with respect to the Lyndon product. Equivalently, a Lyndon word is a finite sequence that is lexicographically strictly less than all of its cyclic rotations. Every finite sequence has a unique (orderless) factorization into Lyndon words, and if these factors are arranged in lexicographically decreasing order, their concatenation is equal to their Lyndon product. For example, (1001) has sorted Lyndon factorization (001)(1).

			The sequence of Lyndon factorizations of the reversed initial terms of A000002 begins:
   1: (1)
   2: (2)(1)
   3: (2)(2)(1)
   4: (122)(1)
   5: (1122)(1)
   6: (2)(1122)(1)
   7: (12)(1122)(1)
   8: (2)(12)(1122)(1)
   9: (2)(2)(12)(1122)(1)
  10: (122)(12)(1122)(1)
  11: (2)(122)(12)(1122)(1)
  12: (2)(2)(122)(12)(1122)(1)
  13: (122)(122)(12)(1122)(1)
  14: (112212212)(1122)(1)
  15: (2)(112212212)(1122)(1)
  16: (12)(112212212)(1122)(1)
  17: (1121122122121122)(1)
  18: (2)(1121122122121122)(1)
  19: (2)(2)(1121122122121122)(1)
  20: (122)(1121122122121122)(1)
For example, the reversed first 13 terms of A000002 are (1221221211221), with Lyndon factorization (122)(122)(12)(1122)(1), so a(13) = 5.

Row-lengths of A329316.
The non-reversed version is A329315.
Cf. A000002, A000031, A001037, A027375, A059966, A060223, A088568, A102659, A211100, A288605, A296372, A296658, A329314, A329325.

lynQ[q_]:=Array[Union[{q,RotateRight[q,#]}]=={q,RotateRight[q,#]}&,Length[q]-1,1,And];
lynfac[q_]:=If[Length[q]==0,{},Function[i,Prepend[lynfac[Drop[q,i]],Take[q,i]]][Last[Select[Range[Length[q]],lynQ[Take[q,#]]&]]]];
kolagrow[q_]:=If[Length[q]<2,Take[{1,2},Length[q]+1],Append[q,Switch[{q[[Length[Split[q]]]],q[[-2]],Last[q]},{1,1,1},0,{1,1,2},1,{1,2,1},2,{1,2,2},0,{2,1,1},2,{2,1,2},2,{2,2,1},1,{2,2,2},1]]]
kol[n_Integer]:=Nest[kolagrow,{1},n-1];
Table[Length[lynfac[Reverse[kol[n]]]],{n,100}]

A329316 Irregular triangle read by rows where row n gives the sequence of lengths of components of the Lyndon factorization of the reversed first n terms of A000002.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 1, 4, 1, 1, 4, 1, 2, 4, 1, 1, 2, 4, 1, 1, 1, 2, 4, 1, 3, 2, 4, 1, 1, 3, 2, 4, 1, 1, 1, 3, 2, 4, 1, 3, 3, 2, 4, 1, 9, 4, 1, 1, 9, 4, 1, 2, 9, 4, 1, 16, 1, 1, 16, 1, 1, 1, 16, 1, 3, 16, 1, 1, 3, 16, 1, 5, 16, 1, 6, 16, 1, 1, 6, 16, 1, 2, 6

Offset: 0

Gus Wiseman, Nov 11 2019

There are no repeated rows, as row n has sum n.
We define the Lyndon product of two or more finite sequences to be the lexicographically maximal sequence obtainable by shuffling the sequences together. For example, the Lyndon product of (231) with (213) is (232131), the product of (221) with (213) is (222131), and the product of (122) with (2121) is (2122121). A Lyndon word is a finite sequence that is prime with respect to the Lyndon product. Equivalently, a Lyndon word is a finite sequence that is lexicographically strictly less than all of its cyclic rotations. Every finite sequence has a unique (orderless) factorization into Lyndon words, and if these factors are arranged in lexicographically decreasing order, their concatenation is equal to their Lyndon product. For example, (1001) has sorted Lyndon factorization (001)(1).
It appears that some numbers (such as 10) never appear in the sequence.

			Triangle begins:
   1: (1)
   2: (1,1)
   3: (1,1,1)
   4: (3,1)
   5: (4,1)
   6: (1,4,1)
   7: (2,4,1)
   8: (1,2,4,1)
   9: (1,1,2,4,1)
  10: (3,2,4,1)
  11: (1,3,2,4,1)
  12: (1,1,3,2,4,1)
  13: (3,3,2,4,1)
  14: (9,4,1)
  15: (1,9,4,1)
  16: (2,9,4,1)
  17: (16,1)
  18: (1,16,1)
  19: (1,1,16,1)
  20: (3,16,1)
For example, the reversed first 13 terms of A000002 are (1221221211221), with Lyndon factorization (122)(122)(12)(1122)(1), so row 13 is (3,3,2,4,1).

Row lengths are A329317.
The non-reversed version is A329315.
Cf. A000002, A000031, A001037, A027375, A059966, A060223, A088568, A102659, A211100, A288605, A296372, A296658, A329314, A329325.

lynQ[q_]:=Array[Union[{q,RotateRight[q,#]}]=={q,RotateRight[q,#]}&,Length[q]-1,1,And];
lynfac[q_]:=If[Length[q]==0,{},Function[i,Prepend[lynfac[Drop[q,i]],Take[q,i]]][Last[Select[Range[Length[q]],lynQ[Take[q,#]]&]]]];
kolagrow[q_]:=If[Length[q]<2,Take[{1,2},Length[q]+1],Append[q,Switch[{q[[Length[Split[q]]]],q[[-2]],Last[q]},{1,1,1},0,{1,1,2},1,{1,2,1},2,{1,2,2},0,{2,1,1},2,{2,1,2},2,{2,2,1},1,{2,2,2},1]]]
kol[n_Integer]:=Nest[kolagrow,{1},n-1];
Table[Length/@lynfac[Reverse[kol[n]]],{n,100}]

cp's OEIS Frontend

A288605 Position of first appearance of each integer in A088568 (number of 1's minus number of 2's in first n terms of A000002).

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A294449 a(n) = index where A088568 (or equally A294448) first reaches or exceeds n in magnitude.

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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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A054353 Partial sums of Kolakoski sequence A000002.

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A054354 First differences of Kolakoski sequence A000002.

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A296658 Length of the standard Lyndon word factorization of the first n terms of A000002.

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A329315 Irregular triangle read by rows where row n gives the sequence of lengths of components of the Lyndon factorization of the first n terms of A000002.

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A376604 Second differences of the Kolakoski sequence (A000002). First differences of A054354.

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