cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

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

Views

Author

Keywords

Comments

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

Examples

			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_
		

References

  • 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.

Crossrefs

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)]).

Programs

  • Haskell
    a = 1:2: drop 2 (concat . zipWith replicate a . cycle $ [1,2]) -- John Tromp, Apr 09 2011
    
  • 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+((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
  • 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 *)
    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 *)
  • PARI
    my(a=[1,2,2]); for(n=3,80, for(i=1,a[n],a=concat(a,2-n%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, 2-m%2)); m++); an[n])};
    
  • 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()
    print([next(K) for  in range(100)]) # _David Eppstein, Oct 15 2016

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(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.

Extensions

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

A066983 a(n+2) = a(n+1) + a(n) + (-1)^n, with a(1) = a(2) = 1.

Original entry on oeis.org

1, 1, 1, 3, 3, 7, 9, 17, 25, 43, 67, 111, 177, 289, 465, 755, 1219, 1975, 3193, 5169, 8361, 13531, 21891, 35423, 57313, 92737, 150049, 242787, 392835, 635623, 1028457, 1664081, 2692537, 4356619, 7049155, 11405775, 18454929, 29860705, 48315633, 78176339
Offset: 1

Views

Author

Benoit Cloitre, Jan 27 2002

Keywords

Comments

Length of strings given by a successive substitution of a "modified" Kolakoski-(3, 1) sequence. Starting with 1, using the rule "string begins with 1 if previous string ends with 3, string begins with 3 if previous string ends with 1" then applying the classical Kolakoski-(3,1) rule. This gives: 1 -> 3 -> 111 -> 313 -> 1113111 -> 313111313 -> 11131113131113111 and the length of string are 1, 1, 3, 3, 7, 9, 17, ... At step n, length = a(n+1). This substitution leads to two sequences: 1, 1, 1, 3, 1, 1, 1, 3, 1, 3, 1, 1, 1, ... and 3, 1, 3, 1, 1, 1, 3, 1, 3, 1, 1, 1, 3, ... - Benoit Cloitre, Jun 01 2004
Lengths of comparators in subsequent layers of correction network F_n. - Grzegorz Stachowiak (gst(AT)ii.uni.wroc.pl), Nov 28 2004
Convolution of F(n+1) and A105812(n). Action of inverse of sequence array for F(n-1)*(-1)^n on F(n+1). - Paul Barry, Oct 29 2006

References

  • Omur Deveci, The Pell-Padovan sequences and the Jacobsthal-Padovan sequences in finite groups, Utilitas Mathematica, 98 (2015), 257-270.

Crossrefs

Programs

  • GAP
    a:=[1,1];; for n in [3..40] do a[n]:=a[n-1]+a[n-2]+(-1)^n; od; a; # Muniru A Asiru, Aug 09 2018
    
  • Magma
    [n le 2 select 1 else Self(n-1)+Self(n-2)+(-1)^n: n in [1..50]]; // Vincenzo Librandi, Aug 13 2018
    
  • Maple
    seq(coeff(series(x*(1+x-x^2)/((1+x)*(1-x-x^2)), x,n+1),x,n),n=1..40); # Muniru A Asiru, Aug 09 2018
  • Mathematica
    Table[ Floor[ GoldenRatio^(k-1) ] - Floor[ GoldenRatio^(k-1) / Sqrt[5] ], {k, 1, 100} ]  (* Federico Provvedi, Mar 26 2013 *)
    LinearRecurrence[{0, 2, 1}, {1, 1, 1}, 40] (* Vincenzo Librandi, Aug 13 2018 *)
  • PARI
    { for (n=1, 250, if (n>2, a=a1 + a2 + (-1)^n; a2=a1; a1=a, a=a1=1; a=a2=1); write("b066983.txt", n, " ", a) ) } \\ Harry J. Smith, Apr 15 2010
    
  • PARI
    vector(40, n, 2*fibonacci(n-2) + (-1)^n) \\ G. C. Greubel, Dec 26 2019
    
  • Python
    from sympy import fibonacci
    def A066983(n): return (fibonacci(n-2)<<1)+(-1 if n&1 else 1) # Chai Wah Wu, May 05 2025
  • Sage
    [2*fibonacci(n-2) + (-1)^n for n in (1..40)] # G. C. Greubel, Dec 26 2019
    

Formula

For n > 4, a(n-2) = floor(2 * phi^n/sqrt(5)) + (1 + (-1)^n)/2.
a(n) = 2 * Fibonacci(n-2) + (-1)^n. - Vladeta Jovovic, Mar 19 2003
G.f.: x*(1+x-x^2)/((1+x)*(1-x-x^2)). - Paul Barry, Oct 29 2006
a(n) = A066629(n-2) - A066629(n-3), n > 2. - R. J. Mathar, Jan 14 2009
a(n) = floor(phi^(n-1)) - floor(phi^(n-1)/sqrt(5)). - Federico Provvedi, Mar 26 2013
a(1) = a(2) = a(3) = 1; for n > 3, a(n) = 2*a(n-2) + a(n-3). - Taras Goy, Aug 03 2018
a(n) = (-1)^n + (-1 - 3/sqrt(5))*((1/2)*(1 - sqrt(5)))^n + (-1 + 3/sqrt(5))*((1/2)*(1 + sqrt(5)))^n. - Stefano Spezia, Jul 22 2019

Extensions

Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

A042942 From substitutional generation of Kolakoski sequence (A000002).

Original entry on oeis.org

1, 2, 4, 6, 9, 14, 22, 33, 49, 74, 112, 169, 254, 381, 573, 862, 1292, 1936, 2902, 4352, 6525, 9788, 14687, 22028, 33050, 49576, 74378, 111579, 167387, 251089, 376630, 564931, 847375, 1271058, 1906627, 2859983, 4289952, 6434942, 9652396
Offset: 1

Views

Author

Keywords

Comments

Generate A000002 via 2 -> 22 -> 2211 -> 221121 -> 221121221 -> ...; sequence gives lengths of successive strings.
a(n) appears to be asymptotic to c*(3/2)^n where c=1.3094... - Benoit Cloitre, Dec 18 2002
A more accurate estimate is c=1.309346948, probably correct to one unit in the last place. - Richard P. Brent, Dec 30 2016

Crossrefs

Formula

a(n) = A001083(n-2) - 1. - Andrey Zabolotskiy, Jan 10 2022

A111123 Number of 2's in n-th "Kolakoski" string defined in A054349.

Original entry on oeis.org

1, 2, 2, 3, 5, 8, 11, 16, 25, 38, 57, 85, 127, 192, 289, 430, 644, 966, 1450, 2173, 3263, 4899, 7341, 11022, 16526, 24802, 37201, 55808, 83702, 125541, 188301, 282444, 423683, 635569, 953356, 1429969, 2144990, 3217454, 4826176, 7239129, 10858479, 16287972, 24431890
Offset: 0

Views

Author

Benoit Cloitre, Oct 16 2005

Keywords

Comments

Also the number of terms in n-th string (starting from n=3) when representing A000002 as a tree. Each branch of this tree is a string. Starting from n=3, each 1 in n-th string generates either 1 or 2 in (n+1)-th string and each 2 in n-th string generates either 11 or 22 in (n+1)-th string based on the previously generated term of either 2 or 1. Hence, the number of terms in (n+1)-th string is the sum of all terms in n-th string. - Rakesh Khanna A, May 24 2020

Crossrefs

Cf. A001083, A042942, A054349, A111124 (number of 1's).

Programs

  • Mathematica
    l = { (*terms in A042942*) }; For[i = 2, i <= Length[l], i++, Print[l[[i]] - l[[i - 1]]]]

Formula

a(0) + a(1) + ... + a(n) = A042942(n+2) - 1.
a(n) = A001083(n+4) - A001083(n+3). - Benoit Cloitre, Nov 07 2010

Extensions

More terms from and offset changed to 0 by Jinyuan Wang, Apr 03 2020

A111124 Number of 1's in n-th "Kolakoski" string defined in A054349.

Original entry on oeis.org

0, 0, 2, 3, 4, 6, 11, 17, 24, 36, 55, 84, 127, 189, 284, 432, 648, 970, 1452, 2179, 3262, 4889, 7346, 11006, 16524, 24774, 37177, 55771, 83685, 125548, 188329, 282487, 423692, 635489, 953271, 1430014, 2144962, 3217488, 4826220, 7239443, 10859222, 16288208, 24432262
Offset: 0

Views

Author

Benoit Cloitre, Oct 16 2005

Keywords

Crossrefs

Cf. A001083, A042942, A054349, A111123 (number of 2's).

Extensions

More terms from and offset changed to 0 by Jinyuan Wang, Apr 03 2020
Showing 1-5 of 5 results.