A054352 Lengths of successive generations of the Kolakoski sequence A000002.

1, 2, 4, 7, 11, 18, 28, 43, 65, 99, 150, 226, 340, 511, 768, 1153, 1728, 2590, 3885, 5826, 8742, 13116, 19674, 29514, 44280, 66431, 99667, 149531, 224306, 336450, 504648, 756961, 1135450, 1703197, 2554846, 3832292, 5748474, 8622646, 12933971, 19400955, 29101203

Offset: 0

Author

N. J. A. Sloane, May 07 2000

Keywords

• nonn
• easy

Comments

Starting with a(0) = 1, the first term of A000002, the n-th generation is the run of figures directly generated from the preceding generation completed with a single last figure which begins the next run. Thus a(0) = 1 -> 1-2 -> 1-22-1 -> 1-2211-2-1 etc. - Jean-Christophe Hervé, Oct 26 2014

It seems that the limit (c =) lim_{n -> oo} a(n)/(3/2)^n exists, with c = 2.63176..., so a(n) ~ (3/2)*a(n-1) ~ c * (3/2)^n, for large n. - A.H.M. Smeets, Apr 12 2024

Links

• Michael S. Branicky, Table of n, a(n) for n = 0..61

Crossrefs

Cf. A054348, A054349, A054350, A054351, A054353, A042942, A000002.

Partial sums of A329758.

Programs

Mathematica

A2 = {1, 2, 2}; Do[If[Mod[n, 10^5] == 0, Print["n = ", n]]; m = 1 + Mod[n - 1, 2]; an = A2[[n]]; A2 = Join[A2, Table[m, {an}]], {n, 3, 10^6}]; A054353 = Accumulate[A2]; Clear[a]; a[0] = 1; a[n_] := a[n] = A054353[[a[n - 1]]] + 1; Table[a[n], {n, 0, 33}] (* Jean-François Alcover, Oct 30 2014, after Jean-Christophe Hervé *)

Python

def aupton(nn):
  alst, A054353, idx = [1], 0, 1
  K = Kolakoski()  # using Kolakoski() in A000002
  for n in range(2, nn+1):
    target = alst[-1]
    while idx <= target:
      A054353 += next(K)
      idx += 1
    alst.append(A054353 + 1)  # a(n) = A054353(a(n-1))+1
  return alst
print(aupton(36))  # Michael S. Branicky, Jan 12 2021

Formula

a(0) = 1, and for n > 0, a(n) = A054353(a(n-1))+1. - Jean-Christophe Hervé, Oct 26 2014

Extensions

a(7)-a(32) from John W. Layman, Aug 20 2002

a(33) from Jean-François Alcover, Oct 30 2014

a(34) and beyond from Michael S. Branicky, Jan 12 2021