A054353 Partial sums of Kolakoski sequence A000002.

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

Author

N. J. A. Sloane, May 07 2000

Keywords

• nonn
• easy

Comments

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

Links

• 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.

Crossrefs

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

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

Programs

Haskell

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

Mathematica

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 *)

Python

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

Formula

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