A079730 Kolakoski variation using (1,2,3,4) starting with 1,2.
1, 2, 2, 3, 3, 4, 4, 4, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 1, 2, 3, 4, 4, 1, 1, 2, 2, 3, 3, 4, 4, 4, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 1, 2, 3, 3, 4, 4, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 1, 1, 1, 1, 2, 3, 4, 1, 1
Offset: 1
Keywords
Examples
Sequence begins: 1,2,2,3,3,4,4,4,1,1,1,2,2,2,2,3,3,3,3, read it as: (1),(2,2),(3,3),(4,4,4),(1,1,1),(2,2,2,2),(3,3,3,3),... then count the terms in parentheses to get: 1,2,2,3,3,4,4,.. which is the same sequence.
Links
- Ivan Neretin, Table of n, a(n) for n = 1..10000
- Ulrich Reitebuch, Henriette-Sophie Lipschütz, and Konrad Polthier, Visualizing the Kolakoski Sequence, Bridges Conf. Proc.; Math., Art, Music, Architecture, Culture (2023) 481-484.
Crossrefs
Cf. A000002.
Programs
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Mathematica
seed = {1, 2, 3, 4}; w = {}; i = 1; Do[ w = Join[w, Array[seed[[Mod[i - 1, Length[seed]] + 1]] &, If[i > Length[w], seed, w][[i]]]]; i++ , {n, 41}]; w
Formula
Partial sum sequence is expected to be asymptotic to 5/2*n.
Extensions
Corrected by Ivan Neretin, Apr 01 2015
Comments