A156257 Digit of runs of length 2 in the Kolakoski sequence A000002: a(n) = A000002(A078649(n)).

2, 1, 2, 2, 1, 1, 2, 1, 2, 1, 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, 2, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 2, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2

Offset: 1

Author

Benoit Cloitre, Feb 07 2009

Keywords

• nonn

Comments

Often equal to A074292 (at the beginning), but not always (see comments in A074292). First differences between the two sequences are at n = 47, 48, 56, 57, 128, 129, 137, 139, 147, 148,176, 177,... (see A248345 = A156257 - A074292). - Jean-Christophe Hervé, Oct 11 2014

As in the Kolakoski sequence, runs in this sequence are of length 1 or 2: a run XX in this sequence implies YXXYX in OK for the first X, and this cannot be continued by a single Y (because XYXYX is not possible), thus we have YXXYXXY, which can be continued by YXXYXXYY or by YXXYXXYXYY, but not by YXXYXXYXX (because this would imply an impossible 21212 in OK). However, words of the form YXYXY appear in this sequence, but they don't in A000002. - Jean-Christophe Hervé, Oct 12 2014

Applying Lenormand's "raboter" transformation (see A318921) to A000002 leads to this sequence. - Rémy Sigrist, Nov 11 2020

Examples

			Kolakoski sequence begins (1),(2,2),(1,1),(2),(1),(2,2),(1),(2,2), so this one begins 2,1,2,2.
		

Links

• Jean-Christophe Hervé, Table of n, a(n) for n = 1..5000

Crossrefs

Cf. A000002, A074292, A318921.

Programs

Maple

A156257 := proc(n)
    A000002(A078649(n)) ;
end proc:
seq(A156257(n),n=1..50) ; # R. J. Mathar, Nov 15 2014

Mathematica

OK = {1, 2, 2}; Do[OK = Join[OK, {1+Mod[n-1, 2]}], {n, 3, 1000}, {OK[[n]]}]; Select[Split[OK], Length[#] == 2&][[All, 1]] (* Jean-François Alcover, Nov 13 2014 *)

Formula

a(n) = A000002(A078649(n)) = A000002(A078649(n)+1).

Strictly positive terms of (A000002(n)-1)*(mod(n-1, 2)+1). - Jean-Christophe Hervé, Oct 11 2014

Strictly positive terms of (1-abs(A000002(n+1)-A000002(n)))*A000002(n). - Jean-Christophe Hervé, Oct 11 2014

Extensions

Definition revised by Jean-Christophe Hervé, Oct 11 2014