A074292 Dominant digit in successive groups of 3 from the Kolakoski sequence (A000002).

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

Offset: 1

Author

Jon Perry, Sep 21 2002

Keywords

• nonn

Comments

This appears to be the same as a sequence studied by Claude Lenormand in a letter dated Nov 17 2003: break up the Kolakoski sequence (A000002) into runs of identical symbols and omit one symbol from each run.

The sequence studied by Claude Lenormand is A156257 and is not equal to this one: see A248805 = A156257 - A074292. Differences between the two sequences are at n = 47, 48, 56, 57, 128, 129, 137, 139, 147, 148, 176, 177,... - Jean-Christophe Hervé, Oct 11 2014

As in the Kolakoski sequence, runs in this sequence are of length 1 or 2, because a run XX implies the repetition of exactly the same 3-group in the Kolakoski sequence: -YXX-YXX- or -XXY-XXY- or -XYX-XYX-, and this is not possible 3 times. However, words of the form YXYXY appear in this sequence, but don't appear in the Kolakoski sequence. - Jean-Christophe Hervé, Oct 12 2014

Examples

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

Links

• Nathaniel Johnston, Table of n, a(n) for n = 1..10001
• Claude Lenormand, Deux transformations sur les mots, Preprint, 5 pages, Nov 17 2003. Apparently unpublished. This is a scanned copy of the version that the author sent to me in 2003. - _N. J. A. Sloane_, Oct 02 2018

Crossrefs

Cf. A000002, A074293, A074295, A156257, A248805, A318921.

Programs

Maple

A074292 := proc(n)
    A000002(3*n-2)+A000002(3*n-1)+A000002(3*n)-3 ;
end proc:
seq(A074292(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]]}]; If[Count[#, 1] > 1, 1, 2]& /@ Partition[OK, 3]  (* Jean-François Alcover, Nov 13 2014 *)

Formula

a(n)=A000002(3n-2)+A000002(3n-1)+A000002(3n)-3. - Benoit Cloitre, Nov 15 2003

Extensions

More terms from Ray Chandler, Nov 16 2003

Offset corrected by Jean-Christophe Hervé, Oct 11 2014