A006928 a(n) = length of (n+1)st run, with initial terms 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, 2, 2, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 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, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• nice

Examples

			Start with [ 1,2 ]. a(1)=1, so the second run has length 1, so a(3) must be 1. a(2)=2, so the third run has length 2, so a(4) must also be 1 and a(5) must be 2. a(3)=1, so the 4th run has length 1, so a(6) must be 1; etc. - _Labos Elemer_
		

References

• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Eric Weisstein's World of Mathematics, Kolakoski Sequence.

Crossrefs

Essentially the same as Kolakoski sequence A000002.

Programs

Mathematica

a = {1, 2}; Do[AppendTo[a, 1+Mod[n, 2]], {n, 2, 80}, {i, a[[n]]}]; a (* Jean-François Alcover, Aug 09 2016, adapted from PARI *)

PARI

a=[ 1,2 ]; for(n=2,80, for(i=1,a[ n ],a=concat(a,1+(n%2)))); a

Formula

a(n) = A000002(n+1), n > 1.