

A022300


The sequence a of 1's and 2's starting with (1,1,2,1) such that a(n) is the length of the (n+2)nd run of a.


15



1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 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, 2, 1, 2, 2, 1, 1
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OFFSET

1,3


COMMENTS

It appears that various properties and unsolved problems associated with the Kolakoski sequence, A000002, apply also to A022300.


LINKS



EXAMPLE

a(1) =1, so the 3rd run has length 1, so a(5) must be 2.
a(2) = 1, so the 4th run has length 1, so a(6) = 1.
a(3) = 2, so the 5th run has length 2, so a(7) = 1 and a(8) = 2.
a(4) = 1, so the 6th run has length 1, so a(9) = 1.
Globally, the runlength sequence of a is 2,1,1,1,2,1,2,1,1,2,1,1,2,...., and deleting the first two terms leaves a = A022300.


MATHEMATICA

a = {1, 1, 2}; Do[a = Join[a, ConstantArray[If[Last[a] == 1, 2, 1], {a[[n]]}]], {n, 200}]; a


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



