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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
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
(* Peter J. C. Moses, Apr 01 2016 *)
CROSSREFS
KEYWORD
nonn
EXTENSIONS
Clarified and augmented by Clark Kimberling, Apr 02 2016
STATUS
approved