OFFSET
1,3
COMMENTS
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..20000
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
AUTHOR
EXTENSIONS
Clarified and augmented by Clark Kimberling, Apr 02 2016
STATUS
approved