OFFSET
1,5
COMMENTS
Guide to related sequences (with adjustments for initial terms):
1, 1, 1, 1; a(n) = length of (n + 1)st run of a; A270641
1, 1, 1, 2; a(n) = length of (n + 2)nd run of a; A270641
1, 1, 2, 1; a(n) = length of (n + 3)rd run of a; A270641
1, 1, 2, 2; a(n) = length of (n + 2)nd run of a; A270642
1, 2, 1, 1; a(n) = length of (n + 3)rd run of a; A022300
1, 2, 1, 2; a(n) = length of (n + 4)th run of a; A270641
1, 2, 2, 1; a(n) = length of (n + 3)rd run of a; A270643
1, 2, 2, 2; a(n) = length of (n + 2)nd run of a; A270644
2, 1, 1, 1; a(n) = length of (n + 2)nd run of a; A270645
2, 1, 1, 2; a(n) = length of (n + 3)rd run of a; A222300
2, 1, 2, 1; a(n) = length of (n + 4)th run of a; A270641
2, 1, 2, 2; a(n) = length of (n + 3)rd run of a; A000002 (Kolakoski)
2, 2, 1, 1; a(n) = length of (n + 2)nd run of a; A270646
2, 2, 1, 2; a(n) = length of (n + 3)rd run of a; A270647
2, 2, 2, 1; a(n) = length of (n + 2)nd run of a; A270644
2, 2, 2, 2; a(n) = length of (n + 1)st run of a; A270648
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..10000
EXAMPLE
a(1) = 1, so the 2nd run has length 1, so a(5) must be 2 and a(6) = 1.
a(2) = 1, so the 3rd run has length 1, so a(7) = 2.
a(3) = 1, so the 4th run has length 1, so a(8) = 1.
a(4) = 1, so the 5th run has length 1, so a(9) = 2.
a(5) = 2, so the 6th run has length 2, so a(10) = 2 and a(11) = 1.
Globally, the runlength sequence of a is 4,1,1,1,1,2,1,2,1,2,2,1,...., and deleting the first term leaves a = A270641.
MATHEMATICA
a = {1, 1, 1, 1};
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,easy
AUTHOR
Clark Kimberling, Apr 05 2016
STATUS
approved