A270641 The sequence a of 1's and 2's starting with (1,1,1,1) such that a(n) is the length of the (n+1)st run of a.
1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 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, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2
Offset: 1
Examples
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.
Links
- Clark Kimberling, Table of n, a(n) for n = 1..10000
Programs
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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 *)
Comments