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%I #6 Mar 30 2012 18:39:18
%S 1,2,1,2,1,1,2,1,2,1,2,1,2,1,2,1,1,2,1,2,1,2,1,1,2,1,2,1,1,2,1,2,1,2,
%T 1,2,1,2,1,1,2,1,2,1,2,1,2,1,2,1,2,1,1,2,1,2,1,2,1,2,1,2,1,1,2,1,2,1,
%U 2,1,1,2,1,2,1,1,2,1,2,1,2,1,2,1,2,1,1,2,1,2,1,2,1,2,1,1,2,1,2,1,1,2,1,2,1
%N Start with (1,2) and then concatenate 2^n+1 previous terms, n>=0 and add 2 if a(2^n+1)=1 or add 1 if a(2^n+1)=2.
%e The first 2^2+1 = 5 terms are 1,2,1,2,1. Concatenate those 5 terms gives 1,2,1,2,1,1,2,1,2,1; the last term a(5) is 1 hence we add 2 and 2^3+1 first terms are 1,2,1,2,1,1,2,1,2,1,2
%Y Cf. A083922 (partial sums).
%K nonn
%O 1,2
%A _Benoit Cloitre_, May 09 2003
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