%N Let f(1)=f(2)=1, f(k)=f(k-1)+f(k-2)+ (k (mod n)). Then f(k)=floor(r(n)*F(k))+g(k) where F(k) denotes the k-th Fibonacci number and g(k) a function becoming periodic. Sequence depends on r(n) which is the largest positive root of : a(3n-2)*X^2-a(3n-1)*X+a(3n)=0.
%C Usually a(3n-2)=A001350(n)
%F It seems that limit n-->infinity r(n)=(9+sqrt(5))/2
%e If n=3 f(k)=floor(r(3)*F(k))+g(k) where r(3)=(9-sqrt(5))/4 is the root of 4*X^2-18*X+19=0 and g(k) is the 6-periodic sequence (0,0,-1,-1,0,-1)
%A _Benoit Cloitre_, Apr 20 2003