%I
%S 0,1,1,1,1,1,4,18,19,5,25,31,11,64,89,4,24,31,29,184,236,45,285,319,
%T 76,486,499,121,759,639,199,1230,855,20,120,59,521,3038,916,841,4727,
%U 341,1364,7386,1189,2205,11445,4889
%N Let f(1)=f(2)=1, f(k)=f(k1)+f(k2)+ (k (mod n)). Then f(k)=floor(r(n)*F(k))+g(k) where F(k) denotes the kth Fibonacci number and g(k) a function becoming periodic. Sequence depends on r(n) which is the largest positive root of : a(3n2)*X^2a(3n1)*X+a(3n)=0.
%C Usually a(3n2)=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)=(9sqrt(5))/4 is the root of 4*X^218*X+19=0 and g(k) is the 6periodic sequence (0,0,1,1,0,1)
%K nonn
%O 1,7
%A _Benoit Cloitre_, Apr 20 2003
