A081420 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.
0, 1, 1, 1, 1, 1, 4, 18, 19, 5, 25, 31, 11, 64, 89, 4, 24, 31, 29, 184, 236, 45, 285, 319, 76, 486, 499, 121, 759, 639, 199, 1230, 855, 20, 120, 59, 521, 3038, 916, 841, 4727, 341, 1364, 7386, 1189, 2205, 11445, 4889
Offset: 1
Keywords
Examples
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)
Formula
It seems that limit n-->infinity r(n)=(9+sqrt(5))/2
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