Riccati Meets Fibonacci

with α(x) = f(x)/f0(x), β(x) = f1(x)/f0(x), and γ(x) = f2(x)/f0(x). Let G(x) generate the number sequence {Gn}0 , i.e. G(x) = ∑∞ n=0Gnx . Because G(x) is the generating function for the convolution of the sequence {Gn}0 with itself, i.e. of G (1) n := ∑n k=0GkGn−k, one can use eq. 2 in order to express the convolution numbers G n in terms of {Gk} 0 and the numbers {αk}0 , {βk}0 , and γn, which are generated by the functions α(x), β(x), and γ(x), respectively, as follows.