Deducing Inequality from an Equation

First all all, let $f_n=g_n-1$ to make $$g_n=g_{n-1}+g_{n-3} \qquad \text{with} \quad g_0=2,g_1=3,g_2=4$$ The real solution of $r^3=r^2+1$ is given by $$r_1=\frac{1}{3} \left(1+2 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{29}{2}\right)\right)\right)$$ Write $$r^3-r^2-1=(r-r_1)(r^2+a r+b)$$ to get $$a=\frac{2}{3} \left(\cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{29}{2}\right)\right)-1\right) \quad\text{and} \quad b=\frac{3}{1+2 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{29}{2}\right)\right)}$$ So, solving the quadratic $r^2+ar+b=0$, you have the complex roots $r_2$ and $r_3$.

Now, use the conditions to get $(c_1,c_2,c_3)$ and I suppose that this will again involve a cubic equation with only one real root to generate real value of $g_n$ and $f_n$.

It could probably be easier to identify the generating function which is not very complex.

In view of the numbers $f(n) > \big[\sqrt 2\big]^n$