How should I find $a_n$ knowing that $a_n = a_{n-1} + a_{n-3}$ I tried using a quadratic formula by using the constants of the recursive formula. Then when I get the solutions of the quadratic function, I would insert the $x$ values gotten to $a_n = a_1 \cdot (x_1)^n+a_2 \cdot (x_2)^n$. After, I would get some initial values such as $a_0$ and $a_3$ and make a system to solve. Unfortunately the $a_n$ formula gotten didn’t work.
 A: It is not so bad.
The characteristic equation being
$$r^3-r^2-1=0$$ just follow the steps given here.
Since $\Delta=-31$, only one real root; using  the hyperbolic method, then
$$r_1=\frac{1}{3} \left(1+2 \cosh \left(\frac{1}{3} \cosh
   ^{-1}\left(\frac{29}{2}\right)\right)\right)$$ Deflating the cubic, $r_2$ and $r_3$ are the solutions of the quadratic
$$r^2+(r_1-1) r+\frac 1 {r_1}=0$$
A: Here's an elementary way to find the roots of your characteristic equation.
$$
\begin{align}
r^3 - r^2 -1 &= 0 \\
\implies (y + \frac{1}{3})^3 - (y + \frac{1}{3})^2 - 1 &= 0 \quad \text{ where $y := r - \frac{1}{3}$} \\
\implies y^3 - \frac{y}{3} - \frac{29}{27} & = 0 \\
\implies 27y^3 - 9y - 29 &= 0 \\
\implies t^3 - 3t - 29 &= 0 \quad \text{ where t := 3y}
\end{align}
$$
Now let us find two real numbers $u$ and $v$ that satisfy $u + v = t$. Notice that $t^3 - 3uv\cdot t - (u^3 + v^3) = 0$. So we have
$$
\begin{align}
u^3 + v^3 & = 29 \quad \text{and,} \\
3uv & = 3 \\~\\
\implies u^3(29 - u^3) &= 1 \\
\implies u^6 - 29u^3 + 1 &= 0 \\~\\
\implies u^3 = \frac{29 \pm \sqrt{837}}{2} &= \frac{-29 \pm 3\sqrt{93}}{2}
\end{align}
$$
Without loss of generality, let $u = \sqrt[3]{\frac{29 + 3\sqrt{93}}{2}}$ and $v = \sqrt[3]{\frac{29 - 3\sqrt{93}}{2}}$. So,
$$
\begin{align}
t = t_1 := \sqrt[3]{\frac{29 + 3\sqrt{93}}{2}} + \sqrt[3]{\frac{29 - 3\sqrt{93}}{2}}
\end{align}
$$
Now, we can generate the other two roots of $t^3 - 3t - 29 = 0$ from $u$ and $v$.
Let $\omega = \frac{-1 + i\sqrt{3}}{2}$. Notice that $\omega$ and $\omega^2$ are the primitive cube roots of unity. Also notice that if we let $t = (u\omega + v\omega^2)$ or $t = (u\omega^2 + v\omega)$, we still get the same relation $t^3 - 3uv\cdot t - (u^3 + v^3) = 0$.
Hence the other two roots of $t^3 - 3t - 29 = 0$ are $t_2 := u\omega + v\omega^2$ and $t_3 := u\omega^2 + v\omega$.
We can find the corresponding value of $r$ for each of the roots since $r = \frac{t}{3} + \frac{1}{3}$.
For the root $t_1$, the corresponding value of $r$ is:
$$
\begin{align}
r = r_1 := \frac{1}{3} \left( 1 + \sqrt[3]{\frac{29 + 3\sqrt{93}}{2}} + \sqrt[3]{\frac{29 - 3\sqrt{93}}{2}} \right ) \approx 1.46557
\end{align}
$$
