Explicit formula from for interesting reccurence $$t_0=-5, t_1=\frac{11}{5} $$
$$t_n=1-\frac{2}{t_{n-1}} + \frac{-4}{t_{n-2}*t_{n-1}} $$
I have never seen so uncommon formula for recurrence like this before. I have no idea how to solve it. Please, can anyone help me ?
 A: The recurrence can be re-written as $t_n t_{n-1} t_{n-2} = t_{n-1}t_{n-2} - 2t_{n-2} -4$.
Let $x_n = t_0 t_1 t_2 \dotsb t_n$. Then
$$\frac{x_n}{x_{n-3}} = \frac{x_{n-1}}{x_{n-3}} - 2\frac{x_{n-2}}{x_{n-3}} -4$$
Hence $x_{n} = x_{n-1} - 2x_{n-2} - 4x_{n-3}$ and $x_0 = -5$, $x_1 = -11$, $x_2 = -5$. You can find $x_n$ from this (one solution of the characteristic equation is $r=-1$), and then $t_n = \frac{x_{n}}{x_{n-1}}$. 
A: I propose a solution utilizing the very interesting auxiliary sequence called $x_n$ introduced by @Catalin Zara, verifying
$$x_{n} = x_{n-1} - 2x_{n-2} - 4x_{n-3}$$
The characteristic equation $r^3 - r^2 + 2r + 4 = 0$  of this recurrence relationship has the following roots $r=-1$, $r=1\pm i \sqrt{3}$.
Therefore, its general term can be expressed, using calculations that we do not reproduce here, in the following way 
$$x_n=-\dfrac{1}{7}\left( A(-1)^n + B 2^n \cos(n\pi/3)+C 2^n \sin(n\pi/3)\right) \ \ \text{with} \ \ \ A=3, B=32, C=16 \sqrt{3}$$
(taking initial conditions into account). Therefore:
$$t_n=\dfrac{x_n}{x_{n-1}}=\dfrac{A(-1)^n + B 2^n \cos(n\pi/3)+C 2^n \sin(n\pi/3)}{A(-1)^{n-1} + B 2^{n-1} \cos((n-1)\pi/3)+C 2^{n-1} \sin((n-1)\pi/3)}$$
Division of the numerator and the denominator by $2^{n-1}$ prove that this sequence is arbitrarily close to the following sequence:
$$u_n:=2\dfrac{B \cos(n\pi/3) + C \sin(n\pi/3)}{B \cos((n-1)\pi/3)+C \sin((n-1)\pi/3)}$$
which has a cycle of length three: $2.5, 0.4,-8$. 
For example, $u_{30}=\dfrac{2 B}{\dfrac{B}{2}-C\dfrac{\sqrt{3}}{2}}=-8$
Remark: this period-3 cyclic behavior persists when changing initial conditions.
