Solving the recurrence $ T(n) = \dfrac{T(n-1) + T(n-3)} {T(n-2)} $
The solution depends on the initial conditions. For example, if $T(0) = T(1) = T(2) = 2$ then $T(n) = 2$ for all $n$, but in general the recurrence converges to a pattern of length $4$. The limiting patterns are of the form
$$
a, \frac{ab\pm\sqrt{(ab)^2-4(a+b)}}{2}, b, \frac{ab\mp\sqrt{(ab)^2-4(a+b)}}{2}
$$
For example, if $T(0)=T(1)=T(2)=1$ then $a ≈ 3.13200592382654$ and $b ≈ 1.35282312408545$. For large $n$, the values $T(4n),T(4n+1),T(4n+2),T(4n+3)$ are close to
$$ 3.13200592382654, 2.06091243965066, 1.35282312408545, 2.17613759887451. $$
(So we choose $-$ in the $\pm$.)
The above was determined empirically. It would be nice to come up with a proof.