Evaluating the recurrence $a_{n+1}=a_n\cdot a_{n-1}-1$ Let the sequence $a_n$ be defined recursively by $a_0=3$, $a_1=1$ and $a_{n+1}=a_n\cdot a_{n-1}-1$ for $n=1,2,\cdots$. Find $a_6$, $a_7$, $a_8$, and $a_{200}$
Here is my attempt as an answer 
$a_0 = 3,a_1=1,a_2=2,a_3=1,a_4=1,a_5=0,a_6=-1,a_7=-1,a_8=0$
I do not know how to find $a_{200}$
 A: Since $a_5=0$ and $a_6=-1$, and we see this again happens at $a_8=0$ and $a_9=-1$, we know this sequence must repeat; a term only depends on the previous two values. Thus, we see the sequence is (from the fifth term on) simply $0,-1,-1$ repeated. 
It is now straightforward to compute $a_{200}$.
A: Perhaps a different point of view would help. Suppose we define a mapping $f(x,y) := (y,xy-1)$ and $P_{n+1} := f(P_n)$ where $P_0 = (a_0,a_1)$. It is easy to see, by induction, that $P_n = (a_n,a_{n+1})$ for all nonnegative $n$. Such a simple mapping $f$ has complicated behavior, but it may have limit cycles. That is, if we can find $i<j$ such that $P_i = P_j$, then the sequence $P$ will repeat indefinitely from $i$ on. That is, going forward, from $i$ on, the only values of $P_n$ will be $P_i,P_{i+1},...,P_{j-1}$ cyclicly repeated. Thus, to find $P_n$, from $i$ on, you only need to know where in the cycle $n$ corresponds. In your problem, I suggest looking at $P_0, P_3, P_6,P_9,\dots$ and I hope this explanation is more understandable.
