Certainly we're getting $(a_{n+3}-a_{n+1})(a_{n+2}+1)=a_{n+2}-a_n$. Now suppose there are infinitely many $n$ such that $a_{n+3}\neq a_{n+2}$ then we get $|a_3-a_1|=|\prod_{i=1}^{k}(a_i+1)(a_{k+1}-a_{k-1})|$. Now clearly there must be finitely many $i$ such that $a_i\neq 0,-2$.So after some stage all of $a_i$ must be $0,-2$.But certainly it's not true.Noe so there are finitely many $n$ such that $a_{n+3}\neq a_{n+2}$.So after some stage for all $n>N$ we'll get there are infinitely many $n$ such that $a_{n+3}=a_{n+2}$. And so for all $n>N$ we get $a_na_{n+1}=2006=ab$.Now suppose $a_N\neq=x$ then certainly either $b=\frac{x+2006}{a+1}\implies x=b$ or $a=\frac{x+2006}{b+1}\implies x=a$.Thus we conclude all of $a_i\in\{a,b\}$ where $ab=2006$ and $a_n=a_{n+2}$ for all $n$.

The above is riddled with typo(s) and doubtful inferences, culminating with omitting to forbid $a_n=-1$. But the idea is good, the solution is easily ironed out, and the answer comes out to being $14$ such sequences.