\(a_1=3, a_2=3\)
\(n\geq2,a_{n+1} a_{n-1}={a^2_n}+2007\)
Find \(\lfloor \frac{{a^2_{2007}}+{a^2_{2006}}}{{a_{2007}}{a_{2006}}}\rfloor\)
The answer is 210.
+150 
