Solve the third-order homogeneous recurrence an=2an−1+an−2−2an−3a_n = 2a_{n-1} + a_{n-2} - 2a_{n-3}an=2an−1+an−2−2an−3 with a0=1a_0 = 1a0=1, a1=0a_1 = 0a1=0, a2=1a_2 = 1a2=1. What is a5a_5a5?
-9
-4
4
9