For the third-order recurrence an=6an−1−11an−2+6an−3a_n = 6a_{n-1} - 11a_{n-2} + 6a_{n-3}an=6an−1−11an−2+6an−3 with a0=0,a1=2,a2=5a_0 = 0, a_1 = 2, a_2 = 5a0=0,a1=2,a2=5, find a3a_3a3.
4
8
12
16