Consider the recurrence an=an−1+an−2+⋯+a0a_n = a_{n-1} + a_{n-2} + \dots + a_0an=an−1+an−2+⋯+a0 for n≥1n \geq 1n≥1. If a0=1a_0 = 1a0=1, find a3a_3a3.
222
444
888
161616