Solve the recurrence relation an−4an−1+3an−2=0a_n - 4a_{n-1} + 3a_{n-2} = 0an−4an−1+3an−2=0 with initial conditions a0=2a_0 = 2a0=2 and a1=4a_1 = 4a1=4. What is a4a_4a4?
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