+876 $a_0 = 54 \div 3^3 = \frac{2}{3} = 2 \cdot 3^{-1}$
$a_1 = \frac{2}{3} \cdot 3 = 2 = 2 \cdot 3^0$
$a_2 = 2 \cdot 3 = 6 = 2 \cdot 3^1$
$a_3 = 6 \cdot 3 = 18 = 2 \cdot 3^2$
$a_4 = 18 \cdot 3 = 54 = 2 \cdot 3^3$
It should be clear by now that for $a_n$, it will be $\boxed{2 \cdot 3^{n-1}}.$
