We claim that for nonnegative integers $m$ and for $0 \leq n<3 \cdot 2^{m}, a_{3 \cdot 2^{m}+n}=\left(3 \cdot 2^{m}+n\right)(m+2)+2 n$. We will prove this by induction; the base case for $a_{3}=6$ (when $m=0$, $n=0$) is given in the problem statement. Now, suppose that this is true for some pair $m$ and $n$. We will divide this into two cases: Case 1: $n<3 \cdot 2^{m}-1$. Then, we want to prove that this is true for $m$ and $n+1$. In particular, writing $a_{3 \cdot 2^{m}+n}$ in base $3 \cdot 2^{m}+n$ results in the digits $m+2$ and $2 n$. Consequently, reading it in base $3 \cdot 2^{m}+n+1$ gives $a_{3 \cdot 2^{m}+n+1}=2+\left(3 \cdot 2^{m}+n+1\right)(m+2)+2 n=\left(2 \cdot 2^{m}+n+1\right)(m+2)+2(n+1)$, as desired. Case 2: $n=3 \cdot 2^{m}-1$. Then, we want to prove that this is true for $m+1$ and 0. Similarly to the previous case, we get that $a_{3 \cdot 2^{m}+n+1}=a_{3 \cdot 2^{m+1}}=2+\left(3 \cdot 2^{m}+n+1\right)(m+2)+2 n=2+\left(3 \cdot 2^{m+1}\right)(m+2)+2\left(3 \cdot 2^{m}-1\right)=\left(3 \cdot 2^{m+1}+0\right)((m+1)+2)+2(0)$, as desired. In both cases, we have proved our claim.