, 2013

Solution:

Answer: 23097

We claim that for nonnegative integers $m$ and for $0 \leq n < 3 \cdot 2^m$, $a_{3 \cdot 2^m + n} = (3 \cdot 2^m + n)(m+2) + 2n$. 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 $2n$. Consequently, reading it in base $3 \cdot 2^m + n + 1$ gives $a_{3 \cdot 2^m + n + 1} = 2 + (3 \cdot 2^m + n + 1)(m+2) + 2n = (3 \cdot 2^m + n + 1)(m+2) + 2(n+1)$, 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 $a_{3 \cdot 2^m + n + 1} = a_{3 \cdot 2^{m+1}} = 2 + (3 \cdot 2^m + n + 1)(m+2) + 2n = 2 + (3 \cdot 2^{m+1})(m+2) + 2(3 \cdot 2^m - 1) = (3 \cdot 2^{m+1} + 0)((m+1)+2) + 2(0)$, as desired.

In both cases, we have proved our claim.