Let $a_0, a_1, a_2, ...$ be a sequence such that: $a_0 = 2$; each $a_n = 2$ or $3; a_n =$the number of $3$s between the $n$th and $n+1$th $2$ in the sequence. So the sequence starts: $233233323332332 ...$ . Show that we can find $\alpha$ such that $a_n = 2$ iff $n = [\alpha m]$ for some integer $m \geq 0$.
1. **Define the sequence and function:**
   Let \( a_0, a_1, a_2, \ldots \) be a sequence such that \( a_0 = 2 \) and each \( a_n = 2 \) or \( 3 \). The sequence is defined such that \( a_n \) is the number of 3's between the \( n \)-th and \( (n+1) \)-th 2 in the sequence. The sequence starts as \( 233233323332332 \ldots \).

2. **Define the function \( f(m) \):**
   Let \( f(m) \) be the position of the \( (m+1) \)-th appearance of 2. Thus, \( f(0) = 1 \), \( f(1) = 4 \), \( f(2) = 8 \), and so on. We aim to prove that \( f(m) = \lceil (2 + \sqrt{3})m \rceil \).

3. **Establish the recurrence relation for \( f \):**
   For \( f \), we have:
   - \( f(k+1) = f(k) + 3 \) if \( k+1 = f(a) \) for some \( a = 0, \ldots, k \)
   - \( f(k+1) = f(k) + 4 \) otherwise.

4. **Summing the recurrence relations:**
   Adding these for \( k = 0, \ldots, f(m) - 1 \), we get:
   \[
   f(f(m)) = 4f(m) - m
   \]
   Extending this, we also get:
   \[
   f(f(m+1) - 1) = 4f(m) - m + 4(f(m+1) - 1 - f(m)) = 4f(m+1) - m - 4
   \]

5. **Inductive proof for the lower bound:**
   We prove inductively that \( f(k) > (2 + \sqrt{3})k \). It suffices to verify this for values \( k = f(m) \) where \( f \) increases by 3. To induct, we need:
   \[
   4f(m) - m = f(f(m)) > (2 + \sqrt{3})f(m)
   \]
   Simplifying, we get:
   \[
   (2 - \sqrt{3})f(m) > m
   \]
   Since \( f(m) > (2 + \sqrt{3})m \), the inequality holds.

6. **Inductive proof for the upper bound:**
   We prove inductively that \( f(k) < (2 + \sqrt{3})k + 1 \). It suffices to verify this for \( k = f(m) - 1 \) where in the next step \( f \) increases by 3. We need:
   \[
   4f(m) - m - 3 = f(f(m) - 1) < (2 + \sqrt{3})(f(m) - 1) + 1
   \]
   Simplifying, we get:
   \[
   (2 - \sqrt{3})f(m) < m + 2 - \sqrt{3}
   \]
   Since \( f(m) < (2 + \sqrt{3})m + 1 \), the inequality holds.

7. **Conclusion:**
   Having proved that:
   \[
   (2 + \sqrt{3})k < f(k) < (2 + \sqrt{3})k + 1
   \]
   we conclude that \( f(m) = \lceil (2 + \sqrt{3})m \rceil \).

The final answer is \( \boxed{ f(m) = \lceil (2 + \sqrt{3})m \rceil } \).