Let $n$ be a positive integer. 
Determine the number of sequences $a_0, a_1, \ldots, a_n$ with terms in the set $\{0,1,2,3\}$ such that $$n=a_0+2a_1+2^2a_2+\ldots+2^na_n.$$
1. **Define the problem and initial conditions:**
   We need to determine the number of sequences \(a_0, a_1, \ldots, a_n\) with terms in the set \(\{0,1,2,3\}\) such that 
   \[
   n = a_0 + 2a_1 + 2^2a_2 + \ldots + 2^n a_n.
   \]
   Let \(b_n\) be the number of such sequences. We aim to prove that \(b_n = \left\lfloor \frac{n}{2} \right\rfloor + 1\).

2. **Base cases:**
   - For \(n = 0\), the only sequence is \(a_0 = 0\), so \(b_0 = 1\).
   - For \(n = 1\), the sequences are \((1, 0)\) and \((0, 1)\), so \(b_1 = 2\).

3. **Inductive hypothesis:**
   Assume that for all \(m \leq k\), \(b_m = \left\lfloor \frac{m}{2} \right\rfloor + 1\).

4. **Inductive step:**
   We need to show that \(b_{k+1} = \left\lfloor \frac{k+1}{2} \right\rfloor + 1\).

5. **Define auxiliary sequences:**
   - Let \(c_n\) be the number of sequences in \(b_n\) where \(a_1 = 0\).
   - Let \(d_n\) be the number of sequences in \(b_n\) where \(a_1 = 3\).

6. **Claim 1:**
   \[
   b_{k+1} = b_{k-1} - d_{k-1} + c_{k+1}.
   \]
   - If \(a_1 \neq 0\), reducing \(a_1\) by 1 produces sequences in \(b_{k-1}\) except those where \(a_1 = 3\).

7. **Claim 2:**
   \[
   c_{k+1} - d_{k-1} = 1.
   \]
   - Consider \(k+1\) in the form \(n = a_0 + 2a_1 + 2^2a_2 + \ldots + 2^n a_n\) with \(a_1 = 0\). Since \(a_0\) can be \(\{0,1,2,3\}\) and \(k+1 \equiv 0,1,2,3 \pmod{4}\), there is a unique \(i\) such that \(a_0 = i\).
   - Similarly, for \(k-1\) with \(a_1 = 3\), there is a unique \(j\) such that \(a_0 = j\).
   - Since \(k+1 - (k-1) = 2 \pmod{4}\), solving \(i - (2 \times 3 + j) = 2 \pmod{4}\) gives \(i = j\).

8. **Relate \(c_{k+1}\) and \(d_{k-1}\):**
   \[
   b_{\frac{k+1 - (0 \times 2 + i)}{4}} = c_{k+1} \quad \text{and} \quad b_{\frac{k-1 - (3 \times 2 + j)}{4}} = d_{k-1}.
   \]
   - Since \(i = j\), we have:
     \[
     \frac{k+1 - (0 \times 2 + i)}{4} - \frac{k-1 - (3 \times 2 + j)}{4} = 2.
     \]
   - By the inductive hypothesis, \(b_{n+2} - b_n = \left\lfloor \frac{n+2}{2} \right\rfloor + 1 - \left\lfloor \frac{n}{2} \right\rfloor + 1 = 1\).

9. **Conclusion:**
   \[
   b_{k+1} = b_{k-1} + 1.
   \]
   - By induction, \(b_{k+1} = \left\lfloor \frac{k+1}{2} \right\rfloor + 1\).

The final answer is \( \boxed{ \left\lfloor \frac{n}{2} \right\rfloor + 1 } \).