On Day $1$, Alice starts with the number $a_1=5$. For all positive integers $n>1$, on Day $n$, Alice randomly selects a positive integer $a_n$ between $a_{n-1}$ and $2a_{n-1}$, inclusive. Given that the probability that all of $a_2,a_3,\ldots,a_7$ are odd can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers, compute $m+n$.


[i]2021 CCA Math Bonanza Lightning Round #1.4[/i]
1. **Initial Condition**: Alice starts with \( a_1 = 5 \), which is an odd number.
2. **Selection Rule**: For \( n > 1 \), Alice selects \( a_n \) randomly between \( a_{n-1} \) and \( 2a_{n-1} \), inclusive.
3. **Odd Number Probability**: We need to determine the probability that \( a_2, a_3, \ldots, a_7 \) are all odd.

   - If \( a_{n-1} \) is odd, the range from \( a_{n-1} \) to \( 2a_{n-1} \) includes \( a_{n-1} \) itself and \( 2a_{n-1} \), which is even.
   - The range \( [a_{n-1}, 2a_{n-1}] \) contains \( a_{n-1} \) numbers. Since \( a_{n-1} \) is odd, the range will have an equal number of odd and even numbers because it spans an even number of integers.

4. **Probability Calculation**:
   - For each \( n \) from 2 to 7, the probability that \( a_n \) is odd given that \( a_{n-1} \) is odd is \( \frac{1}{2} \).
   - This is because half of the numbers in the range \( [a_{n-1}, 2a_{n-1}] \) are odd.

5. **Combined Probability**:
   - The probability that \( a_2, a_3, \ldots, a_7 \) are all odd is the product of the individual probabilities:
   \[
   \left( \frac{1}{2} \right)^6 = \frac{1}{64}
   \]

6. **Simplification**:
   - The fraction \( \frac{1}{64} \) is already in its simplest form, where \( m = 1 \) and \( n = 64 \).
   - Therefore, \( m + n = 1 + 64 = 65 \).

The final answer is \( \boxed{65} \).