A fair coin is to be tossed $10$ times. Let $i/j$, in lowest terms, be the probability that heads never occur on consecutive tosses. Find $i+j$.
1. **Define the problem in terms of sequences:**
   We need to find the number of sequences of length 10 consisting of heads (H) and tails (T) such that no two heads are consecutive.

2. **Use dynamic programming to count valid sequences:**
   Let \( a_n \) be the number of valid sequences of length \( n \) ending in T, and \( b_n \) be the number of valid sequences of length \( n \) ending in H. The total number of valid sequences of length \( n \) is \( a_n + b_n \).

3. **Establish recurrence relations:**
   - If a sequence of length \( n \) ends in T, the previous \( n-1 \) sequence can end in either H or T. Thus, \( a_n = a_{n-1} + b_{n-1} \).
   - If a sequence of length \( n \) ends in H, the previous \( n-1 \) sequence must end in T (to avoid consecutive heads). Thus, \( b_n = a_{n-1} \).

4. **Initial conditions:**
   - For \( n = 1 \), we have \( a_1 = 1 \) (sequence: T) and \( b_1 = 1 \) (sequence: H).

5. **Compute the values up to \( n = 10 \):**
   \[
   \begin{align*}
   a_2 &= a_1 + b_1 = 1 + 1 = 2, \\
   b_2 &= a_1 = 1, \\
   a_3 &= a_2 + b_2 = 2 + 1 = 3, \\
   b_3 &= a_2 = 2, \\
   a_4 &= a_3 + b_3 = 3 + 2 = 5, \\
   b_4 &= a_3 = 3, \\
   a_5 &= a_4 + b_4 = 5 + 3 = 8, \\
   b_5 &= a_4 = 5, \\
   a_6 &= a_5 + b_5 = 8 + 5 = 13, \\
   b_6 &= a_5 = 8, \\
   a_7 &= a_6 + b_6 = 13 + 8 = 21, \\
   b_7 &= a_6 = 13, \\
   a_8 &= a_7 + b_7 = 21 + 13 = 34, \\
   b_8 &= a_7 = 21, \\
   a_9 &= a_8 + b_8 = 34 + 21 = 55, \\
   b_9 &= a_8 = 34, \\
   a_{10} &= a_9 + b_9 = 55 + 34 = 89, \\
   b_{10} &= a_9 = 55.
   \end{align*}
   \]

6. **Total number of valid sequences of length 10:**
   \[
   a_{10} + b_{10} = 89 + 55 = 144.
   \]

7. **Total number of possible sequences:**
   \[
   2^{10} = 1024.
   \]

8. **Probability that heads never occur on consecutive tosses:**
   \[
   \frac{144}{1024} = \frac{9}{64}.
   \]

9. **Sum of the numerator and denominator:**
   \[
   i + j = 9 + 64 = 73.
   \]

The final answer is \( \boxed{73} \).