Problem 1292

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 $\boxed{73}$.