Answer: 73

Small Hint:

Count valid length-$n$ toss strings according to whether they end in tails or heads

Big Hint:

The resulting recurrence is Fibonacci-like, with initial counts $2$ and $3$

Solution:

Let $u_n$ be the number of length-$n$ toss strings with no consecutive heads. A valid string ending in tails is obtained by appending $\mathrm{T}$ to any valid length-$(n-1)$ string, while one ending in heads is obtained by appending $\mathrm{TH}$ to any valid length-$(n-2)$ string. Hence $u_n=u_{n-1}+u_{n-2},$ with $u_1=2$ and $u_2=3.$ This gives $u_{10}=144.$ The probability is $\frac{144}{2^{10}}=\frac{9}{64},$ so $i+j=9+64=73.$