Probability that no two consecutive heads occur?

A fair coin is tossed $10$ times. What is the probability that no two consecutive tosses are heads?

Possibilities are (dont mind the number of terms):

$H TTTTTTH$, $HTHTHTHTHTHTHT$.

But except for those,

let $y(n)$ be the number of sequences that start with $T$

$T _$, there are two options, $T$ and $H$ so,

$y(n) = y(n - 1) + x(n-1) = y(n - 1) + y(n - 2)$

Let $x(n)$ be the number of sequences that start with $H$,

$H _$ the next option is only $T$ hence,

$x(n) = y(n - 1)$

The total number of sequences $F(n)$ is:

$$F(n) = y(n) + x(n) = 2y(n - 1) + y(n - 2)$$

We are after $F(10)$,

$$F(10) = 2y(9) + y(8)$$

$$F(3) = 2y(2) + y(1) = 1 + 4 = 5$$

$$F(4) = 2y(3) + y(2) = 6 + 2 = 8$$

But I'm not quite sure where this will take me though.

Let $f(n)$ be the noumber of sequences without two consecutive $H$, you want $\frac{f(10)}{2^{10}}$.

$f(1)=2$ and $f(2)=3$ by inspection.

We obtain the recursion $f_n=f_{n-1}+f_{n-2}$ why? every sequence of length $n$ without consecutive $H$ can be obtained uniquely by taking a sequence of length $n-2$ and adding $TH$ or by taking a sequence of length $n-1$ and adding $T$ at the end.

Using this we have $f(3)=5,f(4)=8,f(5)=13,f(6)=21,f(7)=34,f(8)=55,f(9)=89$ $f(10)=144$

So we want $\frac{144}{2^{10}}\approx0.14$

$F (n)$ is in fact a fibonacci sequence: \begin{align} F (n+1) &= 2 y (n) + y (n-1)\\ &= 2 [y (n-1) +y (n-2)] + y (n-2) + y (n-1)\\ &= 2 y (n-1) + y (n-2) + 2 y(n-2) +y (n-3)\\ &= F (n) +F (n-1)\\ \end{align}

Now $F (1) = |\{H,T\}|=2$

And $F (2) = |\{HT,TH,TT\}|=3$

So $ F (10) = 144$

There are $ 2^{10}=1024$ possible events.

Thus the required probability is $\frac {144}{1024} $.