Occurrence of 5 consecutive tails before occurrence of 2 consecutive heads In repeated tossing of a fair coin find the probability that $5$ consecutive tails occur before occurrence of $2$ consecutive heads.
My attempt:
I tried to find the probability of non-occurrence of two consecutive heads in $n$ throws.
Let $a_{n}$ be the number of possibilities in which $2$ consecutive heads do not occur in $n$ throws.
I managed to find the recursion formula.
$a_{1}=2$
$a_{2}=3$
$a_{n}=a_{n-1}+a_{n-2}$
But I am not able to get a closed form of $a_{n}$.
Once $a_{n}$ gets determined it may be possible then to find probability of occurrence of $5$ consecutive heads

Here is a  slightly different proof which may be  of interest although
it is not as elegant as the one by @leonbloy.
Suppose we treat the problem of $t$ tails before $h$ heads.
Encoding this in  a generating function with $u$  marking sequences of
tails of length at  least $t$ and $v$ sequences of  heads of length at
least $h$  and finally $w$ marking  the final occurrence of  $h$ heads
and introducing
$$G_t(z) = z+z^2+\cdots +z^{t-1}+uz^t\frac{1}{1-z}
\quad\text{and}\quad
G_h(z) = z+z^2+\cdots +z^{h-1}+vz^h\frac{1}{1-z}$$
we obtain
$$H(z) = (1+G_t(z))
\left(\sum_{k\ge 0} G_h(z)^k G_t(z)^k\right)
\left(1+z+\cdots+z^{h-1} + wz^h + z^{h+1}\frac{1}{1-z}\right).$$
Observe that when we remove the three markers $u,v$ and $w$ we obtain
$$Q(z) = \frac{1}{1-z} 
\left(\sum_{k\ge 0} \frac{z^k}{(1-z)^k} \frac{z^k}{(1-z)^k}\right)
\frac{1}{1-z}
\\ = \frac{1}{(1-z)^2} \frac{1}{1-z^2/(1-z)^2}
= \frac{1}{(1-z)^2-z^2} = \frac{1}{1-2z}$$
which  is good  news because  it means  we have  enumerated all  $2^n$
possible bit strings of length $n.$
Now extracting  coefficients we  are interested in  the series  on $w$
which yields
$$H_1(z) = z^h (1+G_t(z))
\left(\sum_{k\ge 0} G_h(z)^k G_t(z)^k\right)$$
The next step is to discard those terms that have $v\ge 1$ (meaning an
internal occurrence of $h$ heads) which yields on setting $v=0$
$$H_2(z) = z^h (1+G_t(z))
\left(\sum_{k\ge 0} 
\left(z\frac{1-z^{h-1}}{1-z}\right)^k G_t(z)^k\right).$$
Finally we need to compute
$$H_3(z) = 
\left. H_2(z)\right|_{u=1} - \left. H_2(z)\right|_{u=0}$$
to remove those terms not containing a run of at least $t$ tails.
This yields
$$H_3(z) = z^h \frac{1}{1-z}
\left(\sum_{k\ge 0} 
\left(z\frac{1-z^{h-1}}{1-z}\right)^k 
\left(\frac{z}{1-z}\right)^k\right)
\\ - z^h \frac{1-z^t}{1-z}
\left(\sum_{k\ge 0} 
\left(z\frac{1-z^{h-1}}{1-z}\right)^k 
\left(z\frac{1-z^{t-1}}{1-z}\right)^k\right).$$
This finally produces
$$H_3(z) = z^h\frac{1}{1-z}
\frac{1}{1-z^2 (1-z^{h-1})/(1-z)^2}
\\ - z^h\frac{1-z^t}{1-z}
\frac{1}{1- z^2(1-z^{h-1})(1-z^{t-1}))/(1-z)^2}
\\ = z^h
\frac{1-z}{(1-z)^2-z^2 (1-z^{h-1})}
\\ - z^h (1-z^t)
\frac{1-z}{(1-z)^2- z^2 (1-z^{h-1})(1-z^{t-1})}
\\ = z^h
\frac{1-z}{1 - 2z + z^{h+1}}
- z^h (1-z^t)
\frac{1-z}{1 - 2z + z^{h+1} + z^{t+1} - z^{h+t}}.$$
We obtain the probability by setting $z=1/2$
which yields
$$\frac{1}{2^{h+1}} 2^{h+1} 
- \frac{1}{2^{h+1}} \left(1-\frac{1}{2^t}\right)
\frac{1}{1/2^{h+1}+1/2^{t+1}-1/2^{h+t}}
\\ = 1
- \frac{2^t-1}{2^{h+t+1}}
\frac{1}{1/2^{h+1}+1/2^{t+1}-1/2^{h+t}}
\\ = 1
- (2^t-1)
\frac{1}{2^t+2^h-2}
= \frac{2^t+2^h-2-(2^t-1)}{2^t+2^h-2}
\\ = \frac{2^h-1}{2^t+2^h-2}.$$
