Find a recursion (combinatorial) 
Consider sequences that consist entirely of $ A$'s and $ B$'s and that have the property that every run of consecutive $ A$'s has even length, and every run of consecutive $ B$'s has odd length. Examples of such sequences are $ AA$, $ B$, and $ AABAA$, while $ BBAB$ is not such a sequence. How many such sequences have length 14?

Let $A(n)$ be the $n$-lettered sequences beginning with an $A$. Let $B(n)$ be the $n$-lettered sequences beginning with a $B$. 
I am very confused how to set a recursion pattern. 
$A$_ _ _ _ _ _ _ _ 
There is one option to put an $A$. So: $A(n) = A(n-1)$.  But then we have:
$AA$_ _ _ _ _ _ _ Now there are two options, $A$ or $B$ but both with strings attached:  If $A$ is chosen, then the next one must be $A$, if $B$ is chosen then the next must be $BB$ or $A$.
It is very confusing what to do here? 
Hints and help only please!
 A: I'll use functions $A(n), B(n)$ as defined in the question.
A string beginning with $A$ must be followed by a second $A$ and then by any valid string of length $n-2$ beginning with either $A$ or $B$.
A string beginning with $B$ must be either (i) followed by a second $B$ and then by any valid string of length $n-2$ beginning with $B$, so we keep an odd number of $B$s at the start, OR (ii) followed by any valid string of length $n-1$ beginning with $A$.
This results in the pair of recurrences:
$$A(n)=A(n-2) + B(n-2) \\
  B(n)=A(n-1) + B(n-2).$$
A: Hint: Start with A is simple, for then we need to have another A, and we append a good string of length $n-2$.
Start with B is more complicated. (i) If we  have an A next, then we need another, then a good string of length $n-3$.
(ii) If we have a B next, then $\dots$.
Added: We expand on the hint. Let $F(n)$ be the number of good strings of length $n$. There are $F(n-2)$ good strings of length $n$ that begin with A. For good strings of length $n$ that begin with B, they start with BAA, or BBBAA, or BBBBBAA, and so on, with a good string of the appropriate length appended,  That gives a count of $F(n-3)+F(n-5)+F(n-7)+\cdots$. This gives the recurrence
$$F(n)=F(n-2)+F(n-3)+F(n-5)+F(n-7)+\cdots.$$
Now we work our way up to $14$. 
