Let $a_{n}$ be the number of ways to form $n$-letter strings made up of As and Bs such that no more than $3$ adjacent letters are identical.
Note that, at the end of each $n$-letter string, there are $3$ possibilities for the last letter chain: it must be either $1$, $2$, or $3$ letters long. Removing this last chain will make a new string that is $n-1$, $n-2$, or $n-3$ letters long, respectively.
Therefore we can deduce that $a_{n}=a_{n-1}+a_{n-2}+a_{n-3}$.
We can see that
\[a_{1}=2\]
\[a_{2}=2^{2}=4\]
\[a_{3}=2^{3}=8\]
so using our recursive relation we find
\[a_{4}=14\]
\[a_{5}=26\]
\[a_{6}=48\]
\[a_{7}=88\]
\[a_{8}=162\]
\[a_{9}=298\]
\[a_{10}=\boxed{548}\]
There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical.