Problem: 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.

Solution: Let us define a "run" as a set of consecutive letters that are all the same and let a "maximal run" be a run that is not the proper subset of any other run, in other words, it cannot be expanded.
From now on, we consider all runs to be maximal.
Note that the minimum number of runs in the $10$ -letter string is $4$ (or else the Pigeonhole Principle tells us that at least $1$ run has $4$ or more letters in it, contradiction), and the maximum number of runs is clearly $10.$
Since an arbitrary run can have anywhere from $1$ to $3$ letters in it, it follows that the number of $10$ -letter strings with $n$ runs (where $4 \leq n \leq 10$ ) is \[2[x^{10}] (x+x^2+x^3)^n,\] i.e. twice the coefficient of $x^{10}$ in the expansion of $(x+x^2+x^3)^n.$ (Note that we multiplied by $2$ because there are two choices for which letter the first run has and then the rest are fixed).
Hence, we wish to find \[2\sum_{n=4}^{10} [x^{10}] (x+x^2+x^3)^n.\] First, we can rewrite this as
\begin{align*}
\sum_{n=4}^{10} [x^{10}] (x+x^2+x^3)^n &= \sum_{n=4}^{10} [x^{10}] \ x^n \cdot (1+x+x^2)^n \\
&= \sum_{n=4}^{10} [x^{10-n}] \left(\frac{1-x^3}{1-x}\right)^n \\
&= \sum_{n=4}^{10} [x^{10-n}] \frac{(1-x^3)^n}{(1-x)^n}. \\
\end{align*}
Now, we proceed case by case, utilizing the Binomial Theorem for the numerator in all of the cases:
For $n=4,$ we have
\begin{align*}
[x^6] \frac{(1-x^3)^4}{(1-x)^4} &= \binom{4}{0} [x^6] (1-x)^{-4} - \binom{4}{1} [x^3] (1-x)^{-4} + \binom{4}{2} [x^0] (1-x)^{-4} \\
&= 1\cdot\binom{9}{3} - 4\cdot\binom{6}{3} + 6\cdot\binom{3}{3} \\
&= 1\cdot84-4\cdot20+6\cdot1 \\
&= 84-80+6 \\
&= \underline{10}.
\end{align*}
For $n=5,$ we have
\begin{align*}
[x^5] \frac{(1-x^3)^5}{(1-x)^5} &= \binom{5}{0} [x^5] (1-x)^{-5} - \binom{5}{1}[x^2] (1-x)^{-5} \\
&= 1\cdot\binom{9}{4}-5\cdot\binom{6}{4} \\
&= 1\cdot126-5\cdot15 \\
&= 126-75 \\
&= \underline{51}.
\end{align*}For $n=6,$ we have
\begin{align*}
[x^4] \frac{(1-x^3)^6}{(1-x)^6} &= \binom{6}{0} [x^4](1-x)^{-6} - \binom{6}{1}[x^1](1-x)^{-6} \\
&= 1 \cdot \binom{9}{5} - 6 \cdot \binom{6}{5} \\
&= 1 \cdot 126 - 6 \cdot 6 \\
&= 126-36 \\
&= \underline{90}.
\end{align*}For $n=7,$ we have
\begin{align*}
[x^3] \frac{(1-x^3)^7}{(1-x)^7} &= \binom{7}{0}[x^3](1-x)^{-7} - \binom{7}{1}[x^0](1-x)^{-7} \\
&= 1\cdot \binom{9}{6} - 7\cdot\binom{6}{6} \\
&= 1\cdot84-7\cdot1\\
&= 84-7 \\
&= \underline{77}.
\end{align*}For $n=8,$ we have
\begin{align*}
[x^2] \frac{(1-x^3)^8}{(1-x)^8} &=\binom{8}{0} [x^2](1-x)^{-8} \\
&= 1\cdot\binom{9}{7} \\
&= 1\cdot36 \\
&= \underline{36}.
\end{align*}For $n=9,$ we have
\begin{align*}
[x^1] \frac{(1-x^3)^9}{(1-x)^9} &= \binom{9}{0} [x^1] (1-x)^{-9}\\
&= 1\cdot\binom{9}{8} \\
&= 1\cdot9 \\
&= \underline{9}.
\end{align*}For $n=10,$ we have
\begin{align*}
[x^0] \frac{(1-x^3)^{10}}{(1-x)^{10}} &= [x^0] (1-x)^{-10} \\
&= 1\cdot\binom{9}{9} \\
&= 1\cdot1 \\
&= \underline{1}.
\end{align*}Hence, the answer is
\begin{align*}
2\sum_{n=4}^{10} [x^{10-n}]  \frac{(1-x^3)^n}{(1-x)^n} &= 2(10+51+90+77+36+9+1) \\
&= 2\cdot274 \\
&= \boxed{548}.
\end{align*}

Answer:  548