Call a string satisfying the stated constraints a good string and let $a_n$ and $b_n$ be the number of good strings of length $n$ which start with letter $A$ and $B$, respectively.

For $n=1$, $2$, and $3$, we have $$a_1=b_1=1,\ a_2=b_2=2,\ a_3=b_3=4$$

When $n\ge 4$, a good string starting with $A$ must start with $AB$, $AAB$ or $AAAB$. The letter after the first appeard $B$ has no restriction. This means that, after having removed leading $A$s, a good string starting with $A$ will become a good string starting with $B$. For example, a good string of length $n$ starting with $AB$ will correspond a good string of length $(n-1)$ starting with $B$ whose count is $b_{n-1}$. Therefore,  we shall have $$a_n=b_{n-1}+b_{n-2} + b_{n-3}$$

Meanwhile, by the principle of symmetry, it must hold that $a_k = b_k$ for all $k=1$, $2$, $\cdots$, $n$. Consequently, we can conclude that $$a_n=b_{n-1}+b_{n-2} + b_{n-3}\implies a_n=a_{n-1}+ a_{n-2}+ a_{n-3}$$

Using the initial values computed early yields $$\begin{array}{rll} a_4&= 4+2+1&=7\\a_5&=7+4+2&=13\\a_6&=13+7+4&=24\\a_7&=24+13+7&=44\\a_8&=44+24+13&=81\\a_9&=81+44+24&=149\\a_{10}&=149+81+44&=274\\b_{10}&=a_{10}\end{array}$$

Hence, the final answer is $2\times 274=\boxed{548}$.