Solution:
Let $a_n$ be the number of $n$-letter Vitas words that start with $B$ and end with $L$.

Let us generalize and define:
- $a_n$: number of $n$-letter words starting with $B$ and ending with $L$.
- $b_n$: number of $n$-letter words starting with $B$ and ending with $R$.

We do not consider words ending with $B$ since the last letter cannot be $B$ (as the word must end with $L$).

For $n = 2$:
- The word must be $BL$ (since it starts with $B$ and ends with $L$), so $a_2 = 1$.
- The word $BR$ is also possible, so $b_2 = 1$.

For $n > 2$:
To form an $n$-letter word ending with $L$, the $(n-1)$-th letter must be $R$ (since no two consecutive letters are the same), so we append $L$ to any $(n-1)$-letter word ending with $R$.
Thus,
$$a_n = b_{n-1}$$

Similarly, to form an $n$-letter word ending with $R$, the $(n-1)$-th letter must be $L$, so
$$b_n = a_{n-1}$$

But, at each step, the letter can be $L$ or $R$ (but not the same as the previous letter), so we must also consider that after the first $B$, the rest of the letters alternate between $L$ and $R$.

But, since the alphabet is only $B$, $L$, $R$, and the first letter is $B$, the rest of the letters (positions $2$ to $n$) are $L$ or $R$, with no two consecutive letters the same.

Let us define $a_n$ as above, and $b_n$ as above.

From above:
$$a_n = b_{n-1}$$
$$b_n = a_{n-1}$$

With initial conditions:
$a_2 = 1$ (word $BL$), $b_2 = 1$ (word $BR$).

So,
$a_3 = b_2 = 1$
$b_3 = a_2 = 1$

$a_4 = b_3 = 1$
$b_4 = a_3 = 1$

So, for all $n \geq 2$, $a_n = 1$, $b_n = 1$? But this cannot be, since the number should grow with $n$.

Wait: Let's reconsider.

Let us define $A_n$ as the number of $n$-letter words starting with $B$ and ending with $L$.
Let $B_n$ as the number of $n$-letter words starting with $B$ and ending with $R$.

For $n = 2$:
Possible words: $BL$, $BR$.
So $A_2 = 1$, $B_2 = 1$.

For $n = 3$:
Words: $BLR$, $BRL$.
So $A_3 = 1$ ($BRL$), $B_3 = 1$ ($BLR$).

For $n = 4$:
Words: $BLRL$, $BRLR$.
So $A_4 = 1$ ($BLRL$), $B_4 = 1$ ($BRLR$).

So, for each $n$, $A_n = 1$, $B_n = 1$.

But for $n = 5$:
Words: $BLRLR$, $BRLRL$.
So $A_5 = 1$ ($BRLRL$), $B_5 = 1$ ($BLRLR$).

So, for each $n$, $A_n = 1$.

But this cannot be, since the number should be larger for $n = 11$.

Let us try to generalize differently.

Let us define $f(n, x)$ as the number of $n$-letter words starting with $B$ and ending with $x$, where $x$ is $L$ or $R$.

Let us consider the sequence of letters:
- The first letter is $B$.
- The next $n-2$ letters are each $L$ or $R$, with no two consecutive letters the same.
- The last letter is $L$.

So, the word is $B x_2 x_3 \ldots x_{n-1} L$, where each $x_i$ is $L$ or $R$, and $x_i \neq x_{i-1}$ for $i \geq 2$.

Let us define $g(k, y)$ as the number of sequences of length $k$ (positions $2$ to $n-1$), starting with $y$ ($L$ or $R$), with no two consecutive letters the same.

But perhaps it's easier to let $S_k$ be the number of sequences of length $k$ (positions $2$ to $n$), with letters $L$ or $R$, no two consecutive letters the same, and ending with $L$.

Let $a_k$ be the number of such sequences of length $k$ ending with $L$.
Let $b_k$ be the number of such sequences of length $k$ ending with $R$.

We want $a_{n-1}$, since the first letter is $B$, and the rest is a sequence of $n-1$ letters, with the last letter $L$, and no two consecutive letters the same.

The recurrence is:
- To get a sequence of length $k$ ending with $L$, the previous letter must be $R$, so $a_k = b_{k-1}$.
- To get a sequence of length $k$ ending with $R$, the previous letter must be $L$, so $b_k = a_{k-1}$.

With initial conditions:
For $k = 1$:
- The sequence is just one letter, which can be $L$ or $R$.
So $a_1 = 1$ (sequence $L$), $b_1 = 1$ (sequence $R$).

Then:
$a_2 = b_1 = 1$
b_2 = a_1 = 1
$a_3 = b_2 = 1$
b_3 = a_2 = 1

So, for all $k$, $a_k = 1$, $b_k = 1$.

But this cannot be, since the number should grow with $n$.

Wait, perhaps the error is in the initial conditions.

Let us try to write out all possible sequences for small $n$.

For $n = 2$:
- $BL$

For $n = 3$:
- $BLR$
- $BRL$

For $n = 4$:
- $BLRL$
- $BRLR$

For $n = 5$:
- $BLRLR$
- $BRLRL$

So, for $n = 2$, $1$ word.
For $n = 3$, $2$ words.
For $n = 4$, $2$ words.
For $n = 5$, $2$ words.

Wait, but for $n = 6$:
- $BLRLRL$
- $BRLRLR$

So, always $2$ words for $n \geq 3$.

But for $n = 11$:
- $BLRLRLRLRLR$
- $BRLRLRLRLRL$

So, the answer is $2$.

Therefore, the number of Vitas words with $11$ letters is $2$.