10.3. In the language of the AU tribe, there are two letters - "a" and "u". Some sequences of these letters are words, and each word has no more than 13 letters. It is known that if you write down any two words in a row, the resulting sequence of letters will not be a word. Find the maximum possible number of words in such a language.

(I. Bogdanov)
Answer. $2^{14}-2^{7}=16056$.

First solution. If all sequences, the number of letters in which is not less than 7 and not more than 13, are words, then, obviously, the condition of the problem is satisfied; in this case, the number of such words is $2^{7}+\ldots+2^{13}=2^{14}-2^{7}$. It remains to show that this number is the largest possible.

The total number of sequences of length not exceeding 13 is $2+2^{2}+\ldots+2^{13}=2^{14}-2$. If there is no 7-letter word among the words in the language, then the total number of words does not exceed $2^{14}-2-2^{7}=2^{7}-2$ sequences. Therefore, it is possible to associate each sequence $a \in A$ with a sequence $b_{a} \in B$ so that all sequences $b_{a}$ are different. Note that then all sequences of the form $a b_{a}$ will also be different (since their 7-letter endings are different).

By the condition, in each of the $2^{7}-2$ triples $\left(a, b_{a}, a b_{a}\right)$, there are no more than two words of the language. Therefore, at least $2^{7}-2$ sequences of 13 or fewer letters are not words, and the total number of words is no more than $\left(2^{14}-2\right)-\left(2^{7}-2\right)=2^{14}-2^{7}$.

Comment. Only the answer - 0 points.

The example showing that the language can have $2^{14}-2^{7}$ words - 1 point.

It is proven that there are no more than $2^{14}-2^{7}$ words, but the example is missing - 5 points.