In the dog dictionary the words are any sequence of letters $A$ and $U$ for example $AA$, $UAU$ and $AUAU$. For each word, your "profundity" will be the quantity of subwords we can obtain by the removal of some letters.
For each positive integer $n$, determine the largest "profundity" of word, in dog dictionary, can have with $n$ letters.
Note: The word $AAUUA$ has "profundity" $14$ because your subwords are $A, U, AU, AA, UU, UA, AUU, UUA, AAU, AUA, AAA, AAUU, AAUA, AUUA$.
1. **Define the problem and initial conditions:**
   - We need to determine the largest "profundity" of a word in the dog dictionary with \( n \) letters.
   - The profundity of a word is the number of subwords that can be obtained by removing some letters.
   - We denote \( A_n \) as the value of the greatest profundity of an \( n \)-letter word in the dog dictionary.
   - We aim to prove by induction that \( A_n = F_{n+3} - 3 \), where \( F_n \) is the Fibonacci sequence.

2. **Base cases:**
   - For \( n = 2 \):
     - The word \( AA \) has profundity 1 (subword: \( A \)).
     - The word \( AU \) has profundity 2 (subwords: \( A, U \)).
     - Thus, \( A_2 = 2 = F_5 - 3 \).
   - For \( n = 3 \):
     - The word \( AAA \) has profundity 2 (subwords: \( A, AA \)).
     - The word \( AAU \) has profundity 4 (subwords: \( A, U, AA, AU \)).
     - The word \( AUA \) has profundity 5 (subwords: \( A, U, AU, UA, AA \)).
     - The word \( AUU \) has profundity 4 (subwords: \( A, U, AU, UU \)).
     - Thus, \( A_3 = 5 = F_6 - 3 \).

3. **Inductive step:**
   - Assume \( A_k = F_{k+3} - 3 \) for all \( k \leq n \).
   - We need to show \( A_{n+1} = F_{n+4} - 3 \).

4. **Statement 1:**
   - Consider the word \( AUAU \ldots \) (with \( n \) letters).
   - Every subword starting with \( A \) is either \( A \) or of the form \( AX \), where \( X \) is a subword of \( UAUA \ldots \) (with \( n-1 \) letters).
   - By hypothesis, there are \( 1 + A_{n-1} \) such subwords.
   - Every subword starting with \( U \) is either \( U \), \( UAUA \ldots \) (with \( n-1 \) letters), or of the form \( UY \), where \( Y \) is a subword of \( AUAU \ldots \) (with \( n-2 \) letters).
   - By hypothesis, there are \( 2 + A_{n-2} \) such subwords.
   - Therefore, the total number of subwords is:
     \[
     (1 + A_{n-1}) + (2 + A_{n-2}) = 3 + A_{n-1} + A_{n-2}
     \]
   - By the induction hypothesis:
     \[
     3 + (F_{n+2} - 3) + (F_{n+1} - 3) = F_{n+3} - 3
     \]
   - Thus, \( AUAU \ldots \) (with \( n \) letters) has profundity \( F_{n+3} - 3 \).

5. **Statement 2:**
   - Let \( T = X_1 X_2 \ldots X_n \) be an \( n \)-letter word with profundity \( A_n \).
   - Suppose \( X_1 = A \).
   - If all \( X_i = A \), then \( A_n = n-1 < F_{n+3} - 3 \), which is a contradiction.
   - Therefore, there exists \( c \in \{2, \ldots, n\} \) such that \( X_c = U \) and \( c \) is minimal.
   - Every subword starting with \( A \) is either \( A \) or of the form \( AX \), where \( X \) is a subword of \( X_2 \ldots X_n \).
   - By hypothesis, there are at most \( 1 + A_{n-1} \) such subwords.
   - Every subword starting with \( U \) is either \( U \), \( X_c \ldots X_n \), or of the form \( X_c Y \), where \( Y \) is a subword of \( X_{c+1} \ldots X_n \).
   - By hypothesis, there are at most \( 2 + A_{n-c} \) such subwords.
   - Therefore:
     \[
     A_n \leq (1 + A_{n-1}) + (2 + A_{n-c}) = 3 + A_{n-1} + A_{n-c}
     \]
   - Since \( A_n \) is non-decreasing:
     \[
     A_n \leq 3 + A_{n-1} + A_{n-2}
     \]

6. **Conclusion:**
   - By the induction hypothesis and Statement 2:
     \[
     A_n \leq 3 + (F_{n+2} - 3) + (F_{n+1} - 3) = F_{n+3} - 3
     \]
   - Since we have an example with profundity \( F_{n+3} - 3 \), the result follows by induction.

\(\blacksquare\)

The final answer is \( \boxed{ A_n = F_{n+3} - 3 } \)