Given a language with an alphabet of n letters, prove that there exists a finite sequence of letters in this language such that no two adjacent blocks of it are the same, but if any letter is added to the beginning or the end of the sequence, this property no longer holds. (A block consists of one or more consecutive letters.)

To prove the existence of such a finite sequence of letters, we will construct an example and show that it has the desired properties.

Consider the language with an alphabet of \( n \) letters. Let's denote these letters by \( a_1, a_2, \ldots, a_n \).

### Construction of the Sequence

We will construct a sequence \( S \) by considering all possible blocks of lengths from 1 up to \( k \), where \( k \) is the maximum length such that the total number of blocks is less than or equal to \( n \). Specifically, we will use the de Bruijn sequence of order 2 over the alphabet of \( n \) letters. A de Bruijn sequence of order \( k \) over an alphabet of size \( n \) is a cyclic sequence in which every possible substring of length \( k \) appears exactly once.

In our case, we are interested in a de Bruijn sequence of order 2, which ensures that every possible pair of letters appears exactly once. This sequence has length \( n^2 \).

### Verifying the Properties

1. **No Two Adjacent Blocks are the Same:**
   - In a de Bruijn sequence of order 2, every possible pair of letters from the alphabet appears exactly once. Therefore, no pair of letters (block of length 2) can be repeated consecutively.

2. **Adding a Letter to the Beginning or End:**
   - If we add any letter to the beginning or end of the de Bruijn sequence, it will either repeat an existing pair or form a new pair that wasn't in the original sequence, breaking the property of every pair appearing exactly once.

### Detailed Explanation

The de Bruijn sequence \( B(2, n) \) of order 2 over \( n \) letters ensures that:
- Every possible combination of two letters (from the alphabet of size \( n \)) appears exactly once in the sequence.

Since \( B(2, n) \) is cyclic, for our purposes, we can consider a non-cyclic version of this sequence by listing the first \( n^2 - 1 \) characters. This ensures that we don't overlap the beginning and end pairs.

1. **Constructing \( S \):**
   - Let \( S \) be the de Bruijn sequence \( B(2, n) \) minus its last character to avoid the cyclic overlap.

2. **No Two Adjacent Blocks are the Same:**
   - By construction, each pair (block of length 2) in \( S \) is unique and appears exactly once.

3. **Adding a Letter Breaks the Property:**
   - If we add any letter \( x \) to the beginning of \( S \), the pair formed by \( x \) and the first letter of \( S \) will either be a repetition of an existing pair or disrupt the sequence's unique pair structure.
   - Similarly, adding any letter \( y \) to the end of \( S \) will form a pair with the last letter of \( S \), either repeating an existing pair or breaking the sequence's property.

Therefore, the sequence \( S \) constructed from the de Bruijn sequence \( B(2, n) \) minus its last character has the desired properties: no two adjacent blocks are the same, but adding any letter to the beginning or end violates this property.

### Conclusion

We have constructed a sequence \( S \) with the required properties using the de Bruijn sequence of order 2 over an alphabet of \( n \) letters. This sequence demonstrates that there exists a finite sequence of letters in the language such that no two adjacent blocks are the same, and adding any letter to the beginning or end disrupts this property.

Prove the existence of a finite sequence in a language with an alphabet of n letters where no two adjacent blocks are the same. Explore the construction of a sequence using combinatorial principles and de Bruijn sequences. Suitable for advanced high school students and beyond.

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