Find a recurrence relation for the number of ternary

strings of length n that contain two consecutive sym-

bols that are the same.
answer: "Let an be the number of ternary strings that contain two consecutive symbols that are the same. We will

develop a recurrence relation for an by exploiting the symmetry among the three symbols. In particular, it

must be the case that an/3 such strings start with each of the three symbols. Now let us see how we might

specify a string of length n satisfying the condition. We can choose the first symbol in any of three ways.

We can follow this by a string that starts with a different symbol but has in it a pair of consecutive symbols;

by what we have just said, there are 2an−1/3 such strings. Alternatively, we can follow the initial symbol

by another copy of itself and then any string of length n − 2; there are clearly 3n−2 such strings. Thus the

recurrence relation is an = 3 · ((2an−1/3) + 3n−2) = 2an−1 + 3n−1 . It is valid for all n ≥ 2."
explain answer

### Explanation of the Recurrence Relation:

The problem is to find the number of **ternary strings** of length \( n \) that contain **two consecutive symbols that are the same**. Let's break the explanation into steps:

---

### 1. **Defining the Recurrence Relation**
Let:
- \( a_n \): The number of ternary strings of length \( n \) that contain at least two consecutive identical symbols.

We aim to derive a recurrence relation for \( a_n \).

---

### 2. **Symmetry Among the Symbols**
Ternary strings consist of three symbols (let's call them \( 0, 1, 2 \)). Each of the symbols plays a symmetric role in constructing the strings, meaning we don't have to treat them differently when counting possibilities. This symmetry will simplify our analysis.

---

### 3. **Case Analysis for Strings of Length \( n \)**
To construct a ternary string of length \( n \) that contains at least one pair of consecutive identical symbols, consider the following:

#### (a) **First Symbol is Fixed**:
We can choose the first symbol in **any of the three ways** (\( 0, 1, \) or \( 2 \)).

#### (b) **What Comes After the First Symbol?**
There are two cases:
1. **Case 1**: The second symbol is **different** from the first symbol, but the remaining part of the string (\( n-1 \)) must contain a pair of consecutive identical symbols.
   - The number of such strings is proportional to \( a_{n-1} \), the number of valid strings of length \( n-1 \) containing a pair of identical consecutive symbols.
   - There are **2 choices** for the second symbol (since it must be different from the first). Therefore, there are \( 2a_{n-1}/3 \) such strings (the division by \( 3 \) accounts for the symmetry, as one-third of all \( a_{n-1} \) strings start with the first fixed symbol).

2. **Case 2**: The second symbol is **the same** as the first symbol.
   - This immediately creates a valid pair of consecutive identical symbols.
   - After this, the remaining \( n-2 \) symbols can form **any ternary string** (because the condition of having at least one pair of identical consecutive symbols is already satisfied).
   - The number of such strings is \( 3^{n-2} \), as there are 3 choices for each of the \( n-2 \) positions.

---

### 4. **Combining the Two Cases**
The total number of valid strings of length \( n \) is the sum of the two cases:
\[
a_n = 3 \cdot \left( \frac{2a_{n-1}}{3} \right) + 3^{n-2}
\]
Simplify:
\[
a_n = 2a_{n-1} + 3^{n-1}
\]

---

### 5. **Validity of the Recurrence Relation**
- The recurrence relation is valid for \( n \geq 2 \), as for \( n=1 \), there cannot be any pairs of consecutive symbols. Typically, we set \( a_1 = 0 \).

---

### Recap:
- **Initial Condition**: \( a_1 = 0 \) (a string of length 1 has no consecutive symbols).
- **Recurrence Relation**: 
  \[
  a_n = 2a_{n-1} + 3^{n-1}, \quad \text{for } n \geq 2
  \]

This recurrence relation models the number of ternary strings of length \( n \) containing at least two consecutive identical symbols.

---

Would you like further details on how the terms are derived, or an example calculation of \( a_n \) for specific \( n \)?

### Related Questions:
1. How do we compute \( a_n \) for specific values of \( n \)?
2. What is the behavior of \( a_n \) as \( n \to \infty \)?
3. How can we use this recurrence relation to solve for \( a_n \) programmatically?
4. Can this method be generalized to other bases (e.g., binary, quaternary)?
5. What does the solution look like for small \( n \) (e.g., \( n=2, 3, 4 \))?

---

**Tip:** When working with recurrence relations, always verify the initial condition and ensure the logic covers all cases for the transition from \( n-1 \) to \( n \).

Find a recurrence relation for the number of ternary strings of length n that contain two consecutive symbols that are the same.

Learn how to derive the recurrence relation for the number of ternary strings of length n that contain two consecutive identical symbols. The problem involves combinatorics and recurrence relations, with a detailed solution using symmetry and case analysis. Suitable for advanced high school students.

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