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:

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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 $a_n$.

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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.

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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.

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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: $a_n = 2a_{n-1} + 3^{n-1}$

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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 $a_1 = 0$.

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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.

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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., $n=2, 3, 4$)?

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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 $n$.