To understand the answer, let’s break it down step by step and analyze the logic used to find the recurrence relation for the number of ternary strings $a_n$ of length $n$ that contain either two consecutive 0's or two consecutive 1's.

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Step 1: Define the problem

We are tasked with counting ternary strings (strings consisting of $0, 1,$ and $2$) of length $n$ that contain at least one occurrence of either "00" or "11." Let $a_n$ represent this count.

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Step 2: Build the recurrence relation

The answer derives $a_n$ by systematically constructing all possible strings of length $n$ that meet the condition.

Case 1: Strings that start with 2

• If the first character of the string is $2$, then the remaining $n-1$ characters can form any ternary string of length $n-1$ that satisfies the condition (contains "00" or "11").
• Thus, there are $a_{n-1}$ such strings.

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Case 2: Strings that alternate 0's and 1's, followed by 2, followed by $k$-length strings

• Suppose the string alternates between 0 and 1 for the first $n-1-k$ positions, followed by a $2$, and ends with a string of length $k$ that satisfies the condition.
• The number of such strings is: $2a_k$ (The factor $2$ arises because the alternating portion can either start with $0$ or $1$).

Summing over all possible $k$ values (from $0$ to $n-2$), we get: $\sum_{k=0}^{n-2} 2a_k.$

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Case 3: Strings with no 2's at all

• These strings consist of alternating 0's and 1's for $n-k-2$ positions, followed by a pair of "00" or "11," followed by any ternary string of length $k$.
• The number of such strings is: $2 \cdot 3^k$ (The factor $2$ accounts for the two possible ending pairs: "00" or "11").

Summing over all possible $k$ values (from $0$ to $n-2$), we get: $\sum_{k=0}^{n-2} 2 \cdot 3^k.$

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Step 3: Summing all contributions

Combining all cases, we derive: $a_n = a_{n-1} + \sum_{k=0}^{n-2} 2a_k + \sum_{k=0}^{n-2} 2 \cdot 3^k.$

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Step 4: Simplify the second summation

The summation $\sum_{k=0}^{n-2} 2 \cdot 3^k$ is a geometric progression: $\sum_{k=0}^{n-2} 3^k = \frac{3^{n-1} - 1}{2}.$ Thus: $\sum_{k=0}^{n-2} 2 \cdot 3^k = 3^{n-1} - 1.$

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Step 5: Substitute into the recurrence

Plugging everything into the recurrence, we have: $a_n = a_{n-1} + \sum_{k=0}^{n-2} 2a_k + (3^{n-1} - 1).$

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Step 6: Derive the closed form

To simplify further, observe that subtracting $a_n$ from $a_{n-1}$ in the recurrence relation leads to the closed form: $a_n = 2a_{n-1} + a_{n-2} + 2 \cdot 3^{n-2}.$

This closed form directly expresses $a_n$ in terms of $a_{n-1}$ and $a_{n-2}$, making it easier to compute values iteratively.

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Summary

The recurrence relation combines contributions from:

1. Strings starting with $2$,
2. Alternating strings with $k$-length suffixes, and
3. Strings with no $2$ at all.

The simplified closed form is: $a_n = 2a_{n-1} + a_{n-2} + 2 \cdot 3^{n-2}.$

Let me know if you have any questions or need further clarification.

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Follow-up Questions

1. How does the alternating portion contribute to the recurrence relation?
2. Why is $\sum_{k=0}^{n-2} 3^k$ a geometric progression?
3. Can this recurrence be solved explicitly for small values of $n$?
4. What is the base case for $a_0$ and $a_1$?
5. How would the recurrence change if ternary strings had additional restrictions?

Tip: Understanding recurrence relations often involves tracing a few values to verify the logic!