Let’s delve into the explanation for the recurrence relation provided in the problem:

1. Understanding the Problem

We want to find a recurrence relation for the number of ternary strings of length $n$ that contain either:

• Two consecutive 0's (00), or
• Two consecutive 1's (11).

Ternary strings use the digits $\{0, 1, 2\}$, so the structure of such strings depends on how the conditions (two consecutive 0's or 1's) are satisfied.

---

2. Breaking Down the String Construction

Case 1: The string starts with 2.

• The rest of the string (of length $n-1$) must still satisfy the condition (contain consecutive 0's or 1's).
• There are $a_{n-1}$ ways to construct such a string.

Case 2: The string starts with a sequence of alternating 0's and 1's, followed by a 2.

• Let the alternating sequence have length $n-1-k$. For example:
  • If $k = 0$, the sequence ends with 2 immediately (e.g., 2).
  • If $k = 1$, the sequence might look like $0, 1, 2$ or $1, 0, 2$, and so on.
• After the alternating sequence, there is a string of length $k$ that must contain either consecutive 0's or 1's. There are $a_k$ ways to construct such strings.
• Since the alternating sequence can start with 0 or 1, there are $2a_k$ total ways for this case.

Case 3: The string contains no 2's at all.

• In this case, the string is entirely made up of $0's$ and $1's$. It consists of:

  • $n-k-2$ alternating $0's$ and $1's$, followed by
  • A pair of consecutive $0's$ or $1's$ to satisfy the condition, followed by
  • A string of length $k$ that satisfies the condition.

• The number of such strings is $2 \cdot 3^k$, where:

  • $2$ accounts for the choice of whether the pair is $00$ or $11$,
  • $3^k$ accounts for the $k$-length ternary string.

---

3. Summing Up

By combining these cases, the recurrence relation is:

$a_n = a_{n-1} + \sum_{k=0}^{n-2} 2a_k + \sum_{k=0}^{n-2} 2 \cdot 3^k.$

---

4. Simplifying

The summation $\sum_{k=0}^{n-2} 2 \cdot 3^k$ is a geometric progression with a common ratio of 3. Its sum is:

$\sum_{k=0}^{n-2} 2 \cdot 3^k = 2 \cdot \frac{3^{n-1} - 1}{3 - 1} = 3^{n-1} - 1.$

Thus, the recurrence becomes:

$a_n = a_{n-1} + 2(a_{n-2} + a_{n-3} + \dots + a_0) + (3^{n-1} - 1).$

---

5. Deriving the Closed Form

By subtracting this recurrence from the same relation for $n-1$, you can simplify to:

$a_n = 2a_{n-1} + a_{n-2} + 2 \cdot 3^{n-2}.$

This is the ultimate recurrence relation for the problem.

---

Key Insights

• The recurrence considers the multiple ways a string can start and satisfy the conditions.
• The $3^{n-2}$ term captures the contribution of strings that are purely alternating and then forced to satisfy the conditions at the end.

Do you have further questions or want examples for clarification?

---

Relative Questions:

1. Can you explain how the geometric progression $3^{n-1} - 1$ arises in this recurrence relation?
2. Why does the recurrence subtract $n-1$ in the geometric progression step?
3. How do the alternating sequences (Case 2) affect the overall recurrence relation?
4. Why is the closed form simpler than the full recurrence relation derived earlier?
5. What would the recurrence look like if we added an additional condition (e.g., no two consecutive 2’s)?

Tip:

Geometric progressions are often key when summing over repeated cases—mastering their derivation can simplify complex recurrence relations!