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Let's address the three parts of the problem in detail.

### Part (a): Recurrence relation for the number of \( n \)-digit binary sequences with no consecutive 1s.

We need to find a recurrence relation for the number of binary sequences of length \( n \) that do not have two consecutive 1s. Let's denote this number by \( a_n \).

- A sequence of length \( n \) can either end with a 0 or a 1.
- If it ends with a 0, then the first \( n-1 \) digits form any valid sequence of length \( n-1 \), so there are \( a_{n-1} \) such sequences.
- If it ends with a 1, the second-to-last digit must be a 0 (to avoid consecutive 1s), so the first \( n-2 \) digits form any valid sequence of length \( n-2 \), giving \( a_{n-2} \) sequences.

Thus, the recurrence relation is:
\[
a_n = a_{n-1} + a_{n-2}
\]
This is similar to the Fibonacci sequence but with different initial conditions:
- \( a_1 = 2 \) (the sequences are 0 and 1)
- \( a_2 = 3 \) (the sequences are 00, 01, and 10)

### Part (b): Recurrence relation for \( n \)-digit ternary sequences.

For a ternary sequence (using digits 0, 1, and 2), there are no restrictions. So, each digit has 3 possible choices.

Thus, the number of \( n \)-digit ternary sequences is simply:
\[
b_n = 3^n
\]
No recurrence relation is needed here since this is a straightforward calculation.

### Part (c): Recurrence relation for \( n \)-digit ternary sequences with no consecutive 1s or consecutive 2s.

Let \( c_n \) represent the number of ternary sequences of length \( n \) with no consecutive 1s or 2s.

- If the sequence ends with a 0, the first \( n-1 \) digits form any valid sequence of length \( n-1 \), giving \( c_{n-1} \) possibilities.
- If it ends with a 1, the second-to-last digit must be a 0 (to avoid consecutive 1s), so the first \( n-2 \) digits form any valid sequence of length \( n-2 \), giving \( c_{n-2} \) possibilities.
- If it ends with a 2, the second-to-last digit must also be a 0 (to avoid consecutive 2s), so the first \( n-2 \) digits again form any valid sequence of length \( n-2 \), giving \( c_{n-2} \) possibilities.

Thus, the recurrence relation is:
\[
c_n = c_{n-1} + 2c_{n-2}
\]
The initial conditions are:
- \( c_1 = 3 \) (the sequences are 0, 1, 2)
- \( c_2 = 7 \) (the sequences are 00, 01, 02, 10, 12, 20, 21)

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Would you like more details or have any questions?

Here are 5 related questions to consider:
1. How can you derive the closed-form solution of the recurrence in part (a)?
2. How does the Fibonacci sequence relate to the solution in part (a)?
3. What is the impact of increasing the base for sequences, as seen in the jump from binary to ternary?
4. Can you find a generating function for the sequence in part (c)?
5. How would the recurrence relation change in part (a) if more restrictions were added?

**Tip**: Recurrence relations often have connections to combinatorial problems, such as counting specific paths or patterns, which helps in understanding them better.

Find a recurrence relation for the number of n-digit binary sequences with no pair of consecutive 1s. Repeat for n-digit ternary sequences. Repeat for n-digit ternary sequences with no consecutive 1s or consecutive 2s.

Explore the recurrence relations for binary sequences with no consecutive 1s, ternary sequences, and ternary sequences with no consecutive 1s or 2s. Learn the step-by-step process of deriving these relations, including a Fibonacci-like relation and exponential growth for unrestricted sequences.

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