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### Problem Breakdown:

We are tasked with analyzing the number of bit strings of length \( n \) that do **not** contain the pattern "101". Let’s denote the number of such bit strings by \( a_k \), where \( k \geq 0 \).

#### Part (a) Proof

We need to show:
\[
a_k = a_{k-1} + a_{k-3} + a_{k-4} + \cdots + a_0 + 2
\]
for every integer \( k \geq 3 \).

#### Part (b) Proof

We are required to use the result of part (a) to show that for \( k \geq 3 \):
\[
a_k = 2a_{k-1} - a_{k-2} + a_{k-3}.
\]

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Let's first approach part (a).

### **Part (a): Understanding and Proof**

We need to determine how to generate bit strings of length \( k \) that do not contain the pattern "101". We can break the string down based on whether it starts with "1" or "0", and build a recurrence relation accordingly. Given that the pattern "101" is forbidden, there are constraints when choosing the next bits after a "1".

- For \( a_k \), we can construct valid strings in two main ways:
  1. **Start with '0':** This reduces the problem to finding a string of length \( k-1 \) that avoids "101", hence there are \( a_{k-1} \) such strings.
  2. **Start with '1':** We must be cautious here, as we can't directly follow with "01" (since "101" is forbidden). Valid subsequences depend on more detailed analysis of previous bit lengths (using prior terms like \( a_{k-3}, a_{k-4} \), etc.).

This structure gives rise to a recurrence relation that sums over earlier values of \( a \), plus a base case or constant.

### **Part (b): Recurrence Relation**

Once we derive a sum expression like the one in part (a), we can then manipulate it (through algebraic techniques like telescoping sums) to simplify it into a linear recurrence relation of the form:
\[
a_k = 2a_{k-1} - a_{k-2} + a_{k-3}.
\]
This form captures dependencies between current and previous values of \( a_k \).

Would you like me to walk you through the algebraic steps for both parts in detail?

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Here are 5 related questions to this problem:

1. How can we generate the initial values \( a_0, a_1, a_2, \dots \) for this recurrence?
2. How do linear recurrence relations relate to combinatorial problems like this one?
3. Could generating functions help in finding a closed form for \( a_k \)?
4. How would the recurrence change if we forbid a different pattern (e.g., "110")?
5. How could we use dynamic programming to compute \( a_k \) for large \( k \)?

**Tip:** When dealing with recurrence relations, always check for patterns in small values to confirm your recurrence formula matches expected results.

For each integer n >= 0, let a_k be the number of bit strings of length n that do not contain the pattern 101. a) Show that a_k = a_(k-1) + a_(k-3) + a_(k-4) + ... + a_0 + 2, for every integer k >= 3. b) Use the result of part (a) to show that if k >= 3, then a_k = 2a_(k-1) - a_(k-2) + a_(k-3).

Learn how to derive recurrence relations for the number of bit strings that do not contain the pattern 101. This problem explores advanced combinatorics concepts using recurrence relations.

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