Problem 1391

To solve this problem, we need to analyze the behavior of the sequence $\{x_n\}$ given the piecewise recurrence relation. We will determine the fundamental period $k$ for which $x_k = a$ and find the values of $a$ that satisfy this condition.

1. Analyzing the recurrence relation:
The sequence $\{x_n\}$ is defined by:
$$x_{n+1} = \begin{cases} 1 + 1009x_n & \text{if } x_n \leq 2 \\ 2021 - x_n & \text{if } 2 < x_n \leq 1010 \\ 3031 - 2x_n & \text{if } 1010 < x_n \leq 1011 \\ 2020 - x_n & \text{if } 1011 < x_n \end{cases}$$

2. **Case 1: $x_n \leq 2$**
$$x_{n+1} = 1 + 1009x_n$$
This is a linear recurrence relation. If $x_0 = a \leq 2$, then:
$$x_1 = 1 + 1009a$$
For $x_1$ to be equal to $a$ again, we need:
$$1 + 1009a = a \implies 1009a = -1 \implies a = -\frac{1}{1009}$$
However, $a$ must be in the range $1 \leq a \leq 2019$, so this case does not provide a valid solution.

3. **Case 2: $2 < x_n \leq 1010$**
$$x_{n+1} = 2021 - x_n$$
This is a simple reflection about the point $1010.5$. If $x_0 = a$ and $2 < a \leq 1010$, then:
$$x_1 = 2021 - a$$
For $x_1$ to be equal to $a$ again, we need:
$$2021 - a = a \implies 2a = 2021 \implies a = 1010.5$$
However, $a$ must be an integer, so this case does not provide a valid solution.

4. **Case 3: $1010 < x_n \leq 1011$**
$$x_{n+1} = 3031 - 2x_n$$
This is another linear recurrence relation. If $x_0 = a$ and $1010 < a \leq 1011$, then:
$$x_1 = 3031 - 2a$$
For $x_1$ to be equal to $a$ again, we need:
$$3031 - 2a = a \implies 3a = 3031 \implies a = \frac{3031}{3} \approx 1010.33$$
Again, $a$ must be an integer, so this case does not provide a valid solution.

5. **Case 4: $1011 < x_n$**
$$x_{n+1} = 2020 - x_n$$
This is another reflection about the point $1010$. If $x_0 = a$ and $1011 < a \leq 2019$, then:
$$x_1 = 2020 - a$$
For $x_1$ to be equal to $a$ again, we need:
$$2020 - a = a \implies 2a = 2020 \implies a = 1010$$
However, $a$ must be greater than $1011$, so this case does not provide a valid solution.

6. Conclusion:
Based on the analysis, we find that the sequence $\{x_n\}$ does not have a fundamental period $k$ for any $a$ in the given range $1 \leq a \leq 2019$. Therefore, there are no integers that can be a fundamental period of the sequence.