Problem 1493

To solve the problem, we need to analyze the behavior of the sequence $\{x_m\}$ defined by $x_0 = a$ and $x_{m+1} = f(x_m)$ for $m = 0, 1, 2, 3, \ldots$, where $f(n) = k \cdot n \cdot \left(1-\frac{1}{p_1}\right) \cdot \left(1-\frac{1}{p_2}\right) \cdots \left(1-\frac{1}{p_r}\right)$ and $p_1, p_2, \ldots, p_r$ are all distinct prime divisors of $n$.

1. **Express $f(n)$ in terms of Euler's Totient Function:**
$$f(n) = k \cdot n \cdot \left(1-\frac{1}{p_1}\right) \cdot \left(1-\frac{1}{p_2}\right) \cdots \left(1-\frac{1}{p_r}\right) = k \cdot \phi(n)$$
where $\phi(n)$ is Euler's Totient Function, which counts the number of integers up to $n$ that are coprime with $n$.

2. **Behavior of $\phi(n)$:**
Euler's Totient Function $\phi(n)$ is always less than $n$ for $n > 1$. Specifically, $\phi(n) = n \prod_{i=1}^r \left(1 - \frac{1}{p_i}\right)$, where $p_1, p_2, \ldots, p_r$ are the distinct prime divisors of $n$.

3. Boundedness of the sequence:
We need to determine for which values of $k$ the sequence $\{x_m\}$ is bounded for all integers $a > 1$.

4. **Case $k = 2$:**
$$f(n) = 2 \cdot \phi(n)$$
For sufficiently large $i$, all prime divisors of $x_i$ are less than or equal to the largest prime divisor of $k$, which is 2. Thus, $x_i$ will eventually be a power of 2, and $\phi(2^m) = 2^m \left(1 - \frac{1}{2}\right) = 2^{m-1}$. Therefore, $f(2^m) = 2 \cdot 2^{m-1} = 2^m$, and the sequence becomes constant and hence bounded.

5. **Case $k = 3$:**
$$f(n) = 3 \cdot \phi(n)$$
For sufficiently large $i$, all prime divisors of $x_i$ are less than or equal to the largest prime divisor of $k$, which is 3. Thus, $x_i$ will eventually be of the form $2^{m_i}3^{l_i}$. For such $x_i$, $\phi(2^{m_i}3^{l_i}) = 2^{m_i-1}3^{l_i-1}(2-1)(3-1) = 2^{m_i-1}3^{l_i-1}2 = 2^{m_i}3^{l_i-1}$. Therefore, $f(2^{m_i}3^{l_i}) = 3 \cdot 2^{m_i}3^{l_i-1} = 2^{m_i}3^{l_i}$, and the sequence becomes constant and hence bounded.

6. **Case $k > 3$:**
For $k > 3$, consider the product of the terms:
$$k \prod_{p \le p_0} \left(1 - \frac{1}{p}\right)$$
where $p_0$ is the largest prime divisor of $k$. If $k > 3$, then:
$$k \prod_{p \le p_0} \left(1 - \frac{1}{p}\right) > 1$$
This implies that $f(n) = k \cdot \phi(n)$ will eventually increase $x_i$ rather than decrease it, making the sequence unbounded.

Conclusion:
The sequence $\{x_m\}$ is bounded for all integers $a > 1$ if and only if $k = 2$ or $k = 3$.

The final answer is $\boxed{ k = 2 }$ or $k = 3$.