Let $c \ge 2$ be a fixed integer. Let $a_1 = c$ and for all $n \ge 2$ let $a_n = c \cdot \phi (a_{n-1})$. What are the numbers $c$ for which sequence $(a_n)$ will be bounded?

$\phi$ denotes Euler’s Phi Function, meaning that $\phi (n)$ gives the number of integers within the set $\{1, 2, . . . , n\}$ that are relative primes to $n$. We call a sequence $(x_n)$ bounded if there exist a constant $D$ such that $|x_n| < D$ for all positive integers $n$.
1. **Initial Setup and Definitions:**
   - Let \( c \ge 2 \) be a fixed integer.
   - Define the sequence \( \{a_n\} \) such that \( a_1 = c \) and for all \( n \ge 2 \), \( a_n = c \cdot \phi(a_{n-1}) \).
   - Here, \(\phi\) denotes Euler’s Phi Function, which gives the number of integers within the set \(\{1, 2, \ldots, n\}\) that are relatively prime to \( n \).

2. **Divisibility and Sequence Behavior:**
   - Since \( a \mid b \) implies \( \phi(a) \mid \phi(b) \) and \( a_1 \mid a_2 \), it follows that \( a_1 \mid a_2 \mid a_3 \mid \ldots \).
   - This implies that the sequence \( \{a_n\} \) is non-decreasing and each term divides the next.

3. **Boundedness and Constancy:**
   - The sequence \( \{a_n\} \) is bounded if and only if it eventually becomes constant. This means there exists some \( N \) such that for all \( n \ge N \), \( a_n = a_{N} \).

4. **Form of \( c \):**
   - For the sequence to be eventually constant, \( c \) must be of the form \( \frac{N}{\phi(N)} \) for some integer \( N \).
   - We need to find the integer values of \( c \) that satisfy this condition.

5. **Analyzing the Form \( \frac{N}{\phi(N)} \):**
   - Let \( N \) have prime factors \( p_1 < p_2 < \ldots < p_k \).
   - The quotient \( \frac{N}{\phi(N)} \) can be written as:
     \[
     \frac{p_1 p_2 \ldots p_k}{(p_1 - 1)(p_2 - 1) \ldots (p_k - 1)}
     \]
   - The denominator \((p_1 - 1)(p_2 - 1) \ldots (p_k - 1)\) is divisible by \( 2^{k-1} \), implying \( k \le 2 \).

6. **Case Analysis:**
   - **Case \( k = 1 \):**
     - \( c = \frac{p}{p-1} \) is an integer if and only if \( p = 2 \).
     - Thus, \( c = 2 \).

   - **Case \( k = 2 \):**
     - \( c = \frac{pq}{(p-1)(q-1)} \) must be an integer.
     - The denominator \((p-1)(q-1)\) is even, so one of the primes must be 2.
     - Let \( p = 2 \), then:
       \[
       c = \frac{2q}{(2-1)(q-1)} = \frac{2q}{q-1} = 2 + \frac{2}{q-1}
       \]
     - For \( c \) to be an integer, \( \frac{2}{q-1} \) must be an integer, implying \( q-1 \) divides 2.
     - The possible values for \( q \) are 2 and 3. Since \( q \) must be a prime greater than 2, \( q = 3 \).
     - Thus, \( c = 3 \).

7. **Verification:**
   - For \( c = 2 \):
     - The sequence is \( 2, 2, 2, \ldots \), which is constant and hence bounded.
   - For \( c = 3 \):
     - The sequence is \( 3, 6, 6, 6, \ldots \), which becomes constant after the second term and hence is bounded.

Conclusion:
The only integer values of \( c \) for which the sequence \( \{a_n\} \) is bounded are \( 2 \) and \( 3 \).

The final answer is \( \boxed{ c = 2 } \) or \( c = 3 \).