A function $f$ defined on integers such that

$f (n) =n + 3$ if $n$ is odd
$f (n) = \frac{n}{2}$ if $n$ is even

If $k$ is an odd integer, determine the values for which $f (f (f (k))) = k$.
Given the function \( f \) defined on integers such that:
\[ f(n) = 
\begin{cases} 
n + 3 & \text{if } n \text{ is odd} \\
\frac{n}{2} & \text{if } n \text{ is even}
\end{cases}
\]

We need to determine the values of \( k \) (where \( k \) is an odd integer) for which \( f(f(f(k))) = k \).

1. **First Application of \( f \):**
   Since \( k \) is odd, we use the first case of the function:
   \[ f(k) = k + 3 \]
   Here, \( k + 3 \) is even because the sum of an odd number and 3 (an odd number) is even.

2. **Second Application of \( f \):**
   Since \( k + 3 \) is even, we use the second case of the function:
   \[ f(f(k)) = f(k + 3) = \frac{k + 3}{2} \]
   Here, \(\frac{k + 3}{2}\) is an integer because \( k + 3 \) is even.

3. **Third Application of \( f \):**
   We need to consider whether \(\frac{k + 3}{2}\) is odd or even:
   - If \(\frac{k + 3}{2}\) is odd, then:
     \[ f(f(f(k))) = f\left(\frac{k + 3}{2}\right) = \frac{k + 3}{2} + 3 \]
   - If \(\frac{k + 3}{2}\) is even, then:
     \[ f(f(f(k))) = f\left(\frac{k + 3}{2}\right) = \frac{\frac{k + 3}{2}}{2} = \frac{k + 3}{4} \]

4. **Setting \( f(f(f(k))) = k \):**
   We need to solve for \( k \) such that:
   \[ f(f(f(k))) = k \]

   Let's consider the case where \(\frac{k + 3}{2}\) is odd:
   \[ \frac{k + 3}{2} + 3 = k \]
   Simplifying this equation:
   \[ \frac{k + 3}{2} + 3 = k \]
   \[ \frac{k + 3 + 6}{2} = k \]
   \[ \frac{k + 9}{2} = k \]
   \[ k + 9 = 2k \]
   \[ 9 = k \]
   This solution is not valid because 9 is not an odd integer.

   Now, consider the case where \(\frac{k + 3}{2}\) is even:
   \[ \frac{k + 3}{4} = k \]
   Simplifying this equation:
   \[ k + 3 = 4k \]
   \[ 3 = 3k \]
   \[ k = 1 \]

   Since \( k = 1 \) is an odd integer, it satisfies the condition.

The final answer is \( \boxed{ k = 1 } \)