$f(n)$ is the least number that there exist a $f(n)-$mino that contains every $n-$mino.
Prove that $10000\leq f(1384)\leq960000$.
Find some bound for $f(n)$
1. **Understanding the Problem:**
   - We need to find the least number \( f(n) \) such that there exists an \( f(n) \)-mino containing every \( n \)-mino.
   - An \( n \)-mino is a shape made up of \( n \) equal squares connected edge-to-edge.

2. **Given Bounds:**
   - We are given that \( 10000 \leq f(1384) \leq 960000 \).

3. **Lower Bound Analysis:**
   - To find a lower bound for \( f(n) \), consider the number of distinct \( n \)-minos.
   - The number of distinct \( n \)-minos grows exponentially with \( n \).
   - For \( n = 1384 \), the number of distinct \( n \)-minos is very large.
   - Each \( n \)-mino must fit within the \( f(n) \)-mino, implying \( f(n) \) must be at least as large as the number of distinct \( n \)-minos.
   - Therefore, \( f(1384) \) must be significantly large, supporting the lower bound \( 10000 \).

4. **Upper Bound Analysis:**
   - To find an upper bound for \( f(n) \), consider the maximum possible size of an \( n \)-mino.
   - An \( n \)-mino can be arranged in a grid of size \( \sqrt{n} \times \sqrt{n} \).
   - For \( n = 1384 \), the grid size is approximately \( \sqrt{1384} \approx 37.2 \).
   - Thus, an \( f(n) \)-mino must be able to contain a grid of size \( 37.2 \times 37.2 \).
   - However, since \( f(n) \) must contain all possible \( n \)-minos, it must be much larger.
   - The upper bound \( 960000 \) is reasonable given the exponential growth of distinct \( n \)-minos.

5. **Conclusion:**
   - The bounds \( 10000 \leq f(1384) \leq 960000 \) are reasonable based on the analysis of the number of distinct \( n \)-minos and the size of the grid required to contain them.

The final answer is \( \boxed{ 10000 \leq f(1384) \leq 960000 } \).