Eulerian Number Calculation

The Eulerian number \( A(n, m) \) is a mathematical concept that counts the number of permutations of the numbers \( 1, 2, \ldots, n \) in which exactly \( m \) elements are greater than the previous element. It can be defined recursively as:

\[
A(n, m) = (n - m) \cdot A(n-1, m-1) + (m + 1) \cdot A(n-1, m)
\]

with the base cases:

\[
A(n, m) = 0 \text{ if } m \geq n \text{ or } n = 0
\]
\[
A(n, 0) = 1 \text{ for all } n > 0
\]

Write a function `eulerian_num(n, m)` that computes the Eulerian number \( A(n, m) \) using the recursive definition.

#### Example Usage
```python [main.nopy]
print(eulerian_num(3, 1))  # Output: 4
print(eulerian_num(4, 1))  # Output: 11
print(eulerian_num(5, 3))  # Output: 26
```

#### Constraints
- The function should handle inputs where \( n \) and \( m \) are non-negative integers.
- The function should use recursion to compute the result.
def eulerian_num(n, m):
    """
    Calculate the Eulerian number A(n, m) using its recursive definition.

    Args:
        n (int): The number of elements.
        m (int): The number of ascents.

    Returns:
        int: The Eulerian number A(n, m).
    """
    # Base cases
    if m >= n or n == 0:
        return 0
    if m == 0:
        return 1

    # Recursive calculation
    # Placeholder for the recursive formula
    pass
Advanced recursionmath
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