Collatz conjecture testing code

Because the main post would've been too crowded otherwise, the Python code is hidden here. Obviously, this isn't production-quality code -- no comments and shorthand for if statements! -- but it's still fun to play with. :)

def f(n):
    """
    Return the number of recursive calls of this function on n needed to
    produce a value of 0.
    """
    return _f(n, 0)

def _f(n, calls):
    result = (n / 2) if n % 2 == 0 else ((3 * n) + 1)
    calls += 1
    return calls if result == 1 else _f(result, calls)

if __name__ == '__main__':

    for n in range(1, 1000):
        print "n = %d: %d" % (n, f(n))