The double factorial function

P2.4.6The double factorial function

The factorial function, $n! = 1\cdot 2\cdot 3\cdot \cdots (n-1)n$ is the product of the first $$n$$ positive integers and is provided by the math module's factorial method. The double factorial function, $$n!!$$ , is the product of the positive odd integers up to and including $$n$$ (which must itself be odd):

n!! = \prod_{i=1}^{(n+1)/2} (2i-1) = 1\cdot 3\cdot 5\cdots (n-2) \cdot n.

Write a routine to calculate $$n!!$$ in Python.

As a bonus exercise, extend the formula for to allow for even $$n$$ as follows:

n!! = \prod_{i=1}^{n/2} (2i) = 2\cdot 4\cdot 6\cdots (n-2)\cdot n.

Show solution

To calculate $$n!!$$ for odd $$n$$ , simply loop over the odd integers up to and including $$n$$ , multiplying them together:

# Calcualate n!! = 1.3.5.7...(n-2)n for odd n
n = 11
dfactorial = 1
for i in range(3, n+1, 2):
    dfactorial *= i
print('{:d}!! = {:d}'.format(n, dfactorial))

For even $$n$$ , note that

n!! = \prod_{i=1}^{n/2} (2i) = 2^{n/2} (n/2)!,

so:

# Calcualate n!! = 2.4.6.8...(n-2)n for even n
import math
n = 10
dfactorial = 2**(n // 2) * math.factorial(n // 2)
print('{:d}!! = {:d}'.format(n, dfactorial))