**Double factorial** of a non-negative integer n, is the product of all the integers from 1 to n that have the same parity (odd or even) as n. It is also called as **semifactorial** of a number and is denoted by **!!**. For example, double factorial of 9 is 9*7*5*3*1 which is 945. Note that, a consequence of this definition is 0!! = 1.

Examples:

Input:6Output:48 Note that 6*4*2 = 48Input:7Output:105 Note that 7*5*3 = 105

For even n, the double factorial is:

For odd n, the double factorial is:

**Recursive Solution:**

Double factorial can be calculated using following recursive formula.

n!! = n * (n-2)!! n!! = 1 if n = 0 or n = 1

Following is C implementation of double factorial.

#include<stdio.h> // function to find double factorial of given number unsigned int doublefactorial(unsigned int n) { if (n == 0 || n==1) return 1; return n*doublefactorial(n-2); } int main() { printf("Double factorial is %d", doublefactorial(5)); return 0; }

Output:

Double factorial is 15

**Iterative Solution:**

Double factorial can also be calculated iteratively as recursion can be costly for large numbers.

#include<stdio.h> // function to find double factorial of given number unsigned int doublefactorial(unsigned int n) { int res = 1; for (int i=n; i>=0; i=i-2) { if (i==0 || i==1) return res; else res *= i; } } int main() { printf("Double factorial is %d", doublefactorial(5)); return 0; }

Output:

Double factorial is 15

Time complexity of the above solutions is O(n).

**Important Points :**

- Double factorial and factorial are related using below formula.
**Note :**n!! means double factorial. If n is even, i.e., n = 2k n!! = 2^{k}k! Else (n = 2k + 1) n!! = (2k)! / 2^{k}k! - Double factorial is frequently used in combinatorics. Refer wiki for list of applications. An example application is count of perfect matchings of a complete graph K
_{n+1}for odd n.

**References:**

https://en.wikipedia.org/wiki/Double_factorial

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