I saw this question, but was unable to answer because of my inexistant knowledge of Rust, so I decided to try and write an algorithm on my own and put it for review.There is the casual way to compute factorials:\$n! = \prod\limits_{k=1}^n k\$public static BigInteger fact(BigInteger n)  => n == 0 ? 1 : n*fact(n - 1);The problem with this one is that with big factorials, it get's slow (or, in this case, throw a StackOverflowException.Now, reading this paper, I figured I'd try to implement a faster algorithm to compute factorials.From what I understood, the equation would look something like this :\$\prod\limits_{i=1}^{\frac{n}{2}}( \sum\limits_{z=0}^i (n - z\times2))((\lceil{\frac{n}{2}\rceil})\$  (if n is odd)\$)\$The idea is to reduce the number of multiplications by half.public static BigInteger Factorial(int n){    BigInteger sum = n;    BigInteger result = n;    for (int i = n - 2; i > 1; i -= 2)    {        sum = (sum + i);        result *= sum;    }    if (n % 2 != 0)        result *= (BigInteger)Math.Round((double)n / 2, MidpointRounding.AwayFromZero);    return result;}I decided to keep a sum variable so I don't need to redo the sum computing at each multiplication's iteration.I'd like to see if I missed something or if this can be optimized. (I'm doing this as a mathematical exercise, so there's no need for performance, I'm looking at ways to make it more performant just because I want to understant the maths behind it).
Faster(?) factorial calculations
c#;mathematics
\$0! = 1\$, which is a case not handled by the code.The method should throw an ArgumentOutOfRangeException if n is negative.You could also replace thisif (n % 2 != 0)    result *= (BigInteger)Math.Round((double)n / 2, MidpointRounding.AwayFromZero);with thisif (n % 2 != 0)    result *= n / 2 + 1;if you find it clearer.