I want to make an algorithm for $n!$ which will divide the number in half and call the algorithm again until n is so close to $0$ that a value of $1$ can be safely returned, and use the value of each $\tfrac{n}{2}!$ to compute the value of $n!$ to be returned to the previous call, and so on, without using iterated multiplication. Here is an example of what I don't want to use:$$\begin{align}(2n)!&=\left(\prod_{n+1}^{2n}{k}\right)n!\\&=\binom{2n}{n}(n!)^2\end{align}$$ In fact, I want something that will work on any positive number, not just integers. Is there a recurrence relation that meets my needs in this regard?
Just to be clear, I'm not asking what you think is the most efficient way, as an approximation may be, and I'm not looking for a divide-and-conquer algorithm. I simply want to find a recurrence relation to test my theory with. This question is still open; if you know such a recurrence relation, please share it.
this is not a question of a recurrence relation, instead its a question of divide and conquer technique to find the factorial, which wont be useful btw, complexity will remain same, as the numbers are eventually have to be multiplied. (This is not the same as merge sort, in merge sort one can gain advantage because of it is easier to sort already sorted lists)
Try asking it in computer programming section.
and yes, Recurrence relation will be used to find the complexity of the program, which should be
$$O(n)$$ even after using D&C.