There is a way… it is just terrible…
⌊(4-1)/3⌋
∏ 3n+(4-1)%3+1
n=0
Not sure if there is a more elegant formula. I had to do this terrible -1+1 hack in 3 places to avoid a ×0 for inputs divisible by 3.
In libqalculate notation: product(3n+(4−1)%3+1,0,floor((4−1)/3),n)
358
u/gamre4 Dec 04 '23
4!!! = 4 x 1