What is the triple factorial of a negative number, e.g., $-2$?

Question

The triple factorial of a positive integer is computed as $7!!! = 7\cdot 4\cdot 1$.

I'm interested in the value of $$(-2)!!!$$

I tried to find this value by using the Wolfram, but I found the Wolfram doesn't support this form.

Answer 1

The "triple factorial" is defined for non-negative integers $n$ as $$ n!!! = n (n-3)(n-6) \cdots (n \bmod 3) \, , $$ compare "Multifactorials" in Wikipedia. For $n \ge 0$ and $j > 0$ $$ n!!! = \frac{(n+3j)!!!}{(n+3)(n+6)\cdots (n+3j)} \tag{1} $$ holds. This relation can be used to define the function for negative integers as well, as long as they are not a multiple of $(-3)$: Just choose $j$ large enough such that $n + 3j\ge 0$. It is easy to see that the definition does not depend on the actual value of $j$.

For $n=-2$ we get $$ (-2) !!! = \frac {(-2 + 3)!!!}{(-2 + 3)} = \frac {1!!!}{1} = 1 \, . $$

Actually it can even be extended to all complex numbers except for the negative integers which are multiple of $(-3)$, and an explicit formula is given in the referenced Wikipedia article in terms of the Gamma function (choose $k=3$ in the "Alternative extension of the multifactorial"): $$ z!!! = 3^{(z-1)/3} \frac{\Gamma(\frac z3 + 1)}{\Gamma(\frac 13 + 1)} $$

For $z = -2$ this gives the same result $$ (-2)!!! = 3^{-1}\frac{\Gamma(\frac 13)}{\Gamma(\frac 43)} = 1 \, . $$

Answer 2

The value of $(-2)!!!=1$ according to $$n!!!=n(n-3)!!!$$ $$1!!!=1(1-3)!!!$$ $$1=(-2)!!!$$