What is the triple factorial of a negative number, e.g., $-2$? 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. 
 A: 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
$$
(-2)!!! = 3^{-1}\frac{\Gamma(\frac 13)}{\Gamma(\frac 43)} = 1 \, .
$$
A: The value of $(-2)!!!=1$ according to 
$$n!!!=n(n-3)!!!$$
$$1!!!=1(1-3)!!!$$
$$1=(-2)!!!$$
