The double factorial $z!!$, like the normal factorial function, can be extended to the complex plane using

$$ z!! = 2^{z/2} \left(\frac{\pi}{2}\right)^{(\cos\pi z-1)/4} \left(\frac{z}{2}\right)! $$

which is a slightly nicer-looking variant from Wolfram Mathworld and is what you get when you type in "double factorial" in Wolfram Alpha. (NB $z! = \Gamma(z+1)$ just for consistency here.)

The function can take complex numbers, and satisfies the empty product, $0!! = 1$ as well as the double factorial for negative numbers using the identity

$$ x!! = x(x-2)!! \implies x!! = \frac{(x+2)!!}{x+2} $$

thus the poles at even negative numbers and the intermediate values $(-1)!! = 1$, $(-3)!! = -1$, $(-5)!! = 1/3$, etc.

My question is that is there a similar extension for the triple factorial and further multifactorials? I know there are extensions for the general multifactorial, but they don't cover all the premises raised here.

For example, this function is one possible extension:

$$ z\overset{\alpha}{\overbrace{!\cdots !}} = \alpha^{(z-1)/\alpha}\frac{(z/\alpha)!}{(1/\alpha)!}$$

but here $0!\cdots ! \neq 1$.

Has there been found a meromorphic function $z\overset{\alpha}{\overbrace{!\cdots !}}, \alpha\gt 0$ (probably in the form $f(z)(z/\alpha)!$) to the multifactorial such that they agree with the following...?

$$z\overset{\alpha}{\overbrace{!\cdots !}} = \prod_{k=0}^{\left\lfloor\frac{z-1}{\alpha}\right\rfloor}(z-\alpha k)\qquad z\in\mathbb{N} \tag{1} $$

$$ \left.\begin{aligned} z\overset{\alpha}{\overbrace{!\cdots !}} = \frac{(z+\alpha) \overset{\alpha}{\overbrace{!\cdots !}}}{z+\alpha} \\ \implies (-n\alpha)\overset{\alpha}{\overbrace{!\cdots !}} = \pm\infty \end{aligned} \ \right\} \ n\in\mathbb{N}\backslash 0 \tag{2} $$

$$ 0\overset{\alpha}{\overbrace{!\cdots !}} =1 \tag{3}$$

  • $\begingroup$ Look at $F_m(z)\sum_{k=0}^{m-1}c_k e^{2i \pi zk/m} \frac{\Gamma'}{\Gamma}(z/m+k/m), G_m'(z)=F_m(z),e^{G_m(z)}$ $\endgroup$ – reuns Apr 18 at 4:15
  • $\begingroup$ @reuns Is that to say $$ \exp\int\sum_{k=0}^{m-1} \left(c_k e^{2i\pi kz/m} \frac{\Gamma'(z/m + k/m)}{\Gamma(z/m + k/m)} \right) dz$$ ? How did you arrive there? $\endgroup$ – Infiaria Apr 18 at 5:26
  • $\begingroup$ @reuns your comment doesn't make much sense as written... what do you intend with the last exponential for example, and where are the arguments for the $\Gamma$ functions? Infiaria's guess at what you mean is as good as mine. $\endgroup$ – Brevan Ellefsen Apr 19 at 8:53

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