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…?
$$z\overset{\alpha}{\overbrace{!\cdots !}} = \prod_{k=0}^{\left\lfloor\frac{z-1}{\alpha}\right\rfloor}(z-\alpha k)\qquad z\in\mathbb{N} \tag{1} $$
$$ \left.\begin{aligned} z\overset{\alpha}{\overbrace{!\cdots !}} = \frac{(z+\alpha) \overset{\alpha}{\overbrace{!\cdots !}}}{z+\alpha} \\ \implies (-n\alpha)\overset{\alpha}{\overbrace{!\cdots !}} = \pm\infty \end{aligned} \ \right\} \ n\in\mathbb{N}\backslash 0 \tag{2} $$
$$ 0\overset{\alpha}{\overbrace{!\cdots !}} =1 \tag{3}$$
Best Answer
I will answer you briefly by summarizing only the formulas that I have already given in 3 answers and questions:
1.1 Compact form
$$z!_{(\alpha)}=\alpha^{\frac{z}{\alpha}}\Gamma\left(1+\frac{z}{\alpha}\right)\prod_{j=1}^{\alpha-1}\left(\frac{\alpha^{\frac{\alpha-j}{\alpha}}}{\Gamma\left(\frac{j}{\alpha}\right)}\right)^{C_{\alpha}(z-j)}$$ Where $$C_{\alpha}(z)=\frac{1}{\alpha}\left(1+2\sum_{k=1}^{\left\lfloor\frac{\alpha-1}{2}\right\rfloor}\cos\left(\frac{2k\pi}{\alpha}z\right)+\delta_{\alpha,2}\cos(\pi z)\right)$$ And $$\delta_{\alpha,2}=\text{mod}(\alpha-1,2)=\mathbf{1}_{2\mathbb{Z}}(\alpha)=\begin{cases}1&\alpha\text{ is even}\\0&\alpha\text{ is odd}\end{cases}$$
1.2 More efficient definition
Let $\beta:=\left\lfloor\frac{\alpha-1}{2}\right\rfloor$ $$z!_{(\alpha)}=\alpha^{\frac{z+z_0(z)}{\alpha}}\Gamma\left(1+\frac{z}{\alpha}\right)\frac{1}{\pi^{p(z)}}\prod_{j=1}^{\beta}\frac{\sin\left(\frac{j\pi}{\alpha}\right)^{C_{\alpha}(z+j)}}{\Gamma\left(\frac{j}{\alpha}\right)^{\gamma_j(z)}}$$
Where:
1.3 First cases
For $\color{red}{\alpha=1}$ the function is:
$$\color{blue}{z!_{(1)}=\Gamma(1+z)}$$
For $\color{red}{\alpha=2}$ the function is:
$$\color{blue}{z!_{(2)}=2^{\frac{z}{2}}\Gamma\left(1+\frac{z}{2}\right)\left(\frac{2}{\pi}\right)^{\frac{1-\cos(\pi z)}{4}}}$$
For $\color{red}{\alpha=3}$ the function is:
$$\color{blue}{z!_{(3)}=\frac{3^{\frac{z}{3}}\Gamma\left(1+\frac{z}{3}\right)}{\Gamma\left(\frac{1}{3}\right)^{\frac{2}{\sqrt{3}}\sin\left(\frac{2\pi z}{3}\right)}}\frac{3^{\frac{1}{2}-\frac{1}{2}\cos\left(\frac{2\pi z}{3}\right)-\frac{\sqrt{3}}{18}\sin\left(\frac{2\pi z}{3}\right)}}{\left(2\pi\right)^{\frac{1}{3}-\frac{2}{3}\cos\left(\frac{2\pi z}{3}-\frac{\pi}{3}\right)}}}$$
For $\color{red}{\alpha=4}$ the function is: $$\color{blue}{z!_{(4)}=2^{\frac{z}{2}}\frac{\Gamma\left(1+\frac{z}{4}\right)}{\Gamma\left(\frac{1}{4}\right)^{\sin\left(\frac{\pi z}{2}\right)}}\frac{\pi^{\frac{\cos(\pi z)-3}{8}+\frac{1}{2}\sin\left(\frac{\pi z}{2}\right)+\frac{1}{4}\cos\left(\frac{\pi z}{2}\right)}}{2^{\frac{\cos(\pi z)-5}{8}-\frac{3}{4}\sin\left(\frac{\pi z}{2}\right)+\frac{1}{2}\cos\left(\frac{\pi z}{2}\right)}}}$$
For $\color{red}{\alpha=5}$ the formula is:
$$\color{blue}{z!_{(5)}=5^{\frac{z+z_{0}(z)}{5}}\cdot\frac{\Gamma\left(1+\frac{z}{5}\right)}{\Gamma\left(\frac{1}{5}\right)^{\frac{\gamma_{1}(z)}{5}}\Gamma\left(\frac{2}{5}\right)^{\frac{\gamma_{2}(z)}{5}}}\cdot\left(2\pi\right)^{\frac{c_{1}(z)}{5}(z)}\phi^{\frac{c_{2}(z)}{5}}}$$
For $\color{red}{\alpha=6}$ the function is: $$\color{blue}{\displaystyle z!_{(6)}=\frac{\sqrt[12]{2}\sqrt{3}6^{\frac{z}{6}}\Gamma\left(1+\frac{z}{6}\right)}{\left(\frac{\sqrt[12]{3}}{\sqrt{2\pi}}\Gamma\left(\frac{1}{3}\right)\right)^{\sqrt{3}\sin\left(\frac{\pi z}{3}\right)+\frac{1}{\sqrt{3}}\sin\left(\frac{2\pi z}{3}\right)}}\cdot\frac{2^{\frac{\sqrt{3}}{9}\sin\left(\frac{\pi z}{3}\right)}\left(2\pi\right)^{\frac{1}{6}\cos\left(\frac{2\pi z}{3}\right)}}{3^{\frac{1}{4}\left(\cos\left(\frac{\pi z}{3}\right)+\cos\left(\frac{2\pi z}{3}\right)\right)}\pi^{\frac{5}{12}}}\cdot\left(\frac{\pi}{2}\right)^{\left(\frac{\cos\left(\pi z\right)}{12}+\frac{1}{6}\cos\left(\frac{\pi z}{3}\right)\right)}}$$
For $\color{red}{\alpha=8}$ the function is:
$$\color{blue}{z!_{(8)}=8^{\frac{z+z_0(z)}{8}}\cdot\frac{\Gamma\left(1+\frac{z}{8}\right)}{\Gamma\left(\frac{1}{8}\right)^{\gamma_{1}(z)}\Gamma\left(\frac{1}{4}\right)^{\gamma_{2}(z)}}\pi^{c_{1}(z)}\left(\frac{1}{2}\right)^{c_{2}(z)}\left(\sqrt{2}+1\right)^{c_{3}(z)}}$$
Etc...
2.1 Fourier expansion
$$x!_{(\alpha)}\approx \alpha^{\frac{x}{\alpha}}\Gamma\left(1+\frac{x}{\alpha}\right)\sum_{j=1}^{\alpha}\frac{\alpha^{\frac{\alpha-j}{\alpha}}}{\Gamma\left(\frac{j}{\alpha}\right)}C_{\alpha}\left(x-j\right)$$
2.2 Max relative error
Here I bring you the results of the maximum relative percentage error that you can obtain if you use the Fourier series $$\begin{array}{|c|c|c|c|}\hline \alpha&\text{max. rel. err.}&\alpha&\text{max. rel. err.}\\\hline 1&=0.000\%&11&\approx 2.688\%\\ 2&\approx 0.639\%&12&\approx 2.752\%\\ 3&\approx 1.530\%&13&\approx 2.796\%\\ 4&\approx 1.814\%&14&\approx 2.848\%\\ 5&\approx 2.056\%&15&\approx 2.883\%\\ 6&\approx 2.225\%&16&\approx 2.925\%\\ 7&\approx 2.350\%&17&\approx 2.954\%\\ 8&\approx 2.465\%&18&\approx 2.989\%\\ 9&\approx 2.546\%&19&\approx 3.013\%\\ 10&\approx 2.630\%&20&\approx 3.043\%\\ ...&...&...&...\\ 100&\approx 3.562\%&500&\approx 3.736\%\\ 200&\approx 3.663\%&1000&\approx 3.766\%\\\hline \end{array}$$
3.1 At infinity
For $\alpha\to\infty$ the function is $$z!_{(\infty)}=\exp\left(\sum_{j=1}^{\infty}\text{sinc}(z-j)\ln(j)\right)$$ Where $$\text{sinc}(z)=\begin{cases}\frac{\sin(\pi z)}{\pi z}&\text{if }z\neq 0\\1&\text{if }z=0\end{cases}$$