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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...?

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

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

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

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  • $\begingroup$ 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, 2019 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, 2019 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. $\endgroup$ Apr 19, 2019 at 8:53
  • 1
    $\begingroup$ Does this answer your question? How to evaluate the limit of multifactorial $\lim_{n\to 0} \sqrt[n]{n!!!!\cdots !}$ $\endgroup$
    – TheSimpliFire
    Apr 9, 2021 at 14:16
  • $\begingroup$ Look at this answer. $\endgroup$ Sep 17, 2023 at 22:58

1 Answer 1

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I will answer you briefly by summarizing only the formulas that I have already given in 3 answers and questions:

  1. Definition of multifactorial
  2. Error of Fourier series of the multifactorial
  3. $\infty$-multifactorial

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:

  • $\displaystyle C_{\alpha}\left(z\right)=\frac{1}{\alpha}\left(1+2\sum_{k=1}^{\beta}\cos\left(\frac{2k\pi}{\alpha}z\right)+\delta_{\alpha,2}\cos\left(\pi z\right)\right)\qquad$ (as defined above)
  • $\displaystyle z_{0}\left(z\right)=\frac{\alpha-1}{2}+\sum_{k=1}^{\beta}\csc\left(\frac{k\pi}{\alpha}\right)\sin\left(\frac{2k\pi}{\alpha}z-\frac{k\pi}{\alpha}\right)-\delta_{\alpha,2}\frac{\cos\left(\pi z\right)}{2}$
  • $\displaystyle p\left(z\right)=\frac{\alpha-1}{2\alpha}+\frac{2}{\alpha}\sum_{k=1}^{\beta}\csc\left(\frac{k\pi}{\alpha}\right)\sin\left(\frac{k\pi}{\alpha}\beta\right)\cos\left(\frac{2k\pi}{\alpha}z+\left(\beta+1\right)\frac{k\pi}{\alpha}\right)+\frac{\delta_{\alpha,2}}{\alpha}\left(\sum_{k=1}^{\beta}\left(-1\right)^{k}\cos\left(\frac{2k\pi}{\alpha}z\right)-\frac{\cos\left(\pi z\right)}{2}\right)$
  • $\displaystyle \gamma_j\left(z\right)=\frac{4}{\alpha}\sum_{k=1}^{\beta}\sin\left(\frac{2k\pi}{\alpha}j\right)\sin\left(\frac{2k\pi}{\alpha}z\right)$

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:

$$\color{blue}{z!_{(5)}=5^{\frac{z+z_{0}(z)}{5}}\cdot\frac{\Gamma\left(1+\frac{z}{5}\right)}{\Gamma\left(\frac{1}{5}\right)^{\frac{\gamma_{1}(z)}{5}}\Gamma\left(\frac{2}{5}\right)^{\frac{\gamma_{2}(z)}{5}}}\cdot\left(2\pi\right)^{\frac{c_{1}(z)}{5}(z)}\phi^{\frac{c_{2}(z)}{5}}}$$

Where:

  • $\phi$ is the golden ratio
  • $z_{0}(z):=\frac{5}{2}-\frac{5}{4}\cos\left(\frac{2\pi z}{5}\right)+\frac{1}{4}\sqrt{5+\frac{2}{\sqrt{5}}}\sin\left(\frac{2\pi z}{5}\right)-\frac{5}{4}\cos\left(\frac{4\pi z}{5}\right)+\frac{1}{4}\sqrt{5-\frac{2}{\sqrt{5}}}\sin\left(\frac{4\pi z}{5}\right)$
  • $\gamma_{1}(z):=\sqrt{2\left(5+\sqrt{5}\right)}\sin\left(\frac{2\pi z}{5}\right)+\sqrt{2\left(5-\sqrt{5}\right)}\sin\left(\frac{4\pi z}{5}\right)$
  • $\gamma_{2}(z):=\sqrt{2\left(5-\sqrt{5}\right)}\sin\left(\frac{2\pi z}{5}\right)-\sqrt{2\left(5+\sqrt{5}\right)}\sin\left(\frac{4\pi z}{5}\right)$
  • $c_{1}(z):=-2+\cos\left(\frac{2\pi z}{5}\right)+\sqrt{5+2\sqrt{5}}\sin\left(\frac{2\pi z}{5}\right)+\cos\left(\frac{4\pi z}{5}\right)-\sqrt{5-2\sqrt{5}}\sin\left(\frac{4\pi z}{5}\right)$
  • $c_{2}(z):=-\frac{\sqrt{5}}{2}\cos\left(\frac{2\pi z}{5}\right)+\frac{\sqrt{5-2\sqrt{5}}}{2}\sin\left(\frac{2\pi z}{5}\right)+\frac{\sqrt{5}}{2}\cos\left(\frac{4\pi z}{5}\right)+\frac{\sqrt{5+2\sqrt{5}}}{2}\sin\left(\frac{4\pi z}{5}\right)$

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 $\color{red}{\alpha=7}$ it is not worth writing the explicit formula since the goniometric functions with argument $\frac{k\pi}{7}$ with $k\in\mathbb{Z}\setminus 7\mathbb{Z}$ cannot be expressed in simpler terms. You can use the definition in the last part of the answer.


For $\color{red}{\alpha=8}$ the function is:

$$\color{blue}{z!_{(8)}=8^{\frac{z+z_0(z)}{8}}\cdot\frac{\Gamma\left(1+\frac{z}{8}\right)}{\Gamma\left(\frac{1}{8}\right)^{\gamma_{1}(z)}\Gamma\left(\frac{1}{4}\right)^{\gamma_{2}(z)}}\pi^{c_{1}(z)}\left(\frac{1}{2}\right)^{c_{2}(z)}\left(\sqrt{2}+1\right)^{c_{3}(z)}}$$

Where

  • $z_{0}(z):=\frac{7}{2}-\cos\left(\frac{\pi z}{4}\right)+\left(1+\sqrt{2}\right)\sin\left(\frac{\pi z}{4}\right)-\cos\left(\frac{\pi z}{2}\right)+\sin\left(\frac{\pi z}{2}\right)-\cos\left(\frac{3\pi z}{4}\right)+\left(\sqrt{2}-1\right)\sin\left(\frac{3\pi z}{4}\right)-\frac{\cos\left(\pi z\right)}{2}$
  • $\gamma_{1}(z):=\frac{\sqrt{2}}{2}\left(\sin\left(\frac{\pi}{4}z\right)+\sin\left(\frac{3\pi}{4}z\right)\right)$
  • $\gamma_{2}(z):=\frac{2-\sqrt{2}}{4}\sin\left(\frac{\pi}{4}z\right)+\frac{1}{2}\sin\left(\frac{\pi z}{2}\right)-\frac{2+\sqrt{2}}{4}\sin\left(\frac{3\pi z}{4}\right)$
  • $c_{1}(z):=-\frac{7}{16}+\frac{1}{8}\cos\left(\frac{\pi z}{4}\right)+\frac{2+\sqrt{2}}{8}\sin\left(\frac{\pi z}{4}\right)+\frac{1}{8}\cos\left(\frac{\pi z}{2}\right)+\frac{1}{4}\sin\left(\frac{\pi z}{2}\right)+\frac{1}{8}\cos\left(\frac{3\pi z}{4}\right)-\frac{2-\sqrt{2}}{8}\sin\left(\frac{3\pi z}{4}\right)+\frac{1}{16}\cos\left(\pi z\right)$
  • $c_{2}(z):=\frac{1}{4}-\frac{2+3\sqrt{2}}{16}\sin\left(\frac{\pi z}{4}\right)-\frac{1}{8}\cos\left(\frac{\pi z}{2}\right)-\frac{3\sqrt{2}}{16}\sin\left(\frac{3\pi z}{4}\right)+\frac{1}{8}\sin\left(\frac{3\pi z}{4}\right)-\frac{1}{8}\cos\left(\pi z\right)$
  • $c_{3}(z)=-\frac{1}{4}\cos\left(\frac{\pi}{4}z+\frac{\pi}{4}\right)+\frac{1}{4}\cos\left(\frac{3\pi z}{4}-\frac{\pi}{4}\right)$

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 $$\text{sinc}(z)=\begin{cases}\frac{\sin(\pi z)}{\pi z}&\text{if }z\neq 0\\1&\text{if }z=0\end{cases}$$

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