I will answer you briefly by summarizing only the formulas that I have already given in 3 answers and questions:
- Definition of multifactorial
- Error of Fourier series of the multifactorial
- $\infty$-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
$$\text{sinc}(z)=\begin{cases}\frac{\sin(\pi z)}{\pi z}&\text{if }z\neq 0\\1&\text{if }z=0\end{cases}$$
Let's "decipher" the notation of "multifactorial". Assume $k>0$. Any $n>0$ can be written in the form $n=kq+r$ with $q\geqslant 0$ and $1\leqslant r\leqslant k$; then $n\underbrace{!\ldots!}_{k}:=\prod_{j=0}^q(kj+r)$, and $0\underbrace{!\ldots!}_{k}:=1$. But $$\prod_{j=0}^q(kj+r)=k^{q+1}\prod_{j=0}^q\left(j+\frac rk\right)=k^{q+1}\frac{\Gamma(q+1+r/k)}{\Gamma(r/k)}=\frac{k^{q+1}q!}{\mathrm{B}(r/k,q+1)}$$ using the gamma and beta functions. Using the integral representation of the latter, we get
\begin{align*}
m(k)&=1+\sum_{r=1}^k\sum_{q=0}^\infty\frac{1}{(kq+r)\underbrace{!\ldots!}_{k}}
\\&=1+\sum_{r=1}^k\sum_{q=0}^\infty\frac{1}{k^{q+1}q!}\mathrm{B}\left(\frac rk,q+1\right)
\\&=1+\sum_{r=1}^k\sum_{q=0}^\infty\frac{1}{k^{q+1}q!}\int_0^1 t^{r/k-1}(1-t)^q\,dt
\\&=1+\frac1k\sum_{r=1}^k\int_0^1 t^{r/k-1}\sum_{q=0}^\infty\frac1{q!}\left(\frac{1-t}{k}\right)^q dt
\\&=1+\frac1k\sum_{r=1}^k\int_0^1 t^{r/k-1}e^{(1-t)/k}\,dt\quad\color{gray}{[t=kx]}
\\&=1+\frac{e^{1/k}}k\sum_{r=1}^k k^{r/k}\int_0^{1/k}x^{r/k-1}e^{-x}\,dx,
\end{align*}
which is the first of the two closed forms. The second one is then easy to obtain: at $r=k$ $$k^{r/k}\int_0^{1/k}x^{r/k-1}e^{-x}\,dx=k\int_0^{1/k}e^{-x}\,dx=k(1-e^{-1/k}).$$
Best Answer
The main task in this problem is to construct an analytic continuation of the multifactorial function $$n!^{(k)}=n(n-k)(n-2k)\cdots$$ over the real numbers. On the set of integers, we can take advantage of modulo arithmetic to arrive at $$n!^{(k)}_i=i\cdot k^{(n-i)/k}\frac{\Gamma(1+n/k)}{\Gamma(1+i/k)}$$ which is defined whenever $n\equiv i\pmod k$. Therefore, we need to find a function which interpolates the values of $$1,\frac{k^{-1/k}}{\Gamma(1+1/k)},\frac{2k^{-2/k}}{\Gamma(1+2/k)},\cdots,\frac{(k-1)k^{-(k-1)/k}}{\Gamma(1+(k-1)/k)}$$ so we consider the ansatz $$f(x)=\prod_{i=1}^{k-1}\left(\frac{ik^{-i/k}}{\Gamma(1+i/k)}\right)^{g(x)}$$ where $g(x)\equiv1$ when $x-i\equiv0\pmod k$ and is zero otherwise. This suggests the analytic function $$\frac{\sin(\pi(x-i))}{k\sin(\pi(x-i)/k)}$$ which is unfortunately negative when $(x-i)/k$ is an odd integer. We can circumvent this by introducing a factor of $\cos(\pi(x-i)/k)$ as its sign agrees with that of the above function when $x-i\equiv0\pmod k$. It follows that we can extend the multifactorial function to the reals through $$x!^{(k)}=k^{x/k}\Gamma\left(1+\frac xk\right)\prod_{i=1}^{k-1}\left(\frac{ik^{-i/k}}{\Gamma(1+i/k)}\right)^{\sin(\pi(x-i))\cot(\pi(x-i)/k)/k}\tag1$$ It is worth noting that this is not unique, as we can multiply by a factor of $\cos(j\pi(x-i)/k)$ for some integer $j$ — here we just took $j=1$. We can now determine the limit by considering each term separately \begin{align}\lim_{x\to0}x!^{(k)/x}&=k^{1/k}\lim_{x\to0}\exp\left(\frac{\log\Gamma(1+x/k)}x\right)\lim_{x\to0}\prod_{i=1}^{k-1}\left(\frac{ik^{-i/k}}{\Gamma(1+i/k)}\right)^{\sin(\pi(x-i))\cot(\pi(x-i)/k)/kx}\\&=k^{1/k}e^{-\gamma/k}\prod_{i=1}^{k-1}\left(\frac{ik^{-i/k}}{\Gamma(1+i/k)}\right)^{h(k)}\end{align} where \begin{align}h(k)&=\lim\limits_{x\to0}\frac{\sin(\pi(x-i))\cot(\pi(x-i)/k)}{kx}=-\cot\frac{\pi i}k\lim\limits_{x\to0}\frac{\sin\pi x\cos\pi i}{kx}\\&=-(-1)^i\frac\pi k\cot\frac{\pi i}k.\end{align} Using this definition of $g(x)$, the limit evaluates to $$\lim_{x\to0}x!^{(k)/x}=\left[\frac k{e^\gamma}\prod_{i=1}^{k-1}\left(\frac{\Gamma(1+i/k)}{ik^{-i/k}}\right)^{\pi(-1)^i\cot\frac{\pi i}k}\right]^{1/k}.$$ Note that the multifactorial function in $(1)$ can be extended onto the complex plane. It is holomorphic everywhere except at negative integer multiples of $k$, similar to the gamma function.