Some years ago I calculated the equation of the multifactorial:
$$z!_{(\alpha)}=\alpha^{\frac{z}{\alpha}}\Gamma\left(1+\frac{z}{\alpha}\right)\prod_{j=1}^{\alpha-1}\left(\frac{\alpha^{\frac{\alpha-j}{\alpha}}}{\Gamma\left(\frac{j}{\alpha}\right)}\right)^{C_{\alpha}(z-j)}$$
Where
$$C_{\alpha}(z)=\frac{1}{\alpha}\left(1+2\sum_{k=1}^{\left\lfloor\frac{\alpha-1}{2}\right\rfloor}\cos\left(\frac{2k\pi}{\alpha}z\right)+\text{mod}(\alpha-1,2)\cos(\pi z)\right)$$
Alternative representation of $C_{\alpha}(z)$
Let $U_n(z)$ the Chebyshev polynomial of the second kind
- $\displaystyle C_{\alpha}\left(x\right)=\frac{1}{\alpha}U_{\alpha-1}\left(\cos\left(\frac{\pi x}{\alpha}\right)\right)\cos\left(\frac{\pi x}{\alpha}\right)^{\text{mod}\left(\alpha-1,2\right)}$
- $\displaystyle C_{\alpha}(x)=\frac{1}{\alpha}\sum_{k=1}^{\alpha}\cos\left(\theta_{k}\ x\right)\qquad\text{ where }\theta_{k}=\arccos\left(\cos\left(\frac{2k\pi}{\alpha}\right)\right)$
This function respects these relationship:
$\displaystyle\bullet\; z!=\prod_{i=0}^{\alpha-1}(z-i)!_{(\alpha)}$
$\displaystyle\bullet\; z!_{(\alpha)}=z(z-\alpha)!_{(\alpha)}$
Also corresponds to the known cases in which $\alpha=1$ and $\alpha=2$:
For $\color{red}{\alpha=1}$ the function is:
$$z!_{(1)}=1^{\frac{z}{1}}\Gamma\left(1+\frac{z}{1}\right)\prod_{j=1}^{0}\left(\frac{1^{\frac{1-j}{\alpha}}}{\Gamma\left(\frac{j}{1}\right)}\right)^{C_{1}(z-j)}$$
$$\color{blue}{z!_{(1)}=\Gamma(1+z)}$$
Pay attention to the fact that
$$\prod_{i=1}^{0}f(z):=1\text{ for every function, you also have that}\sum_{i=1}^{0}f(z):=0$$
For $\color{red}{\alpha=2}$ the function is:
$$z!_{(2)}=2^{\frac{z}{2}}\Gamma\left(1+\frac{z}{2}\right)\prod_{j=1}^{2-1}\left(\frac{2^{\frac{2-j}{2}}}{\Gamma\left(\frac{j}{2}\right)}\right)^{C_{2}(z-j)}=2^{\frac{z}{2}}\Gamma\left(1+\frac{z}{2}\right)\left(\frac{2^{\frac{2-1}{2}}}{\Gamma\left(\frac{1}{2}\right)}\right)^{C_{2}(z-1)}$$
$$C_2(z)=\frac{1+\cos(\pi z)}{2}\quad\Rightarrow\quad C_2(z-1)=\frac{1-\cos(\pi z)}{2}\qquad \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$$
$$z!_{(2)}=2^{\frac{z}{2}}\Gamma\left(1+\frac{z}{2}\right)\left(\frac{\sqrt{2}}{\sqrt{\pi}}\right)^{\frac{1-\cos(\pi z)}{2}}$$
$$\color{blue}{z!_{(2)}=2^{\frac{z}{2}}\Gamma\left(1+\frac{z}{2}\right)\left(\frac{2}{\pi}\right)^{\frac{1-\cos(\pi z)}{4}}}$$
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...
So in your case
$$12.1!_{(6)}= 72.7394522427...$$
Bearing in mind that $12!_{(6)}:=12\cdot(12-6)=72$ you can see that the values are also quite similar.
Fourier expansion
If you only need an approximate value you can also use its Fourier expansion:
$$x!_{(\alpha)}\approx \alpha^{\frac{x}{\alpha}}\Gamma\left(1+\frac{x}{\alpha}\right)\sum_{j=1}^{\alpha}\frac{\alpha^{\frac{\alpha-j}{\alpha}}}{\Gamma\left(\frac{j}{\alpha}\right)}C_{\alpha}\left(x-j\right)$$
which is a very similar form to
$$z!_{(\alpha)}=\alpha^{\frac{z}{\alpha}}\Gamma\left(1+\frac{z}{\alpha}\right)\prod_{j=1}^{\alpha-1}\left(\frac{\alpha^{\frac{\alpha-j}{\alpha}}}{\Gamma\left(\frac{j}{\alpha}\right)}\right)^{C_{\alpha}(z-j)}$$
With the difference that:
- product from 1 to $\alpha-1$ $\mapsto$ sum from 1 to $\alpha$
- power $\mapsto$ product
Using this approximation you get $12.1!_{(6)}\approx 72.8179454437$.
More efficient definition
it is possible to give a more efficient formulation of the multifactorial, which is essentially the one that comes out when all the summations and all the products are carried out (exploiting the symmetry properties of the Gamma function, of the powers and of the goniometric functions):
For brevity, I indicate $\beta:=\left\lfloor\frac{\alpha-1}{2}\right\rfloor$
$$z!_{(\alpha)}=\alpha^{\frac{z+z_0(z)}{\alpha}}\Gamma\left(1+\frac{z}{\alpha}\right)\frac{1}{\pi^{p(z)}}\prod_{j=1}^{\beta}\frac{\sin\left(\frac{j\pi}{\alpha}\right)^{C_{\alpha}(z+j)}}{\Gamma\left(\frac{j}{\alpha}\right)^{\gamma_j(z)}}$$
Where:
- $\displaystyle C_{\alpha}\left(z\right)=\frac{1}{\alpha}\left(1+2\sum_{k=1}^{\beta}\cos\left(\frac{2k\pi}{\alpha}z\right)+\operatorname{mod}\left(\alpha-1,2\right)\cos\left(\pi z\right)\right)\qquad$ (as defined above)
- $\displaystyle z_{0}\left(z\right)=\frac{\alpha-1}{2}+\sum_{k=1}^{\beta}\csc\left(\frac{k\pi}{\alpha}\right)\sin\left(\frac{2k\pi}{\alpha}z-\frac{k\pi}{\alpha}\right)-\operatorname{mod}\left(\alpha-1,2\right)\frac{\cos\left(\pi z\right)}{2}$
- $\displaystyle p\left(z\right)=\frac{\alpha-1}{2\alpha}+\frac{2}{\alpha}\sum_{k=1}^{\beta}\csc\left(\frac{k\pi}{\alpha}\right)\sin\left(\frac{k\pi}{\alpha}\beta\right)\cos\left(\frac{2k\pi}{\alpha}z+\left(\beta+1\right)\frac{k\pi}{\alpha}\right)+\frac{\operatorname{mod}\left(\alpha-1,2\right)}{\alpha}\left(\sum_{k=1}^{\beta}\left(-1\right)^{k}\cos\left(\frac{2k\pi}{\alpha}z\right)-\frac{\cos\left(\pi z\right)}{2}\right)$
- $\displaystyle \gamma_j\left(z\right)=\frac{4}{\alpha}\sum_{k=1}^{\beta}\sin\left(\frac{2k\pi}{\alpha}j\right)\sin\left(\frac{2k\pi}{\alpha}z\right)$
This formulation is equivalent and immediately returns the "final form". Therefore (except in particular cases where $\Gamma\left(\frac{j}{\alpha}\right)$ or $\sin\left(\frac{j\pi}{\alpha}\right)$ can be written in different terms) there is no need to further develop the formula once $\alpha$ has been set.
Note that in this case half of the products are made and the terms involving the powers of $\pi$ and $\alpha$ are taken out of the production, so there is no longer any need to perform two summations to find the final terms.
