For the double factorial and even numbers $(2k)$ is well known that:
$$(2k)!!=2^k k!$$
Which is easy to see just from the definition. From here we can get for odd numbers $(2k-1)$:
$$(2k-1)!!=\dfrac{(2k)!}{2^k k!}.$$
Then we can express double factorial in terms of simple factorials and powers of $2$. I wonder if there is some well known formula for triple factorial (or multifactorial in general). Is always easy to see that for $3k$:
$$(3k)!!!=3^k k!$$
But I don't have idea of how to find a formula for $3k-1$ or $3k-2$:
$$(3k-1)!!!=?$$
$$(3k-2)!!!=?$$
I also would love to see if there is a general way of find these formulas for any multifactorial. Thanks.
Triple factorial in terms of simple factorial
combinatoricselementary-number-theoryfactorialprobabilitysequences-and-series
Related Solutions
In the $k=4$ case, $$\frac{\left(\sum_{i=n}^{n+k-1}i\right)!}{\sum_{i=n}^{n+k-1}i!} = \frac{(4n+6)!}{n!(n+2)(n^2+5n+5)}.$$
For a lot of (likely infinitely many) values of $n$, $n^2+5n+5$ is a prime larger than $4n+6$, and so the expression is not an integer for those $n$.
(In the $k=3$ case, we get $$ \frac{(3n+3)!}{n!(n+2)^2} $$ and since $2(n+2)<3n+3$, this simplifies to an integer for $n\ge1$. Similarly, with $k=2$, we have $$ \frac{(2n+1)!}{n!(n+2)} $$ which again simplifies to an integer for $n\ge1$.)
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:
- $\displaystyle C_{\alpha}\left(z\right)=\frac{1}{\alpha}\left(1+2\sum_{k=1}^{\beta}\cos\left(\frac{2k\pi}{\alpha}z\right)+\delta_{\alpha,2}\cos\left(\pi z\right)\right)\qquad$ (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 $$\text{sinc}(z)=\begin{cases}\frac{\sin(\pi z)}{\pi z}&\text{if }z\neq 0\\1&\text{if }z=0\end{cases}$$
Related Question
- Define the triple factorial, $n!!!$, as a continuous function for $n \in \mathbb{C}$
- Is there a name for a factorial-like falling product which uses an arbitrary step $x \in \mathbb{R}$ instead of 1.
- Asymptotic Approximations for Higher-Order Factorials (e.g. triple factorial) and the Gamma Function
- Sum of Squares and Factorial – Is There a Hidden Relation?
Best Answer
Unfortunately, the examples you mentioned are the only "nice extensions" for multifactorials. Anything else requires expansions with the Gamma function, which this answer and the corresponding thread explain pretty well.
As you would expect, separating double factorials by parity allowed for the simplifications above, but triple factorials and higher level multifactorials lack this capability.