How to get decimal answers of a factorial I have had a familiar understanding of how factorials work within mathematics for some time now. However, I have recently discovered that you are able to get decimal factorials.
So far under my research and calculations, I have been able to deduce that:
$\left(t+\frac{1}{2}\right)!=\frac{\sqrt{\pi }}{2}\cdot \prod _{n=1}^(t\frac{2n+1}{2}$

t is a variable

Additionally:
$\left(t-\frac{1}{2}\right)!=\frac{\sqrt{\pi }}{2}\cdot \prod _{n=1}^{t-1}\frac{2n+1}{2}$

t is a variable

This so far will only work for half values and therefore I have not found a way for it to work with other fractions e.g. 
$\frac{1}{3},\ \frac{1}{4},\ \frac{1}{5},\ ...$
$\left(t+\frac{1}{v}\right)!= ?$

t and v are both variables

If anyone has any intel on how this would work please free to leave a comment below or any solutions. It would be as always, much appreciated. I am aware that there is a calculus solution but I am looking for a non-calculus solution. Anyone interested in the calculus solution it is:
$Γ\left(t\right)=\int _0^∞x^{\left(t-1\right)}\cdot e^{-x}dx$

t is a variable

 A: First, let's seek $\Gamma(x)$ where $x=n+\varepsilon$ where $n$ is a large integer and $|\varepsilon|<1$.
Let $f(x) = \log(\Gamma(x))$. Since $f(x+1) = f(x) + \log(x)$ we have
$$ f'(x) \sim \log x $$
$\log x$ doesn't change very quickly compared to its magnitude when $x$ is large, so let's approximate
$$ f(n+\varepsilon) \approx f(n) + \varepsilon f'(n) \approx f(n) + \varepsilon \log n $$
which is the same as
$$ \Gamma(n+\varepsilon) \approx \Gamma(n) \cdot n^\varepsilon = (n-1)!\cdot n^\varepsilon$$
This approximation is of course exact for $\varepsilon=0$ and $\varepsilon=1$. To get an idea of the error we risk, let's set $\varepsilon=-1$ and compare:
$$ \frac{\Gamma(n-1)_{\rm approx}}{\Gamma(n-1)} =
\frac{\Gamma(n) / n }{\Gamma(n) / (n-1) } = 1 + \frac{1}{n-1} $$
So now we can compute $\Gamma(n+\varepsilon)$ to a relative precision of about $\frac1n$.
If we want better precision than that, choose a larger $n$ (but the same $\varepsilon$) and work our way down to the original $n$ by repeatedly applying
$$ \Gamma(x-1) = \frac{\Gamma(x)}{x-1} $$
which preserves the relative precision all the way down (assuming the divisions are exact).

A slightly more careful derivation yields that
$$ \Gamma(n+\varepsilon) \approx (n-1)! (n-\tfrac12+\tfrac\varepsilon2)^\varepsilon
\quad \text{with }\varepsilon\in[-\tfrac12,\tfrac12] $$
gives a relative error of at most $\frac{1}{8n^2}$.
A: For the exact values of $x!$ for $x=a/b$ and $120\equiv0\pmod b$, see this paper.  However, the solutions are extraordinarily complicated.
There are a few approaches, however, to dealing with the factorials of fractional values.  The first is the multiplication formula:
$$(2\pi)^{\frac{k-1}2}k^{\frac12-kz}\Gamma(kz)=\prod_{i=0}^{k-1}\Gamma\left(z+\frac ik\right)$$
For example, with $k=2$, we get
$$\sqrt{2\pi}k^{\frac12-2z}\frac{\Gamma(2z)}{\Gamma(z)}=\Gamma\left(z+\frac12\right)$$
However, this formula does not really provide closed forms for any other fractional values.  At best, it will give you $\Gamma(z+\frac13)$ in terms of $\Gamma(z+\frac23)$, for example..
At $1/4$, we happen to have Gauss' constant,
$$\Gamma\left(\frac14\right)=\sqrt{G\sqrt{8\pi^3}}$$
$G$ can be easily calculated using the AM-GM form:
$$G=\frac1{\operatorname{agm}(1,\sqrt2)}$$
As a quick example,
$a_0=1,g_0=\sqrt2$
$a_1=\frac{1+\sqrt2}2\approx1.207,g_1=\sqrt[4]{2}\approx1.189$
$a_{n+1}=\frac{a_n+g_n}2,g_{n+1}=\sqrt{a_ng_n}$
Thus. we know that $G\approx1/1.195=0.83$
From there, it is easy to find $\Gamma\left(n+\frac14\right)$ from $\Gamma(x+1)=x\Gamma(x)$.
Values for $\Gamma\left(n+\frac34\right)$ may then be calculated from the $\Gamma\left(z+\frac12\right)$ formula.
Straight away approximations may also be calculated from the full Stirling approximation:
$$\ln(\Gamma(x))=x\ln(x)-x-\frac12\ln\left(\frac x{2\pi}\right)+\sum_{k=2}^\infty\frac{B_{k+1}}{k(k+1)z^k}$$
where $B_k$ are the Bernoulli numbers.
A: In this answer, it is shown that
$$
\frac1{\Gamma(1+z)}=e^{z\gamma}\prod_{k=1}^\infty\left(1+\frac{z}{k}\right)e^{-\frac{z}{k}}
$$
Therefore, we have
$$
z!=\Gamma(1+z)=e^{-z\gamma}\prod_{k=1}^\infty e^{\frac{z}{k}}\left(1+\frac{z}{k}\right)^{-1}
$$
where $\gamma$ is the Euler-Mascheroni Constant.
