Binomial Coefficients with fractions: $\binom {m-\frac 12}m=\frac 1{2^{2m}}\binom {2m}m$. From my earlier question here and the interesting solutions posted, we find interesting equivalents converting binomial coefficients with fractions to those without, e.g. 
$$\binom {m-\frac 12}m=\frac 1{2^{2m}}\binom {2m}m$$
and
$$\binom {n+\frac 12}n=\frac {n+1}{2^{2n+1}}\binom {2n+2}{n+1}$$

Are there any "rules of thumb" for quickly converting a binomial coefficient with fractions into a binomial coefficient without fractions, adjusted with a coefficient as necessary?

Further edit:
The purpose of this question is not to derive the above (that has already been done elsewhere) but to ask if there is a handy rule of thumb for converting one form to another (with basis provided, of course). 
Another example might be
$$\binom {m-\frac 34}m$$
Perhaps one could consider a fractional binomial coefficient of the form 
$$\binom {m-\frac pq}m$$
and see if that can be converted into a binomial coefficient of integer parameters. 
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
{m - 1/2 \choose m} & =
{\pars{m - 1/2}! \over m!\pars{-1/2}!}=
{\Gamma\pars{m + 1/2} \over m!\,\Gamma\pars{1/2}}
\\[5mm] & =
{1 \over m!\,\root{\pi}}\,\
\overbrace{{\root{2\pi}2^{1/2 - 2m}\,\Gamma\pars{2m} \over \Gamma\pars{m}}}
^{\ds{\color{#f00}{\large\S}}\,,\ \Gamma\pars{m + 1/2}}\,,\quad
\pars{~\Gamma\pars{1 \over 2} = \root{\pi}~}
\\[5mm] & =
{1 \over 2^{2m - 1}}\,{\pars{2m - 1}! \over m!\pars{m - 1}!} =
{1 \over 2^{2m - 1}}\,{\pars{2m}!/\pars{2m} \over m!\pars{m!/m}}
\\[5mm] & =
{1 \over 2^{2m}}\,{\pars{2m}! \over m!\, m!} =
\color{#f00}{{1 \over 2^{2m}}{2m \choose m}}
\end{align}
$\ds{\color{#f00}{\large\S}:\ \Gamma\!-\!Duplication\ Formula}$.
See $\ds{\mathbf{6.1.18}}$
in Abramowitz & Stegun Table.

Note that there are several useful ways to express $\ds{2m \choose m}$:

$$
{2m \choose m} =
2^{2m}{m - 1/2 \choose m} =
2^{2m}\bracks{{-1/2 \choose m}\pars{-1}^{m}} =
{-1/2 \choose m}\pars{-4}^{m} =
{-1/2 \choose -1/2 - m}\pars{-4}^{m}
$$

The other one is quite similar to this one.

A: In equation $(6)$ of this answer, the Euler-Maclaurin Sum Formula was used to derive the asymptotic formula
$$
\binom{n+\alpha}{n}=\frac{n^\alpha}{\Gamma(1+\alpha)}\left(1+\frac{\alpha+\alpha^2}{2n}-\frac{2\alpha+3\alpha^2-2\alpha^3-3\alpha^4}{24n^2}+O\!\left(\frac1{n^3}\right)\right)\tag{1}
$$
by applying the idenity
$$
\binom{n+\alpha}{n}=\prod_{k=1}^n\left(1+\frac\alpha{k}\right)\tag{2}
$$
Proven in equation $(11)$ of this answer, Gauss' Multiplication Formula says
$$
\prod_{k=0}^{n-1}\Gamma\!\left(x+\frac kn\right)
=\sqrt{n2^{n-1}\pi^{n-1}}\frac{\Gamma(nx)}{n^{nx}}\tag{3}
$$
So that
$$
\begin{align}
\prod_{k=0}^1\binom{m+\frac k2}{m}
&=\prod_{k=0}^1\frac{\Gamma\!\left(m+1+\frac k2\right)}{m!\,\Gamma\left(1+\frac k2\right)}\\
&=\frac{\Gamma(2m+2)}{m!^2\,2^{2m}\,\Gamma(2)}\\
&=\frac1{2^{2m}}\binom{2m+1}{1}\frac{(2m)!}{(m!)^2}\\
&=\frac{2m+1}{4^m}\binom{2m}{m}\tag{4}
\end{align}
$$
and
$$
\begin{align}
\prod_{k=0}^3\binom{m+\frac k4}{m}
&=\prod_{k=0}^3\frac{\Gamma\!\left(m+1+\frac k4\right)}{m!\,\Gamma\left(1+\frac k4\right)}\\
&=\frac{\Gamma(4m+4)}{m!^4\,4^{4m}\,\Gamma(4)}\\
&=\frac1{4^{4m}}\binom{4m+3}{3}\frac{(4m)!}{(m!)^4}\\
&=\frac1{256^m}\binom{4m+3}{3}\binom{4m}{m}\binom{3m}{m}\binom{2m}{m}\tag{5}
\end{align}
$$
so that
$$
\binom{m+\frac14}{m}\binom{m+\frac34}{m}=\frac{(4m+1)(4m+3)}{3\cdot64^m}\binom{4m}{m}\binom{3m}{m}\tag{6}
$$
So we can compute the product $\binom{m+\frac14}{m}\binom{m+\frac34}{m}$ in terms of integer binomimals, but I don't know of a way to compute each in such terms.
