What are the properties of the "unfolded" gamma distribution generalization of a normal distribution? In a prior post, I developed an "unfolded" gamma distribution generalization of a normal distribution as an example of how to relate a gamma distribution to a normal distribution. This yielded
$$
\text{ND}(x;\mu,\sigma^2,a) =
\dfrac{a e^{-2^{-\frac{a}{2}} \left(\frac{1}{\sigma }\right)^a \left| x-\mu\right| ^a}}{2 \sqrt{2} \sigma  \Gamma \left(\frac{1}{a}\right)}
\,,$$
where the mean is $\mu$, the variance is $\sigma^2$, and the shape is $a>0$, where $a=2$ for an ordinary normal distribution. 
This appears to be a different distribution from the generalized error distribution
$$\text{GED}(x;\mu,\alpha,\beta)=\frac{\beta}{2\alpha\Gamma(1/\beta)} \; e^{-(|x-\mu|/\alpha)^\beta}\,,$$
where $\mu$ is the location, $\alpha>0$ is the scale, and $\beta>0$ is the shape and where $\beta=2$ yields a normal distribution. It includes the Laplace distribution when $\beta=1$. As $\beta\rightarrow\infty$, the density converges pointwise to a uniform density on $(\mu-\alpha,\mu+\alpha)$.
Questions Are these distributions the same? If so, how does one convert between them? If not, and $\text{ND}(x;\mu,\sigma^2,a)$ is a different distribution, what are its other properties, for example, does it reduce to a distribution other than a normal distribution?  
 A: Indeed, my notation was bad. Better notation is now used on the post this came from, 
The generalized gamma distribution
$$\text{GD}\left(x;\alpha ,\beta ,\gamma ,\mu \right)=\begin{array}{cc}
  & 
\begin{cases}
 \dfrac{\gamma  e^{-\left(\dfrac{x-\mu }{\beta }\right)^{\gamma }} \left(\dfrac{x-\mu }{\beta }\right)^{\alpha  \gamma -1}}{\beta  \,\Gamma (\alpha )} & x>\mu  \\
 0 & \text{other} \\
\end{cases}
 \\
\end{array}\,,$$
implies that 
$$
\begin{align}
\text{GND}(x;\mu,\alpha,\beta) &=
\frac{1}{2}\text{GD}\left(x;\frac{1}{\beta},\alpha,\beta,\mu \right)+\frac{1}{2}\text{GD}\left(-x;\frac{1}{\beta},\alpha,\beta,\mu \right)\\
&=
\frac{\beta  e^{-\left(\dfrac{\left|x-\mu\right|}{\alpha }\right)^{\mathrm{\Large{\beta}}}}}{2 \alpha  \Gamma \left(\dfrac{1}{\beta }\right)}\\
\end{align}
\,,$$ 
is a generalization of the normal distribution, where $\mu$ is the location, $\alpha>0$ is the scale, and $\beta>0$ is the shape and where $\beta=2$ yields a normal distribution. It includes the Laplace distribution when $\beta=1$. As $\beta\rightarrow\infty$, the density converges pointwise to a uniform density on $(\mu-\alpha,\mu+\alpha)$.
See the original post for the logic behind this.
A: The result from Whuber's comment in an answer:
It seems like both expressions can be related to the function below
$$f(x;c_1,c_2,c_3) = c_1 e^{-c_2 \vert x - \mu \vert^{c_3}}$$

GED: $\frac{\beta}{2\alpha\Gamma(1/\beta)} \; e^{-(|x-\mu|/\alpha)^\beta}$
$$\begin{array}{rcl}
c_1 &=& \frac{\beta}{2\alpha\Gamma(1/\beta)} \\
c_2 &=& \left(\frac{1}{\alpha} \right)^\beta \\
c_3 &=& \beta
\end{array}$$
ND: $\dfrac{a e^{\left(-2^{-\frac{a}{2}} \left(\frac{1}{\sigma }\right)^a \left| x-\mu\right| ^a\right)}}{2 \sqrt{2} \sigma  \Gamma \left(\frac{1}{a}\right)}
$
$$\begin{array}{rcl}
c_1 &=& \frac{a}{2 \sqrt{2} \sigma\Gamma(1/a)} \\
c_2 &=& 2^{-\frac{a}{2}}\left( \frac{1}{\sigma} \right)^a \\
c_3 &=& a
\end{array}$$

and you can say
$$\begin{array}{rcl}
\mu &=& \mu \\
{\alpha}  &=& \sqrt{2}{\sigma}  \\
\beta &=& a
\end{array}$$
