Identifying a distribution from its moments I came across a random variable whose sequence of central standard moments empirically seems to be $0, 1/2, 0, 3/2, \dots$.  (That's as far as I could compute.) Is this a well-known distribution?
 A: Given a finite number of central moments, there are typically a large number of possibilities, and it is also possible to shift the location of the distribution arbitrarily.  
Your particular numbers with zero skew and with excess kurtosis of $\dfrac{3/2}{\left(1/2 \right)^2}-3 = 3$ would be consistent with a Laplace distribution with (since you know the variance) scale parameter $b=\frac12$. 
But there are many other possibilities, not necessarily symmetric despite the zero skewness. 
A: Edit: I desire to give precisions to the answer of @Henry (I overlooked that you gave the precise value of the parameter) as the Laplace distribution is not so well known.
The OP says in particular that $\mu_2=\frac12$ and $\mu_4=\frac32$. Thus, we must have
$$\dfrac{\mu_4}{\mu_2}=3  \ \ \ \ (1)$$
The standard Laplace (sometimes name "double exponential") distribution with pdf $f(x)=\dfrac{1}{2}e^{-|x|}$ is not a direct answer because
$\mu_2=\int_{-\infty}^{\infty}x^2\dfrac{1}{2}e^{-|x|}dx=2$ whereas $\mu_4=\int_{-\infty}^{\infty}x^4\dfrac{1}{2}e^{-|x|}dx=24$, and (1) is not fulfilled.
We have to modulate the distribution and consider a more general Laplace distribution with parameter $b>0$, i.e., with pdf $f_b(x)=\dfrac{1}{2b}e^{-|\frac{x}{b}|}$. 
There is a general formula for the moments of this distribution:
$$\mu_k=\begin{cases}b^k k!&\text{for even } \ k\\0& \text{for odd values of } k\end{cases}$$
Condition (1) gives $\dfrac{24 b^4}{2 b^2}=12 b^2=3$. Then: $b=\frac12$ (the value of the parameter given by Henry).
Therefore, an answer compatible with the given constraints, (1) in particular, is the Laplace distribution with pdf $f(x)=e^{-2|x|}$.
