In an enumeration problem the sequence of number of Dyck paths semilength n having no UUDD's starting at level 0 with generating function $$\frac{2}{(1+2z^2+\sqrt{1-4z})}$$ showed up, see also <https://oeis.org/A114487>.
It is not really important but I wonder whether a nice explicit formula exists for this sequence.


Im not really experienced with such problems but I saw that sometimes very nice formulas exit for such non-rational generating functions.
For example for the Motzkin numbers $M\_n$ with generating function $\frac{ 1 - x - (\sqrt{1-2x-3x^2} )}{ 2x^2}$ the explicit formula $M\_n=\sum\limits\_{k=0}^{[\frac{n}{2}]}{\binom{n}{2k}C\_k}$ exists when $C\_k$ are the Catalan numbers, see <https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5310552/>.



Explicit formula for a generating function
The coefficient of $z^n$ is 
$$\sum\_{0\le k\le n/2} (-1)^k \frac{k+1}{2n-3k+1}\binom{2n-3k+1}{n-2k}.$$


To see this, let $C(z)$ be the Catalan number generating function,
$$C(z) = \frac{1-\sqrt{1-4z}}{2z}=\frac{2}{1+\sqrt{1-4z}}.$$
Then
$$
\frac{2}{1+2z^2+\sqrt{1-4z}} = \frac{C(z)}{1+z^2 C(z)}.
$$
It is well known that 
$$
C(z)^j = \sum\_{m=0}^\infty \frac{j}{2m+j}\binom{2m+j}{m} z^m.
$$
Thus 
$$
\begin{aligned}
\frac{2}{1+2z^2+\sqrt{1-4z}}&=\sum\_{k=0}^\infty (-1)^k z^{2k}C(z)^{k+1}\\
 &=\sum\_{k=0}^\infty (-1)^k z^{2k} \sum\_m \frac{k+1}{2m+k+1}\binom{2m+k+1}{m} z^m\\
 &=\sum\_n z^n \sum\_{0\le k\le n/2} (-1)^k \frac{k+1}{2n-3k+1}\binom{2n-3k+1}{n-2k}.
\end{aligned}
$$


I have added this formula to the OEIS entry.


