<s>[INST] A conjectural closed form for $\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!}$ Let
$$S=\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!},\tag1$$
its numeric value is approximately $S \approx 0.517977853388534047...$${}^{[more\ digits]}$
$S$ can be represented in terms of the generalized hypergeometric function:
$$S={_3F_2}\left(\frac12,1,1;\ \frac43,\frac53;\ \frac4{27}\right)\cdot\frac12.\tag2$$

Let $\sigma$ be the closed-form expression constructed from integers and elementary functions as follows:
$$\sigma=3\,\alpha\,\ln(2\,\alpha+1)-\sqrt{\beta\,}\arccos\gamma,\tag3$$
where
$$\alpha=\frac{\sqrt[3]{3\,}}{6\,\sqrt[3]{2\,}}\left(\sqrt[3]{9-\sqrt{69}}+\sqrt[3]{9+\sqrt{69}}\right),\tag4$$
$$\beta=\frac1{4\,\sqrt[3]{2\,}}\left(\sqrt[3]{25+3\,\sqrt{69}}+\sqrt[3]{25-3\,\sqrt{69}}\right)-\frac12,\tag5$$
$$\gamma=\frac1{6\,\sqrt[3]{2\,}}\left(\sqrt[3]{57\,\sqrt{69}-459}-\sqrt[3]{57\,\sqrt{69}+459}\right)+\frac12\tag6$$
are the unique real roots of the following cubic equations:
$$8\,\alpha^3-2\,\alpha-1=0,\tag7$$
$$64\,\beta^3+96\,\beta^2+36\,\beta-23=0,\tag8$$
$$8\,\gamma^3-12\,\gamma^2+16\,\gamma+11=0.\tag9$$
Equivalently,
$$\sigma = \frac{3\,p}{2}\,\ln\big(p+1\big)-\frac{1}{2}\sqrt{\frac{3-p}{p}}\arccos\Big( \frac{p-6}{6p+2}\Big)\tag{10}$$
where $p$ is the plastic constant or the real root of
$$p^3-p-1=0\tag{11}$$

It can be numerically checked that the following inequality holds:
$$\Big|S-\sigma\Big|<10^{-10^5},\tag{12}$$
I conjecture that the actual difference is the exact zero, and thus $S$ has an elementary closed form:
$$\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!}\stackrel?=3\,\alpha\,\ln(2\,\alpha+1)-\sqrt{\beta\,}\arccos\gamma,\tag{13}$$
I am asking for you help in proving this conjecture.
 [/INST] partial answer 
Use the method of LINK
Since
$$
\int_0^1 t^n (1-t)^{2n}\;dt = \frac{n!(2n)!}{(3n+1)!} ,
$$
If we write
$$
S(x) = \sum_{n=0}^\infty \frac{n!(2n)!}{(3n+2)!}\;x^{3n+2}
$$
then
$$
S(x) = \int_0^1 \left(\sum_{n=0}^\infty t^n(1-t)^{2n} \frac{x^{3n+2}}{3n+2}\right)dt
$$
and $S = S(1)$.
But the derivative with respect to $x$ of $\sum_{n=0}^\infty t^n(1-t)^{2n} \frac{x^{3n+2}}{3n+2}$ is a geometric series.  Its sum is a rational function, which may be integrated (with some work or a CAS).  Plug in $x=1$.  Result (if I copied right):
$$
S = \int_0^1\frac{F(t)}{18(1-t)^{4/3}t^{2/3}}\;dt
$$
where
$$
F(t) = \pi \,\sqrt {3}-6\,\sqrt {3}\arctan
 \left( 2/\sqrt {3}\cdot \left( 1-t \right) ^{2/3}{t}^{1/3}+1/
\sqrt {3} \right) \\
-6\,\ln  \left( ({1-t})^{1/3}
{t}^{2/3} -\left( 1-t \right) t \right) +3\,\ln  \left( t \left( 1-t \right) ^{2}+ \left( 1-t \right) ^{4/3}{t}^{2/3}+ \left( 1-t \right) ^{2/3}{t}^{1/3}
 \right) \\
+6\,\ln  \left( ({1-t})^{1/3}{t}^{2/3} \right) -3\,\ln 
 \left(  \left( 1-t \right) ^{2/3}{t}^{1/3} \right) 
$$
</s>