Are there any natural numbers $n\not\equiv1\bmod3$, and $n\not\equiv2\bmod4$, so that $~\dfrac{\displaystyle{2n\choose n}}{n+2}\not\in\mathbb N$ ?
Since $C_n=\dfrac{\displaystyle{2n\choose n}}{n+1}\in\mathbb N$ for all n, this is equivalent to asking whether there are any Catalan
numbers not divisible by $n+2$, with n neither of the form $3k+1$, nor of the form $4k+2$.
Inspired by the afore-mentioned Catalan numbers, I began investigating integers with the property
that $~\dfrac{\displaystyle{2n\choose n}}{n+2}\in\mathbb N$, and then soon afterwards arrived at the conclusion that there were too many of
them for such a quest to be even remotely interesting, so I negated the above property, and, after a
rather short while, I immediately started noticing that the new results were either of the form $n=$
$=3k+1$, or of the form $n=4k+2$. Updating the search parameters so as to eliminate these two
classes as well, I eventually came to the realization that there were no solutions for $n\le10^5$.
Note that Catalan Numbers are $${2n\choose n}-{2n\choose n+1}=\frac{(2n)!}{n!(n+1)!}(n+1-n)$$ Try a similar thing for these numbers: $$A{2n\choose n}+B{2n\choose n+1}+C{2n\choose n+2}\\ =\frac{(2n)!}{n!(n+2)!}\left[A(n+2)(n+1)+Bn(n+2)+Cn(n-1)\right]\\ A+B+C=0,3A+2B-C=1,2A=1\\ A=1/2,B=-1/3,C=-1/6\\ D_n=\frac1{n+2}{2n\choose n}=\frac12{2n\choose n}-\frac13{2n\choose n+1}-\frac16{2n\choose n+2}\\ =\frac12{2n\choose n}-\frac13\left[{2n\choose n+1}-{2n\choose n+2}\right]-\frac12{2n\choose n+2}$$ ${2n\choose n}=2{2n-1\choose n}$ so the first term is an integer.
$$(n-1)\left[{2n\choose n+1}-{2n\choose n+2}\right]=\frac{3(2n)!}{(n-2)!(n+2)!} $$ which is a multiple of $3$, so when you divide by $n-1$ it is still a multiple of $3$ unless $n=1\mod 3$. That leaves $$\frac12{2n\choose n+2}=\frac n{n+2}{2n-1\choose n+1}$$ which only has an extra 2 in the denominator if $n=2\mod 4$.