Let $c(n)=\frac{1}{n+1}\binom{2n}{n}$ be the Catalan number. It seems that a product $\prod\_{n\in I} c(n)$, where $I\subset\mathbb N\_{>1}$, is never a Catalan number. Is this a (known) fact?



Products of Catalan numbers
It should be possible here to mimic the same argument that Erdos uses in his paper ["On some divisibility properties of $\binom{2n}{n}$"](https://cms.math.ca/10.4153/CMB-1964-047-5).


Suppose that $c(n)=c(a\_1)c(a\_2)\cdots c(a\_k)$ and $n$ is large enough (not sure what constant is exactly needed here, but checking all $n\le 25$ should be enough). Since we can always find a prime $n+1<p<2n$, it must divide $c(n)$. Therefore we must have
$$p|c(a\_i)$$
for some $i$, which in turn implies $a\_i> n/2$. From here we want to show that it is impossible to have $\frac{c(n)}{c(m)}\in \mathbb Z$, with $\max(25,m)< n<2m$ giving us our contradiction.


To show this last part we can look at the fraction
$$\frac{c(m+k)}{c(m)}=\frac{\frac{1}{(m+1)(m+k+1)}\prod\_{i=1}^{2k} (2m+i)}{\prod\_{i=2}^k (m+i)^2}$$
for integers $m>k>0$ with $m+k>25$.


In case $m<\frac{5}{6}(m+k)$ we have a prime $p$ that appears twice in the denominator and once in the numerator. In the opposite case, $m\geq 5k$ is enough to guarantee that the largest prime factor in the denominator is $\geq 2k$. If this largest prime appears with multiplicity $2\alpha$ in the denominator, it only appears with multiplicity $\alpha$ in the numerator. So either way the ratio can not be an integer.


Of course, I would consult with an actual number theorist about what the actual bounds should be, or if there can be more simplifications.


