# Products of Catalan numbers

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?

• It should not be surprising, as Catalan numbers are divisible by almost all primes between n and 2n. Do you need a proof that a product does not produce such a number? Gerhard "Proudly, Proactively Proposing Productive Proposition" Paseman, 2018.12.05. – Gerhard Paseman Dec 6 '18 at 6:43

It should be possible here to mimic the same argument that Erdos uses in his paper "On some divisibility properties of $$\binom{2n}{n}$$".
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 $$n+1, it must divide $$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.
• Great! Experimentally, $n>5$ should be enough. (trivial typo: I think we are happy if $c(n)/c(m) \not\in\mathbb{Z}$ if $\max(25,m) < n < 2m$) – Martin Rubey Dec 6 '18 at 9:07