The diagram shows Pascal's triangle down to row $10$.
I noticed that the product of the blue numbers is divisible by the product of the orange numbers. That is (including the bottom centre number $252$ in both products),
$$\frac{\prod\limits_{k=0}^{5}\binom{10}{k}}{\prod\limits_{k=0}^5\binom{2k}{k}}=\frac{2857680000}{4233600}=675\in\mathbb{Z}.$$
Now let
$$P=\frac{\prod\limits_{k=0}^{n}\binom{2n}{k}}{\prod\limits_{k=0}^n\binom{2k}{k}}$$
According to Wolfram, for $n=1$ to $n=50$, $P$ is an integer only when $n=1,2,5$.
Is the following conjecture true or false:
$P$ is an integer only when $n=1,2,5$.
$P$ can be rewritten as
$$P=\prod\limits_{k=0}^n\frac{(2n)!k!}{(2k)!(2n-k)!}$$
but that doesn't seem to make the question easier to answer.
I did a search on the OEIS using "$1,2,5$ Pascal divisible", and found no sequence with $1,2,5$ and the next number being greater than $50$.
Context: I have been trying to unravel some of the mysteries of Pascal's triangle recently, for example with questions about random walks, three consecutive numbers and killing.