Conjecture: $\prod\limits_{k=0}^{n}\binom{2n}{k}$ is divisible by $\prod\limits_{k=0}^n\binom{2k}{k}$ only if $n=1,2,5$.

Question

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.

Answer 1

Preliminaries

We will use the following

Theorem (Jitsuro Nagura). There exists at least one prime number $p$ such as $$ a_m\leq n < p < \left(1+\frac{1}{m}\right)n\tag1 $$ where $m=1,2,3,4,5$ and $a_m=2,8,9,24,25$, respectively.

Proof. See "On The Interval Containing At Least One Prime Number" by Jitsuro Nagura in Proc. Japan Acad. 28(4): 177-181 (1952). DOI: 10.3792/pja/1195570997. Accessible for example here.$\square$

Proof of conjecture

Theorem. Let $n$ be a positive integer. If $$ P_n=\prod\limits_{k=0}^n\frac{(2n)!k!}{(2k)!(2n-k)!}=\prod\limits_{k=1}^n\frac{(2n)!k!}{(2k)!(2n-k)!}\tag2 $$ is an integer, then $n\in\{1,2,5\}$.

Proof. Calculation shows that if $1\leq n\leq 8$, then $P_n$ is an integer only for $n\in\{1,2,5\}$. Assume $n\geq 9$. By Nagura's theorem, there exists at least one prime $p$ with $n<p<\left(1+\frac{1}{3}\right)n$. Denote by $k_p(a)$ the number of times prime $p$ appears in the prime factorization of an integer $a$ and define $$ \begin{align} A_n&:=\prod_{k=1}^n(2n)!k!\tag3\\ B_n&:=\prod_{k=1}^n(2k)!\tag4\\ C_n&:=\prod_{k=1}^n(2n-k)!\tag5 \end{align} $$ so that $P_n=\frac{A_n}{B_nC_n}$. Now, $n<p<2n$, so $k_p(A_n)=n$. Moreover, $p<\frac{4n}{3}$, so $k_p(B_n)>\frac{n}{3}$ and $k_p(C_n)>\frac{2n}{3}$. Therefore, $k_p(B_nC_n)>n=k_p(A_n)$ and we conclude that if $n\geq 9$, then $P_n\notin\mathbb{Z}$.$\square$