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I found $n=1,3,4,7,10,24$ in a search up to $n=70.$ I didn't consider my software reliable enough to go much higher.

This came up while trying to find the probability of an all black card bridge hand. [If cards drawn with replacement answer would have been $1/2^{13}…$] In particular $n=13$ was not on my short list noted above. I'm wondering whether there can be a decision one way or the other about there being infinitely many $n$ that work.

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    $\begingroup$ Wolfram agrees with you up to e.g. $n=100$, https://www.wolframalpha.com/input/?i=%284n+choose+n+%29+%2F+%282n+choose+n+%29+for+n%3D1...100 $\endgroup$ Sep 2 '19 at 7:09
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    $\begingroup$ No other solution up to 900. $\endgroup$
    – BillyJoe
    Sep 2 '19 at 7:28
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    $\begingroup$ May i say that putting the problem into : When a fraction $\in \mathbb{N}$ ? is helpfull and often seen $\endgroup$
    – Toni Mhax
    Sep 2 '19 at 7:47
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    $\begingroup$ @ToniMhax Good point, given that the accepted answer does that, turning it into ratio of to products of factorials. (+1 on comment...) $\endgroup$
    – coffeemath
    Sep 2 '19 at 12:46
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Claim:

If $n > 24$, then ${\large{\binom{2n}{n}}}$ does not divide ${\large{\binom{4n}{n}}}$.

Proof:

It's easily verified that the claim holds for $n < 38$.

Suppose $n\ge 38$.

Let $p\;$be the least prime such that $p > {\large{\frac{4}{3}}}n$.

Then for $n \ge 2010760$, we have $p \le {\large{\frac{3}{2}}}n$ by Lowell Schoenfeld's generalization of Bertrand's Postulate

$\qquad$https://en.wikipedia.org/wiki/Bertrand%27s_postulate#Better_results

and for $38\le n < 2010760$, we have $p \le {\large{\frac{3}{2}}}n$ by direct evaluation via a Maple test program.

Then from ${\large{\frac{4}{3}}}n < p \le {\large{\frac{3}{2}}}n$, we get \begin{align*} &\bullet\;p{\,\not\mid\,}n!\\[4pt] &\bullet\;p{\,\mid\,}(2n)!\\[4pt] &\bullet\;p^2{\,\mid\,}(3n)!\\[4pt] &\bullet\;p^3{\,\not\mid\,}(4n)!\\[4pt] \end{align*} hence $${\large{\frac{\binom{4n}{n}}{\binom{2n}{n}}}}$$ is not an integer since identically we have $${\large{\frac{\binom{4n}{n}}{\binom{2n}{n}}}}=\frac{n!(4n)!}{(2n)!(3n)!}$$ and for the fraction on the right, $p^3$ divides the denominator, but doesn't divide the numerator.

This completes the proof.

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  • $\begingroup$ Why $p^3{\,\not\mid\,}(4n)!$? $\endgroup$
    – BillyJoe
    Sep 2 '19 at 9:16
  • $\begingroup$ @mbjoe:$\;$Since ${\large{\frac{4}{3}}}n < p \le {\large{\frac{3}{2}}}n$, the only multiples of $p$ which are less than or equal to $4n$ are $p$ and $2p$, hence $p^2{\,||\,}(4n)!$. $\endgroup$
    – quasi
    Sep 2 '19 at 9:35
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    $\begingroup$ It would be interesting to generalize the result to: if $n > m$, then ${\large{\binom{2n}{n}}}$ does not divide ${\large{\binom{2^{k}n}{n}}}$. $\endgroup$
    – BillyJoe
    Sep 2 '19 at 9:44

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