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.

  • 3
    $\begingroup$ 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
  • 2
    $\begingroup$ No other solution up to 900. $\endgroup$
    – BillyJoe
    Sep 2 '19 at 7:28
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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


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


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


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.

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.