# For which $n$ does $\binom{2n}{n}$ divide $\binom{4n}{n}$?

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.

• 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 Sep 2 '19 at 7:09
• No other solution up to 900. Sep 2 '19 at 7:28
• May i say that putting the problem into : When a fraction $\in \mathbb{N}$ ? is helpfull and often seen Sep 2 '19 at 7:47
• @ToniMhax Good point, given that the accepted answer does that, turning it into ratio of to products of factorials. (+1 on comment...) Sep 2 '19 at 12:46

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.

• Why $p^3{\,\not\mid\,}(4n)!$? Sep 2 '19 at 9:16
• @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 $p^2{\,||\,}(4n)!$. Sep 2 '19 at 9:35
• It would be interesting to generalize the result to: if $n > m$, then ${\large{\binom{2n}{n}}}$ does not divide ${\large{\binom{2^{k}n}{n}}}$. Sep 2 '19 at 9:44