TITLE: On computing: $ \gcd \left({2n \choose 1}, {2n \choose 3},\cdots, {2n \choose 2n-1}\right)$
QUESTION [17 upvotes]: I would like to calculate 
$$ d=\gcd \left({2n \choose 1}, {2n \choose 3},\cdots, {2n \choose 2n-1}\right) $$
We have:
$$ \sum_{k=0}^{n-1}{2n \choose 2k+1}=2^{2n-1} $$
$$ d=2^k, 0\leq k\leq2n-1 $$
...
Any idea?

REPLY [4 votes]: Here’s a short argument for the second part of Shaun’s proof.
Suppose that $p$ is an odd prime dividing $2n$, say $2n = mp^k$, where $p\nmid m$. Then I claim that
$$\begin{align*}
\binom{2n}{p^k}&=\frac{(2n)!}{p^k!(2n-p^k)!}\\
&=\frac{(mp^k)!}{p^k!((m-1)p^k)!}\\
&=\frac1{p^k!}\prod_{i=0}^{p^k-1}(mp^k-i)\\
&=m\prod_{i=1}^{p^k-1}\frac{(m-1)p^k+i}i
\end{align*}$$
is not divisible by $p$. Clearly $p\nmid m$. If $1\le i\le p^k-1$, let $p^s$ be the highest power of $p$ dividing $(m-1)p^k+i$. Then $s<k$ and $p^s\mid i$, so the factor $$\frac{(m-1)p^k+i}i$$ contributes no factor of $p$ to $\dbinom{2n}{p^k}$. Thus, $p\nmid d$.
On the other hand, $d\mid 2n$, so any prime factor of $d$ must divide $2n$. It follows that $d$ has no odd prime factors and hence (by the first part of Shaun’s proof) that $d$ is the highest power of $2$ dividing $2n$.