<s>[INST] On computing: $ \gcd \left({2n \choose 1}, {2n \choose 3},\cdots, {2n \choose 2n-1}\right)$ 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?
 [/INST] Let $q = 2^i$ where $2^i | 2n$ and $2^{i+1} \not|2n$.  Claim:  $d=q$.
First we'll show that each term ${2n \choose 2k+1}$, for $0 \leq k \leq n-1$ has $q$ as a factor.  Consider,
$$\begin{align*}
  {2n \choose 2k+1} &= \frac{(2n)(2n-1)(2n-2) \cdots (2n- [2k+1] + 1)}{(2k+1)(2k)(2k-1)\cdots(1)}\\
  &= \frac{2n}{2k+1} \cdot {2n-1 \choose 2k}\\
  &= \frac{2n{2n-1 \choose 2k}}{2k+1} 
\end{align*}$$
Since $k \leq n-1$, $2k < 2n-1$, and the number ${2n-1 \choose 2k}$ is a nonzero integer.  Now since ${2n \choose 2k+1}$ is an integer, we know that $2k+1$ divides the numerator.  There are no factors of $2$ in $2k+1$, therefore, 
$q$ must survive to be a factor of ${2n \choose 2k+1}$.
Second we show that $d \leq q$.  We know at least that $q$ and $d$ share the same highest power of 2, since one of the terms is ${2n \choose 1} = 2n$.
Now as Andre points out in the comments, this means we're done at this point.  Since OP proved that $d$ is a power of 2, we have $d=q$.
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