TITLE: How can we find the gcd for elements (binomial coefficient)?
QUESTION [9 upvotes]: $\gcd\left(\binom{2n}1,\binom{2n}3,\binom{2n}5,\ldots,\binom{2n}{2n-1}\right)$
i want to know what is specialty of such a series.I am not able to generalize the problem solution.Is there any rule for such gcd computation.

REPLY [2 votes]: Following on user94529's answer, since the sum $\sum_k \binom{2n}{2k+1} = 2^{2n-1}$ is a power of two, we will be done if we look only at the greatest power of two dividng them all. By experimenting a bit, it seems that $\binom{2n}{1} = 2n$ has the smallest power of 2 of all. 
This is indeed the case, looking at $\binom{2n}{2k+1} = \frac{2n}{2k+1} \times \binom{2n-1}{2k}$ since $2k+1$ is odd and $\binom{2n-1}{2k}$ is an integer, we get that the greatest power of 2 dividing $\binom{2n}{2k+1}$ is greater or equal to that of $2n$.
Hence, if $n=2^j \times m$ where $m$ is odd, then the answer is
$$ \gcd\left( \binom{2n}{1},\ldots,\binom{2n}{2n-1} \right) = 2^{j+1} $$