TITLE: Why is ${2n \choose n}/2 $ odd if and only if $n=2^k$?
QUESTION [6 upvotes]: If we consider the numbers ${2n \choose n}/2 $ for $n$ small:  

[ 1, 1 ][ 2, 3 ][ 3, 10 ][ 4, 35 ][ 5, 126 ][ 6, 462 ][ 7, 1716 ][ 8,
  6435 ] [ 9, 24310 ][ 10, 92378 ][ 11, 352716 ][ 12, 1352078 ][ 13,
  5200300 ] [ 14, 20058300 ][ 15, 77558760 ][ 16, 300540195 ]

we observe that ${2n \choose n}/2 $ is odd if and only if $n$ is a power of $2$.     
We can easily check with a computer that it is true for $n \le 2^{10}$, so that we can expect it is true in general.  
Question: What's the proof?

REPLY [5 votes]: By Kummer's theorem, the number of prime factors $p$ in a binomial coefficient $\binom {a+b}a=\binom {a+b}b$ is the number of carries arising when adding $a+b$ represented as numbers in base$~p$. Taking $p=2$ and $a=b=n$, it is easy to see that you get one carry (and so one factor$~2$) for every bit '$1$' in the binary representation of $n$; you get one carry precisely when the representation has a single bit '$1$', which is when $n$ is a power of $2$.