Let $k$ be a given positive integer (fixed). I want to determine whether $$ 2n-k\choose n $$ is even or odd, for each positive integer $n$. Is there any general result?
My attempt:
Case (1). $k=0$.
Then the power of 2 in $2n\choose n$ is $\sum _{i=1}^\infty [\frac{2n}{2^i}]-2[\frac{n}{2^i}]$. It is zero if and only if $n\geq 1$. Hence $2n\choose n$ is even if and only if $n=0$.
Case (2). $k=1$.
Then the power of 2 in $2n-1\choose n$ is $\sum _{i=1}^\infty [\frac{2n-1}{2^i}]-[\frac{n}{2^i}]-[\frac{n-1}{2^i}]$. It is zero if and only if $n=0,1,2,\cdots,2^j$. Hence $2n-1\choose n$ is even if and only if $n=0$ or $n$ is $2^j$, $j\geq 0$.
Is my computation correct? I do not know how to solve the general case...