TITLE: Is this an integer?
QUESTION [10 upvotes]: The following expression arose in a binomial coefficient
summation :
$$f(n,k) = \frac{(2n + 1)!}{n!k!(n-k)!(2k + 1)} $$
where  $k$ ranges from  $0$ to $n$.
This fraction is clearly an integer for $k = 0$ and $k = n$.
It can also be proved to be integral in certain other cases
including when $2k + 1$ is prime .
Question:  Is $f(n,k)$ an integer for all values of $k$?
Thanks

REPLY [2 votes]: If we consider darij's comment, it is enough to prove that 
$$
r(a, b, c) = \frac{(2a+2b+c)!(a+c)!}{(a+b)!b!(2a+c)!c!}
$$
is an integer for all $a, b, c\geq 0$. (I just slightly changed the definition for convenience.) To prove this, we will prove the following lemma :
$$
\lfloor 2x+2y+z\rfloor+\lfloor x+z\rfloor\geq \lfloor x+y\rfloor +\lfloor y\rfloor + \lfloor 2x+z \rfloor+\lfloor z \rfloor
$$
for all $x, y, z\in \mathbb{R}$. If we can show this, our previous claim follows from the Lagrange's identity
$$
\mathrm{ord}_{p}(n!)=\sum_{k=1}^{\infty}\bigg\lfloor\frac{n}{p^{k}}\bigg\rfloor.
$$
By the way, since the equation in the lemma remains same if we change $x$ to $x+n$ for any $n\in \mathbb{Z}$ (and same for $y$ and $z$), we only need to check for $0\leq x, y, z<1$, which is an easy exercise.