The number $ \frac{(m)^{(k)}(m)_k}{(1/2)^{(k)} k!}$ For a real number $a$ and a positive integer $k$, denote by $(a)^{(k)}$ the number $a(a+1)\cdots (a+k-1)$ and $(a)_k$ the number
$a(a-1)\cdots (a-k+1)$. Let $m$ be a positive integer $\ge k$. Can anyone show me, or point me to a reference,  why the number
$$ \frac{(m)^{(k)}(m)_k}{(1/2)^{(k)} k!}= \frac{2^{2k}(m)^{(k)}(m)_k}{(2k)!}$$ is always an integer?
 A: \begin{eqnarray*}& &\frac{2^{2k}(m)^{(k)}(m)_k}{(2k)!}\\
&=&\frac{2^{2k}(m-k+1)(m-k+2)\cdots (m-1)(m)(m)(m+1)\cdots (m+k-2)(m+k-1)}{(2k)!}
\end{eqnarray*}
Now we write one of the $m$ as $\frac{1}{2}[(m-k)+(m+k)]$ and distribute, and the last expression becomes
$$2^{2k-1}\left[\frac{(m+k-1)(m+k-2)\cdots(m-k) }{(2k)!}+\frac{(m+k)(m+k-1)\cdots (m-k+1)}{(2k)!}\right]$$
which is equal to 
$$2^{2k-1}\left[{m+k-1\choose 2k}+{m+k\choose 2k}\right],$$ an integer.
A: We use the idea- product of n consecutive integers is divisible by $n!$.
The numerator = $$2^{2k}(m-k+1)(m-k+2)\cdots (m-1)(m)(m)(m+1)\cdots (m+k-2)(m+k-1)$$
= 2k-1 consecutive integers $\times m$
This must be divisible by $2^{2k}(2k-1)!\cdot m$
A: Using Knuth's notation:
$$
x^{\underline{k}} = x (x - 1) \ldots (x - k + 1) = (x)_{(k)} \qquad
x^{\overline{k}} = x (x + 1) \ldots (x + k - 1) = (x)^{(k)}
$$
What you are looking for is:
$$
\frac{m^{\overline{k}} m^{\underline{k}}}{(1/2)^{\overline{k}} k!}
$$
As if $k > m$ then $m^{\underline{k}} = 0$, while all other factors are positive, the range of interest is $0 \le k \le m$. We also have:
$$
\frac{m^{\underline{k}}}{k!} = \binom{m}{k}
$$
which is an integer, we still need:
$$
m^{\overline{k}} = (m + k - 1)^{\underline{k}}
$$
$$
\begin{align*}
(1/2)^{\overline{k}} &= (1/2) \cdot (3 / 2) \cdot \ldots \cdot (1/2 + k - 1) \\
                     &= \frac{1 \cdot 3 \cdot \ldots \cdot (2 k - 1)}{2^k} \\
                     &= \frac{1 \cdot 2 \cdot \ldots \cdot 2k}
                             {2^k \cdot 2^k \cdot k!} \\
                     &= \frac{(2 k)!}{2^{2 k} k!}
\end{align*}
$$
Thus:
$$
\frac{m^{\overline{k}} m^{\underline{k}}}{(1/2)^{\overline{k}} k!}
 = \frac{m (m +  k - 1)^{\underline{2 k-1}} 2^{2 k} }{(2 k)!}
$$
