Show that $2n\choose n$ is divisible by $2n-1$ I have found many questions asking to prove $2n\choose n$ is divisible $2$, but I also observed by trying the first few "$n$" that $2n\choose n$ is divisible  by $2n-1$. 
It sure seems true when $2n-1$ is prime, but is it true in general for all $n$?
 A: Since
$$
\binom{2n}{n}=2\frac{2n-1}n\binom{2n-2}{n-1}
$$
we get
$$
2n\binom{2n}{n}=4(2n-1)\binom{2n-2}{n-1}
$$
and therefore,
$$
\bbox[5px,border:2px solid #C0A000]{\binom{2n}{n}=\left[4\binom{2n-2}{n-1}-\binom{2n}{n}\right](2n-1)}
$$
A: The case when $2n-1$ is prime is trivial. This is true, as we know that $\binom{2n}{n}$ is integer, so then we have that
$$\binom{2n}{n} = \frac{(2n)!}{(n!)^2} = \frac{(n+1) \cdots (2n-1)\cdot 2n}{n!}$$
and $2n-1$ doesn't appear in the prime factorization in the denominator.
Now assume that $2n-1$ is composite and $p$ is a prime divisor of it. Obviously we must have $p \le \frac{2n-1}{2} \implies p < n$. Then if $p \mid m \le n$ we get that $m$ contributes in the prime factorization of $(n!)^2$ twice, but similarly $m$ and $2m$ contribute by at least the same amount in $(2n)!$. Obviously $2n-1$ can't be equal to $m$ ($m \le n$) nor to $2m$ ($2n-1$ is odd), so the numerator has at least one more contributor of $p$. So we get that $p^k \mid \binom{2n}{n}$, where $k$ is the highest power of $p$ dividing $2n-1$. This is true for every prime divisor of $2n-1$, so we get
$$2n-1 \mid \binom{2n}{n} $$
Hence the proof.
A: Catalan's numbers $\{ C_n\}_{n \ge 1}$ are defined by
$$C_n=\frac{1}{n+1}\binom{2n}{n}$$
these are known to be integers for all $n \ge 1$.
Now, for $n \ge 2$,
$$\binom{2n}{n}=\frac{2n \cdot (2n-1) \cdot ((2n-2)!)}{n \cdot ((n-1)!) \cdot n \cdot ((n-1)!)} = 2 \cdot \frac{1}{n} \binom{2n-2}{n-1} \cdot (2n-1) = 2C_{n-1} \cdot (2n-1)$$
This shows that $(2n-1)$ always divides $\binom{2n}{n}$, and the quotient is $2C_{n-1}$.
A: $$\frac{\binom{2n-1}{n-1}}{2n-1}\in\mathbb N.$$
See here: To prove $n\binom{2n-1}{n-1}$ is divisible by $n(2n-1)$
Thus, $$\frac{\binom{2n}{n}}{2n-1}=2\cdot\frac{\binom{2n-1}{n-1}}{2n-1}\in\mathbb N$$ 
A: Note that $\dbinom{2n}{n}$ is an integer. Note also that $1$ divides $\binom  21 $ and assume $n>1$ hereafter.
$\begin{align}
\dbinom{2n}{n} &= \dfrac{2n!}{n!\cdot n!} \\ &= \dfrac{2n\cdot (2n-1)\cdots (n+1)}{n\cdot (n-1)\cdots 1}\\
&= \dfrac{(2n)(2n-1)}{n\cdot n}\dbinom{2(n-1)}{n-1}\\
&= 2\dfrac{2n-1}{n}\dbinom{2(n-1)}{n-1}\\
\end{align}$
Now $\gcd(2n-1,n)=1$ so $n$ must divide  $2\dbinom{2(n-1)}{n-1}$, that is, $2\dbinom{2(n-1)}{n-1}= kn$ for some integer $k$.
Then $\dbinom{2n}{n} = k(2n-1)$ as required.
A: Simply expand  and seperate the factorial in the numerator. 

Clearly,  is a factor in the numerator. Therefore,  is indeed divisible by .
