# Thread: Divisibility of central binomial coefficient

1. Yes, you are right.

2. Originally Posted by chiph588@
I was hoping with the addition of $\left\{\frac np\right\}=0$ and a constraint on the size of $p$, this was an if and only if statement, i.e. "if $0<\left\{\frac np\right\}<\frac12$, then $p\not|\binom{2n}{n}$.

This turns out to be false as $3\mid\binom{14}{7}$ and $\left\{\frac73\right\}=\frac13$.
Hello chiph588@, I think I've found what you're looking for, namely a sufficient and necessary condition of the divisibility of $\binom{2n}{n}$ by a prime $p$ in general!

A prime $p$ divides $\binom{2n}{n}$ if and only if the fractional part $\left \{ \frac{n}{p^k} \right \}\geq\frac{1}{2}$ for at least one $k\in \mathbb{N}$. In particular, when $k=1$.

For example, $3\not|\binom{8}{4}$ and indeed $\left \{ \frac{4}{3^1} \right \}<\frac{1}{2}$; $\left \{ \frac{4}{3^2} \right \}<\frac{1}{2}$; and so on...

Another example, the one that you gave here is $3\mid\binom{14}{7}
$
and as you noted $\left\{\frac73\right\}=\frac13$, but if you continue $\left\{\frac{7}{3^2}\right\}\geq\frac12$.

3. Originally Posted by melese
Hello chiph588@, I think I've found what you're looking for, namely a sufficient and necessary condition of the divisibility of $\binom{2n}{n}$ by a prime $p$ in general!

A prime $p$ divides $\binom{2n}{n}$ if and only if the fractional part $\left \{ \frac{n}{p^k} \right \}\geq\frac{1}{2}$ for at least one $k\in \mathbb{N}$. In particular, when $k=1$.

For example, $3\not|\binom{8}{4}$ and indeed $\left \{ \frac{4}{3^1} \right \}<\frac{1}{2}$; $\left \{ \frac{4}{3^2} \right \}<\frac{1}{2}$; and so on...

Another example, the one that you gave here is $3\mid\binom{14}{7}
$
and as you noted $\left\{\frac73\right\}=\frac13$, but if you continue $\left\{\frac{7}{3^2}\right\}\geq\frac12$.
Yes, I came to the same conclusion last night too!

4. Me too.

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