# Divisibility of central binomial coefficient

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• Jul 11th 2010, 02:56 PM
Ben92
Yes, you are right. (Happy)
• Jul 11th 2010, 06:10 PM
melese
Quote:

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

This turns out to be false as $\displaystyle 3\mid\binom{14}{7}$ and $\displaystyle \left\{\frac73\right\}=\frac13$.

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

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

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

Another example, the one that you gave here is $\displaystyle 3\mid\binom{14}{7}$ and as you noted $\displaystyle \left\{\frac73\right\}=\frac13$, but if you continue $\displaystyle \left\{\frac{7}{3^2}\right\}\geq\frac12$.
• Jul 11th 2010, 06:12 PM
chiph588@
Quote:

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 $\displaystyle \binom{2n}{n}$ by a prime $\displaystyle p$ in general!

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

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

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

Yes, I came to the same conclusion last night too!
• Jul 11th 2010, 08:43 PM
Ben92
Me too.
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