# Maximum $r$ such $p^r$ divides ${2n \choose n}$.

It is known that the number $n!$ contains the prime factor $p$ exactly $$\displaystyle\sum_{k\geq1}\left\lfloor\dfrac{n}{p^k}\right\rfloor$$

Then, if for a fixed prime $p$, define $R(p,n)$ to be the largest natural number $r$ such that $p^r$ divides ${2n \choose n}$, then: $$R(p,n)=\displaystyle\sum_{j\geq1} \left \lfloor \dfrac{2n}{p^j} \right \rfloor -\color{blue}{2\displaystyle\sum_{j\geq1} \left \lfloor \dfrac{n}{p^j} \right \rfloor}$$ But each term of the last summation can either be zero or 1, (if $n/p^j \bmod 1< 1/2$) or $1$ (if $n/p^j \bmod 1\ge 1/2$) and all terms with $j>\log_p(2n)$ are zero. Therefore $$R(p,n)\leq \log_{p}(2n)$$

I don't understand the $\color{blue}{\mbox{blue}}$ lines, how can I get it?

• If I want $\max r$, why I rest the blue summation? the blue sumation is the max power which divides $n!\times n!$, if I want the maximum would be better not to take the minimum exponent wich divides $n!\times n!$ to maximize the sum? – Luis Felipe May 18 '15 at 1:18
• $x\bmod 1\ge 1/2$ is an excessively fancy way of saying that the fractional part of $x$ (the part after the decimal point) is $\ge 1/2$. – André Nicolas May 18 '15 at 1:22
• Thank you, now it has color black. – Luis Felipe May 18 '15 at 1:24
• You are welcome. The $x\bmod r$ notation sort of makes sense. For $x=(q)(1)+r$, for some integer $q$, and some $r$ with $0\le r\lt 1$. So the fractional part is (sort of) the "remainder" on division by $1$. – André Nicolas May 18 '15 at 1:32
• sure, I read the $mod$ notation only with an integer modulus. I'm reading about Bertrand postulate proof from Erdös. – Luis Felipe May 18 '15 at 1:34

If $j \gt \log_p(2n)$, then $p^j \gt 2n$. It follows that $\frac{2n}{p^j}\lt 1$, and therefore $\left\lfloor \frac{2n}{p^j}\right\rfloor=0$. For $j \gt \log_p(2n)$, all the terms in the "infinite" sums of your second displayed formula are $0$.