Which primes divide C(2n, n)?

1. $\binom{2n}n=\frac{(2n)!}{n!\,n!}$. The prime $p\le2n$ divides $(2n)!$, but $p>n$ does not divide $n!$. So $p$ divides the numerator and not the denominator.

2. Legendre's formula gives
$$v_p\binom{2n}n=\sum_{k\ge1}\left(\left\lfloor\frac{2n}{p^k}\right\rfloor-2\left\lfloor\frac n{p^k}\right\rfloor\right).$$

• For $k\ge2$: $p^k\ge p^2>2n$, so both floors are $0$.
• For $k=1$: from $\frac{2n}3<p\le n$ we get $1\le\frac np<\frac32$, so $\lfloor\frac np\rfloor=1$ and $2\le\frac{2n}p<3$, so $\lfloor\frac{2n}p\rfloor=2$. The term is $2-2\cdot1=0$.

So $v_p\binom{2n}n=0$.

Check. $n=10$: $\binom{20}{10}=184756=2^2\cdot11\cdot13\cdot17\cdot19$. Primes in $(10,20]$: $11,13,17,19$ all divide it ✓. Primes in $(\frac{20}3,10]$ with $p^2>20$: only $7$, and $7\nmid184756$ ✓.

Remark. These two facts are the heart of Erdős's proof of Bertrand's postulate.

Lesson. Express $v_p$ of a binomial coefficient via Legendre's formula as a sum of terms $\lfloor2x\rfloor-2\lfloor x\rfloor\in\{0,1\}$ and control each term.