Power of 2

The standard notation for the highest power of $2$ that divides $k$ is $\upsilon_2(k)$ . You can see it in, e.g., the LTE paper .

$\upsilon_2((n+1)(n+2)\cdots (2n))=\upsilon_2\left(\frac{(2n)!}{n!}\right)=\upsilon_2((2n)!)-\upsilon_2(n!)$

See Legendre's formula .

$=\left(\lfloor\frac{2n}{2}\rfloor+\lfloor\frac{2n}{4}\rfloor+\lfloor\frac{2n}{8}\rfloor+\cdots\right)-\left(\lfloor\frac{n}{2}\rfloor+\lfloor\frac{n}{4}\rfloor+\lfloor\frac{n}{8}\rfloor+\cdots\right)=\lfloor n\rfloor = n$

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Alternative solution: $(n+1)(n+2)\cdots (2n)=\frac{(2n)!}{n!}=\frac{(1\cdot 3\cdot 5\cdots (2n-1))\cdot (2\cdot 4\cdot 6\cdots 2n)}{n!}$

$=\frac{(1\cdot 3\cdot 5\cdots (2n-1))\cdot ((2\cdot 1)\cdot (2\cdot 2)\cdot (2\cdot 3)\cdots (2\cdot n))}{n!}=\frac{(1\cdot 3\cdot 5\cdots (2n-1))\cdot (2^nn!)}{n!}=(1\cdot 3\cdot 5\cdots (2n-1))\cdot 2^n$ .