Exercise 4.1.5 (Compute $\nu_p(2\cdot 4 \cdot 6 \cdots (2n))$ and $\nu_p(1\cdot 3 \cdot 5 \cdots (2n+1))$.)

Proof.

(a)

The product of the first $n$ even numbers is $$P=2\cdot 4\cdot 6\cdots (2n)=2^{n}n!.$$

Using the de Polignac’s formula, we obtain

$${\nu }_p(P)=\{\begin{array}{ccc}{\nu }_p(n!)\hfill & =\underset{i=1}{\overset{\infty }{\sum }}\lfloor \frac{n}{p^i}\rfloor \hfill & \text{if }p\ne 2,\hfill \\ n+{\nu }_2(n!)\hfill & =n+\underset{i=1}{\overset{\infty }{\sum }}\lfloor \frac{n}{2^i}\rfloor \hfill & \text{if }p=2.\hfill \end{array}$$

(b)

The product of the first $n$ odd numbers is $$Q=1\cdot 3\cdot 5\cdots (2n+1)=\frac{(2n+1)!}{2^{n}n!}.$$

If $p\ne 2$,

$$\begin{array}{llll}\hfill {\nu }_p(Q)& ={\nu }_p((2n+1)!)-{\nu }_p(n!)& \hfill & \\ \hfill & =\underset{i=1}{\overset{\infty }{\sum }}\left(\lfloor \frac{2n+1}{p^i}\rfloor -\lfloor \frac{n}{p^i}\rfloor \right).& \hfill & \end{array}$$

If $p=2$, since all factors in $Q$ are odd,

$${\nu }_2(Q)=0.$$