Problem 1453

To prove that
$$\frac{(a_1 + a_2 + \cdots + a_n)!!}{a_1!! a_2!! \cdots a_n!!}$$
is an integer, we will use properties of factorials and Legendre's formula for the valuation of factorials at a prime.

1. Definition and Setup:
- Define $n!! = (n!)!$.
- We need to show that the given fraction is an integer.

2. Prime Factorization Insight:
- Consider a prime $p$. We will use Legendre's formula to count the power of $p$ in the factorials involved.
- Legendre's formula states that for any integer $m$,
$$v_p(m!) = \sum_{i=1}^{\infty} \left\lfloor \frac{m}{p^i} \right\rfloor$$
where $v_p(m!)$ denotes the exponent of the prime $p$ in the prime factorization of $m!$.

3. Applying Legendre's Formula:
- For the numerator $(a_1 + a_2 + \cdots + a_n)!!$, we have:
$$v_p((a_1 + a_2 + \cdots + a_n)!!) = v_p(((a_1 + a_2 + \cdots + a_n)!)!)$$
Using Legendre's formula again for $(a_1 + a_2 + \cdots + a_n)!$, we get:
$$v_p((a_1 + a_2 + \cdots + a_n)!) = \sum_{i=1}^{\infty} \left\lfloor \frac{a_1 + a_2 + \cdots + a_n}{p^i} \right\rfloor$$
Therefore,
$$v_p((a_1 + a_2 + \cdots + a_n)!!) = \sum_{i=1}^{\infty} \left\lfloor \frac{(a_1 + a_2 + \cdots + a_n)!}{p^i} \right\rfloor$$

4. Denominator Analysis:
- For the denominator $a_1!! a_2!! \cdots a_n!!$, we have:
$$v_p(a_1!! a_2!! \cdots a_n!!) = v_p(a_1!!) + v_p(a_2!!) + \cdots + v_p(a_n!!)$$
Each $v_p(a_i!!) = v_p((a_i!)!)$, and using Legendre's formula for $a_i!$, we get:
$$v_p(a_i!) = \sum_{j=1}^{\infty} \left\lfloor \frac{a_i}{p^j} \right\rfloor$$
Therefore,
$$v_p(a_i!!) = \sum_{j=1}^{\infty} \left\lfloor \frac{a_i!}{p^j} \right\rfloor$$
Summing over all $i$,
$$v_p(a_1!! a_2!! \cdots a_n!!) = \sum_{i=1}^{n} \sum_{j=1}^{\infty} \left\lfloor \frac{a_i!}{p^j} \right\rfloor$$

5. Comparison:
- We need to show that:
$$v_p((a_1 + a_2 + \cdots + a_n)!!) \geq v_p(a_1!! a_2!! \cdots a_n!!)$$
- Since $(a_1 + a_2 + \cdots + a_n)! > a_1! + a_2! + \cdots + a_n!$, it follows that:
$$\left\lfloor \frac{(a_1 + a_2 + \cdots + a_n)!}{p^i} \right\rfloor \geq \sum_{j=1}^{n} \left\lfloor \frac{a_j!}{p^i} \right\rfloor$$
for all $i \geq 1$.

6. Conclusion:
- Since the power of any prime $p$ in the numerator is at least as large as the power of $p$ in the denominator, the fraction is an integer.

$\blacksquare$