Let’s break it down
There is an integer $a_1$ and then its double $a_2$
The largest value is $n$ and its half is $n/2$
The second largest value is $n-2$ and its half is $\frac{n-2}{2}$
We subtract 2 since odd numbers have no halves
This pattern tells us that starting from n, every other value decrementing results in a double and integer
When we rephrase this statement, we’re saying half of the values from n to 2 are doubles of some values in the set
We can take the integer and its double and place it at the front
This means there are possibly $ \frac{n}{2} $ pairs
The remaining $n-2$ integers can be arranged in $ (n-2)! $ ways
In total, there are $ \frac{n}{2} \cdot (n-2)! $ awesome permutations of the set ${1, 2, …, n}$.