Median divisor of even perfect numbers

Question

I noticed that when divisors of even perfect numbers are listed in ascending order, the middle divisor (I guess the median), is always of the form $2^n$, some power of 2. If true is there a proof for this, or does it happen all the time? I only checked up to the 8th perfect number. Thank you and apologies for the possible silliness of the question.

Answer 1

We know that all even perfect numbers are of the form $(2^n-1)2^{n-1}$ where $2^n-1$ is prime, so there are $2n$ divisors. Ignoring the divisor that is the perfect number itself (so that the remaining $2n-1$ divisors sum to the perfect number), it is easy to see that those divisors have a simple ordering too: $$\underbrace{1,2,\dots,2^{n-1}}_{n},2^n-1,2(2^n-1),\dots,2^{n-2}(2^n-1)$$ The median is the $n$th divisor in this ordering, which we see is $2^{n-1}$. Therefore your claim is correct.

Answer 2

Every even perfect number is of the form $n=2^{p-1}(2^p-1)$ with $2^p-1$ prime.

Therefore, the factors of $n$ are

$1, 2, 2^2, 2^3, ..., 2^{p-1}, 2^p-1, 2(2^p-1), 2^2(2^p-1), 2^3(2^p-1), ...2^{p-2}(2^p-1),$

and the middle one is $2^{p-1}.$