s(n) = kn or s(n) = n/k? [closed]

Question

This is not an important question, just for fun.

Definition:

$\sigma (n)$ = sum of the positive divisors of $n$.
$s(n)$ = sum of the proper positive divisors of $n$.

For $s(n) = kn$ , where $k$ is a natural number:

When $k = 1,$ then $n$ is a perfect number which has been discussed a lot.

How about $k = 2,3,4,5,\ldots$?

Based on some computations ( $n < 1.5\cdot 10^9$), I haven't found

1. Any odd number satisfying $s(n) = kn$.
2. $s(n) \geq 5n$
3. $s(n) = 4n$

for $s(n) = n/k$ with natural $ k >1$:

$n$ must be prime (and thus $k=n$.)

Conjecture:

1. If $s(n) = kn$, then $n$ must be even.
2. $s(n) < 5n$
3. $s(n) = n/k$ for a natural $ k >1$ $\iff n$ is prime .

Question: Could you provide a counterexample or prove it?

Answer 1

This is related to some open questions in number theory. See http://oeis.org/A134639 and references/links there.

Conjecture 2 is incorrect, the smallest counterexample is given by $$n=154345556085770649600.$$

Conjecture 3 holds trivially if one notice that for a composite $n$, $s(n)$ is larger than the largest proper divisor of $n$, and thus $s(n)=n/k$ implies $k=1$.