It seems like something that should be pretty obvious but I don't quite get why would it be true. For example, in the case of 2-fold perfect numbers, or simply perfect numbers, it is evident because

$$\sigma(n) - n = n$$

so you just add all the divisors except itself.

We call a positive integer $n>1$ multi-perfect , if $k:=\frac{\sigma(n)}{n}$ is an integer , in the case $k=2$ the number is called perfect.

What about the multi-perfect numbers that are not perfect (the case $k>2$) ? Are they all pseudoperfect ?

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    $\begingroup$ Pseudoperfect numbers, sometimes also called semiperfect, are numbers for which the sum of all or some of its divisors is the number itself. $\endgroup$ – aaac991 Dec 13 '20 at 22:45
  • $\begingroup$ All multiples of $6$ are pseudoperfect, and so far all multiperfect numbers I got with PARI/GP (the perfect numbers excluded) are multiples of $6$. There could however be counterexamples. But at first sight, it seems to be the case. $\endgroup$ – Peter Dec 14 '20 at 14:28
  • $\begingroup$ $459818240$ is not a multiple of $6$, but it is a multiple of $20$. Such numbers are pseudoperfect as well. $\endgroup$ – Peter Dec 14 '20 at 14:44
  • $\begingroup$ Thanks for the help, I appreciate it. $\endgroup$ – aaac991 Dec 15 '20 at 2:37
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    $\begingroup$ The multi-perfect numbers in the above OEIS-link that correspond with $k>2$ , are all divisible by $6$ or by $20$ , hence are pseudoperfect. If we can trust this entry, all multi-perfect numbers upto $10^{300}$ are pseudoperfect. $\endgroup$ – Peter Feb 21 at 13:46

Here I consider the case of $k$-perfect numbers where $k \geq 3$. (The case $k=2$ was considered in the OP.)

It has been conjectured that $k$-perfect numbers ought to be divisible by $k!$. From Peter's comment, this then implies that $k$-perfect numbers are also pseudoperfect. (Note that the conjecture mentioned also implies that there are no odd $k$-perfect numbers for $k \geq 2$.)

Update: As correctly pointed out by Peter, $459818240$ is $3$-perfect but not divisible by $6$. It is however, divisible by $20$, which still makes it pseudoperfect, as had already been pointed by Peter in a comment underneath the OP.

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    $\begingroup$ $459818240$ is $3$-perfect, but not divisible by $6$. $\endgroup$ – Peter Feb 21 at 10:36

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