# Is every multiperfect number also pseudoperfect?

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 ?

• Pseudoperfect numbers, sometimes also called semiperfect, are numbers for which the sum of all or some of its divisors is the number itself. – aaac991 Dec 13 '20 at 22:45
• 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. – Peter Dec 14 '20 at 14:28
• $459818240$ is not a multiple of $6$, but it is a multiple of $20$. Such numbers are pseudoperfect as well. – Peter Dec 14 '20 at 14:44
• Thanks for the help, I appreciate it. – aaac991 Dec 15 '20 at 2:37
• 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. – 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.
• $459818240$ is $3$-perfect, but not divisible by $6$. – Peter Feb 21 at 10:36