Multiply perfect number

The sum of the divisors of 120 is

1 + 2 + 3 + 4 + 5 + 6 + 8 + 10 + 12 + 15 + 20 + 24 + 30 + 40 + 60 + 120 = 360

which is 3 × 120. Therefore 120 is a 3-perfect number.

The following table gives an overview of the smallest known k-perfect numbers for k ≤ 11 (sequence A007539 in the OEIS):

It can be proven that:

• For a given prime number p, if n is p-perfect and p does not divide n, then pn is (p + 1)-perfect. This implies that an integer n is a 3-perfect number divisible by 2 but not by 4, if and only if n/2 is an odd perfect number, of which none are known.
• If 3n is 4k-perfect and 3 does not divide n, then n is 3k-perfect.

It is unknown whether there are any odd multiply perfect numbers other than 1. However if an odd k-perfect number n exists where k > 2, then it must satisfy the following conditions:[2]

• The largest prime factor is ≥ 100129
• The second largest prime factor is ≥ 1009
• The third largest prime factor is ≥ 101

In little-o notation, the number of multiply perfect numbers less than x is ${\displaystyle o(x^{\varepsilon })}$ for all ε > 0.[2]

The number of k-perfect numbers n for n ≤ x is less than ${\displaystyle cx^{c'\log \log \log x/\log \log x}}$, where c and c' are constants independent of k.[2]

Under the assumption of the Riemann hypothesis, the following inequality is true for all k-perfect numbers n, where k > 3

${\displaystyle \log \log n>k\cdot e^{-\gamma }}$

where ${\displaystyle \gamma }$ is Euler's gamma constant. This can be proven using Robin's theorem.

The number of divisors τ(n) of a k-perfect number n satisfies the inequality[3]

${\displaystyle \tau (n)>e^{k-\gamma }.}$

The number of distinct prime factors ω(n) of n satisfies[4]

${\displaystyle \omega (n)\geq k^{2}-1.}$

If the distinct prime factors of n are ${\displaystyle p_{1},p_{2},\ldots ,p_{r}}$, then:[4]

${\displaystyle r\left({\sqrt[{r}]{3/2}}-1\right)<\sum _{i=1}^{r}{\frac {1}{p_{i}}}<r\left(1-{\sqrt[{r}]{6/k^{2}}}\right),~~{\text{if }}n{\text{ is even}}}$

${\displaystyle r\left({\sqrt[{3r}]{k^{2}}}-1\right)<\sum _{i=1}^{r}{\frac {1}{p_{i}}}<r\left(1-{\sqrt[{r}]{8/(k\pi ^{2})}}\right),~~{\text{if }}n{\text{ is odd}}}$

• Hemiperfect number

• The Multiply Perfect Numbers page
• The Prime Glossary: Multiply perfect numbers
• The Six Triperfect Numbers on YouTube