Perfect Number is Ore Number

Theorem

Let $n \in \Z_{>0}$ be a perfect number.

Then $n$ is an Ore number.

Proof

From Harmonic Mean of Divisors in terms of Divisor Count and Divisor Sum, the harmonic mean of the divisors of $n$ is given by:

$\map H n = \dfrac {n \map {\sigma_0} n} {\map {\sigma_1} n}$

where:

$\map {\sigma_0} n$ denotes the divisor count function: the number of divisors of $n$

$\map {\sigma_1} n$ denotes the divisor sum function: the sum of the divisors of $n$.

Let $n$ be a perfect number.

By definition of perfect number:

$\dfrac {\map {\sigma_1} n} n = 2$

From Divisor Count Function is Odd Iff Argument is Square:

$\map {\sigma_0} n = 2 k$

for some $k \in \Z$.

Hence:

$\map H n = \dfrac {2 k} 2 = k$

Hence the result.

$\blacksquare$

Sources

• 1948: Øystein Ore: On the averages of the divisors of a number (Amer. Math. Monthly Vol. 55, no. 10: pp. 615 – 619)  www.jstor.org/stable/2305616

• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $140$