Perfect Number is Ore Number

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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 Tau and Sigma, the harmonic mean of the divisors of $n$ is given by:

$H \left({n}\right) = \dfrac {n \tau \left({n}\right)} {\sigma \left({n}\right)}$

where:

$\tau \left({n}\right)$ denotes the $\tau$ (tau) function: the number of divisors of $n$
$\sigma \left({n}\right)$ denotes the $\sigma$ (sigma) function: the sum of the divisors of $n$.


Let $n$ be a perfect number.

By definition of perfect number:

$\dfrac {\sigma \left({n}\right)} n = 2$

From Tau Function Odd Iff Argument is Square:

$\tau \left({n}\right) = 2 k$

for some $k \in \Z$.

Hence:

$H \left({n}\right) = \dfrac {2 k} 2 = k$

Hence the result.

$\blacksquare$


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