# On identities and congruences involving the harmonic mean of odd perfect numbers

I would like to know if my calculations were rights and if you want to deduce other different interesting congruences following my approach.

On assumption (notice that it isn't the usually accepted conjecture, I am saying that the belief is that there exists no odd perfect numbers; but it is required suppose such hypothesis to try deduce something about odd perfect numbers and also in an hypothetical proof by contradiction) that there exists an odd perfect number $n$, then its harmonic mean

$$H(n)=\frac{n\sigma_{0}(n)}{\sigma(n)},$$ is integer (see the genuine definition from previous Wikipedia's article) where $\sigma(n)=\sum_{d\mid n}d$ is the sum of divisors function and $\sigma_0(n)=\sum_{d\mid n}1$ the number of such positive divisors. Please, notice thus from this factorization that this arithmetic function is also multiplicative.

Claim 1. If there are no mistakes in my calculations I can deduce by substitution of $H(n)$ in the particular value of $H(2^an)$, where $a\geq 1$ is an integer, the following identity that holds on assumption that $n$ is an odd perfect number $$(2^{a+1}-1)H(2^an)=2^a(a+1)H(n).$$

And also using Euler-Fermat Theorem with specialization $a=\phi(n)$, where $\phi(m)$ is the Euler's totient function, one gets the following first statement, and secondly with the specialization $a=n$ and dividing (without Euler-Fermat) the second one, of the following

Claim 2. On assumption that there exists an odd perfect number $n$, then $$(2^{\phi(n)+1}-1)H(2^{\phi(n)}n)\equiv(\phi(n)+1)H(n)\text{ mod }n,$$ and $$(2^{n+1}-1)H(2^n n)\equiv 2^nH(n)\text{ mod }n.$$

(Also was fixed a typo.)

Then

Question 1. What's about previous claims? Can you provide us detatils to show that my calculations were rights, or is there a mistake? Also you are welcome to discuss in comments if those claims are interesting.

The following is to encourage to study this kind of questions, isn't required a full answer (only a contribution as an answer, a reference, hints...) for the following

Question 2. As a secondary question: can you provide us different congruences from same approach? I am saying if it is possible deduce more interesting recurrences, relationships or congruences for the harmonic mean of an odd perfect number (maybe you can assume different primes that don't divide $n$, or to use that the harmonic mean is multiplicative in a different way, I was thinking what's about two odd perfect numbers, being coprime...).

Many thanks users!

• Also it's possible deduce $$(H(n))^2H(n+1)H(n-1)\sigma(n-1)\sigma(n+1)=(\frac{n^2-1}{2})(\sigma_0(n))^2\sigma_0(n+1)\sigma_0(n-1),$$ where $\frac{n^2-1}{2}$ is integer. – user243301 Sep 5 '16 at 22:24
• I am waiting the proof verification from some user and if he/she can deduce some interesting identities, and why his/her identities could be potentially useful. I've deduced one more: let $N=p^{4\lambda+1}m^2$ the eulerian form of an odd perfect number then $$2\sigma_0(2^{p-1})H(p^{4\lambda }m^2)=\sigma_0(p^{4\lambda}m^2).$$ On the other hand I've found the equivalence: $n\geq 1$ is perfect if and only if $$\sum_{d\mid n}(d+n/d)^2=2\sigma_2(n)+4nH(n),$$ where $\sigma_2(n)=\sum_{d\mid n}d^2$. (One can write $\sigma_0(n)\sigma(n)$ instead of $4nH(n)$ in previous equivalence). – user243301 Sep 7 '16 at 18:30
• Many thanks to you and other users for this edit @2015 – user243301 Sep 22 '16 at 11:34