Let $\sigma(n)$ the sum of divisors function, and $\varphi(n)$ the Euler's totient function. And $p\mid n$ means that $p$ runs over primes (dividing an integer $n\geq 1$).

I am interested if was in the literature some relevant theorem or conjecture about the equation $$2n\varphi(\sigma(n))=\sigma(n)\varphi(n).\tag{E}$$ I am interested especially in odd (integer) solutions having the form $12\lambda+1$ or $36\lambda+9$. I didn't numerical experiments with my computer, but I have searched the string phi(sigma(n)) in the OEIS. I am interested in this Question when I was combining simple hypothesis and similar equations concerning odd perfect nubmers.

Motivation. If $n$ is an odd perfect number and $7\nmid n$ then one has $$\varphi(\sigma(4n))=3\sigma(n)\prod_{p\mid n}\left(1-\frac{1}{p}\right).\tag{1}$$ On the other hand, if an odd integer $m\geq 1$ satisfies $7\nmid \sigma(m)$ and previous equation $(1)$ then $$2m\varphi(\sigma(m))=\sigma(m)\varphi(m).\tag{2}$$ As remark, even perfect numbers satisfy an equation similar than $(2)$.

Question. Is it kwnon from the literature the sequence of odd integers $n\geq 1$ satisfying $$2n\varphi(\sigma(n))=\sigma(n)\varphi(n)?$$ If was in the literature, refer papers with relevant theorems or related conjectures. If you know that it isn't in the literature, what simple facts or conjectures (should be arbitrarly large solutions?) can we set for the equation, especially concerning odd solutions $n\geq 1$, that is simplified as $$2\varphi\left(\prod_{p\mid n}\frac{p^{e_p+1}-1}{p-1}\right)=\prod_{p\mid n}\frac{p^{e_p+1}-1}{p}$$ (in terms of the factorization of integers $n=\prod_{p\mid n}p^{e_p}$ satisfying such equation)? Many thanks.


I did a numerical search and found $$ 135,\ 891,\ 200655,\ 307125,\ 544635 $$ This is sequence A238330 in OEIS.

  • $\begingroup$ Many thanks! Many thanks for your help, my problem was that I (I've deleted in the past) lost my programs written in pascal and when I need those, well, I have no those and I can not numerical calculations. As you see if $n$ is an odd perfect number, if there exist such $n$, also satisfies A238330, because the radical of an integer $\operatorname{rad}(n)$ is a multiplicative function. $\endgroup$ – user243301 May 18 '17 at 10:52

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