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Are there infinitely many postive integers $\ n\ $ satisfying $$\varphi(n)\mid \sigma(n)$$ where $\varphi(n)$ is Euler’s totient function, and $\sigma(n)$ is the sum of divisors of $n$? If yes, can it be proven?

The first few such $\ n\ $ are $$n=1, 2, 3, 6, 12, 14, 15, 30, 35, 42, 56, 70, 78, 105, 140, 168, 190.$$

Ideas for a proof of the existence of infinitely many such positive integers (like families of such numbers probably being infinite) are appreciated.

I posted this question to Math.StackExchange, but I couldn't get reaction. That is why I'm asking this here. If this is not suitable for this place, please answer at Math.StackExchange: Are there infinite many positive integers $\ n\ $ satisfying $\ \varphi(n)|\sigma(n)\ $?

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    $\begingroup$ This is A020492 in the Online Encyclopedia of Integer Sequences. Most likely the references there will answer your question. oeis.org/A020492 $\endgroup$
    – Aeryk
    Commented Mar 25, 2022 at 15:55
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    $\begingroup$ @Aeryk: Did you look at these references? $\endgroup$
    – markvs
    Commented Mar 26, 2022 at 3:01
  • $\begingroup$ Note that both of these are multiplicative functions. Now $\varphi(p^k)=p^k(1-1/p)$ whereas $\sigma(p^k)=(p^k-1)/(p-1)$. If $p\geq 3$ and $k\geq 2$, then $\varphi(p^k)>\sigma(p^k)$. Hence, one might imagine that $n$ cannot be divisible by $p^2$ for $p\geq 3$. However, $n=270, 594, ...$ are in the sequence. In other words, the multiplicative properties alone are not enough to resolve this issue! $\endgroup$
    – Kapil
    Commented Mar 26, 2022 at 6:14

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