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For any positive integer $n$, let $\sigma(n)$ be the sum of all positive divisors of $n$. Clearly, $\sigma(n)\ge n\ge \varphi(n)$ for all $n\in\mathbb{Z}_{\geq 1}$, where $\varphi$ is Euler's totient function.

For any $\varepsilon>0$, we may take a prime $p>1+2/\varepsilon$ so that $$1<\frac{\sigma(p)}{\varphi(p)}=\frac{p+1}{p-1}=1+\frac2{p-1}<1+\varepsilon.$$ Let us consider the set $$S=\left\{\frac{\sigma(n)}{\varphi(n)}:\ n\in\mathbb{Z}_{\geq 1}\right\}.$$

Question. Is it true that $S=\{r\in\mathbb Q:\ r\ge1\}$?

Recently I found that all those rational numbers $a/b$ with $36\ge a\ge b\ge1$ belong to the set $S$. This led me to conjecture that $S$ indeed coincides with $\{r\in\mathbb Q:\ r\ge1\}$. For example, $41/2\in S$ since $$25604040 = 2^3\times3\times5\times7\times11\times17\times163 \ \ \text{and}\ \ \frac{\sigma(25604040)}{\varphi(25604040)}=\frac{102021120}{4976640} =\frac{41}2.$$ Also, $41/19\in S$ since $$121473159=3\times31\times179\times7297 \ \ \text{and}\ \ \frac{\sigma(121473159)}{\varphi(121473159)}=\frac{168145920}{77921280} =\frac{41}{19}.$$

Your comments are welcome!

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    $\begingroup$ The standard way to define a set is $\{a\in A:\ P(a)\}$, where $A$ is a set and $P(a)$ is a first-order formula in ZFC. Hence I changed $\{r\ge1:\ r\in\mathbb Q\}$ to $\{r\in\mathbb Q:\ r\ge1\}$. $\endgroup$
    – GH from MO
    Commented Aug 8 at 17:21
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    $\begingroup$ @YaakovBaruch I think he meant that $\sigma(1)=\varphi(1)=1$. $\endgroup$
    – GH from MO
    Commented Aug 8 at 17:23
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    $\begingroup$ This may be true even if we restrict $n$ to squarefree integers. I have verified that $\frac{\sigma(n)}{\varphi(n)}=\frac{a}{b}$ is solvable for squarefree $n$ for all $b<a\leq 100$. For example: $\frac{\sigma(n)}{\varphi(n)}=\frac{41}2$ holds for squarefree $n=2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 29\cdot 163$. $\endgroup$
    – Max Alekseyev
    Commented Aug 9 at 3:16
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    $\begingroup$ Is it clear that every positive integer can be expressed as the quotient $\sigma(n)/\varphi(n)$? That might be a good start. $\endgroup$
    – Stanley Yao Xiao
    Commented Aug 10 at 2:09
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    $\begingroup$ @StanleyYaoXiao Up to $n=10000000$ one gets integers from $1$ to $22$, represented with the following frequencies:$$ \begin{array}{r|r} 1 & 1 \\ 2 & 26 \\ 3 & 38 \\ 4 & 90 \\ 5 & 73 \\ 6 & 78 \\ 7 & 75 \\ 8 & 83 \\ 9 & 89 \\ 10 & 86 \\ 11 & 51 \\ 12 & 79 \\ 13 & 54 \\ 14 & 57 \\ 15 & 49 \\ 16 & 33 \\ 17 & 6 \\ 18 & 22 \\ 19 & 12 \\ 20 & 12 \\ 21 & 12 \\ 22 & 2 \\ \end{array}$$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Aug 13 at 17:12

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