Perfect numbers of the form $12m+1$ and $\sum_{d\mid n}\frac{1}{\phi(d)}$, where $\phi(m)$ is Euler's totient function

Question

If there are no mistakes combining Exercise 9 a) (Chapter 3, page 71) and Exercise (Chapter2, page 47) of Apostol's Introduction to Analytic Number Theory we can prove easily

Lemma. If $n$ is a perfect number then $$2<\sum_{d\mid n}\frac{1}{\phi(d)},$$ where $\phi(m)$ is the Euler's totient function.

On the other hand, we know the so called Touchard's theorem.

Question. Can you give a proof that all integer $n\geq 3$ of the form $n=12m+1$ satisfies $$\sum_{d\mid n}\frac{1}{\phi(d)}\leq 2?$$ Or well can you find a counterexample (I say with a computer), an integer $n\geq 3$ such that $n\equiv 1\mod 12$ satisfying $\sum_{d\mid n}\frac{1}{\phi(d)}> 2?$ Thanks in advance.

Answer 1

The smallest counterexamples are $37182145$, $56581525$, $61686625$, $92317225$, $96521425$, $107632525,\dots$.

For example, for $37182145=5\cdot7\cdot11\cdot13\cdot17\cdot19\cdot23$, your sum has value $\frac{676039}{331776}$ which is about 2.0376.