For integers $n\geq 1$ with $$\operatorname{rad}(n)=\prod\_{\substack{p\mid n\\p\text{ prime}}}p$$ we denote the squarefree kernel or radical of an integer $n$ (see if you want this [Wikipedia](https://en.wikipedia.org/wiki/Radical_of_an_integer)). And $\varphi(n)$ denotes the Euler's totient funciton. Then while I was stuying the equation $$\operatorname{rad}(\varphi(2\cdot n))=2$$
in the context of constructible polygons with compass and straightedge, see this section of this [Wikipedia](https://en.wikipedia.org/wiki/Fermat_number#Relationship_to_constructible_polygons), I wondered next variation 
$$\operatorname{rad}(\varphi(3\cdot n))=2\cdot 3.\tag{1}$$


Ater I search the resulting sequence of solutions of our equation $(1)$, the first few terms are $3,6,7,9,12,13,14,\ldots$, in The On-Line Encyclopedia of Integer Sequences, I would like to do a comparison to the sequence [*A135412*](https://oeis.org/A135412). I know that there are terms in *A135412*, but don't satisfy our equation (1). You can read the definition of Heronian mean from this [Wikipedia.](https://en.wikipedia.org/wiki/Heronian_mean)



> 
> **Question.** I would like to know\* if it is possible to prove or refute that: 
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> *If* $n$ *satisfies* $$\operatorname{rad}(\varphi(3 n))=6$$ *then our solution* $n$ *equals three times the Heronian mean of two positive integers.* 
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> 
> **Many thanks.**
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> 
> 


I think that this question is curious. If a full answer isn't feasible I'm interested to know what work can be done.


\*Additionally if you want to add remarks in your answer about different equations involving the Euler's totient function and/or the squarefree kernel, showing some remarkable relationship to the sequence of the Heronian means in some way feel free to provide me such feedback.



The Euler's totient function and the product of distinct primes dividing $n$ versus the Heronian means
Your conclusion is valid, and for this we only need to assume that $\varphi(3n)$ is divisible by $3$. Indeed, this weaker condition is equivalent to the existence of a prime divisor $p\mid n$ that is either $3$ or congruent to $1$ modulo $3$. It is known that $p$ is the norm of some [Eulerian integer](https://en.wikipedia.org/wiki/Eisenstein_integer) $z=x+y\frac{1+i\sqrt{3}}{2}$. Multiplying $z$ with a sixth root of unity, we can achieve that $0<\arg(z)<\pi/3$. Then, $x$ and $y$ are positive integers satisfying $p=x^2+xy+y^2$, and hence $n$ is three times the Heronian mean of $(n/p)x^2$ and $(n/p)y^2$.


