Asymptotic behavior $\sum_{n=1}^x\phi_k(n)$, a variant of Euler's Totient function Let $$\phi_k(x)=\sum_{1\le n \le x \\(n,x)=1} n^k$$
What's the asymptotic behavior of 
$$\sum_{n=1}^x\phi_k(n)?$$
According to the wikipedia $\sum^x_{n=1} \phi_0 (n) \approx  \frac{3}{\pi^2}x^2 $. It also appears in page $69$ and $70$ which are $30$ and $31$ of this pdf.  
The possible routes 
Route 1 (For someone who wants some practice with Abel Summations): There should be an approach which is an analog to the techniques shown here: sum of the divisor functions and I think that $\sum_{n=1}^x \frac{\phi_k(n)}{n^{k+1}}$ is always on the order of a linear function. So that might be the place to start. 
If no one takes this route I will almost certainly post my own answer in 2 or 3 weeks and ask this community for help verifying my proof. This is the most obvious route for me to take to make progress on this.  
Route 2: Also it would be particularly interesting to see an argument which isn't an analog of the linked post and which exploits what we already know about the asymptotic behavior of $\sum \sigma_k(n)$ to make claims about $\sum \phi_k(n)$. I am not sure this possible but it may be a route forward. 
 A: Reuns asks: What is $$\sum_{d|m} \mu(d)\sum_{nd\le m} (dn)^k$$
I dunno. Maybe: 
$$\sum_{d|m} \mu(d)d^k\sum_{nd\le m} n^k$$
$$\sum_{d|m} \mu(d)d^kf_k(\lfloor m/d \rfloor)$$
Where $f_k(x)=\sum_{n=1}^x n^k $ is a Faulhaber sum. This may not be what is being sought though... 
A: Not an answer. 
$$\sum_{n=1}^\infty\frac{\phi_k(n)}{n^s}=\frac{1}{\zeta(s-k)}\sum_{l=0}^{k+1}c(k,l)\zeta(s-l)$$
Where $c(k,k+1)=\frac{1}{k+1}, c(k,k)=\frac{1}{2}$ and $c(k, k-l+1) = \frac{B_lk!}{l!(k-l+1)!}$ 
where $B_k$ is the $k$th Bernoulli number which we define in terms of Stirling numbers of the second kind.   
$$B_k=\sum_{m=0}^k \frac{(-1)^mm!}{m+1}S(k,m), \text{ and } S(k,m)=\frac{1}{k!}\sum_{j=1}^k(-1)^{k-j} {k \choose j} j^m$$
$c(a,b)$ are coefficients which we can find in Faulhaber's triangle. In particular, let's consider $k=4$, 
$$1^4+2^4+3^4+ \dots x^k \\=c(4,5)x^{5}+c(4,4)x^4+c(4,3)x^3+c(4,2)x^2+c(4,1)x
\\ = \frac{1}{5}x^5+\frac{1}{2}x^4+\frac{1}{3}x^3-\frac{1}{30}x $$
$$\sum_{n=1}^\infty \frac{\phi_4(n)}{n^s} = \frac{1}{\zeta(s-4)}\bigg[ \frac{1}{5} \zeta(s-5)+\frac{1}{2}\zeta(s-4)+\frac{1}{3}\zeta(s-3)- \frac{1}{30} \zeta(s-1) \bigg]$$
Here's a graph of this lining up! Lovely. 
Then we can somehow look at the poles of this to find the asymptotic behavior. 
