# 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.

• – Mason Nov 24 '18 at 17:40
• @Reuns. You are confident about the formula above? Taking this $k=0$. We have here that $\phi = \mu * d$ where $d(n)=\sigma_0(n)$ but the wiki says that $\phi = \mu * id$. – Mason Nov 30 '18 at 5:32
• @reuns. Are you confident about the formula above? – Mason Dec 4 '18 at 1:12
• $\sum_{d | m} \mu(d) \sum_{dn \le m} (dn)^k = ?$ – reuns Dec 4 '18 at 1:17
• – Mason Dec 18 '18 at 19:19

## 2 Answers

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...

• Sorry for my sloppy old comment $\phi_k(n) = \sum_{m \le n, (m,n)=1} m^k$ is a sum of multiplicative functions (related to $\zeta(s)$) because $\sum_{d | n} d^k \phi_k(n/d)=\sum_{d | n} \sum_{m \le n, (m,n)=d} m^k= \sum_{m=1}^n m^k = \sum_{l=0}^{k+1} c_l n^l$ (for some $c_l= c_l(k)$ not depending on $n$) so $\phi_k(n) =\sum_{d | n} \mu(d) \sum_{l=0}^{k+1} c_l (n/d)^l= \sum_{l=0}^{k+1} c_l \sum_{d | n} \mu(d)(n/d)^l$ and hence $$\sum_{n=1}^\infty \phi_k(n) n^{-s} = \sum_{l=0}^{k+1} c_l\frac{\zeta(s-l)}{\zeta(s)}$$ Looking at the poles shows the possible asymptotic of $\sum_{n \le N}\phi_k(n)$ – reuns Dec 9 '18 at 23:05
• Awesome Awesome Awesome! This looks good and definitely something I can work with. – Mason Dec 10 '18 at 23:23
• I meant $\phi_k(n) =\sum_{d | n} \mu(d)d^{k} \sum_{l=0}^{k+1} c_l (n/d)^l$ so it is $\sum_l c_l \zeta(s-l)/\zeta(s-k)$ – reuns Dec 11 '18 at 17:04

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.

• @reuns. Now equipped with a dirichlet series we can associate with $\phi_4$ is it possible to make any progress on the asympotic behavior of $\sum^x_{n=1}\phi_4(n)$? I suppose the poles mean when $\zeta(s-4)=0$? This seems like it should be as hard as the RH... Anyway: I wanted to comment that the approach above works for whole numbered $k$ but in general this techinique won't work arbitrary $k$. – Mason Dec 17 '18 at 21:26