# On a generalization for $\sum_{d|n}rad(d)\phi(\frac{n}{d})$ and related questions

Let $\phi(m)$ Euler's totient function and $rad(m)$ the multiplicative function defined by $rad(1)=1$ and, for integers $m>1$ by $rad(m)=\prod_{p\mid m}p$ the product of distinct primes dividing $m$ (it is obvious that it is a multiplicative funtion since in the definition is $\prod_{p\mid m}\text{something}$ and since empty products are defined by $1$).

Denoting $r_1(n)=rad(n)$, and $$R_1(n)=\sum_{d|n}rad(d)\phi(\frac{n}{d}),$$ I claim that it is possible to proof that this Dirichlet product of multiplicative functions (thus is multiplicative) is computed as $$\frac{n}{rad(n)}\prod_{p\mid n}(2p-1).$$

Question 1. Can you prove or refute that $$R_k:=\sum_{d|n}r_k(d)\phi(\frac{n}{d})=\frac{n}{rad(n)}r_{k+1}(n)$$ for $$r_{k+1}(n)=\prod_{p\mid n}((k+1)p-k),$$ with $k\geq 1$? Thanks in advance.

I excuse this question since I've obtain the first examples and I don't know if I have mistakes. I know that the proof should be by induction. Since computations are tedious I would like see a full proof. In this ocassion if you are sure in your computations, you can provide to me a summary answer. The following is to obtain a more best post, in other case I should to write a new post.

I know the theorem about Dirichlet product versus Dirichlet series that provide us to write

$$\sum_{n=1}^\infty\frac{\frac{n}{rad(n)}r_2(n)}{n^2}=\left(\sum_{n=1}^\infty\frac{rad(n)}{n^s}\right)\left(\sum_{n=1}^\infty\frac{\phi(n)}{n^s}\right)=\sum_{n=1}^\infty\frac{\sum_{d\mid n}rad(d)\phi(n/d)}{n^s},$$ for $\Re s=\sigma>2$ (I've read notes in Apostol's book about this and follows ). By a copy and paste from  we can write $$\frac{\zeta(s)^2}{\zeta(2s)}<R(s)<\frac{\zeta(s)\zeta(s-1)}{\zeta(2s-2)},$$ where $R(s)$ is the Dirichlet series for $rad(n)$, and I believe that previous inequality holds for $\sigma>2$.

Question 2. Can you write and claim the convergence statement corresponding to Dirichlet series for $r_k(n)$? I say if Question 1 is true, and looking to compute these Dirichlet series for $r_k(n)$ as values, or inequalities involving these values, of the zeta function. Thanks in advance.

I excuse this Question 2 to encourage to me read and understand well, previous references  and .

 Ethan's answer, this site Arithmetical Functions Sum, $\sum_{d|n}\sigma(d)\phi(\frac{n}{d})$ and $\sum_{d|n}\tau(d)\phi(\frac{n}{d})$

 LinusL's question, this site, Average order of $\mathrm{rad}(n)$

• Too I believe that it is possible to prove inductively $\prod_{p\mid n}\left(1+\sum_{k=1}^{e_p}2^{k-1}R_1(p^k)\right)=2^{\eta(n)}n$, where $\eta(n)=\sum_{p\mid n}e_p$, the sum of exponents corresponding to the factorization of $n=\prod_{p\mid n}p^{e_p}$. And by the same method of first question that multiplicative function $n/rad(n)$ satisfies $\sum_{d\mid n}\frac{d}{rad(d)}\phi(n/d)=\frac{n}{(rad(n))^2}\prod_{p\mid n}(p^2+(e_p-1)(p-1))$. I leave this comment, because if you want read this notes. – user243301 Nov 27 '15 at 10:45

About your first question, you just have to observe that $R_k(n)$ is multiplicative beeing a Dirichlet product of multiplicative functions, and as a consequence so is $r_{k}(n)$, so you can compute it for a prime power a then multiply,

$$R_k(n) = \prod_{p^j\vert\vert n} R_k(p^j)$$ For computing $R_k(p^j)$ we treat separately the divisor 1 for the rest of divisors of $p^j$, ($p,p^2,\dots,p^j$) obtaining: $$R_k(p^j) = p^j-p^{j-1} + \sum_{i=1}^j (kp-(k-1))\phi(p^{j-i}) = \\ p^j-p^{j-1} + (kp-(k-1)) \left( (p^{j-1}-p^{j-2})+ \dots + (p-1) + 1\right) = \\ (k+1)p^j-kp^{j-1} = \frac{p^{j}}{rad(p^{j})}((k+1)p-k)$$

And you are done.

I'm not entirely sure what you are asking in the second question, if I understand you are interested in the convergence of the Dirichlet series $$\sum_n \frac{r_k(n)}{n^s}$$ suppose it converges for some $s=\sigma+it$, then it is easy to show that it converges for any $s$ with real part $>\sigma$, know in that hypothesis it will have also an expresion as an Euler product: $$\sum_n \frac{r_k(n)}{n^s} = \prod_p \left( 1 + (kp-(k-1))(p^{-s} + p^{-2s}+\dots) \right)= \\ \prod_p \left( \frac{1+p^{-s}(kp-k))}{1-p^{-s}} \right)=\zeta(s)A(s)$$ Where $A(s)$ has the Euler product $$A(s) = \prod_p (1+p^{-s}(kp-k))$$ if $s$ is real then all the factors are positive so you can bound it above by $$A(s) < \prod_p (1+p^{-s+1})^k = \left(\sum_n \frac{\vert \mu(n) \vert}{n^{s-1}}\right)^k$$ this implies that the original series converges for $\sigma >2$, to see that it diverges for $\sigma < 2$ it is easy for $k > 1$ as we have $kp-k >= p$, and so again for $s$ real $$A(s) > \prod_p(1+ p^{-s+1})=\sum_n \frac{\vert \mu(n) \vert}{n^{s-1}}$$ and the right hand series diverges for $s=2$. It still remains to prove that it diverges for $k=1$ I can't see now any simple proof but a limiting argument should work.

I hope this is what you were looking for.

• Very thanks much, this is more clear and concise. I vote up your good answer, but I will wait an answer for second question. Thanks @EstebanCrespi – user243301 Nov 27 '15 at 19:36
• I will study this in detail, very thanks much @EstebanCrespi – user243301 Nov 28 '15 at 17:27
• Yesterday I was a few tired, now I take notes from your computations, and study this. Is recognized that yours is a good an answer.Thanks @EstebanCrespi – user243301 Nov 29 '15 at 10:09