Arithmetical Functions Sum, $\sum\limits_{d|n}\sigma(d)\phi(\frac{n}{d})$ and $\sum\limits_{d|n}\tau(d)\phi(\frac{n}{d})$

Solution 1:

Here I can give you three quick different proofs, first in terms of Dirichlet multiplication we have: $$\sigma*\phi=(\text{id}*1)*\phi=\text{id}*1*\mu*\text{id}=\text{id}*\text{id}=\text{id}\tau$$ $$\tau*\phi=(1*1)*\phi=1*1*\mu*\text{id}=1*\text{id}=\sigma$$

Or note that if $f$ and $g$ are multiplictive with $v_p(n)$ the p-adic order of $n$ then,

$$\sum_{d\mid n}f(d)g\left(\frac{n}{d}\right)=\prod_{p\mid n}\left(\sum_{k=0}^{v_p(n)} f(p^k)g(p^{v_p(n)-k})\right)$$

Thus if $f=\sigma,\tau$ and $g=\phi$ then substituting there values in at prime powers gives both results.

Alternatively if you know their Dirichlet generating functions then multiplying them out yields:

$$\sum_{n=1}^\infty \frac{n\tau(n)}{n^s}=\zeta(s-1)^2=\left(\sum_{n=1}^\infty\frac{\sigma(n)}{n^s}\right)\left(\sum_{n=1}^\infty\frac{\phi(n)}{n^s}\right)=\sum_{n=1}^\infty \frac{\sum_{d\mid n}\sigma(d)\phi(n/d)}{n^s}$$

$$\sum_{n=1}^\infty \frac{\sigma(n)}{n^s}=\zeta(s)\zeta(s-1)=\left(\sum_{n=1}^\infty\frac{\tau(n)}{n^s}\right)\left(\sum_{n=1}^\infty\frac{\phi(n)}{n^s}\right)=\sum_{n=1}^\infty \frac{\sum_{d\mid n}\tau(d)\phi(n/d)}{n^s}$$

Therefore we have both $n\tau(n)=\sum_{d\mid n}\tau(d)\phi(n/d)$ and also $\sigma(n)=\sum_{d\mid n}\tau(d)\phi(n/d)$ as required.

Lastly for future reference I would advise not using $\tau$ to refer to the divisor function, I only did so in this post to reduce any confusion for you. This is because in analytic number theory $\tau$ is often used to refer to Ramanujan's tau function. I would use either of these notations $d(n)=\sigma_0(n)=\sum_{d\mid n}1$.

Solution 2:

There is a request for Pillai's paper or proof. The paper is a bit hard to find and quite old, so I scanned it. It's called "On An Arithmetic Function" and was published at Annamalai University.

I've made the scan available at my website. [Presumably I'll host it somewhere else sometime - I do not know of another online copy]. The first desired equation is the combination of his Theorem II and Theorem III, for instance. The second desired equation isn't an explicit result, but instead is a byproduct of his other proofs and results.