Determine $f(n)$ such that for all $n\geq 1$,

$$\frac{1}{\varphi (n)}=\sum_{d\vert n}\left(\frac{1}{d}\right)f\left(\frac{n}{d}\right)$$

This is not a homework question, just a question I stumbled upon.

I have tried writing $\varphi (n)$ as

$$\varphi (n)=\sum_{d\vert n}^{}{\mu(d) \frac{n}{d}},$$ where $\mu$ is the Möbius function.

I am not sure if this is the right approach, but I was stuck here.

Sincere thanks for any help!

  • $\begingroup$ Have you checked to see whether Mobius inversion gives you anything? $\endgroup$ – Gerry Myerson Nov 19 '12 at 11:28
  • $\begingroup$ there is a somewhat different formula for $\frac{n}{\varphi(n)}$ at wikipedia (with 15 at the right and the link to Dineva's paper at the end). $\endgroup$ – Raymond Manzoni Nov 19 '12 at 13:02

Assume that $f$ is multiplicative. This will ensure that we can determine $f(n)$ as the product of its values at prime powers $f(p^v).$ We set $f(1)=1.$

The values of $f(p^v)$ can be determined recursively. Start with $f(p)$, which produces the equation $$ \frac{1}{p-1} = f(p) + \frac{1}{p} f(1) $$ or $$ f(p) = \frac{1}{p} \frac{1}{p-1}.$$

Now claim that $f(p^v) = 0$ when $v\ge 2.$ Reasoning inductively, we find $$ \frac{1}{p^v-p^{v-1}} = f(p^v) + \frac{1}{p^{v-1}} f(p) + \frac{1}{p^v} f(1)$$ which implies $$ f(p^v) = \frac{1}{p^v-p^{v-1}} - \frac{1}{p^v} \frac{1}{p-1} - \frac{1}{p^v} = \frac{p - 1 - (p-1)}{p^{v+1}-p^v} = 0.$$ This shows that $$f(p^v) = \begin{cases} 1 & \text{if} \quad v=0 \\ \frac{1}{p} \frac{1}{p-1} & \text{if} \quad v=1 \\ 0 & \text{otherwise.} \end{cases}$$ To conclude we now identify this function. It must be zero if the square of a prime divides $n$, and positive otherwise, hence it is a multiple of $\mu^2(n).$ The denominator is simply $n\varphi(n)$, so that the end result is $$ f(n) = \frac{\mu^2(n)}{n\varphi(n)}.$$

The above process reflects the Dirichlet convolution $$ f \star \frac{1}{n} = \frac{1}{\varphi}.$$ This would suggest a possibility to compute a closed form of the function $G(s)$ from this post. However we have the Euler product $$ \sum_{n\ge 1} \frac{1/\varphi(n)}{n^s} = \prod_p \left( 1 + \frac{1}{p-1} \frac{1}{p^s} + \frac{1}{p}\frac{1}{p-1} \frac{1}{p^{2s}} + \frac{1}{p^2}\frac{1}{p-1} \frac{1}{p^{3s}} + \cdots \right)$$ which is $$ \prod_p \left( 1 + \frac{p}{p} \frac{1}{p-1} \frac{1}{p^s} + \frac{p}{p^2}\frac{1}{p-1} \frac{1}{p^{2s}} + \frac{p}{p^3}\frac{1}{p-1} \frac{1}{p^{3s}} + \cdots \right)$$ which yields in turn $$\prod_p \left( 1 + \frac{p}{p-1} \frac{1/p^{s+1}}{1-1/p^{s+1}} \right)$$ Now we examine the roots and the singularities of this expression as in $$ 1 + \frac{p}{p-1} \frac{1/z/p}{1-1/z/p}$$ getting $$ z = \frac{1}{p(1-p)},$$ so no closed form appears possible.

  • $\begingroup$ Is your derivation of a formula for $f(n)$ not just Mobius inversion carried out by hand, as it were? $\endgroup$ – Gerry Myerson Nov 19 '12 at 22:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.