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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!

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Have you checked to see whether Mobius inversion gives you anything? – Gerry Myerson Nov 19 '12 at 11:28
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). – 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.

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

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