Reciprocal-totient function, in term of the totient function?

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I've been looking at a function $\phi_{-1}:\mathbb N\mapsto\mathbb Q$, similar to Euler's totient function $\phi$, defined recursively as $$\phi_{-1}(1):=1$$ $$\phi_{-1}(n):=\frac{1}{n}-\sum_{d|n, d\lt n}\phi_{-1}(d)$$ which can be contrasted with the recursive definition of the totient function: $$\phi(n):=n-\sum_{d|n, d\lt n}\phi(d)$$ Some of its special values include $$\phi_{-1}(p)=\frac{1-p}{p}, \space\space p\space\space\text{prime}$$ $$\phi_{-1}(p^m)=\frac{1-p}{p^m}, \space\space p\space\space\text{prime}$$ $$|\phi_{-1}(c)|=\frac{\phi(c)}{c},\space\space \text{if each prime factor of } c\space\space\text{has a multiplicity of 1}$$ I am looking for some other properties of this function, especially a representation of it in terms of other common number-theoretical function, or a formula for the sum $$\sum_{d|n}\phi_{-1}(d)$$ I am also trying to find a rule about when $\phi_{-1}$ takes on positive values.

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If $\phi_{-1}:\mathbb{N}^*\to\mathbb{Q}$ is defined through $\phi_{-1}(1)=1$ and $$ \phi_{-1}(n) = \frac{1}{n}-\sum_{\substack{d\mid n\\ d<n}}\phi_{-1}(d) $$ then $\sum_{d\mid n}\phi_{-1}(d)=\frac{1}{n}$. Since $\frac{1}{n}$ is a multiplicative function, by Moebius inversion it follows that $\phi_{-1}$ is a multiplicative function too. We know its values over prime powers, hence

$$ \phi_{-1}(n) = \prod_{p\mid n}\frac{1-p}{p^{\nu_p(n)}}=\frac{(-1)^{\omega(n)}}{n}\prod_{p\mid n}(p-1)=\color{red}{\frac{(-1)^{\omega(n)}\,\text{rad}(n)\,\varphi(n)}{n^2}} $$ here $\varphi(n)$ is the usual Euler's totient function, $\omega(n)$ is the number of distinct prime factors of $n$ and $\text{rad}(n)=\prod_{p\mid n}p$ is the square-free part of $n$.