I'm looking for some $f(x)$ that has the following property:


$\sum\_{x=1}^\infty f(kx) = r^k$


for some real $0 < r < 1$, and at least for strictly positive integer $k$.


Does such an $f(x)$ exist?


This could also be thought of in terms of some sequence of real numbers $f[n]$.


I posted this at MSE but got no answer, so thought I would try MO.



Looking for infinite series resembling an exponential
You can solve this system of equations explicitly in terms of the function
$$F(z)=\sum\_{m=1}^{\infty} \mu(m)z^m=z-z^2-z^3-z^5+z^6+\cdots$$
where $\mu(m)$ is the [Möbius function](https://en.wikipedia.org/wiki/M%C3%B6bius_function).
The function $F(z)$ can be shown to be transcendental (see the paper ["Transcendence of Power Series for Some Number Theoretic Functions"](https://arxiv.org/abs/0806.1563), by Coons and Borwein). With this notation your sequence can be succinctly expressed as
$$f(n)=F(r^n).$$
Indeed, one can check that 
$$\sum\_{n=1}^{\infty} f(kn)=\sum\_{n=1}^{\infty} F(r^{kn})=\sum\_{m,n=1}^{\infty} \mu(m)r^{nmk}=\sum\_{m=1}^{\infty}\sum\_{d|m}\mu(d)r^{mk}=r^k$$
as desired.


