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.


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. The function $F(z)$ can be shown to be transcendental (see the paper "Transcendence of Power Series for Some Number Theoretic Functions", 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.


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