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I'm looking for a function $f(x)$ 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 a function exist?

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

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  • $\begingroup$ Do you really want a function $\;f\;$ such that $\;f(k)+f(2k)+f(3k)+\ldots=r^k\;$ ? $\endgroup$ – DonAntonio Jan 19 at 0:06
  • $\begingroup$ Right. Is there something wrong with that? $\endgroup$ – Mike Battaglia Jan 19 at 0:07
  • $\begingroup$ @Mi No, not really...It is just that it looks pretty weird to me, that's all. It isn't a power series nor a series of functions, but just the sum of the values of $\;f\;$ on integral multiples of some number $\;k\;$ ....and that must equal $\;r^k\;$ ...It'll be interesting if someone can come up with something like that. $\endgroup$ – DonAntonio Jan 19 at 0:09
  • $\begingroup$ That's true. Would it be clearer if I rewrote this in terms of some sequence f[n] instead of f(x)? $\endgroup$ – Mike Battaglia Jan 19 at 0:10
  • $\begingroup$ Now posted to MO, mathoverflow.net/questions/321432/… $\endgroup$ – Gerry Myerson Jan 23 at 20:00
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We have that $\;r^k=e^{k\log r}\;$ , so

$$e^{k\log r}=\sum_{n=0}^\infty\frac{k^n(\log r)^n}{n!}$$

Yet the above doesn't have the form you seem to be searching...

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