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We all know about the Riemann zeta function, where $s$ is a complex number:

$$ \sum^\infty_{n=1}\frac{1}{n^s} $$

But what about replacing all natural $n$s with $p_n$ primes? Is there anything on a function like this out there and/or is it worth exploring?

What about only composite numbers for $n$?

What about using some other set of numbers, for example using $F_n$ Fibonacci numbers?

Can these functions be expressed using the zeta function?

In other words, I'm looking for everything or at least anything out there that explores or mentions a function like a zeta function but the natural numbers are replaced with some other "famous" set of numbers, for example primes, or composites, or fibonaccis...

Edit

Looks like Prime zeta function and Fibonacci zeta function were "a google away", my mistake for trying to search for "zeta functions" which brought me to some of the generalizations only, instead of the specific functions.

The questions follows, did anyone ever somewhere tried to explore a similar function like these two but with some other set of numbers and found something interesting?

This might seem like a vague question so I won't be suprised if it gets voted to be closed, but it is worth a shot if someone reading this knows of something.

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    The sets of integers you are mentionning are not multiplicative, so the Dirichlet series doesn't have an Euler product anymore. Also you'd like to know that $\log \zeta(s) = \sum_p \sum_{m=1}^\infty \frac{p^{-sm}}{m}$ means that $\sum_p p^{-s} = \sum_{k=1}^\infty \frac{\mu(k)}{k} \log \zeta(ks)$ – 2017-01-11
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    :P sorry for being slow, I added in the Fibonacci numbers. – 2017-01-11
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    And... I updated to your edit...? – 2017-01-11
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    Maybe related (not completely of course...) https://math.stackexchange.com/questions/1415062 – 2017-01-11
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    @Watson It is related in the sense that it is important for the OP to realize that $\prod_{p \equiv 1 \bmod 4} \frac{1}{1-p^{-s}}$ doesn't have a simple expression as $\sum_n a_n n^{-s}$, while $\sum_{n \equiv 1 \bmod 4} n^{-s}$ doesn't have an Euler product (even if $\sum_{n \equiv 1 \bmod 4} n^{-s} = \frac{(1-2^{-s})\zeta(s)+\beta(s)}{2}$ where $\beta(s) = \sum_{n=0}^\infty (2n+1)^{-s} (-1)^n = \prod_{p \ge 3} \frac{1}{1-(-1)^{(p-1)/2}p^{-s}}$ the Dirichlet beta function) – 2017-01-11

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Lucky for you, there happens to be a prime zeta function, and if you want composite numbers, just subtract the prime zeta from the Riemann zeta.

$$P(s)=\sum_p\frac1{p^s}$$

Some basic asymptotes:

$$P(s)\sim\log\zeta(s)\sim\log\frac1{s-1}$$

As $s\to1$.

We could also have

$$\sum_{n=1}^\infty\frac{F_n}{n^s}=\frac{\operatorname{Li}_s(\phi)-\operatorname{Li}_s((-\phi)^{-1})}{\sqrt5}$$

Which are polylogarithms and golden ratios. But I'm not really sure about $\sum\limits_{f}\frac1{f^s}$.

In general,

$$\sum_{n=1}^\infty\frac{a_n}{n^s}$$

Is called a dirichlet series.

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    Just FYI, I don't think there exists a super general version, but you may want to look up a Dirichlet series anyways. – 2017-01-11
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    (you mean $\sum_{n=1}^\infty F_n n^{-s}$ not $\sum_{n=1}^\infty (F_n)^{-s}$ as the OP was thinking to) – 2017-01-11
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    @user1952009 Oh, you are right... – 2017-01-11