What is the Möbius analoge for Ihara's $\zeta$ function? 
The Dirichlet series that generates the Möbius function is the (multiplicative) inverse of the Riemann zeta function; if s is a complex number with real part larger than 1 we have
  $$\sum_{n=1}^\infty \frac{\mu(n)}{n^s}=\frac{1}{\zeta(s)}. $$
  ... from Wiki

What is the analoge of the right hand side of the above equation, if we put Ihara's $\zeta$ function instead of Riemann's?
Multiplicatively they go along same lines:
$$    \frac{1}{\zeta(s)} = \prod_{p\in \mathbb{P}}{\left(1-\frac{1}{p^{s}}\right)}= \left(1-\frac{1}{2^{s}}\right)\left(1-\frac{1}{3^{s}}\right)\left(1-\frac{1}{5^{s}}\right)\cdots \tag{Riemann}
$$
compared with
$$
    \frac{1}{\zeta_G(u)} = \prod_{p} ({1 - u^{L(p)}}). \tag{Ihara}
$$
This product is taken over all prime walks $p$ of the graph $G = (V, E)$ - that is, closed cycles $p = (u_0, \cdots, u_{L(p)-1}, u_0)$ such that
$$
    (u_i, u_{(i+1)\bmod L(p)}) \in E~; \quad u_i \neq u_{(i+2) \bmod L(p)~}, 
$$
and $L(p)$ is the length of cycle $p$.
What is the Möbius analoge for Ihara's $\zeta$ function? And what is $n$? The paths composed of certain prime cycles like natual numbers in number theory?
Would this mean that $\mu(n_G)=0$ when the composed closed path $n_G$ contains a certain vertex twice, like $\mu(n)=0$, when $n=p_k^2\cdot K$?
I found this but I don't feel my questions answered there...
 A: for a graph $G$ :  $p$ are the prime cycles, that is a cycle in the graph that doesn't pass through the same vertex two times (I think they did a mistake about that on wiki). $L(c)$ is the length of a cycle.
two cycles are considered the same if they are the concatenation of the same  prime cycles, each the same number of times, but in different order.
because each cycle is a concatenation of prime cycles, there length will be $L(p_i) + L(p_j) + \cdots$ and your zeta function computes :
$$\zeta_G(u) = \sum_{c \in G} u^{L(C)} = \sum_{n} b_G(n) u^n$$ 
where $c$ are the cycles of $G$. the coefficients of this analytic function are $b_G(n) = \sum_{L(c) = n} 1$ which depend on the graph. there is an infinity of cycles because once you got one, you can concanate it with itself anytimes you want. so $\zeta_G(u)$ converges at most for $|u| < 1$. it means that $\lim_{n \to \infty} b_G(n) \ne 0$.
now we look at it's inverse, which is a simple polynomial if the graph is finite :
$$\frac{1}{\zeta_G(u)} = \sum_{c \in G} \mu(c) u^{L(C)} = \sum_{n} a_G(n) u^n$$ 
where $\mu(c) = (-1)^k$ if $c$ is composed of $k$ different prime  cycles, and $\mu(c) = 0$ if there is a prime cycle appearing two times or more in it. 
the sequence $a_G(n) = \sum_{L(c) = n} \mu(c)$ is the convolution inverse of the sequence $b_G(n)$ :
$$( a_G \ast b_G)(n) = \sum_{k=0}^n a_G(k) b_G(n-k) = \delta(n)$$
note that it is normal convolution, additive, not the multiplicative one as for the Riemann zeta function with $\sum_{d | n}$. your $\zeta$  is additive because we considered addition of cycle lengths, not multiplication. that makes a huge difference between additive and multiplicative zeta function.
now if it's a normal graph, your $\frac{1}{\zeta_G(u)} = \sum_{n} a_G(n) u^n$ is a polynomial because there is only a finite number of different prime cycles.
thus $\frac{1}{\zeta_G(u)}$ is holomorphic on the whole complex plane, and $\zeta_G(u)$ is holomorphic on the whole complex plane except at a finite number of poles (the zeros of the polynomial $\frac{1}{\zeta_G(u)}$).
I guess you can also consider a $\zeta$ function for a non-finite graph (Riemann surface topology ?), which maybe (I don't know because I didn't understand it) could take you into the wonderful world of Selberg zeta functions for hyperbolic surfaces...
