Let $G$ be a finite group, $S \subset G$ a generating set, $|g|:=|g|\_S=$ word-length with respect to $S$. Let $\phi(g,h)=|g|+|h|-|gh| \ge 0$ be the "defect-function" of $S$. The set $\mathbb{Z}\times G$ builds a group for the following operation:


$$(a,g) \oplus (b,h) = (a+b+\phi(g,h),gh)$$


On $\mathbb{N}\times G$ is the "norm": $|(a,g)| := |a|+|g|$ additive, which means that $|a \oplus b| = |a|+|b|$. Define the multiplication with $n \in \mathbb{N\_0}$ to be:


$$ n \cdot a := a \oplus a \oplus \cdots \oplus a$$


(if $n=0$ then $n \cdot a := (0,1) \in \mathbb{Z} \times G$).


A word $w := w\_{n-1} w\_{n-2} \cdots w\_0$ is mapped to an element of $\mathbb{Z} \times G$ as follows:


$$\zeta(w) := \oplus\_{i=0}^{n-1} (m^i \cdot (0,w\_i))$$


where $m := \min\_{g,h\in G, \phi(g,h) \neq 0} \phi(g,h)$.


We let $|w|:=|\zeta(w)|$ and $w\_1 \oplus w\_2:=\zeta(w\_1)\oplus \zeta(w\_2)$


Then we have $|w\_1 \oplus w\_2| = |w\_1|+|w\_2|$.


For instance for the Klein four group $\{0,a,b,c=a+b\}$ generated by $S:=\{a,b\}$, we get sorting the words $w$ by their word-length:


$$0,a,b,c,a0,aa,ab,ac,b0,ba,bb,bc,c0,ca,cb,cc,a00,a0a,a0b,a0c$$


corresponding to the following $\mathbb{Z}\times K\_4$ elements $\zeta(w)$:


$$(0,0),(0,a),(0,b),(0,c),(2,0),(2,a),(2,b),(2,c),(2,0),(2,a),(2,b),(2,c),(4,0),(4,a),(4,b),(4,c),(4,0),(4,a),(4,b),(4,c)$$


corresponding to the the following "norms" of words $|w| = |\zeta(w)|$:


$$0,1,1,2,2,3,3,4,2,3,3,4,4,5,5,6,4,5,5,6$$


Let $a\_n, n\ge 0$ be the sequence of numbers generated by the Klein four group.


1) Is 
$$\sum\_{n=1}^\infty \frac{1}{a\_n^s} = \sum\_{n=1}^\infty \frac{n+1}{n^s} = \zeta(s-1) + \zeta(s)$$
where $\zeta$ denotes the Riemann zeta function?


I have checked this with SAGE math up to a certain degree and it seems plausible, however I have no idea how to prove it.


2) Is every $a\_n$ the product of primes $p=a\_k$ for some $k\le n$?


3) Let $\pi\_{K\_4}(n) = |\{ k : \text{$a\_k$ is prime, $k \le n$}\}|$ be the prime counting function of the sequence. What is the approximate relationship to the usual prime counting function $\pi(n)$?



A zeta function for the Klein Four group?
Consider a word $w=w\_0w\_1\cdots w\_{n-1}$ with each $w\_i\in\{0,a,b,c\}$.
Following your notation,
$$\lvert w\rvert=\lvert\zeta(w)\rvert=\left\lvert\bigoplus\_{i=0}^\infty m^i\cdot(0,w\_i)\right\rvert=\sum\_{i=0}^{n-1}\lvert 2^i\cdot(0,w\_i)\rvert=\sum\_{i=0}^{n-1} 2^i\lvert w\_i\rvert$$
where $\lvert0\rvert=0$, $\lvert a\rvert=1$, $\lvert b\rvert=1$, $\lvert c\rvert=2$.
We now consider the generating function
$$F\_n(x)=\sum\_wx^{\lvert w\rvert}$$
where the sum is over all words $w=w\_0w\_1\cdots w\_{n-1}$.
Then
\begin{align\*}
F\_n(x)&=\sum\_{w\_0}\sum\_{w\_1}\cdots\sum\_{w\_{n-1}}x^{|w\_0|+2|w\_1|+\cdots+2^{n-1}|w\_{n-1}|}\\
&=\left(\sum\_{w\_0}x^{|w\_0|}\right)\left(\sum\_{w\_1}x^{2|w\_1|}\right)\cdots\left(\sum\_{w\_{n-1}}x^{2^{n-1}|w\_{n-1}|}\right)\\
&=(1+2x+x^2)(1+2x^2+x^4)(1+2x^4+x^8)\cdots(1+2x^{2^{n-1}}+x^{2^n})\\
&=(1+x)^2(1+x^2)^2(1+x^4)^2\cdots(1+x^{2^{n-1}})^2\\
&=(1+x+x^2+\cdots+x^{2^n-1})^2\\
&=1+2x+\cdots+(2^n-1)x^{2^n-2}+2^nx^{2^n-1}+(2^n-1)x^{2^n}+\cdots+x^{2^{n+1}-2}.
\end{align\*}




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What this shows is that among the first $4^n$ terms of the sequence:
\begin{align\*}
&0\text{ appears }1\text{ times},\\
&1\text{ appears }2\text{ times},\\
&2\text{ appears }3\text{ times},\\
&\cdots\\
&2^n-2\text{ appears }2^n-1\text{ times},\\
&2^n-1\text{ appears }2^n\text{ times},\\
&2^n\text{ appears }2^n-1\text{ times},\\
&\cdots\\
&2^{n+1}-4\text{ appears }3\text{ times},\\
&2^{n+1}-3\text{ appears }2\text{ times},\\
&2^{n+1}-2\text{ appears }1\text{ times}.\\
\end{align\*}
This resolves question 1.
Since this seems to be purely a combinatorics question, I would not expect the second and third questions to have particularly interesting answers.




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I will remark that
$$\pi\_{K\_4}(4^n)=\sum\_{p\leq2^{n+1}-2}(2^n-\lvert p-2^n+1\rvert)p$$
so you could use some analytic number theory to get asymptotics for $\pi\_{K\_4}(n)$ if you wanted.


