do we know for a given $k > 2$ the Euler product of $\ \displaystyle\sum_{n=0}^\infty (2^k n + 1)^{\textstyle-s} \ $ ?
I saw that every prime numbers will appear in it, as well as some non-prime generators of $2^k \mathbb{N} + 1$ (the set $A_k$) :
$$F_k(s) =\sum (2^k n + 1)^{\textstyle-s} = \left(\prod_{\textstyle p \in \mathcal{P}} \frac{1}{\textstyle 1-p^{\textstyle- e(k,p) s}}\right) \left( \prod_{\textstyle r \in A_k} \frac{1}{1 - r^{\textstyle -s}} \right)$$
of course my problem is to describe precisely how looks that set $A_k$, which is a subset of $2^k \mathbb{N} + 1$, and probably is tightly related to the other generators $p^{\textstyle e(k,p)}$ ?
another way to ask the question would be : how can we predict in some sort the result of the sieve of Eratosthene on the set of integers $\equiv 1 \text{ mod } 2^k $ ?
(some precisions)
given $p$ an odd prime number : $e(k,p)$ is the smallest integer $ \ge 1$ such that $p^{d} \equiv 1 \text{ mod } 2^k$ and there always exists such a $d$, because $p^m \equiv 1 \text{ mod } 2^n \implies$ $p^{2m} \equiv 1 \text{ mod } 2^{n+1}$. this means that every prime appears as a power of itself in the set $2^k \mathbb{N} +1$ and thus it must appear in the Euler product.
Now the set $A_k$ : it is the set of the generators of $2^k \mathbb{N} +1$ which are not prime powers. Indeed, the Euler product is not only onto the primes numbers : for example $21$ is $ \equiv 1 \text{ mod } 4$ but nor $3$ nor $7$ are $ \equiv 1 \text{ mod } 4$ ($3^2$ and $7^2$ are $ \equiv 1 \text{ mod } 4$ but they don't divide $21$) and so $21$ must appear in the Euler product of $\sum_{n=0}^\infty (4n+1)^{-s}$ : $21$ is one of the generators of the set $2^2 \mathbb{N} +1$ . Notice the uniqueness of the generators of $2^k \mathbb{N} + 1$ and thus the uniqueness of the exponents $e(k,p)$ and the set $A_k$: they can be obtained from the sieve of Eratosthene on $2^k \mathbb{N} + 1$.
After all this, the idea would be that having a sieve of Eratosthene (*) and furthermore having every odd prime appearing as a power of itself as generator, the "topology" of $2^k \mathbb{N} +1$ is very similar to the "topology" of $\mathbb{N}^*$, and maybe it would be possible to derive some recurrences formulas for $A_k$, for $e(k,p)$, for the zeros and poles of $\sum (2^k n +1)^{-s}$, the growth of it's Mertens function ... Of course the fact that $2 F_{k+1}(s) = F_k(s) + \sum (2^k n + 1)^{-s} (-1)^n$ is very useful to link the properties of $F_k(s)$ to $\zeta(s)$.
(*) but no simple functional relation as $\zeta(s)$, even if with some recurrences formulas on $k$, one could maybe define a functional relation different from the usual functionnal relation for Hurwitz zeta functions..
I found that $e(k,p) = 2^{k-m}$ where $m$ is the integer such that $p = (2a+1) 2^m +1$
(proof)
consider an odd prime $p \equiv 2^{m}+1 \text{ mod } 2^{m+1}$ :
$p = a 2^m + 1 $ where $a$ is odd $ \implies p^2 = 2^{m+1} \left( a^2 2^{m-1} + a\right)+ 1 \implies p^{\textstyle 2^d}$ is at least $\equiv 1 \text{ mod } 2^{m+d}$.
now if $p^c = a2^l+1$ and $p^d=b2^n+1$ where $a,b$ are odds $\implies p^{c+d} = ab 2^{l+n} + a2^l+b2^n+1$ $ \implies p^{c+d} \equiv 1 \text{ mod } 2^{n+1}$ if and only if $l = n$, so that $p^{\textstyle 2^d}$ is at most $\equiv 1 \text{ mod } 2^{m+d}$.