Let $0<a<m$ be integers with $\gcd(a,m)=1$. For $s\in\mathbb{C}$ with $\Re s>1$, define $$\zeta(m,a;s)=\prod_{\substack{p\text{ is prime}\\p\equiv a\pmod m}}(1-p^{-s})^{-1}.$$
I'm looking for an efficient (arbitrary precision) computation of $\zeta(m,a;s)$. Assume that we have such a computation of Dirichet $L$-functions.
This paper by H. Cohen (which motivated me the most) covers various sums and products over primes, as well as Dirichlet $L$-functions and more, but not the above.
I found a satisfactory solution for some specific cases.
The simplest are $m\in\{3,4,6\}$, with a single nontrivial character $\chi$ modulo $m$. Let \begin{align*}f_1(s)&=\log\zeta(3,1;s),&g_1(s)&=\log[(1-3^{-s})\zeta(s)],\\f_2(s)&=\log\zeta(3,2;s),&g_2(s)&=\log L(s,\chi),\end{align*} then $g_1(s)=f_1(s)+f_2(s)$ and $g_2(s)=f_1(s)+f_2(2s)-f_2(s)$. This gives $$f_2(s)=\sum_{n=0}^\infty 2^{-n-1}g(2^n s),\qquad g(s)=g_1(s)-g_2(s),$$ a series that converges very quickly (its term is $O(2^{-2^n\sigma})$ with $\sigma=\Re s$). Thus, we have an efficient (enough) way to compute $f_2(s)$ and then $f_1(s)=g_1(s)-f_2(s)$.
The case $m=8$ is just a bit harder. Take the following table of characters:
| $a$ | $1$ | $3$ | $5$ | $7$ |
|---|---|---|---|---|
| $\chi_1(a)$ | $1$ | $1$ | $1$ | $1$ |
| $\chi_2(a)$ | $1$ | $1$ | $-1$ | $-1$ |
| $\chi_3(a)$ | $1$ | $-1$ | $-1$ | $1$ |
| $\chi_4(a)$ | $1$ | $-1$ | $1$ | $-1$ |
Then, if we put $g_k(s)=\log L(s,\chi_k)$, we obtain (for $a=3$, and similarly for other values) $$\log\zeta(8,3;s)=\sum_{n=0}^\infty 2^{-n-2}g(2^n s),\qquad g(s)=g_1(s)+g_2(s)-g_3(s)-g_4(s).$$
But already the case $m=5$ gets me confused completely.
