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For $\Re(s) > 1$, the following product is valid (see also this question):

$$\displaystyle \prod _{n=2}^{\infty } \left( \dfrac{1}{1-\frac{1}{n^{s}}} \right)=\prod _{primes}^{\infty } \left( \dfrac{1}{1-\frac{1}{p^{s}}} \right)\cdot\prod _{composites}^{\infty } \left( \dfrac{1}{1-\frac{1}{c^{s}}} \right)$$

The products of primes and composites are only equal at $s= 1.397737620...$.

I wondered whether a similar result could be obtained for arithmetic progressions under Dirichlet's theorem i.e. that for any two positive coprime integers $a$ and $d$, there are infinitely many primes of the form $d\,n+a$, where $n$ is a non-negative integer. So, assuming the Euler product for the progression, this would look like:

$$\displaystyle \prod _{n=2}^{\infty } \left( \dfrac{1}{1-\frac{1}{(d\,n+a)^{s}}} \right)=\prod _{p=d\,n+a}^{\infty } \left( \dfrac{1}{1-\frac{1}{p^{s}}} \right)\cdot\prod _{c=d\,n+a}^{\infty } \left( \dfrac{1}{1-\frac{1}{c^{s}}} \right)$$

I found that both products of primes and composites only for some progressions are equal once. Here is the initial result (products for $\underline{d\,n+a}$ are equal once, the others never):

$\underline{2n+1}$ $4n+1$ $\underline{4n+3}$ $6n+1$ $6n+5$ $\underline{8n+1}$ $\underline{8n+3}$ $8n+5$ $\underline{8n+7}$ $10n+1$ $\underline{10n+3}$ $10n+7$ $\underline{10n+9}$ $12n+1$ $\underline{12n+5}$ $\underline{12n+7}$ $\underline{12n+11}$ $\dots$

Also tested some progressions with higher values for $d$ and $a$ and found again that some are equal and some never are. I guess this has something to do with a temporary bias from a number of lower primes being included in the prime product or not and since primes are thinning out for the higher numbers in the progression, the composites will always 'beat' them.

Q1: Is there any connection to be made between the progression $d\,n+a$ and the prime and composite products crossing once or not?

Q2: Does there exist a closed form for finite products $\displaystyle \prod _{n=2}^{N} \left( \dfrac{1}{1-\frac{1}{(d\,n+a)^{s}}} \right)$ (e.g. in terms of $\Gamma$'s similar to general arithmetic progressions?)

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  • $\begingroup$ why don't you look at the logarithm ? you'll get usual Dirichlet series $\endgroup$ – reuns Aug 27 '16 at 20:44
  • $\begingroup$ and see fr.wikipedia.org/wiki/… how the Dirichlet characters are useful $\endgroup$ – reuns Aug 27 '16 at 20:51

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