Does this Euler product have a name?

Question

Let

$$\displaystyle f(s) = \prod_p \left(1 - \frac{1}{2p^s}\right), \Re(s) > 1.$$

This Euler product converges. Does it have a name?

Answer 1

It has an expression in terms of fractions involving $\zeta(s)$. Start with $$ \frac{1}{\zeta(s)}=\prod_p\left(1-\frac{1}{p^s}\right), $$ and then $$ f(s)=\sum_{n=1}^{\infty}\frac{\mu(n)}{2^{\omega (n)}n^s}. $$ For the plus sign we have $$ \prod_p \left(1+\frac{1}{2p^s}\right)=\sum_{n=1}^{\infty}\frac{2^{\omega(n)}}{n^s}=\frac{\zeta(s)^2}{\zeta(2s)}. $$ Reference: Here.

Answer 2

It could help to visualize this product by rewriting it as $$f(s)=\prod_{p} (1-q^{-s})$$ where $$q=2p$$ Recall that $$\frac{1}{\zeta(s)}=\prod_{p} (1-p^{-s})=\sum_{n=1}^\infty \frac{\mu(n)}{n^s}$$

So if you expand the product into a sum you should get an answer very similar to that, which is $$\sum_{n=1}^\infty \frac{\mu(n)2^{-\omega(n)}}{n^s}$$ Where $\omega(n)$=amount of prime factors of n.

This looks to be similar to the identity $$\frac{\zeta^2(s)}{\zeta(2s)}=\sum_{n=1}^\infty \frac{2^{\omega(n)}}{n^s}$$ but I could not find any way to relate your function to the above sum.