The classical Euler product $\displaystyle \prod_{p \in \mathbb{P}} \left( \frac{1}{1-p^{-s}} \right)$ could be rewritten as:
$$eu_0(s):=\prod_{n=2}^{\infty} \left( \frac{1-\frac{1}{(n+1)^s}}{1-\frac{1}{n^s}}\right)^{\pi(n)}\qquad \qquad \Re(s) > 1$$
where $\pi(n)$ is the prime counting function.
After injecting an alternating "character" as follows:
$$eu_1(s):=\prod_{n=2}^{\infty} \left( \frac{1-\frac{(-1)^n}{(n+1)^s}}{1-\frac{(-1)^n}{n^s}}\right)^{\pi(n)}\qquad \qquad \Re(s) > \frac12 ?$$
the domain of convergence seems to get extended and I am particularly interested in the case $s=1$ where numerical evidence suggests that:
$$\prod_{n=2}^{\infty} \left( \frac{1-\frac{(-1)^n}{(n+1)}}{1-\frac{(-1)^n}{n}}\right)^{\pi(n)}=\frac43$$
Curious whether this is provable?
EDIT: Probably just a coincidence, but also: $$\frac12 \cdot\prod_{n=2}^{\infty} \left( \frac{1-\frac{1}{(n+1)^2}}{1-\frac{1}{n^2}}\right)^{n}=\frac43$$