Divergent Products over Prime Numbers

SPECIES

$\displaystyle \prod_{p\in\mathbb{P}} p = (2\pi)^2 = 4\pi^2$

$\displaystyle \prod_{p\in\mathbb{P}} \sqrt{p} = 2\pi$

$\displaystyle \prod_{p\in\mathbb{P}} (p-1) = \frac{(2\pi)^2}{\zeta(1)} = \frac{4\pi^2}{\gamma}$

$\displaystyle \prod_{p\in\mathbb{P}} \sqrt{p^2-1} = \frac{(2\pi)^2}{\sqrt{\zeta(2)}} = 4\sqrt{6}\,\pi$

$\displaystyle \prod_{p\in\mathbb{P}} (p^2-1) = \frac{(2\pi)^4}{\zeta(2)} = 96\,\pi^2$

$\displaystyle \prod_{p\in\mathbb{P}} (p^3-1) = \frac{(2\pi)^6}{\zeta(3)}$

$\displaystyle \prod_{p\in\mathbb{P}} (p^4-1) = \frac{(2\pi)^8}{\zeta(4)} = 23040\,\pi^4$

$\displaystyle \prod_{p\in\mathbb{P}} (p+1) = 24\,\zeta(1) = 24\,\gamma$

$\displaystyle \prod_{p\in\mathbb{P}} (p^2+1) = 1440\,\zeta(2) = 240\,\pi^2$

$\displaystyle \prod_{p\in\mathbb{P}} (p^3+1) = 60480\,\zeta(3)$

$\displaystyle \prod_{p\in\mathbb{P}} (1+\frac{1}{p}) = \frac{\zeta(1)}{\zeta(2)} = \frac{6\gamma}{\pi^2}$

$\displaystyle \prod_{p\in\mathbb{P}} (1+\frac{1}{p}+\frac{1}{p^2}) = \frac{\zeta(1)}{\zeta(3)} = \frac{\gamma}{\zeta(3)}$

$\displaystyle \prod_{p\in\mathbb{P}} (1+\frac{1}{p}+\frac{1}{p^2}+\frac{1}{p^3}) = \frac{\zeta(1)}{\zeta(4)} = \frac{90\gamma}{\pi^4}$

$\displaystyle \prod_{p\in\mathbb{P}} (1+p+p^2) = \frac{\zeta(1)}{\zeta(3)} (2\pi)^4 = \frac{\gamma}{\zeta(3)} (2\pi)^4$

$\displaystyle \prod_{p\in\mathbb{P}} (1+p+p^2+p^3) = \frac{\zeta(1)}{\zeta(4)} (2\pi)^6 = 5760\,\pi^2 \gamma$

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