Divergent Products over Prime Numbers

SPECIES

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

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

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

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

$\displaystyle \prod_{p\in\mathbb{P}} \, (p^3\!-\!1) = \dfrac{(2\pi)^6}{\zeta(3)} = \dfrac{64\pi^6}{\zeta(3)}$

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

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

$\displaystyle \prod_{p\in\mathbb{P}} \, p\,(p\!-\!1) = (2\pi)^2 \, (2\pi)^2 = (2\pi)^4 = 16\,\pi^4$

$\displaystyle \prod_{p\in\mathbb{P}} \, (p^2\!-\!p\!+\!1) = \zeta(2)\,\dfrac{\zeta(3)}{\zeta(6)}\,(2\pi)^2\,(2\pi)^2 = \dfrac{7!}{2}\,\zeta(3) = 2520\,\zeta(3)$

$\displaystyle \prod_{p\in\mathbb{P}} \, (p^2\!-\!p\!-\!1) = (2\pi)^2\,(2\pi)^2\,C_\text{Artin} = (2\pi)^4\,C_\text{Artin}$

$\displaystyle \prod_{p\in\mathbb{P}} \left(1\!-\!\dfrac{1}{p}\right) = \dfrac{(2\pi)^2}{\zeta(1)}\,\dfrac{1}{(2\pi)^2} = \dfrac{1}{\zeta(1)} = \dfrac{(2\pi)^2}{(2\pi)^2} = 1$

$\displaystyle \prod_{p\in\mathbb{P}} \left(1\!-\!\dfrac{1}{p^2}\right) = \dfrac{(2\pi)^4}{\zeta(2)}\,\dfrac{1}{(2\pi)^4} = \dfrac{1}{\zeta(2)} = \dfrac{24\,(2\pi)^2}{(2\pi)^4} = \dfrac{6}{\pi^2}$

$\displaystyle \prod_{p\in\mathbb{P}} \left(1\!-\!\dfrac{1}{p^3}\right) = \dfrac{(2\pi)^6}{\zeta(3)}\,\dfrac{1}{(2\pi)^6} = \dfrac{1}{\zeta(3)}$

$\displaystyle \prod_{p\in\mathbb{P}} \left(1\!-\!\dfrac{1}{p^4}\right) = \dfrac{(2\pi)^8}{\zeta(4)}\,\dfrac{1}{(2\pi)^8} = \dfrac{1}{\zeta(2)}\,\dfrac{\zeta(2)}{\zeta(4)} = \dfrac{1}{\zeta(4)} = \dfrac{90}{\pi^4}$

$\displaystyle \prod_{p\in\mathbb{P}} \left(1\!-\!\dfrac{1}{p}\!-\!\dfrac{1}{p^2}\right) = C_\text{Artin}$

$\displaystyle \prod_{p\in\mathbb{P}} \left(p\!-\!1\!-\!\dfrac{1}{p}\right) = (2\pi)^2\,C_\text{Artin}$

$\displaystyle \prod_{p\in\mathbb{P}} \left(1\!-\!\dfrac{1}{p(p\!-\!1)}\right) = C_\text{Artin}$

$\displaystyle \prod_{p\in\mathbb{P}} \left(p\!-\!\dfrac{1}{p\!-\!1}\right) = (2\pi)^2\,C_\text{Artin}$

$\displaystyle \prod_{p\in\mathbb{P}} \, (p+\!1) = (2\pi)^2\,\dfrac{\zeta(1)}{\zeta(2)} = 24$

$\displaystyle \prod_{p\in\mathbb{P}} \, (p^2\!+\!1) = (2\pi)^4\,\dfrac{\zeta(2)}{\zeta(4)} = 240\,\pi^2$

$\displaystyle \prod_{p\in\mathbb{P}} \, (p^3\!+\!1) = (2\pi)^6\,\dfrac{\zeta(3)}{\zeta(6)} = 60480\,\zeta(3)$

$\displaystyle \prod_{p\in\mathbb{P}} \, (p^4\!+\!1) = (2\pi)^8\,\dfrac{\zeta(4)}{\zeta(8)} = 26880\,\pi^4$

$\displaystyle \prod_{p\in\mathbb{P}} \, \sqrt{p^2\!+\!1} = (2\pi)^2\,\sqrt{\dfrac{\zeta(2)}{\zeta(4)}} = 4\pi\sqrt{15}$

$\displaystyle \prod_{p\in\mathbb{P}} \, p\,(p\!+\!1) = 24\,(2\pi)^2 = 96\pi^2$

$\displaystyle \prod_{p\in\mathbb{P}} \, (p^2\!+\!p\!+\!1) \\ \\ = 24\,(2\pi)^2\,\dfrac{\zeta(2)}{\zeta(3)} = \dfrac{(2\pi)^6}{\zeta(3)}\,\dfrac{1}{(2\pi)^2} = (2\pi)^4\,\dfrac{\zeta(1)}{\zeta(3)} = \dfrac{(2\pi)^4}{\zeta(3)} = \dfrac{16\pi^4}{\zeta(3)}$

$\displaystyle \prod_{p\in\mathbb{P}} \, (p^4\!+\!p^2\!+\!1) = \dfrac{7!}{2}\,\zeta(3)\,\dfrac{(2\pi)^4}{\zeta(3)} = 2520\,(2\pi)^4 = 40320\,\pi^4$

$\displaystyle \prod_{p\in\mathbb{P}} \, (p^3\!+\!p^2\!+\!p\!+\!1) = \dfrac{(2\pi)^8}{\zeta(4)}\,\dfrac{1}{(2\pi)^2} = (2\pi)^6\,\dfrac{\zeta(1)}{\zeta(4)} = \dfrac{(2\pi)^6}{\zeta(4)} = 5760\,\pi^2$

$\displaystyle \prod_{p\in\mathbb{P}} \, \left(1\!+\!\dfrac{1}{p}\right) = (2\pi)^2\,\dfrac{\zeta(1)}{\zeta(2)}\,\dfrac{1}{(2\pi)^2} = \dfrac{\zeta(1)}{\zeta(2)} = \dfrac{24}{(2\pi)^2} = \dfrac{6}{\pi^2}$

$\displaystyle \prod_{p\in\mathbb{P}} \, \left(1\!+\!\dfrac{1}{p^2}\right) = (2\pi)^4\,\dfrac{\zeta(2)}{\zeta(4)}\,\dfrac{1}{(2\pi)^4} = \dfrac{\zeta(2)}{\zeta(4)} = \dfrac{15}{\pi^2}$

$\displaystyle \prod_{p\in\mathbb{P}} \, \left(1\!+\!\dfrac{1}{p^3}\right) = (2\pi)^6\,\dfrac{\zeta(3)}{\zeta(6)}\,\dfrac{1}{(2\pi)^6} = \dfrac{\zeta(3)}{\zeta(6)} = \dfrac{945}{\pi^6}\,\zeta(3)$

$\displaystyle \prod_{p\in\mathbb{P}} \, \left(1\!+\!\dfrac{1}{p^4}\right) = (2\pi)^8\,\dfrac{\zeta(4)}{\zeta(8)}\,\dfrac{1}{(2\pi)^8} = \dfrac{\zeta(4)}{\zeta(8)} = \dfrac{105}{\pi^4}$

$\displaystyle \prod_{p\in\mathbb{P}} \, \left(1\!+\!\dfrac{1}{p}\!+\!\dfrac{1}{p^2}\right) = \dfrac{\zeta(1)}{\zeta(3)} = \dfrac{1}{\zeta(3)}$

$\displaystyle \prod_{p\in\mathbb{P}} \,\left (1\!+\!\dfrac{1}{p}\!+\!\dfrac{1}{p^2}\!+\!\dfrac{1}{p^3}\right) = \dfrac{\zeta(1)}{\zeta(4)} = \dfrac{1}{\zeta(4)} = \dfrac{90}{\pi^4}$

$\displaystyle \prod_{p\in\mathbb{P}} \, \left(1\!+\!\dfrac{1}{p^2}\!+\!\dfrac{1}{p^4}\right) = \dfrac{2520\,(2\pi)^4}{(2\pi)^8} = \dfrac{2520}{(2\pi)^4} = \dfrac{315}{2\pi^4}$

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