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Peđa Terzić
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$$\Large\color{blue}{\frac{\sqrt{3}\pi}{6}=\left(\displaystyle\prod_{p \equiv 1 \pmod{6} \atop p \in \mathbb{P} } \frac{p}{p-1}\right) \cdot \left(\displaystyle\prod_{p \equiv 5 \pmod{6} \atop p \in \mathbb{P} } \frac{p}{p+1}\right)}$$

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