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##### Activity
Peter Hizalev
@petrohi
[Peter H, sameroom] 11
[Peter Hizalev] BOOO
Peter Hizalev
@petrohi
[Peter H, sameroom] YO
[Peter Hizalev] 1111
[Peter H, sameroom] jaja
[Peter Hizalev] xyz
[Peter H, sameroom] JAJA
[Peter H, sameroom] aaa
[Peter Hizalev] 1
[Peter H, sameroom] 2
[Peter H, sameroom] 3
[Peter Hizalev] 4
Peter Hizalev
@petrohi
[Peter H, sameroom] asdasdsd
[Peter Hizalev] 111111
[Peter Hizalev] 111111
[Peter H, sameroom] 123123 123 123
[Peter Hizalev] 2222
[Peter Hizalev] 2222
Peter Hizalev
@petrohi
[Peter H, sameroom] a
[Peter Hizalev] b
[Peter H, sameroom] c
[Peter Hizalev] d
[Peter H, sameroom] xyz
[Peter Hizalev] /me foo
Peter Hizalev
@petrohi
[Peter H, sameroom] 12312
[Peter H, sameroom] 112
Peter Hizalev
@petrohi
[Peter H, sameroom] vbvb
[Peter H, sameroom] 11
Andrei Soroker
@abs
\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } }
\$ \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } }
$\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } }$
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