Replying to @tomcuchta
There are lots of ways to q it, with vastly different results. And different analytic completions. Euler product prod_{m=1}^{\infty} (1- 1/ [p_{m}]_{q}^{s})^{-1} does something different that sum_{n=1}^{\infty} 1/[n]_{q}^{s}
10:49 AM - 11 Feb 2018
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