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I found this question on $\prod_{n\to \infty}(1-1/p_n)$, played a little at Wolfram's Alpha and found the following:

The series expansion of a related indefinite integral $\int \log (1-1/p_n)dn$ gave a series expansion at $n=0$ of

$$ n \log(1-1/p_0)+O(n^2)+\text{constant} $$

I continued and asked W|A what the $0^\text{th}$ prime is and got $$ p_0 = -\sqrt{10} \lfloor \sqrt{10}\alpha\rfloor +\lfloor 10^1 \alpha\rfloor \tag{*} $$ for $\alpha = \sum_{k=1}^\infty p_k/10^{2^k}\approx 0.0203001$. So $p_0=0$. Fine (for the moment), but where does $(*)$ come from?

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draks ...
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    This is obtained by considering identity $(*)$, valid for every integer $n\geqslant1$, out of its range of validity. (Oh... and zero is not prime.) – Did Apr 01 '15 at 19:43
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    $0$ is not a prime but $(0)$ is a prime ideal of $\mathbb{Z}$ – marwalix Apr 01 '15 at 19:59

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