The $0^{\rm th}$ prime number

I was performing some calculations with the online version of Wolfram Alpha, when I was presented with a formula containing the number $p_0$. Since Wolfram defines $p_n$ as the $n^{\rm th}$ prime, I was naturally intrigued as to what Wolfram meant by $p_0$. Inputting Prime[0] into Wolfram, I was presented with the following expression:

$$ p_0 = -\sqrt{10} \lfloor\sqrt{10} \alpha\rfloor + \lfloor 10^1 \alpha\rfloor, $$

where

$$ \alpha = \sum_{k=1}^{\infty} \frac{p_k}{10^{2^k}}. $$

What does this mean, where does it come from and why does it make sense to talk about a $0^{\rm th}$ prime number?

You're looking at a nonsensical circular definition.

$\alpha$ is a number that contains every prime by adding sufficiently shifted base 10 numbers. E.g.:

$\alpha = 0.02030005\ldots$

The expression above just extracts the primes by unshifting (multiplying by a power of 10) and flooring, subtracting the previous terms.

This expression results from plugging in $n = 0$ into the formula, leading to a non-sensical result (essentially a domain error).

If $\alpha =\sum_{i=1}^{\infty} \frac{p_i}{10^{2^i}}$ then we can definitely get the values of $p_i$ back out by computing:

$$p_i=\left\lfloor 10^{2^i}\alpha\right\rfloor-10^{2^{i-1}}\left\lfloor 10^{2^{i-1}}\alpha\right\rfloor$$

This is just saying that we are encoding $p_i$ in the digits of $\alpha.$ It is essentially a silly formula, which would let you compute $p_{\sqrt{2}}$ and $p_{-\pi}$, but has no actual numeric use.

In particular, the choice of base $10$ is completely arbitrary.