Talk:Prime Enumeration Function is Primitive Recursive

Less than not less than or equals?

Not being an expert I am reluctant to comment on this, but the proof now says:

We can define $p$ recursively by:

$p \left({n + 1}\right) = \text{the smallest } y \in \N \text { such that } y \text { is prime and } p \left({n}\right) \le y$.

Hence we can express it as:

$p \left({n + 1}\right) = \mu y \left({\chi_\Bbb P \left({y}\right) = 1 \land p \left({n}\right) \le y}\right)$

which seems to say one prime and the next prime would be equal. Is this what was meant instead?:

We can define $p$ recursively by:

$p \left({n + 1}\right) = \text{the smallest } y \in \N \text { such that } y \text { is prime and } p \left({n}\right) < y$.

Hence we can express it as:

$p \left({n + 1}\right) = \mu y \left({\chi_\Bbb P \left({y}\right) = 1 \land p \left({n}\right) < y}\right)$

--EntropyReversal 12:53, 16 May 2011 (CDT) (belated sig)

Oo-er, you could be right. I'll need to get back to this. I'll flag it for now while I go and think about it (and hunt my books down). --prime mover 12:33, 16 May 2011 (CDT)