On Mon, 2006-01-23 at 08:08, Brian Beesley wrote:
> Yes. In fact I think there is a formal proof that _every_ simple algorithm
> generating a list of numbers _must_ generate at least some composites. Maybe
> even for sufficiently large n that _all_ subsequent terms must be composite?
> Can't remember where I saw this but it does look intuitively reasonable...
Refutation by counterexample:
The algorithm: "n:=1; repeat {n:=n+2; output n}" is, I hope you agree, a
simple algorithm and generates an infinite number of primes (indeed, all
of them bar 2) and so "_all_ subsequent terms must be composite" is
false.
Paul
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