Re: Sieve needed for k*b1^m*b2^n+1
- --- In firstname.lastname@example.org, "Bryan" <beyastard@...> wrote:
> I've been trying to find a type of prime that has not been explored and
> yet easily
> factored for p+1 or p-1 and came up with the following:
> k*b1^m*b2^n+1 where b1 and b2 are prime bases > 2, bases differing and k
> is even
> for example:
> One thing I have noticed is for every k=2i+1, i>=0, the composite isReally?
> divisble by 3.
That's false for k=0 mod 3; and for k=1, b1=3; and for...