Something interesting happens if instead of starting with a finite set of primes, we start with the set of all primes (call it P). In that situation, the factor reactor that uses f(p) = 2p-1, f(p) = 2p+1, or f(p) = p^2-1 gives back P again, while the factor reactor that uses f(p) = p^2+1 gives just the prime 2 and the primes that are 1 mod 4. That is, the three rules that tend to make big finite sets of primes get smaller leaves the set P alone, while the rule that tends to make nonempty finite sets get bigger makes P get smaller. These assertions are not hard to prove. To show that every prime odd prime q divides some number of the form 2p-1, find k such that 2k is 1 mod q, and note that by Dirichlet's theorem, there are infinitely many primes p (and hence at least one!) such that p is k mod q; then 2p-1 is 0 mod q. The same argument shows that q divides some number of the form 2p+1 and that q divides some number of the form p-1 (which in turn divides p^2-1). The assertion about f(p) = p^2+1 follows from that fact that for every prime q that's 1 mod 4, there exists k such that k^2 is -1 mod q, combined with one more application of Dirichlet's theorem. Jim Propp