Hans, If I get you correctly, the "covering congruences" will not help.
I performed sieving of candidate primes p and filtered out all such that P(p) has a small prime factor less than 1M. The sieving ended up with still quite a large number of candidates. C++ source sieve.cpp and its output sieved.txt can be downloaded at http://www.mytempdir.com/459551 For every prime p from sieved.txt it is guaranteed that P(p) does not have prime divisors smaller than 1M. It may be useful if somebody wants to continue search for p beyond 150000. Max On 2/18/06, Hans Riesel <[EMAIL PROTECTED]> wrote: > Hello Max, > > I think your problem could be solved with the technique of > "covering congruences". It is easy to prove that for the case > p=6k-1 the number P(p) is always divisible by 7. The remaining > numbers p=6k+1 can be split into two classes, p=12k+1 and > p=12k+7. For the case p=12k+7 it is easy to prove that P(p) is > aiways divisible by 13.---I have unfortunately not been able to > do more research on the remaining case p=12k+1, because I have > been given a new computer, which has not yet had all systems > installed. But if you could send me factorizations of the > numbers P(p) for p=13, 25, 37, 49, and 61, say, I'll give it a > try. > > Sincerely yours, > > Hans Riesel > > > _______________________________________________ > Prime mailing list > [email protected] > http://hogranch.com/mailman/listinfo/prime > _______________________________________________ Prime mailing list [email protected] http://hogranch.com/mailman/listinfo/prime
