S(0)=3
 

Mp is prime if it divides S(p-2) with S(0)=4

Some Mp is prime if it divides S(p-2) with S(0)=3
Marcel Martin has checked this up to L(2^4499)


So, is Mp prime if it divides the remainder R of :
[S(p-2) with S(0)=4] = R (mod [S(p-2) with S(0)=3] ) ?


It seems so, and it much smaller than either S(p-2).
There are probably various algorithms to speed up this integer relationship.


Please, comments or disproofs welcome!

When S[0]=3, Mp divides S[p-2] iff Mp is prime [b]and p=3(mod4)[/b]