A115091 Primes p such that p^2 divides m!+1 for some integer m

5, 11, 13, 47, 71, 563, 613

Offset: 1

Author

T. D. Noe, Mar 01 2006

Keywords

• hard
• more
• nonn

Comments

By Wilson's theorem, we know that there is an m=p-1 such that p divides m!+1. Sequence A115092 gives the number of m for each prime. Occasionally p^2 also divides m!+1. These primes seem to be only slightly more plentiful than Wilson primes (A007540). No other primes < 10^6.

There is no prime p < 10^8 such that p^2 divides m!+1 for some m <= 1200. [From F. Brunault (brunault(AT)gmail.com), Nov 23 2008]

For a(n), m = p-A259230(n). - Felix Fröhlich, Jan 24 2016

References

• R. K. Guy, Unsolved Problems in Number Theory, 3rd Ed., New York, Springer-Verlag, 2004, Section A2.

Crossrefs

Cf. A064237 (n!+1 is divisible by a square), A259230.

Programs

Mathematica

nn=1000; lst={}; Do[p=Prime[i]; p2=p^2; f=1; m=1; While[m
Michael De Vlieger, Jan 24 2016, Version 10 *)

PARI

forprime(p=1, , for(k=1, p-1, if(Mod((p-k)!, p^2)==-1, print1(p, ", "); break({1})))) \\ Felix Fröhlich, Jan 24 2016