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 A115091 Primes p such that p^2 divides m!+1 for some integer m
 5, 11, 13, 47, 71, 563, 613 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 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. LINKS MATHEMATICA nn=1000; lst={}; Do[p=Prime[i]; p2=p^2; f=1; m=1; While[m

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