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A115091
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Primes p such that p^2 divides m!+1 for some integer m<p.
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1
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OFFSET
| 1,1
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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]
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REFERENCES
| R. K. Guy, Unsolved Problems in Number Theory, 3rd Ed., New York, Springer-Verlag, 2004, Section A2.
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MATHEMATICA
| nn=1000; lst={}; Do[p=Prime[i]; p2=p^2; f=1; m=1; While[m<p && f+1<p2, m++; f=Mod[f*m, p2]]; If[m<p, AppendTo[lst, p]], {i, PrimePi[nn]}]; lst
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CROSSREFS
| Cf. A064237 (n!+1 is divisible by a square).
Sequence in context: A125742 A116440 A098720 * A034924 A018607 A032481
Adjacent sequences: A115088 A115089 A115090 * A115092 A115093 A115094
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KEYWORD
| hard,more,nonn
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AUTHOR
| T. D. Noe (noe(AT)sspectra.com), Mar 01 2006
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