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

Table of n, a(n) for n=1..7.

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

Select[Prime@ Range@ 1000, Function[p, AnyTrue[Range[p - 1], Divisible[#! + 1, p^2] &]]] (* Michael De Vlieger, Jan 24 2016, Version 10 *)

PROG

(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

CROSSREFS

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

Sequence in context: A269844 A116440 A098720 * A034924 A018607 A032481

Adjacent sequences:  A115088 A115089 A115090 * A115092 A115093 A115094

KEYWORD

hard,more,nonn

AUTHOR

T. D. Noe, Mar 01 2006

STATUS

approved

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Last modified May 30 04:27 EDT 2017. Contains 287305 sequences.