A115092 The number of m such that prime(n) divides m!+1.

1, 1, 1, 2, 2, 1, 1, 2, 3, 2, 1, 1, 1, 2, 2, 1, 4, 4, 3, 7, 1, 4, 4, 1, 1, 1, 3, 1, 2, 1, 2, 2, 4, 2, 2, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 5, 1, 2, 2, 1, 3, 3, 2, 3, 3, 2, 1, 1, 5, 4, 2, 1, 3, 1, 1, 2, 1, 1, 2, 2, 1, 3, 4, 3, 4, 6, 1, 3, 1, 3, 1, 1, 2, 2, 1, 2, 3, 3, 4, 1, 2, 2, 4, 1, 3, 2, 1, 1, 2, 4, 3, 4

Offset: 1

Author

T. D. Noe, Mar 01 2006

Keywords

• nonn

Comments

By Wilson's theorem, we know that for each prime p there is at least one m such that p divides m!+1. The largest such m is p-1. Sequence A073944 lists the smallest m for each prime.

Examples

			a(20)=7 because 71, the 20th prime, divides m!+1 for the seven values m=7,9,19,51,61,63,70. Interesting, note that 7+63=9+61=19+51=70.
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..2000

Programs

Mathematica

Table[p=Prime[i]; cnt=0; f=1; Do[f=Mod[f*m,p]; If[f+1==p,cnt++ ], {m,p-1}]; cnt, {i,150}]

PARI

a(n,p=prime(n))=my(t=Mod(1,p)); sum(k=1,p-1, t*=k; t==-1) \\ Charles R Greathouse IV, May 15 2015