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%I #7 May 15 2015 15:13:33
%S 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,
%T 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,
%U 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
%N The number of m such that prime(n) divides m!+1.
%C 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.
%H T. D. Noe, <a href="/A115092/b115092.txt">Table of n, a(n) for n = 1..2000</a>
%e 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.
%t 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}]
%o (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
%K nonn
%O 1,4
%A _T. D. Noe_, Mar 01 2006
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