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A115092 The number of m such that prime(n) divides m!+1. 5
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,4
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.
LINKS
EXAMPLE
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.
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}]
PROG
(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
CROSSREFS
Sequence in context: A352620 A103343 A085263 * A172281 A304945 A176298
KEYWORD
nonn
AUTHOR
T. D. Noe, Mar 01 2006
STATUS
approved

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Last modified April 23 20:33 EDT 2024. Contains 371916 sequences. (Running on oeis4.)