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Search: id:A115092
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| A115092 |
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The number of m such that prime(n) divides m!+1. |
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+0
5
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| 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; listen)
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OFFSET
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1,4
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COMMENT
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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.
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LINKS
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T. D. Noe, Table of n, a(n) for n=1..2000
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EXAMPLE
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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.
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MATHEMATICA
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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}]
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CROSSREFS
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Sequence in context: A098281 A103343 A085263 this_sequence A011847 A091325 A143974
Adjacent sequences: A115089 A115090 A115091 this_sequence A115093 A115094 A115095
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KEYWORD
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nonn
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AUTHOR
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T. D. Noe (noe(AT)sspectra.com), Mar 01 2006
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