

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
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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 p1. Sequence A073944 lists the smallest m for each prime.


LINKS

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


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, p1}]; cnt, {i, 150}]


PROG

(PARI) a(n, p=prime(n))=my(t=Mod(1, p)); sum(k=1, p1, t*=k; t==1) \\ Charles R Greathouse IV, May 15 2015


CROSSREFS

Sequence in context: A207324 A103343 A085263 * A172281 A304945 A176298
Adjacent sequences: A115089 A115090 A115091 * A115093 A115094 A115095


KEYWORD

nonn


AUTHOR

T. D. Noe, Mar 01 2006


STATUS

approved



