

A166862


Primes p that divide n! + 1 for some n other than p1.


1



2, 7, 11, 19, 23, 29, 43, 47, 59, 61, 67, 71, 79, 83, 103, 109, 127, 131, 137, 139, 149, 163, 179, 191, 193, 199, 227, 233, 239, 251, 257, 263, 269, 271, 277, 293, 307, 311, 317, 347, 359, 367, 379, 383, 389, 397, 401, 419, 431, 443, 449, 461, 463, 467, 479
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OFFSET

1,1


COMMENTS

For n >= p, p is one of the factors of n!, so p cannot divide n! + 1. As a result, only 0 <= n <= p2 needs to be searched.
For n = p1, by Wilson's Theorem, (p1)! = 1 (mod p), so p divides (p1)! + 1.
Since by convention 0! = 1, 2 is included in the sequence as dividing 0!+1 = 2.
The standard heuristic suggests that the fraction of primes in this sequence is 1  1/e or about 63%.  Charles R Greathouse IV, Apr 17 2013


LINKS

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Wilson's Theorem


EXAMPLE

11 is included in the sequence since 11 divides 5! + 1 = 121.
13 is not included in the sequence since the only n for which 13 divides n! + 1 is n = 12.


PROG

(PARI) isA166862(n) = {local(r); r=0; for(i=0, n2, if((i!+1)%n==0, r=1)); r}
(PARI) is(p)=my(m=Mod(1, p)); for(k=2, p2, m*=k; if(m==1, return(isprime(p)))); p==2 \\ Charles R Greathouse IV, Apr 17 2013


CROSSREFS

Cf. A000142, A038507, A051301, A002583.
Sequence in context: A260889 A289994 A040154 * A105881 A060191 A178899
Adjacent sequences: A166859 A166860 A166861 * A166863 A166864 A166865


KEYWORD

nonn


AUTHOR

Michael B. Porter, Oct 22 2009


STATUS

approved



