A166862 Primes p that divide n! + 1 for some n other than p-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

Offset: 1

Author

Michael B. Porter, Oct 22 2009

Keywords

• nonn

Comments

For n >= p, p is one of the factors of n!, so p cannot divide n! + 1. As a result, only 0 <= n <= p-2 needs to be searched.

For n = p-1, by Wilson's Theorem, (p-1)! = -1 (mod p), so p divides (p-1)! + 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

Examples

			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.
		

Links

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

Crossrefs

Cf. A000142, A038507, A051301, A002583.

Programs

PARI

isA166862(n) = {local(r);r=0;for(i=0,n-2,if((i!+1)%n==0,r=1));r}

PARI

is(p)=my(m=Mod(1,p)); for(k=2,p-2,m*=k; if(m==-1, return(isprime(p)))); p==2 \\ Charles R Greathouse IV, Apr 17 2013