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 <= 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
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, 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
CROSSREFS
KEYWORD
nonn
AUTHOR
Michael B. Porter, Oct 22 2009
STATUS
approved