OFFSET
1,1
COMMENTS
Since n! - 1 = 0 for n=1 and n=2, the restriction n > 1 needed to be placed.
For n >= p, p is one of the factors of n!, so p cannot divide n! - 1.
For n = p-1, by Wilson's Theorem, (p-1)! = -1 (mod p), so p divides (p-1)! + 1, and cannot also divide (p-1)! - 1 unless p = 2.
For n = p-2, again by Wilson's Theorem, (p-1)! = (p-1)(p-2)! = (-1)(p-2)! = -1 (mod p), so (p-2)! = 1 (mod p) and p divides (p-2)! - 1. As a result, only 2 <= n <= p-3 needs to be searched.
LINKS
Eric Weisstein's World of Mathematics, Wilson's Theorem
EXAMPLE
17 is included in the sequence since 17 divides 5! - 1 = 119.
19 is not included in the sequence since the only n for which 19 divides n! - 1 is n = 17.
PROG
(PARI) isA166864(n) = {local(r); r=0; for(i=2, n-3, if((i!-1)%n==0, r=1)); r}
CROSSREFS
KEYWORD
nonn
AUTHOR
Michael B. Porter, Oct 22 2009
STATUS
approved