A061006 a(n) = (n-1)! mod n.

0, 1, 2, 2, 4, 0, 6, 0, 0, 0, 10, 0, 12, 0, 0, 0, 16, 0, 18, 0, 0, 0, 22, 0, 0, 0, 0, 0, 28, 0, 30, 0, 0, 0, 0, 0, 36, 0, 0, 0, 40, 0, 42, 0, 0, 0, 46, 0, 0, 0, 0, 0, 52, 0, 0, 0, 0, 0, 58, 0, 60, 0, 0, 0, 0, 0, 66, 0, 0, 0, 70, 0, 72, 0, 0, 0, 0, 0, 78, 0, 0, 0, 82, 0, 0, 0, 0, 0, 88, 0, 0, 0, 0, 0

Offset: 1

Author

Henry Bottomley, Apr 12 2001

Keywords

• nonn
• easy

Comments

It appears that a(n) = (n!*h(n)) mod n, where h(n) = Sum_{k = 1..n} 1/k. - Gary Detlefs, Sep 04 2010

Indeed: It is easy to show n!*h(n) - (n-1)! = n*(n-1)!*h(n-1). Since (n-1)!*h(n-1) is integral, n!*h(n) == (n-1)! mod n. - Franz Vrabec, Apr 08 2017

Examples

			a(4) = 2 since (4-1)! = 6 = 2 mod 4.
a(5) = 4 since (5-1)! = 24 = 4 mod 5.
a(6) = 0 since (6-1)! = 120 = 0 mod 6.
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..1000
• Wikipedia, Wilson's theorem

Crossrefs

Positive for all but the first term of A046022. Cf. A000040, A000142, A061007, A061008, A061009.

Programs

Mathematica

Table[Mod[(n - 1)!, n], {n, 100}] (* Alonso del Arte, Feb 16 2014 *)

PARI

a(n)=if(isprime(n),n-1,if(n==4,2,0)) \\ Charles R Greathouse IV, Mar 31 2014

Formula

a(4) = 2, a(p) = p - 1 for p prime (Wilson's_theorem), a(n) = 0 otherwise. Apart from n = 4, a(n) = (n-1)*A061007(n) = (n-1)*A010051(n).