A211173 (2n)!^n (modulo 2n+1).

0, 2, 1, 6, 0, 10, 1, 0, 1, 18, 0, 22, 0, 0, 1, 30, 0, 0, 1, 0, 1, 42, 0, 46, 0, 0, 1, 0, 0, 58, 1, 0, 0, 66, 0, 70, 1, 0, 0, 78, 0, 82, 0, 0, 1, 0, 0, 0, 1, 0, 1, 102, 0, 106, 1, 0, 1, 0, 0, 0, 0, 0, 0, 126, 0, 130, 0, 0, 1, 138, 0

Offset: 0

Author

Larry Riddle (LRiddle(AT)AgnesScott.edu) and Robert G. Wilson v, Jan 31 2013

Keywords

• nonn
• easy

Comments

a(n)= 0, 1 or 2n. In fact, a(n) = 0 iff n belongs to A047845, a(n) = 1 iff n belongs to A104636 and a(n) = 2n iff n belongs to A104635.

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
• Larry Riddle, Sophie Germain and Fermat's Last Theorem, Agnes Scott College, Math. Dept. Jul, 1999.

Crossrefs

Cf. A047845, A104635, A104636.

Programs

Mathematica

f[n_] := Mod[((2 n)!)^n, 2 n + 1]; Array[f, 70]
Table[PowerMod[(2n)!,n,2n+1],{n,0,70}] (* Harvey P. Dale, Nov 02 2019 *)

PARI

a(n)=if(isprime(2*n+1),if(n%2,2*n,1),0) \\ Charles R Greathouse IV, Feb 07 2013

Formula

a(n) = (2n)!^n (modulo 2n+1).