A212158 - OEIS

A212158

a(n) = ((prime(n) - 1)/2)!, n >= 2.

1, 2, 6, 120, 720, 40320, 362880, 39916800, 87178291200, 1307674368000, 6402373705728000, 2432902008176640000, 51090942171709440000, 25852016738884976640000, 403291461126605635584000000, 8841761993739701954543616000000

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OFFSET

2,2

COMMENTS

a(n)^2 == (-1)^((prime(n) + 1)/2) (mod prime(n)).

Use product(p-j,j=1..(p-1)/2) == (-1)^((p-1)/2)*a(n) (mod p) for p=prime(n), n>=2, hence a(n)*(-1)^((p-1)/2)*a(n) == (p-1)! (mod p), then apply Wilson's theorem. That is, a(n)^2 == -1 (mod prime(n)) for primes of the form 4*k+1 (see A002144) and +1 for primes of the form 4*k+3 (see A002145). See the link with a blog by W. Holsztyński.

See A004055 for a(n) (mod prime(n)), n>=2.

See A212159 for a(n)^2 (mod prime(n)), n>=2.

LINKS

Table of n, a(n) for n=2..17.

Holsztyński Włodzimierz, Congruence x^2==-1 (mod p) (Euler), and a super-Wilson Theorem

FORMULA

a(n) = ((prime(n) - 1)/2)!, n>=2, with prime(n) = A000040(n).

a(n) = A005097(n-1)!, n>=2.

EXAMPLE

a(4) = ((7-1)/2)! = 3! = 6.

a(4)^2 = 36 == +1 (mod 7), because (7 + 1)/2 = 4, and 4 is even.

a(6) = ((13-1)/2)! = 6! = 720.