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%I #10 Jan 24 2021 13:55:46
%S 1,2,6,120,720,40320,362880,39916800,87178291200,1307674368000,
%T 6402373705728000,2432902008176640000,51090942171709440000,
%U 25852016738884976640000,403291461126605635584000000,8841761993739701954543616000000
%N ((prime(n)- 1)/2)!, n >= 2.
%C a(n)^2 == (-1)^((prime(n) + 1)/2) (mod prime(n)).
%C 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.
%C See A004055 for a(n) (mod prime(n)), n>=2.
%C See A212159 for a(n)^2 (mod prime(n)), n>=2.
%H Holsztyński Włodzimierz, <a href="http://wlod.wordpress.com/article/congruence-x-2-1-mod-p-euler-and-a-1jxfhq4x4sw0j-65/">Congruence x^2==-1 (mod p) (Euler), and a super-Wilson Theorem</a>
%F a(n) = ((prime(n)- 1)/2)!, n>=2, with prime(n) = A000040(n).
%F a(n) = A005097(n-1)!, n>=2.
%e a(4) = (7-1)/2)! = 3! = 6.
%e a(4)^2 = 36 == +1 (mod 7), because (7 + 1)/2 = 4, and 4 is even.
%e a(6) = ((13-1)/2)! = 6! = 720.
%e a(6)^2 = 518400 == -1 (mod 13) = 12 (mod 13) because (13+1)/2 = 7, and 7 is odd.
%t ((Prime[Range[2,20]]-1)/2))! (* _Harvey P. Dale_, Jan 24 2021 *)
%K nonn
%O 2,2
%A _Wolfdieter Lang_, May 08 2012
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