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User: Carlo Sanna

Carlo Sanna has authored 1 sequences.

A265643 a(n) = +-1 == ((p - 1)/2)! (mod p), where p is the n-th prime number == 3 (mod 4).

1, -1, -1, -1, 1, 1, -1, -1, 1, -1, 1, -1, 1, -1, 1, -1, -1, 1, 1, -1, 1, -1, -1, -1, 1, 1, -1, 1, 1, -1, 1, 1, 1, 1, 1, -1, 1, -1, 1, -1, -1, -1, 1, -1, 1, 1, -1, 1, -1, 1, -1, -1, 1, -1, -1, 1, -1, -1, -1, -1, 1, 1, 1, -1, -1, 1, -1, -1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, 1, 1, -1, 1, 1, 1, -1, -1, 1, 1, -1, 1, -1, -1, 1, -1, -1, 1, 1, -1

Offset: 1

Carlo Sanna, Dec 11 2015

sign

By Wilson's theorem, ((p - 1)/2)!^2 == (-1)^((p + 1)/2) (mod p) for each prime number p. Hence, if p == 3 (mod 4), then ((p - 1)/2)! == +-1 (mod p).
Michele Elia proved that a(n) = (-1)^((1 + h(-p)) / 2) for n > 1, where p is the n-th prime number == 3 (mod 4), and h(-p) is the class number of the quadratic field Q(sqrt(-p)).
Mordell (1961) proved the same result 52 years earlier in a 2-page note in the Monthly. - Jonathan Sondow, Apr 09 2017

			The second prime number == 3 (mod 4) is 7. Since ((7 - 1)/2)! = 3! = 6 == -1 (mod 7), it follows that a(2) = -1.

Robert Israel, Table of n, a(n) for n = 1..10000
Michele Elia, A note on the sequence ((p-1)/2)! mod p, International Mathematical Forum, 2013, Vol. 8, no. 37, pages 1813-1825.
L.J. Mordell, The congruence (p-1/2)! == +-1 (mod p), Amer. Math. Monthly, 68 (1961), 145-146.

Cf. A000924, A002145, A004055.

map(p -> if isprime(p) then mods(((p-1)/2)!, p) fi, [seq(i,i=3..10000, 4)]); # Robert Israel, Dec 11 2015

Function[p, Mod[((p-1)/2)!, p, -1]] /@ Select[Range[3, 2003, 4], PrimeQ] (* Jean-François Alcover, Feb 27 2016 *)

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User: Carlo Sanna

A265643 a(n) = +-1 == ((p - 1)/2)! (mod p), where p is the n-th prime number == 3 (mod 4).

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