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A265643
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a(n) = +-1 == ((p - 1)/2)! (mod p), where p is the n-th prime number == 3 (mod 4).
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1
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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
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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EXAMPLE
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The second prime number == 3 (mod 4) is 7. Since ((7 - 1)/2)! = 3! = 6 == -1 (mod 7), it follows that a(2) = -1.
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MAPLE
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map(p -> if isprime(p) then mods(((p-1)/2)!, p) fi, [seq(i, i=3..10000, 4)]); # Robert Israel, Dec 11 2015
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MATHEMATICA
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Function[p, Mod[((p-1)/2)!, p, -1]] /@ Select[Range[3, 2003, 4], PrimeQ] (* Jean-François Alcover, Feb 27 2016 *)
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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