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%I #18 Feb 26 2023 03:08:31
%S 3,8,13,17,31,32,30,50,46,55,75,91,76,98,100,105,129,93,162,112,183,
%T 122,144,177,241,187,217,228,155,288,203,189,213,311,269,274,334,381,
%U 266,392,254,382,348,413,301,286,489,439,483,553,516,476,578,423,487,504
%N a(n) = A002144(n) - A002314(n).
%C For the four numbers {1, A002314(n), A152676(n), A152680(n)}, the multiplication table modulo A002144(n) is isomorphic with the Latin square
%C 1 2 3 4
%C 2 4 1 3
%C 3 1 4 2
%C 4 3 2 1
%C and is isomorphic with the multiplication table for {1,i,-i,-1} where i = sqrt(-1), A152680(n) is isomorphic with -1, A002314(n) with i or -i and A152676(n) vice versa -i or i.
%C 1, A002314(n), A152676(n), A152680(n) are a subfield of the Galois Field [A002144(n)].
%C Let p = A002144(n), the n-th prime of the form 4k+1. Then a(n) and A002314(n) are the two square roots of -1 (mod p). Note that a(n) is also the multiplicative inverse of A002314(n) (mod p). - _T. D. Noe_, Feb 18 2010
%H T. D. Noe, <a href="/A152676/b152676.txt">Table of n, a(n) for n = 1..1000</a>
%H A. J. C. Cunningham, <a href="https://archive.org/details/binomialfactoris01cunn/page/1/mode/1up">Binomial Factorisations</a>, Vols. 1-9, Hodgson, London, 1923-1929. See Vol. 1, pp. 1-21.
%t aa = {}; Do[If[Mod[Prime[n], 4] == 1, k = 1; While[ ! Mod[k^2 + 1, Prime[n]] == 0, k++ ]; AppendTo[aa, Prime[n] - k]], {n, 1, 200}]; aa
%Y Cf. A002144, A002314, A152680.
%K nonn
%O 1,1
%A _Artur Jasinski_, Dec 10 2008
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