%I A120457
%S A120457 81,108,162,216,256,324,378,486,504,512,540,648,756,810,896,972,1024,
%T A120457 1080,1280,1512,1620,1764,1792,1944,2048,2268,2520,2560,2916,3136,3240,
%U A120457 3528,3584,3780,4096,4480,4536,4860,5120,5400,5832,6272,6400,7168,7560
%N A120457 Sequence of unique powers from a quaternion generalization of Gaussian 
               quadratic reciprocity ( quaternion quartic reciprocity).
%C A120457 Quaternion[ -1/2, 1/2, 1/2, 1/2] is equivalent here to the Gaussian (-1). 
               I've eliminated all the powers that give the identity matrix. These 
               matrices are all unitary (determinant one). When the matrices of 
               these unique powers are sorted they only make 9 types in the first 
               10^4 products.
%F A120457 a(n) = Sorted[16*Powerof[((Prime[n] + 1)/2)*((Prime[m] + 1)/2)*((Prime[o] 
               + 1)/2)*((Prime[p] + 1)/2)]]
%e A120457 q[ -1/2, 1/2, 1/2, 1/2].q[ -1/2, -1/2, -1/2, -1/2] = {{1,0},{0,1}}
%t A120457 i = {{0, 1}, {-1, 0}}; j = {{0, I}, {I, 0}}; k = {{I, 0}, {0, -I}}; e 
               = IdentityMatrix[2]; q[t_, x_, y_, z_] = e*t + x*i + j*y + k*z; f[n_, 
               m_, o_, p_] = ((Prime[n] + 1)/2)*((Prime[m] + 1)/2)*((Prime[o] + 
               1)/2)*((Prime[p] + 1)/2) a = 16*Union[Flatten[Table[If[MatrixPower[q[ 
               -1/2, 1/2,1/2, 1/2], f[n, m, o, p]] - e == {{0, 0}, {0, 0}}, {}, 
               f[n, m, o, p]], {n, 1,10}, {m, 1, 10}, {o, 1, 10}, {p, 1, 10}], 3]]
%Y A120457 Sequence in context: A104113 A102766 A064828 this_sequence A129151 A039546 
               A053887
%Y A120457 Adjacent sequences: A120454 A120455 A120456 this_sequence A120458 A120459 
               A120460
%K A120457 nonn,uned
%O A120457 0,1
%A A120457 Roger Bagula (rlbagulatftn(AT)yahoo.com), Jun 24 2006