%I
%S 81,108,162,216,256,324,378,486,504,512,540,648,756,810,896,972,1024,
%T 1080,1280,1512,1620,1764,1792,1944,2048,2268,2520,2560,2916,3136,
%U 3240,3528,3584,3780,4096,4480,4536,4860,5120,5400,5832,6272,6400,7168,7560
%N Sorted list of numbers of the form (p+1)*(q+1)*(r+1)*(s+1) that are not multiples of 48, where p, q, r, and s are primes.
%C Original name: Sequence of unique powers from a quaternion generalization of Gaussian quadratic reciprocity (quaternion quartic reciprocity).
%C The 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. [This comment needs to be clarified]
%C This quaternion is a primitive cube root of 1; its powers behave like any other primitive cube root of 1. We are thus looking at products (p+1)*(q+1)*(r+1)*(s+1) modulo 48, where p, q, r, and s are primes. The only possible odd result is with all 4 primes = 2: 81. If no prime = 2, the result must be a multiple of 16, which gives three residues, one of which is the identity. If 2 and another prime are present, the result is a multiple of 6, which produces 8 residues; again, one is the identity.  Franklin T. AdamsWatters, Aug 20 2011
%F a(n) = Sorted[16*Powerof[((Prime[n] + 1)/2)*((Prime[m] + 1)/2)*((Prime[o] + 1)/2)*((Prime[p] + 1)/2)]]
%e q[ 1/2, 1/2, 1/2, 1/2]*q[ 1/2, 1/2, 1/2, 1/2] = {{1,0},{0,1}}
%t 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]]
%K nonn
%O 1,1
%A _Roger L. Bagula_, Jun 24 2006
