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%I #13 Mar 12 2014 16:36:53
%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. Adams-Watters, 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
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