%I #37 Jun 13 2015 00:55:17
%S 3,-1,-2,1,-9,4,7,-3,30,-13,-23,10,-99,43,76,-33,327,-142,-251,109,
%T -1080,469,829,-360,3567,-1549,-2738,1189,-11781,5116,9043,-3927,
%U 38910,-16897,-29867,12970,-128511,55807,98644,-42837
%N List of quadruples (r,s,t,u): the matrix M = [[4,12,9][2,5,3][1,2,1]] is raised to successive negative powers, then (r,s,t,u) are the square roots of M[1,3], M[1,1], M[3,3], M[3,1] respectively.
%C The sequence, in reverse order, comprises numbers to the left of a(0) in A249579, where the terms would be labeled a(-1), a(-2), a(-3), ... .
%C This sequence 'factors' into four sequences with alternating signs. Ignoring signage, they are A052906, A003688, A052924 and A006190 (listed as crossrefs below).
%H Colin Barker, <a href="/A249580/b249580.txt">Table of n, a(n) for n = 1..1000</a>
%H Russell Walsmith, <a href="/A249580/a249580.pdf">A sequence of matrices</a>
%H Russell Walsmith, <a href="/A249580/a249580_1.pdf">DCL-Chemy III: Hyper-Quadratics</a>
%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,-3,0,0,0,1).
%F a(n) = -3*a(n-4)+a(n-8). - _Colin Barker_, Nov 06 2014
%F G.f.: -x*(x^6+x^5+x^3-2*x^2-x+3) / (x^8-3*x^4-1). - _Colin Barker_, Nov 06 2014
%e M^-1 = [[1,-6,9][-1,5,-6][1,-4,4]]. sqrt(M[1,3]) = 3, sqrt(M[1,1]) = -1, sqrt(M[3,3]) = -2, sqrt(M[3,1]) = 1. Then r = 3; s = -1; t = -2; ; u = 1.
%e M^-2 = [[16,-72,81][-12,55,-63][9,-42,49]]. sqrt(M[1,3]) = -9, sqrt(M[1,1]) = 4, sqrt(M[3,3]) = 7, sqrt(M[3,1]) = -3. Then r = -9; s = 4; t = 7; ; u = -3.
%t m[e_] := MatrixPower[{{4, 12, 9}, {2, 5, 3}, {1, 2, 1}}, -e]; g[e_, x_, y_] := (-1)^If[x == y, e, e + 1] Sqrt@ m[e][[x, y]]; f[e_] := {g[e, 1, 3], g[e, 1, 1], g[e, 3, 3], g[e, 3, 1]}; Array[f, 10] // Flatten (* _Robert G. Wilson v_, Dec 19 2014 *)
%o (PARI) Vec(-x*(x^6+x^5+x^3-2*x^2-x+3)/(x^8-3*x^4-1) + O(x^100)) \\ _Colin Barker_, Nov 06 2014
%Y Cf. A249579. Disregarding signage, a(4n) = A052906; a(4n+1) = A003688; a(4n+2) = A052924; a(4n+3) = A006190.
%K sign,easy
%O 1,1
%A _Russell Walsmith_, Nov 02 2014