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A138976 The discriminant of the characteristic polynomial of the O+ and O- submatrix for spin 3 of the nuclear electric quadrupole Hamiltonian is a perfect square for these values.

%I A138976 #20 Apr 14 2024 08:50:37
%S A138976 0,-3,-18,-45,-192,-459,-1914,-4557,-18960,-45123,-187698,-446685,
%T A138976 -1858032,-4421739,-18392634,-43770717,-182068320,-433285443,
%U A138976 -1802290578,-4289083725,-17840837472,-42457551819,-176606084154,-420286434477,-1748220004080,-4160406792963
%N A138976 The discriminant of the characteristic polynomial of the O+ and O- submatrix for spin 3 of the nuclear electric quadrupole Hamiltonian is a perfect square for these values.
%C A138976 Perfect square values for discriminants are used to classify the Galois group of a polynomial. The O+ discriminant component is Sqrt[6(x^2-3x+6)] (used to generate these values) and for the O- discriminant Sqrt[6(x^2+3x+6)].
%C A138976 This sequence is the negative of the O+ sequence. Also, note that if 3*a[n] represents the positive terms, the negative terms are generated from 3 - 3*a[n].
%C A138976 For the O- sequence reverse the O+ sequence and change all of the signs to generate ...-446688, -45126, -18963, -4560, -1917, -462, -195, -48, -21, -6, -3, 0, 3, 18, 45, 192, 459, 1914, 4557, 18960, 45123, 187698, 446685.
%C A138976 Note that the difference equation a[n] generates the above sequence divided by 3 or ...,-148895, -62566, -15041, -6320, -1519, -638, -153, -64, -15, -6, -1, 0, 1, 2, 7, 16, 65, 154, 639, 15 20, 6321, 15042, 148896,...
%C A138976 This sequence, its reverse and the division by 3 form, all appear to be new.
%D A138976 The physics reference is G. W. King, "The Asymmetric Rotor I. Calculation and Symmetry Classification of Energy Levels", Journal of Chemical Physics, Jan 1943, Volume 11, p27-42.
%H A138976 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,10,-10,-1,1).
%F A138976 The difference equation is a[n]=11(a[n-2] - a[n-4])+a[n-6] with a[0]=0, a[1]=1, a[2]=2, a[3]=7, a[4]=16, a[5]=65. The solution is for even n: a[n]=(1/2) - (1/12)*(3+2*Sqrt[6])*(5-2*Sqrt[6])^(n/2)+(1/12)*(-3+2*Sqrt[6])*(5+2*Sqrt[6])^(n/2), for odd n a[n]=(1/2) - (1/12)*(3*Sqrt[2]+Sqrt[3])*(5-2*Sqrt[6])^(n/2)+(1/12)*(3*Sqrt[2]-Sqrt[3])*(5+2*Sqrt[6])^(n/2). Multiply the resultant sequence by 3 to generate the present sequence.
%F A138976 G.f.: -3*x*(1+5*x-x^2-x^3)/((1-x)*(1-10*x^2+x^4)). [_Colin Barker_, Aug 22 2012]
%t A138976 Do[If[IntegerQ[Sqrt[6 (6 - 3 x + x^2)]], Print[{x, Sqrt[6 (6 - 3 x + x^2)]}]], {x, -1000, 1000}]; Do[If[IntegerQ[Sqrt[6 (6 + 3 x + x^2)]], Print[{x, Sqrt[6 (6 + 3 x + x^2)]}]], {x, -1000, 1000}];
%o A138976 (PARI) {a(n) = local(m); m = if( n>0, m = 1+n, -n); 3 * ((n>0) + (-1)^(n>0) * polcoeff( (x + x^2 - 5*x^3 - x^4) / ((1 - x) * (1 - 10*x^2 + x^4)) + x*O(x^m), m))} /* _Michael Somos_, Apr 05 2008 */
%Y A138976 A136331(n) = a(-n).
%K A138976 sign,easy
%O A138976 0,2
%A A138976 _Lorenz H. Menke, Jr._, Mar 27 2008