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%I #23 Jul 25 2021 02:36:45
%S 0,3,6,21,48,195,462,1917,4560,18963,45126,187701,446688,1858035,
%T 4421742,18392637,43770720,182068323,433285446,1802290581,4289083728,
%U 17840837475,42457551822,176606084157,420286434480,1748220004083
%N 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 Perfect square values for discriminants are used to classify the Galois group of a polynomial. The O+ discriminant component is sqrt(6*(x^2-3*x+6)) (used to generate these values) and for the O- discriminant sqrt(6*(x^2+3*x+6)).
%C 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 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 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, 1520, 6321, 15042, 148896, ...
%C This sequence, its reverse, and the division-by-3 form all appear to be new.
%D Mohammad K. Azarian, On the Hyperfactorial Function, Hypertriangular Function, and the Discriminants of Certain Polynomials, International Journal of Pure and Applied Mathematics, Vol. 36, No. 2, 2007, pp. 251-257. Mathematical Reviews, MR2312537. Zentralblatt MATH, Zbl 1133.11012.
%D 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, pp. 27-42.
%F 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 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 even 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) for odd n. Multiply the resultant sequence by 3 to generate the present sequence.
%F G.f.: 3 * (x + x^2 - 5*x^3 - x^4) / (1 - x - 10*x^2 + 10*x^3 + x^4 - x^5). - _Michael Somos_, Apr 05 2008
%F a(n) = A138976(-n) for all n in Z. a(n) = 3 * A129444(n+1).
%e G.f. = 3*x + 6*x^2 + 21*x^3 + 48*x^4 + 195*x^5 + 462*x^6 + 1917*x^7 + ...
%t 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 (PARI) {a(n) = my(m); m = if( n<0, m = 1-n, n); 3*(n<0) + 3*(-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 Cf. A129444, A138976.
%K nonn,uned
%O 0,2
%A _Lorenz H. Menke, Jr._, Mar 27 2008
%E I rather feel that this should be broken up into two sequences, one each for the positive and negative terms, both starting at 0. - _N. J. A. Sloane_, Apr 04 2008
%E More terms from _Michael Somos_, Apr 05 2008
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