A136331 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.
0, 3, 6, 21, 48, 195, 462, 1917, 4560, 18963, 45126, 187701, 446688, 1858035, 4421742, 18392637, 43770720, 182068323, 433285446, 1802290581, 4289083728, 17840837475, 42457551822, 176606084157, 420286434480, 1748220004083
Offset: 0
Examples
G.f. = 3*x + 6*x^2 + 21*x^3 + 48*x^4 + 195*x^5 + 462*x^6 + 1917*x^7 + ...
References
- 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.
- 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.
Programs
-
Mathematica
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}];
-
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 */
Formula
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.
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
Extensions
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
More terms from Michael Somos, Apr 05 2008
Comments