A318245 Scaled g.f. T(v) = Sum_{n>=0} a(n)*(3*v/64)^n satisfies 9*(5*v-4)*T + d/dv(16*v*(v-1)*(3*v-4)*T') = 0, and a(0)=1; sequence gives a(n).

1, 12, 180, 2928, 49860, 875952, 15754704, 288722880, 5373771876, 101334517680, 1932405892560, 37208369165760, 722497419680400, 14132680228175040, 278236490874120000, 5508974545258860288, 109624581377872629156, 2191185332414847848880, 43971545517545956240464

Offset: 0

Author

Bradley Klee, Aug 22 2018

Keywords

• nonn

Comments

The linked document "Proof Certificate" explains that period function T(v) measures precession of the J-vector along an algebraic sphere curve with local cyclic C_4 symmetry (also cf. Examples and A186375).

Examples

			Period function T_{O}(w): Take T_{C3}(u) and T_{C4}(v) from A186375 and A318245 respectively. Set (u,v)=(w-2/3,2-w), with u in [0,1/3], v in [0,1], and w in [2/3,2]. Define piecewise function T_{O}(w) = T_{C3}(w-2/3) if w in [2/3,1] or T_{O}(w) = T_{C4}(2-w) if w in [1,2].
Geometric Singular Points: Construct a family of algebraic sphere curves by intersecting a sphere 1=X^2+Y^2+Z^2 with the octahedral surface w=2*(X^4+Y^4+Z^4). Four cube vertex axes--(x+y+z, -x+y+z, x-y+z, x+y-z)--intersect the sphere in eight circular points with w=2/3. Three octahedron vertex axes--(x, y, z)--intersect the sphere in six circular points with w=2. Six cuboctahedron vertex axes--(x+y, x-y, y+z, y-z, z+x, z-x)--intersect the sphere in twelve hyperbolic points with w=1.
		

Links

• W. G. Harter and C. W. Patterson, Rotational energy surfaces and high-J eigenvalue structure of polyatomic molecules, The Journal of Chemical Physics, 80 (1984), 4252.
• S. Herfurtner, Elliptic surfaces with four singular fibres, Mathematische Annalen, 1991. Preprint.
• Bradley Klee, Proof Certificate.
• Bradley Klee, Checking Weierstrass data, 2023.
• Eric Weisstein's World of Mathematics, Goursat's Surface.

Crossrefs

Periods: A186375, A318417.

Programs

Mathematica

CoefficientList[Expand[Normal@Series[Divide[Sqrt[S],Sqrt[1-4*S*x]*Sqrt[S-8 + 8*Sqrt[1-4*S*x]]], {x, 0, 13}]/.{S->12+4*Q^2}]/.{Q^n_:>(1/2)^n*Binomial[n, n/2]} /.{x->1/3*x}, x]
RecurrenceTable[{3*n^2*a[n] - 4*(28*n^2-28*n+9)*a[n-1] + 64*(4*n-5)*(4*n-3)*a[n-2] == 0, a[0]==1, a[1]==12}, a, {n,0,1000}]

Formula

3*n^2*a(n) - 4*(28*n^2-28*n+9)*a(n-1) + 64*(4*n-5)*(4*n-3)*a(n-2) = 0.

For n > 0, a(n) mod 3 = 0 (conjecture, tested up to n=3*10^6).

From Bradley Klee, May 30 2023: (Start)

The defining ODE can be derived from the following Weierstrass data:

g2 = (3/16)*(256 - 576*x + 405*x^2 - 81*x^3);

g3 = (1/64)*(4096 - 13824*x + 17496*x^2 - 9963*x^3 + 2187*x^4);

which determine an elliptic surface with four singular fibers. (End)