A318417 Scaled g.f. T(u) = Sum_{n>=0} a(n)*(3*u/48)^n satisfies 3*(2*u-1)*T + d/du(4*u*(2*u-1)*(u-1)*T') = 0, and a(0)=1; sequence gives a(n).

1, 12, 228, 5040, 121380, 3093552, 82047504, 2240162496, 62508328740, 1773580002480, 50988042273168, 1481392081181376, 43413834762798864, 1281498837550545600, 38059165854011995200, 1136249610240102992640, 34076899109906247654180, 1026061759878805529676720

Offset: 0

Author

Bradley Klee, Aug 26 2018

Keywords

• nonn

Comments

The defining differential equation determines the period function T(u) along a square-symmetric Hamiltonian family of algebraic plane curves, u=2*H=x^2+y^2-(1/2)*(x^4+y^4). The separatrix with u=1/2 is a product of two congruent ellipses. Transformation u->1-u leaves T(u) invariant.

Examples

			Singular Points of u=x^2+y^2-(1/2)*(x^4+y^4).
   u             (x,y)              type
=============================================
   0             (0,0)            circular
  1/2      (0,+/-1) (+/-1,0)     hyperbolic
   1       +/-(1,1) +/-(1,-1)     circular
G.f. = 1 + 12*x + 228*x^2 + 5040*x^3 + 121380*x^4 + 3093552*x^5 + 82047504*x^6 + ... - _Michael Somos_, Aug 27 2018
		

Links

• Bradley Klee, Contour and Period Graphs.

Crossrefs

Factors: A000984, A098410. Quartic Periods: A113424, A002894, A318245.

Programs

Mathematica

RecurrenceTable[{n^2*a[n]-12*(2*n-1)^2*a[n-1] + 128*(2*n-3)*(2*n-1)*a[n-2] == 0, a[0] == 1, a[1] == 12}, a, {n,0,1000}]
PeriodIntegral[n_]:= CoefficientList[Series[1/Sqrt[1-2*x*((z+y)^4+(z-y)^4)], {x,0,n}], {x, y, z}][[#+1,2*#+1,2*#+1]]&/@Range[0,n];
PeriodIntegral[10]
HadamardProduct[n_]:=Times@@Map[CoefficientList[Normal[Series[#,{x,0,n}]],x,n+1]&, {1/Sqrt[1-12*x+32*x^2], 1/Sqrt[1-4*x]}];
HadamardProduct[10]
a[ n_] := Binomial[2 n, n] SeriesCoefficient[ (1 - 12 x + 32 x^2)^(-1/2), {x, 0, n}]; (* Michael Somos, Aug 27 2018 *)

PARI

{a(n) = if( n<0, 0, binomial(2*n, n) * polcoeff( (1 - 12*x + 32*x^2 + x * O(x^n))^(-1/2), n))}; /* Michael Somos, Aug 27 2018 */

Formula

Define t(u,z) = 1/sqrt(1-2*u*(cos(z)^4+sin(z)^4)), then certify that:

3*(2*u-1)*t + d/du(4*u*(2*u-1)*(u-1)*dt/du) = d/dz((1/4)*sin(4*z)*t^3).

G.f.: G(x) = T(48*x/3), T(u) = 1/(2*Pi)*Integral_{z=0..2*Pi} t(u)*dz.

n^2*a(n) - 12*(2*n-1)^2*a(n-1) + 128*(2*n-3)*(2*n-1)*a(n-2) = 0.

a(n) = A000984(n)*A098410(n).

0 = +a(n)*(+a(n+1)*([phone]*a(n+2) -25165824*a(n+3) +819200*a(n+4)) +a(n+2)*(-6291456*a(n+2) +1998848*a(n+3) -78336*a(n+4)) +a(n+3)*(-29184*a(n+3) +1472*a(n+4))) +a(n+1)*(+a(n+1)*(-6291456*a(n+2) +1179648*a(n+3) -38400*a(n+4)) +a(n+2)*(+327680*a(n+2) -110592*a(n+3) +4448*a(n+4)) +a(n+3)*(+1728*a(n+3) -90*a(n+4))) +a(n+2)*(+a(n+2)*(-1536*a(n+2) +800*a(n+3) -36*a(n+4)) +a(n+3)*(-18*a(n+3) +a(n+4))) for all n in Z. - Michael Somos, Aug 27 2018

G.f.: hypergeom([1/2, 1/2],[1],16*x/(32*x-1))/sqrt(1-32*x). - Mark van Hoeij, Dec 13 2024