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

0,2

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.

Links

Table of n, a(n) for n=0..17.

Bradley Klee, Contour and Period Graphs.

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)*(+134217728*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

Example

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

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 *)

Prog

(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 */

Crossrefs

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

Sequence in context: A358362 A337332 A290754 * A034832 A231448 A221657

Adjacent sequences: A318414 A318415 A318416 * A318418 A318419 A318420

Keyword

nonn

Author

Bradley Klee, Aug 26 2018

Status

approved