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%I #14 Aug 30 2018 22:42:56
%S 1,12,228,5040,121380,3093552,82047504,2240162496,62508328740,
%T 1773580002480,50988042273168,1481392081181376,43413834762798864,
%U 1281498837550545600,38059165854011995200,1136249610240102992640,34076899109906247654180,1026061759878805529676720
%N 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).
%C 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.
%H Bradley Klee, <a href="/A318417/a318417.pdf">Contour and Period Graphs</a>.
%F Define t(u,z) = 1/sqrt(1-2*u*(cos(z)^4+sin(z)^4)), then certify that:
%F 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).
%F G.f.: G(x) = T(48*x/3), T(u) = 1/(2*Pi)*Integral_{z=0..2*Pi} t(u)*dz.
%F n^2*a(n) - 12*(2*n-1)^2*a(n-1) + 128*(2*n-3)*(2*n-1)*a(n-2) = 0.
%F a(n) = A000984(n)*A098410(n).
%F 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
%e Singular Points of u=x^2+y^2-(1/2)*(x^4+y^4).
%e u (x,y) type
%e =============================================
%e 0 (0,0) circular
%e 1/2 (0,+/-1) (+/-1,0) hyperbolic
%e 1 +/-(1,1) +/-(1,-1) circular
%e 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
%t 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}]
%t 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];
%t PeriodIntegral[10]
%t 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]}];
%t HadamardProduct[10]
%t a[ n_] := Binomial[2 n, n] SeriesCoefficient[ (1 - 12 x + 32 x^2)^(-1/2), {x, 0, n}]; (* _Michael Somos_, Aug 27 2018 *)
%o (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 */
%Y Factors: A000984, A098410. Quartic Periods: A113424, A002894, A318245.
%K nonn
%O 0,2
%A _Bradley Klee_, Aug 26 2018
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