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%I #33 Sep 12 2018 02:18:10
%S 1,492,707220,1298204880,2654173160100,5765723073622512,
%T 13021894087331233104,30217387890886676251200,
%U 71532102917478013611243300,171944976047709681477985038000,418347201888204996027087975427920
%N a(n) = binomial(3*n,n)/(2*Pi)*Integral_{x=0..2*Pi} (12*cos^2(x)*sin(x) + 20*sin^3(x))^(2*n) dx.
%C Compare with A295870. The series expansion "T(x)=2*Pi*sqrt(3/5)*Sum_{n>=0} a_n*(x/25)^n" determines the period T of anharmonic oscillation along a contour of the Hamiltonian energy surface "x=2H=(5/3)*p^2+q^2+4*(p^2+q^2)*q,0<x<1/108".
%C The period-energy function T(x) satisfies the Picard-Fuchs equation "(2460+28512*x+2239488*x^2)*T(x)-(125-24840*x-1423008*x^2-20155392*x^3)*T'(x)+(-125*x+1620*x^2+1189728*x^3+10077696 x^4)*T''(x)", also the P.F.Eq. of A295870 under transformation x->x'=1/108-x.
%C A300057 has a similar definition to A005721, with a couple of extra integers appearing in the integrand. This makes a nice analogy between real and complex periods A295870, A300058. Second-order recurrences with polynomial coefficients define both sequences.
%H Bradley Klee, <a href="/A295870/a295870.pdf">Phase Plane Geometry</a>.
%H Brad Klee, <a href="http://demonstrations.wolfram.com/DerivingHypergeometricPicardFuchsEquations/">Deriving Hypergeometric Picard-Fuchs Equations</a>, Wolfram Demonstrations Project (2018).
%H M. Kontsevich and D. Zagier, <a href="http://www.ihes.fr/~maxim/TEXTS/Periods.pdf">Periods</a>, Institut des Hautes Etudes Scientifiques 2001 IHES/M/01/22.
%F a(n) = A005809(n)*A300057(n).
%F a(n) = Sum_{k1=0..2n} Sum_{k2=0..2n} binomial(3*n,n)*binomial(2*n,k2)*binomial(2*n,k1)*binomial(2*n,3*n-k1-k2)*((4+sqrt(15))^(2*n-k1))*((4-sqrt(15))^(2*n-k2))
%F a(0) = 1; a(1) = 492; a(n):=(c1/c3)*a(n-1)+(c2/c3)*a(n-2); with
%F c1 = 12*(-230+2259*n-3933*n^2+1863*n^3);
%F c2 = 5248800*(n-5/3)*(n-4/3)*(n-1/9);
%F c3 = 9*n^2*(n-10/9);
%F a(n) ~ 2^(2*n - 1) * 3^(3*n - 1/2) * 5^(2*n + 1/2) / (Pi*n). - _Vaclav Kotesovec_, Apr 18 2018
%p a := n -> 36^n*(3*n)!/n!^3*hypergeom([-2*n, n+1/2], [n+1], -2/3):
%p seq(simplify(a(n)), n=0..10); # _Peter Luschny_, Apr 19 2018
%t c1=12*(-230+2259*n-3933*n^2+1863*n^3);c2=5248800*(n-5/3)*(n-4/3)*(n-1/9);c3=9*n^2*(n-10/9);a[0]=1;a[1]=492;a[n0_]:=ReplaceAll[(c1/c3)*a[n0-1]+(c2/c3)*a[n0-2],n->n0];
%t b[NN_]:=Expand[Total[Flatten[#]]&/@Table[Binomial[3*n,n]*Binomial[2*n,k2]*Binomial[2*n,k1]*Binomial[2*n,3*n-k1-k2]*((4+Sqrt[15])^(2*n-k1))*((4-Sqrt[15])^(2*n-k2)),{n,0,NN},{k1,0,2*n},{k2,0,2*n}]]; ({#,SameQ[#,a/@Range[0,10]]}&@b[10])[[1]]
%t Table[Binomial[3*n, n] * SeriesCoefficient[(1 + 9*z + 9*z^2 + z^3)^(2*n), {z, 0, 3*n}], {n, 0, 15}] (* _Vaclav Kotesovec_, Apr 18 2018 *)
%Y Cf. A002894, A113424, A006480, A000897. Factors: A005809, A300057. Real Period: A295870.
%K nonn
%O 0,2
%A _Bradley Klee_, Feb 23 2018
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