A307618 A Calabi-Yau period integral: a(n) = C(4*n,2*n)*C(2*n,n)^3.
1, 48, 15120, 7392000, 4414410000, 2956651746048, 2133278987583744, 1621682968820428800, 1281351259836532170000, 1043032815185819858400000, 869343653096068540955685120, 738637974389826550020188712960, 637665137404661719206664998969600
Offset: 0
Keywords
Links
- G. Almkvist et al., Tables of Calabi-Yau Equations, arXiv:math/0507430 [math.AG], 2005-2010, p. 10.
- G. Almkvist et al., Raw Calabi-Yau Period Data, Johannes Gutenberg Universität Mainz, 2019, case 1.6.
Crossrefs
Programs
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Mathematica
Binomial[4*#,2*#]*Binomial[2*#,#]^3&/@Range[0,10]
Formula
G.f.: 4F3({1/4, 3/4, 1/2, 1/2}, {1, 1, 1}, 1024*x).
Define the period integral:
dt(x) = dz1*dz2*dz3/sqrt(1-32*x*cos(z1)*cos(z2)*cos(z3)).
T(x)=1/(2*Pi)^3*Integral_{0..2*Pi,0..2*Pi,0..2*Pi} dt(x),
the Picard-Fuchs coefficients:(c0,c1,c2,c3,c4)=
(768*x, 14592*x^2-1, x*(25344*x^2-7), 2*x^2*(5120*x^2-3), x^3*(32*x-1)*(32*x+1)),
and the certificate function:
G(z1,z2,z3)=(16*sin(z1)*(
48*x*cos(z1)
+ cos(z2)*cos(z3)
+ 48*x*cos(z1)*(cos(z3)^2 + cos(z2)^2)
+ 2304*x^2*cos(z1)^2*cos(z2)*cos(z3)
+ 80*x*cos(z1)*cos(z2)^2*cos(z3)^2
+ 384*x^2*cos(z1)^2*(cos(z2)*cos(z3)^3 + cos(z2)^3*cos(z3))
+ 256*x^2*cos(z1)^2*cos(z2)^3*cos(z3)^3)
)/(3*(1 - 32*x*cos(z1)*cos(z2)*cos(z3))^(7/2)),
Then: 0 = Sum_{n=0..4}cn*d^n/dx^n dt(x) + d/dz1 G(z1,z2,z3) + d/dz2 G(z2,z3,z1) + d/dz3 G(z3,z1,z2), thus: 0 = Sum_{n=0..4} cn*d^n/dx^n T(x).
Furthermore, let (a1,a2,a3)=(c1,c2,c3)/c0, then also: 0 = (1/2)*a2*a3 - (1/8)*a3^3 + d/dx(a2) - (3/4)*a3*d/dx(a3) - (1/2)*d^2/dx^2(a3) - a1.
D-finite with recurrence: n^4*a(n) -16*(4*n-1)*(4*n-3)*(-1+2*n)^2*a(n-1)=0. - R. J. Mathar, Jan 27 2020
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