A275051 Expansion of 3F2([1/9, 4/9, 5/9], [1/3,1], 729*x).
1, 60, 20475, 9373650, 4881796920, 2734407111744, 1605040007778900, 973419698810097000, 604759111060745718000, 382741738086972337402560, 245810413547242455520545552, 159759730493918131135425965280, 104861901534978616465850670348000
Offset: 0
Keywords
Examples
1 + 60*x + 20475*x^2 + 9373650*x^3 + ...
Links
- Gheorghe Coserea, Table of n, a(n) for n = 0..200
- A. Bostan, S. Boukraa, G. Christol, S. Hassani, and J-M. Maillard Ising n-fold integrals as diagonals of rational functions and integrality of series expansions: integrality versus modularity, arXiv:1211.6031 [math-ph], 2012.
- S. Boukraa, S. Hassani, J-M. Maillard, and J-A. Weil, Differential algebra on lattice Green functions and Calabi-Yau operators (unabridged version), arXiv:1311.2470 [math-ph], 2013.
Programs
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Mathematica
CoefficientList[Series[HypergeometricPFQ[{1/9, 4/9, 5/9}, {1/3,1}, 729*x], {x, 0, 15}], x] (* Vaclav Kotesovec, Jul 28 2016 *) a[n_] := FullSimplify[(729^n Cos[Pi/18] Gamma[1/3] Gamma[1/9 + n] Gamma[4/9 + n] Gamma[5/9 + n])/(Pi Gamma[1/9] Gamma[1/3 + n] n!^2)] (* Benedict W. J. Irwin, Aug 05 2016 *) -
PARI
\\ system("wget http://www.jjj.de/pari/hypergeom.gpi"); read("hypergeom.gpi"); N = 21; x = 'x + O('x^N); Vec(hypergeom_sym([1/9, 4/9, 5/9], [1/3,1], 729*x, N)) -
PARI
my(x = 'x + O('x^20)); Vec(hypergeom([1/9, 4/9, 5/9], [1/3,1], 729*x)) \\ Michel Marcus, Apr 11 2023
Formula
G.f.: hypergeom([1/9, 4/9, 5/9], [1/3,1], 729*x).
From Vaclav Kotesovec, Jul 28 2016: (Start)
Recurrence: n^2*(3*n - 2)*a(n) = 3*(9*n - 8)*(9*n - 5)*(9*n - 4)*a(n-1).
a(n) ~ Gamma(1/3) * cos(Pi/18) * 3^(6*n) / (Pi * Gamma(1/9) * n^(11/9)).
(End)
a(n) = 729^n*cos(Pi/18)*Gamma(1/3)*Gamma(1/9+n)*Gamma(4/9+n)*Gamma(5/9+n) /(Pi*Gamma(1/9)*Gamma(1/3+n)*n!^2). - Benedict W. J. Irwin, Aug 05 2016
From Karol A. Penson, Apr 11 2023: (Start)
a(n) = Integral_{x=0..729} x^n*W(x), where
W(x) = W1(x) + W2(x) + W3(x), and
W1(x) = (2*cos(Pi/18)*3^(1/3)*2^(4/9)*sqrt(Pi)*Gamma(13/18)*hypergeom([1/9, 1/9, 7/9], [5/9, 2/3], x/729))/(9*Gamma(2/3)^2*Gamma(1/9)*Gamma(8/9)^2*x^(8/9));
W2(x) = cos(Pi/18)*2^(1/9)*Gamma(2/9)*Gamma(1/18)*hypergeom([4/9, 4/9, 10/9], [8/9, 4/3], x/729)/(162*Gamma(2/3)*Gamma(1/9)*Pi^(3/2)*x^(5/9));
W3(x) = cos(Pi/18)*3^(1/6)*2^(4/9)*Gamma(5/18)*Gamma(-1/18)*hypergeom([5/9, 5/9, 11/9], [10/9, 13/9], x/729)/(324*Gamma(2/3)*Gamma(1/9)*Pi*Gamma(4/9)*x^(4/9)).
This integral representation is unique as W(x) is the solution of the Hausdorff power moment problem. Using only the definition of a(n), W(x) can be proven to be positive. W(x) is singular at x = 0, with the singularity x^(-4/9), and for x > 0 is monotonically decreasing to zero at x = 729. At x = 729 the first derivative of W(x) is infinite. (End)
Comments