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A275051
Expansion of 3F2([1/9, 4/9, 5/9], [1/3,1], 729*x).
11
1, 60, 20475, 9373650, 4881796920, 2734407111744, 1605040007778900, 973419698810097000, 604759111060745718000, 382741738086972337402560, 245810413547242455520545552, 159759730493918131135425965280, 104861901534978616465850670348000
OFFSET
0,2
COMMENTS
"One may consider the following conjecture: all the irreducible factors of the minimal order linear differential operator annihilating a diagonal of a rational function should be homomorphic to their adjoint (possibly on an algebraic extension). [...]
"If our conjecture above was correct, this would be a way to show that the series cannot be the diagonal of a rational function." (See Boukraa link.)
LINKS
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.
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)
EXAMPLE
1 + 60*x + 20475*x^2 + 9373650*x^3 + ...
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Gheorghe Coserea, Jul 19 2016
STATUS
approved