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

0,2

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

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.

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

Cf. A268545-A268555.

Sequence in context: A291912 A001460 A003794 * A068295 A177643 A123482

Adjacent sequences: A275048 A275049 A275050 * A275052 A275053 A275054

Keyword

nonn

Author

Gheorghe Coserea, Jul 19 2016

Status

approved