A208425 Expansion of Sum_{n>=0} (3*n)!/n!^3 * x^(2*n)/(1-x)^(3*n+1).

1, 1, 7, 25, 151, 751, 4411, 24697, 146455, 862351, 5195257, 31392967, 191815339, 1177508515, 7276161907, 45154764025, 281492498455, 1761076827895, 11055132835705, 69600761349175, 439370198255401, 2780265190892641, 17631718101804517, 112038660509078695

Offset: 0

Author

Paul D. Hanna, Feb 26 2012

Keywords

• nonn

Comments

Compare g.f. to: Sum_{n>=0} (3*n)!/n!^3 * x^(2*n)/(1-2*x)^(3*n+1), which is a g.f. of the Franel numbers (A000172).

From Zhi-Wei Sun, Nov 12 2016: (Start)

Conjecture: (i) For any prime p > 3 and positive integer n, the number (a(p*n)-a(n))/(p*n)^3 is always a p-adic integer.

(ii) For any prime p == 1 (mod 3), we have Sum_{k=0..p-1}a(k) == C(2(p-1)/3,(p-1)/3) (mod p^2). For any prime p == 2 (mod 3), we have Sum_{k=0..p-1}a(k) == 2p/C(2(p+1)/3,(p+1)/3) (mod p^2).

We have proved part (i) of this conjecture for n = 1. (End)

Diagonal of rational functions 1/(1 - x*y - y*z - x*z - x*y*z), 1/(1 - x*y + y*z + x*z - x*y*z). - Gheorghe Coserea, Jul 03 2018

Number of paths from (0,0,0) to (n,n,n) using steps (1,1,0), (1,0,1), (0,1,1), and (1,1,1). - William J. Wang, Dec 07 2020

Diagonal of the rational function 1/(1 - (x^2 + y^2 + z^2 + x*y*z)). - Seiichi Manyama, Jul 04 2025

Examples

			G.f.: A(x) = 1 + x + 7*x^2 + 25*x^3 + 151*x^4 + 751*x^5 + 4411*x^6 +...
where
A(x) = 1/(1-x) + 6*x^2/(1-x)^4 + 90*x^4/(1-x)^7 + 1680*x^6/(1-x)^10 + 34650*x^8/(1-x)^13 + 756756*x^10/(1-x)^16 +...
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..1000
• A. Bostan, S. Boukraa, J.-M. Maillard and J.-A. Weil, Diagonals of rational functions and selected differential Galois groups, arXiv preprint arXiv:1507.03227 [math-ph], 2015.
• Hao Pan and Zhi-Wei Sun, Supercongruences for central trinomial coefficients, arXiv:2012.05121 [math.NT], 2020.
• Zhi-Wei Sun, Supercongruences involving Lucas sequences, arXiv:1610.03384 [math.NT], 2016.

Crossrefs

Cf. A000172, A001850, A208426, A244973, A274783.

Cf. A081798, A344560.

Programs

Maple

series(hypergeom([1/3, 2/3], [1], 27*x^2/(1 - x)^3)/(1 - x), x=0, 25): seq(coeff(%, x, n), n=0..23);  # Mark van Hoeij, May 20 2013
a := n -> hypergeom([1/2 - n/2, -n/2, n + 1], [1, 1], 4); seq(simplify(a(n)), n=0..23);  # Peter Luschny, Jan 11 2025

Mathematica

nmax = 20; CoefficientList[Series[Sum[(3*n)!/n!^3 * x^(2*n)/(1-x)^(3*n+1), {n, 0, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jul 05 2016 *)

PARI

{a(n)=polcoeff(sum(m=0,n, (3*m)!/m!^3*x^(2*m)/(1-x+x*O(x^n))^(3*m+1)),n)}
for(n=0,25,print1(a(n),", "))

Formula

Conjecture: n^2*(3*n-5)*a(n) +(-9*n^3+24*n^2-17*n+4) *a(n-1) -(3*n-4) *(24*n^2-56*n+27)*a(n-2) -(3*n-2)*(n-2)^2*a(n-3)=0. - R. J. Mathar, Mar 10 2016

a(n) ~ sqrt(1/2 + sqrt(13)*cos(arctan(53*sqrt(3)/19)/3)/6) * (1 + 6*cos(Pi/9))^n / (Pi*n). - Vaclav Kotesovec, Jul 05 2016

It is easy to show that a(n) = Sum_{k=0..n}C(n,k)*C(n-k,k)*C(n+k,k) = Sum_{k=0..n}C(n+k,k)*C(n,2k)*C(2k,k). By this formula and the Zeilberger algorithm, we confirm the recurrence conjectured by R. J. Mathar. - Zhi-Wei Sun, Nov 12 2016

G.f. y=A(x) satisfies: 0 = x*(x + 2)*(x^3 + 24*x^2 + 3*x - 1)*y'' + (3*x^4 + 56*x^3 + 147*x^2 + 12*x - 2)*y' + (x^3 + 9*x^2 + 42*x + 2)*y. - Gheorghe Coserea, Jul 03 2018

a(n) = hypergeom([1/2 - n/2, -n/2, n + 1], [1, 1], 4). - Peter Luschny, Jan 11 2025