A024485 a(n) = (2/(3*n-1))*binomial(3*n,n).

-2, 3, 6, 21, 90, 429, 2184, 11628, 63954, 360525, 2072070, 12096045, 71524440, 427496076, 2578547760, 15675792072, 95951017602, 590842763469, 3657598059570, 22749427475775, 142096423925610, 890949529108485, 5605635937900320, 35380499289211440, 223951032734902200

Offset: 0

Author

Clark Kimberling

Keywords

• sign

Comments

For n >= 1, a(n) is the number of lattice paths from (0,0) to (2n,n) using only the steps (1,0) and (0,1) and which do not touch the line y = x/2 except at the path's endpoints. - Lucas A. Brown, Aug 21 2020

Examples

			G.f. =  -2 + 3*x + 6*x^2 + 21*x^3 + 90*x^4 + 429*x^5 + 2184*x^6 + ... - _Michael Somos_, Nov 08 2024
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 0..500
• R. J. Mathar, The Eggenberger-Polya urn process: Probabilities of revisited ball ratios, vixra:2502.0097 (2025) Table 4

Crossrefs

Cf. A005809, A006013.

Programs

Maple

[seq((2/(3*n-1))*binomial(3*n,n), n=0..40)];

Mathematica

Table[2/(3n-1) Binomial[3n,n],{n,0,20}] (* Harvey P. Dale, Nov 21 2015 *)

PARI

a(n) = (2/(3*n-1))*binomial(3*n, n); \\ Michel Marcus, May 10 2020

Formula

G.f.: 3*g-2 where g*(1-g)^2 = x. - Mark van Hoeij, Nov 09 2011

a(n) = 2*A005809(n)/(3*n-1). - R. J. Mathar, Apr 27 2020

D-finite with recurrence: 2*n*(2*n-1)*a(n) -3*(3*n-2)*(3*n-4)*a(n-1)=0. - R. J. Mathar, Apr 27 2020

a(n) = A006013(n-1)/3 for n >= 1. - Lucas A. Brown, Aug 21 2020

From Karol A. Penson, Dec 18 2023: (Start)

G.f.: - (sqrt(1-27*z/4)+i*sqrt(27*z/4))^(2/3) - (sqrt(1-27*z/4)-i*sqrt(27*z/4))^(2/3), where i = sqrt(-1).

(G.f.)^3 = G satisfies the cubic equation:

-(27*z - 2)^3 + 3*(27*z + 1)*(27*z - 5)*G + 3*(-27*z+2)*G^2 + G^3 = 0.

a(n) = Integral_{x=0..27/4} x^n*W(x) dx, for n>=1, where

W(x) = -(3^(1/6)*(9+sqrt(3)*sqrt(27-4*x))^(1/3))*(-27*(2^(1/3))*(3^(1/6)) + 3*2^(1/3)*3^(2/3)*sqrt(27-4*x)-2*(9+sqrt(3)*sqrt(27-4*x))^(1/3)*sqrt(27-4*x)*x^(1/3) + 4*2^(1/3)*3^(1/6)*x)/(4*2^(2/3)*Pi*sqrt(27-4*x)*x^(5/3)), for x in (0, 27/4). For n=0, Integral_{x=0..27/4} W(x) dx diverges and is not suited to reproduce a(0).

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, and for x > 0 is monotonically decreasing to zero at x = 27/4. At x = 27/4 the first derivative of W(x) is infinite. (End)

G.f.: -2*hypergeometric2F1([1/3,-1/3],[1/2],27*z/4). - Karol A. Penson, Oct 08 2024

0 = 3*a(n)^2*(405*a(n+1) - 154*a(n+2))*(81*a(n+1) - 70*a(n+2)) + 4*a(n)*a(n+1)*a(n+2)*(111*a(n+1) - 742*a(n+2)) - 4*a(n+1)^2*(5*a(n+1) - 8*a(n+2))*(3*a(n+1) + 2*a(n+2)) for all n in Z. - Michael Somos, Nov 08 2024

Extensions

Terms a(21) and beyond from Andrew Howroyd, May 10 2020