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a(n) = (2/(3*n-1))*binomial(3*n,n).
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%I #38 Oct 10 2024 15:38:08

%S -2,3,6,21,90,429,2184,11628,63954,360525,2072070,12096045,71524440,

%T 427496076,2578547760,15675792072,95951017602,590842763469,

%U 3657598059570,22749427475775,142096423925610,890949529108485,5605635937900320,35380499289211440,223951032734902200

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

%C 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

%H Andrew Howroyd, <a href="/A024485/b024485.txt">Table of n, a(n) for n = 0..500</a>

%F G.f.: 3*g-2 where g*(1-g)^2 = x. - _Mark van Hoeij_, Nov 09 2011

%F a(n) = 2*A005809(n)/(3*n-1). - _R. J. Mathar_, Apr 27 2020

%F 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

%F a(n) = A006013(n-1)/3 for n >= 1. - _Lucas A. Brown_, Aug 21 2020

%F From _Karol A. Penson_, Dec 18 2023: (Start)

%F 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).

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

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

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

%F 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).

%F 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)

%F G.f.: -2*hypergeometric2F1([1/3,-1/3],[1/2],27*z/4). - _Karol A. Penson_, Oct 08 2024

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

%t Table[2/(3n-1) Binomial[3n,n],{n,0,20}] (* _Harvey P. Dale_, Nov 21 2015 *)

%o (PARI) a(n) = (2/(3*n-1))*binomial(3*n, n); \\ _Michel Marcus_, May 10 2020

%K sign

%O 0,1

%A _Clark Kimberling_

%E Terms a(21) and beyond from _Andrew Howroyd_, May 10 2020