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

%I A024485 #43 Feb 28 2025 06:22:03
%S A024485 -2,3,6,21,90,429,2184,11628,63954,360525,2072070,12096045,71524440,
%T A024485 427496076,2578547760,15675792072,95951017602,590842763469,
%U A024485 3657598059570,22749427475775,142096423925610,890949529108485,5605635937900320,35380499289211440,223951032734902200
%N A024485 a(n) = (2/(3*n-1))*binomial(3*n,n).
%C A024485 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 A024485 Andrew Howroyd, <a href="/A024485/b024485.txt">Table of n, a(n) for n = 0..500</a>
%H A024485 R. J. Mathar, <a href="https://vixra.org/abs/2502.0097">The Eggenberger-Polya urn process: Probabilities of revisited ball ratios</a>, vixra:2502.0097 (2025) Table 4
%F A024485 G.f.: 3*g-2 where g*(1-g)^2 = x. - _Mark van Hoeij_, Nov 09 2011
%F A024485 a(n) = 2*A005809(n)/(3*n-1). - _R. J. Mathar_, Apr 27 2020
%F A024485 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 A024485 a(n) = A006013(n-1)/3 for n >= 1. - _Lucas A. Brown_, Aug 21 2020
%F A024485 From _Karol A. Penson_, Dec 18 2023: (Start)
%F A024485 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 A024485 (G.f.)^3 = G satisfies the cubic equation:
%F A024485 -(27*z - 2)^3 + 3*(27*z + 1)*(27*z - 5)*G + 3*(-27*z+2)*G^2 + G^3 = 0.
%F A024485 a(n) = Integral_{x=0..27/4} x^n*W(x) dx, for n>=1, where
%F A024485 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 A024485 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 A024485 G.f.: -2*hypergeometric2F1([1/3,-1/3],[1/2],27*z/4). - _Karol A. Penson_, Oct 08 2024
%F A024485 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
%e A024485 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
%p A024485 [seq((2/(3*n-1))*binomial(3*n,n), n=0..40)];
%t A024485 Table[2/(3n-1) Binomial[3n,n],{n,0,20}] (* _Harvey P. Dale_, Nov 21 2015 *)
%o A024485 (PARI) a(n) = (2/(3*n-1))*binomial(3*n, n); \\ _Michel Marcus_, May 10 2020
%Y A024485 Cf. A005809, A006013.
%K A024485 sign
%O A024485 0,1
%A A024485 _Clark Kimberling_
%E A024485 Terms a(21) and beyond from _Andrew Howroyd_, May 10 2020