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A273055 a(n) = Sum_{k=0..n} binomial(2*k, k) * binomial(2*n+1, 2*k). 6

%I

%S 1,7,51,393,3139,25653,212941,1787607,15134931,128996853,1105350729,

%T 9513228123,82176836301,712070156203,6186675630819,53876592856681,

%U 470139239360787,4109922421017093,35986168879543609,315544068167601787,2770417140954208377,24352194654450483759

%N a(n) = Sum_{k=0..n} binomial(2*k, k) * binomial(2*n+1, 2*k).

%C These are the central coefficients of the trinomial irregular triangle A027907 [Comtet, pp. 77-78, with references] for odd-indexed rows. For the central coefficients of the even-indexed rows see A082758. - _Wolfdieter Lang_, Apr 19 2018

%D L. Comtet, Advanced Combinatorics, Reidel, 1974., pp. 77-78. (In the integral formula on p. 77, a left bracket is missing for the cosine argument.)

%H Indranil Ghosh, <a href="/A273055/b273055.txt">Table of n, a(n) for n = 0..826</a>

%F a(n) = 4^(2*n+1)*JacobiP(2*n+1, -2*n-3/2, -2*n-3/2, -1/2).

%F a(n) = GegenbauerC(2*n+1, -2*n-1, -1/2).

%F a(n) = hypergeom([-n-1/2, -n], [1], 4).

%F a(n) = (2*n+1)!*( [x^(2*n+1)] exp(x)*BesselI(0,2*x) ).

%F a(n) = [x^(2*n+1)] (1-2*x-3*x^2)^(-1/2).

%F a(n) = [x^(2*n+1)] (1+x+x^2)^(2*n+1).

%F a(n) = ((4*n-1)*(20*n^2-10*n-3)*a(n-1)-9*(n-1)*(2*n-1)*(4*n+1)*a(n-2))/(n*(2*n+1)*(4*n-3)) for n>1.

%F a(n) = A002426(2*n+1).

%F a(n) ~ 3^(2*n + 3/2) / sqrt(8*Pi*n). - _Vaclav Kotesovec_, Feb 16 2017

%F From _Peter Bala_, Mar 16 2018: (Start)

%F a(n) = sqrt(-3)^(2*n+1)*P(2*n+1,1/sqrt(-3)), where P(n,x) is the Legendre polynomia1 of degree n.

%F a(n) = 1/C(2*n+1,n)*Sum_{k = 0..n} C(n,k)*C(n+k+1,k)* C(2*n+2*k+1,n+k+1)*(-3)^(n-k). Cf. A082758.

%F a(n) = (-3)^n*hypergeom([-n, n + 3/2], [1], 4/3).

%F (End)

%F From _Wolfdieter Lang_, Apr 19 2018: (Start)

%F a(n) = (2/Pi)*Integral_{phi = 0..Pi/2} (sin(3*phi))/sin(phi))^(2*n+1) [Comtet, p. 77, q = 3, n = k -> 2*n+1] = (2/Pi)*Integral_{x=0..2} (x^2 - 1)^(2*n+1) / sqrt(4-x^2) (with x = 2*cos(phi).

%F a(n) = 3^(2*n+1)*Sum_{k=0..2*n+1} binomial(2*n+1, k)*binomial(2*k, k)*(-1/3)^k = 3^(2*n+1)*hypergeometric([-(2*n+1), 1/2], [1], 4/3). See the version in Bala's formulas, also for the Legendre polynomial version. (End)

%F G.f.: sqrt( 2 / ((1-10*x+9*x^2)*(1-3*x+sqrt(1-10*x+9*x^2))) ). - _Seiichi Manyama_, Aug 25 2020

%p a := n -> GegenbauerC(2*n+1, -2*n-1, -1/2): seq(simplify(a(n)), n=0..21);

%t Table[Hypergeometric2F1[-n-1/2, -n, 1, 4], {n, 0, 21}]

%t Table[GegenbauerC[2 n + 1, -2 n - 1, -1/2], {n, 0, 21}] (* _Michael De Vlieger_, May 14 2016 *)

%o (Sage)

%o def a():

%o a, b, n = 1, 7, 2

%o yield a

%o while True:

%o yield b

%o a, b = b, ((4*n-1)*(20*n^2-10*n-3)*b-9*(n-1)*(2*n-1)*(4*n+1)*a)//(n*(2*n+1)*(4*n-3))

%o n += 1

%o A = a()

%o [next(A) for i in range(22)]

%o (PARI) a(n) = sum(k=0, n, binomial(2*k, k) * binomial(2*n+1, 2*k)); \\ _Michel Marcus_, Mar 01 2020

%o (PARI) N=40; x='x+O('x^N); Vec(sqrt(2/((1-10*x+9*x^2)*(1-3*x+sqrt(1-10*x+9*x^2))))) \\ _Seiichi Manyama_, Aug 25 2020

%Y Cf. A002426, A027907, A082758.

%K nonn,easy

%O 0,2

%A _Peter Luschny_, May 14 2016

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Last modified June 13 11:29 EDT 2021. Contains 344990 sequences. (Running on oeis4.)