|
%I #59 Dec 19 2023 17:44:57
%S 1,6,60,700,8820,116424,1585584,22084920,312869700,4491418360,
%T 65166397296,953799087696,14062422446800,208618354980000,
%U 3111393751416000,46619049708716400,701342468412012900
%N a(n) = (2*n)!*(2*n+1)! /((n+1)! *n!^3).
%C a(n) = (n+1) * A000891(n) = A248045(n+1) / A000142(n). - _Reinhard Zumkeller_, Sep 30 2014
%C This sequence is one half of the odd part of the bisection of A241530. The even part is given in A002894. - _Wolfdieter Lang_, Sep 06 2016
%D E. R. Hansen, A Table of Series and Products, Prentice-Hall, Englewood Cliffs, NJ, 1975, p. 96.
%H Vincenzo Librandi, <a href="/A000894/b000894.txt">Table of n, a(n) for n = 0..180</a>
%H Ling Gao, <a href="http://hdl.handle.net/20.500.12680/h989rb533">Graph assembly for spider and tadpole graphs</a>, Master's Thesis, Cal. State Poly. Univ. (2023). See p. 36.
%H Pedro J. Miana and Natalia Romero, <a href="https://doi.org/10.1016/j.jnt.2010.01.018">Moments of combinatorial and Catalan numbers</a>, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Omega1 Remark 3 p. 1882.
%H Yidong Sun and Fei Ma, <a href="http://arxiv.org/abs/1305.2017">Four transformations on the Catalan triangle</a>, arXiv preprint arXiv:1305.2017 [math.CO], 2013 (see Omega_1).
%H Yidong Sun and Fei Ma, <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v21i1p33">Some new binomial sums related to the Catalan triangle</a>, Electronic Journal of Combinatorics 21(1) (2014), #P1.33.
%F a(n) = C(2*n+1,n)*C(2*n,n) = A001700(n)*A000984(n) = A000984(n)*A000984(n+1)/2, n>=0. - _Zerinvary Lajos_, Jan 23 2007
%F G.f.: (EllipticK(4*x^(1/2)) - EllipticE(4*x^(1/2)))/(4*x*Pi). - _Mark van Hoeij_, Oct 24 2011
%F n*(n+1)*a(n) -4*(2*n-1)*(2*n+1)*a(n-1)=0. - _R. J. Mathar_, Sep 08 2013
%F a(n) = A103371(2*n,n) = A132813(2*n,n). - _Reinhard Zumkeller_, Apr 04 2014
%F 0 = a(n)*(+65536*a(n+2) - 23040*a(n+3) + 1400*a(n+4)) + a(n+1)*(-1536*a(n+2) + 1184*a(n+3) - 90*a(n+4)) + a(n+2)*(-24*a(n+2) - 6*a(n+3) + a(n+4)) for all n in Z. - _Michael Somos_, May 28 2014
%F 0 = a(n+1)^3 * (+256*a(n) - 6*a(n+1) + a(n+2)) + a(n) * a(n+1) * a(n+
%F 2) * (-768*a(n) - 20*a(n+1) - 3*a(n+2)) + 90*a(n)^2*a(n+2)^2 for all n in Z. - _Michael Somos_, Sep 17 2014
%F a(n) = A241530(2n+1)/2, n >= 0. - _Wolfdieter Lang_, Sep 06 2016
%F a(n) ~ 2^(4*n+1)/(Pi*n). - _Ilya Gutkovskiy_, Sep 06 2016
%e G.f. = 1 + 6*x + 60*x^2 + 700*x^3 + 8820*x^4 + 116424*x^5 + ...
%p seq(binomial(2*n+1,n)*binomial(2*n,n), n=0..16); # _Zerinvary Lajos_, Jan 23 2007
%t a[ n_] := Binomial[2 n + 1, n] Binomial[2 n, n]; (* _Michael Somos_, May 28 2014 *)
%t a[ n_] := SeriesCoefficient[ (EllipticK[ 16 x] - EllipticE[ 16 x]) / (4 x Pi), {x, 0, n}]; (* _Michael Somos_, May 28 2014 *)
%t Table[(2 n)!*(2 n + 1)!/((n + 1)!*n!^3), {n, 0, 16}] (* _Michael De Vlieger_, Sep 06 2016 *)
%o (Magma) [Factorial(2*n)*Factorial(2*n+1) /(Factorial(n+1)* Factorial(n)^3): n in [0..20]]; // _Vincenzo Librandi_, Oct 25 2011
%o (Haskell)
%o a000894 n = a132813 (2 * n) n -- _Reinhard Zumkeller_, Apr 04 2014
%o (PARI) {a(n) = binomial( 2*n + 1, n) * binomial( 2*n, n)}; /* _Michael Somos_, May 28 2014 */
%Y First differences of A082578. Cf. A002463.
%Y Cf. A001700, A000984.
%Y Cf. A000142, A000891, A248045, A002894, A241530.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_
| 