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a(n) = binomial(2n,n)*n*(2n+1)/2.
7

%I #28 Sep 08 2022 08:44:59

%S 0,3,30,210,1260,6930,36036,180180,875160,4157010,19399380,89237148,

%T 405623400,1825305300,8143669800,36064823400,158685222960,

%U 694247850450,3022020054900,13095420237900,56517076816200,243023430309660,1041528987041400

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

%H G. C. Greubel, <a href="/A051133/b051133.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n) = (1/2) * A000911(n-1).

%F a(n) = (1/2)*A000984(n+1)*A000217(n). - _Zerinvary Lajos_, May 05 2007

%F a(n) = 3*A002802(n-1). - _Zerinvary Lajos_, Jun 02 2007

%F (-n+1)*a(n) + 2*(2*n+1)*a(n-1) = 0. - _R. J. Mathar_, Feb 05 2013

%F G.f.: 3*x * (1 - 4*x)^(-5/2). - _Michael Somos_, Sep 09 2013

%F Sum_{n>=1} 1/a(n) = 4 - 2*Pi/sqrt(3). - _Amiram Eldar_, Oct 22 2020

%e G.f. = 3*x + 30*x^2 + 210*x^3 + 1260*x^4 + 6930*x^5 + 36036*x^6 + ...

%p seq(binomial(2*n,n)*binomial(n,(n-2))/2, n=1..23); # _Zerinvary Lajos_, May 05 2007

%t a[ n_]:= SeriesCoefficient[ 3x(1-4x)^(-5/2), {x, 0, n}]; (* _Michael Somos_, Sep 09 2013 *)

%t Table[Binomial[2*n, n]*n*(2*n + 1)/2, {n, 0, 22}] (* _Amiram Eldar_, Oct 22 2020 *)

%o (PARI) {a(n) = if( n<1, 0, (2*n + 1)! / (2 * n! *(n-1)!))}; /* _Michael Somos_, Sep 09 2013 */

%o (PARI) {a(n) = 2^(n+2) * polcoeff( pollegendre( n+3), n-1)}; /* _Michael Somos_, Sep 09 2013 */

%o (Magma) [Binomial(2*n,n)*n*(2*n+1)/2: n in [0..25]]; // _G. C. Greubel_, Feb 10 2019

%o (Sage) [binomial(2*n,n)*n*(2*n+1)/2 for n in (0..25)] # _G. C. Greubel_, Feb 10 2019

%Y Cf. A000911, A000984, A000217, A002802.

%K nonn,easy

%O 0,2

%A _Wouter Meeussen_