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A038665 Convolution of A007054 (super ballot numbers) with A000984 (central binomial coefficients).

%I A038665 #30 May 16 2022 02:49:04
%S A038665 3,8,25,84,294,1056,3861,14300,53482,201552,764218,2912168,11143500,
%T A038665 42791040,164812365,636438060,2463251010,9552774000,37112526990,
%U A038665 144410649240,562724141460,2195581527360,8576490341250,33537507830424
%N A038665 Convolution of A007054 (super ballot numbers) with A000984 (central binomial coefficients).
%H A038665 Vincenzo Librandi, <a href="/A038665/b038665.txt">Table of n, a(n) for n = 0..200</a>
%H A038665 Guo-Niu Han, <a href="/A196265/a196265.pdf">Enumeration of Standard Puzzles</a>, 2011. [Cached copy]
%H A038665 Guo-Niu Han, <a href="https://arxiv.org/abs/2006.14070">Enumeration of Standard Puzzles</a>, arXiv:2006.14070 [math.CO], 2020.
%F A038665 a(n) = (n+3)*C(n+1) with C(n) the Catalan numbers A000108.
%F A038665 G.f.: c(x)*(4 - c(x))/sqrt(1 - 4*x) with c(x) the g.f. for the Catalan numbers.
%F A038665 From _Amiram Eldar_, May 16 2022: (Start)
%F A038665 Sum_{n>=0} 1/a(n) = 41/6 - 64*Pi/(9*sqrt(3)) + 2*Pi^2/3.
%F A038665 Sum_{n>=0} (-1)^n/a(n) = 57/10 - 256*log(phi)/(5*sqrt(5)) + 24*log(phi)^2, where phi is the golden ratio (A001622). (End)
%p A038665 seq((n+3)*binomial(2*n+2, n+1)/(n+2), n=0..24); # _Zerinvary Lajos_, Dec 08 2008
%t A038665 Table[(n + 3) (CatalanNumber[n + 1]), {n, 0, 30}] (* _Vincenzo Librandi_, Sep 11 2016 *)
%o A038665 (Magma) [(n+3)*Catalan(n+1): n in [0..30]]; // _Vincenzo Librandi_, Sep 11 2016
%Y A038665 Cf. A000108, A000777, A000984, A001622, A007054.
%K A038665 easy,nonn
%O A038665 0,1
%A A038665 _Wolfdieter Lang_