%I #76 Nov 12 2022 10:47:36
%S 1,2,9,48,275,1638,9996,62016,389367,2466750,15737865,100975680,
%T 650872404,4211628008,27341497800,177996090624,1161588834303,
%U 7596549816030,49772989810635,326658445806000,2147042307851595,14130873926790390,93115841412899760
%N Central coefficients T(2n,n) of the Catalan triangle A033184.
%C A033184 is the Riordan array (c(x), x*c(x)), c(x) the g.f. of A000108.
%C Number of standard Young tableaux of shape [2n, n]. Also the number of binary words with 2n 1's and n 0's such that for every prefix the number of 1's is >= the number of 0's. The a(2) = 9 words are: 101011, 101101, 101110, 110011, 110101, 110110, 111001, 111010, 111100. - _Alois P. Heinz_, Aug 15 2012
%C Number of lattice paths from (0,0) to (2n,n) not above y=x. - _Ran Pan_, Apr 08 2015
%H Alois P. Heinz, <a href="/A174687/b174687.txt">Table of n, a(n) for n = 0..370</a>
%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL16/Barry1/barry242.html">On the Central Coefficients of Riordan Matrices</a>, Journal of Integer Sequences, 16 (2013), #13.5.1.
%H D. Kruchinin and V. Kruchinin, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL15/Kruchinin/kruchinin5.html">A Method for Obtaining Generating Function for Central Coefficients of Triangles</a>, Journal of Integer Sequence, Vol. 15 (2012), article 12.9.3.
%H Ran Pan, <a href="http://www.math.ucsd.edu/~projectp/warmups/eL.html">Exercise L</a>, Project P
%H Anssi Yli-Jyrä and Carlos Gómez-Rodríguez, <a href="https://arxiv.org/abs/1706.03357">Generic Axiomatization of Families of Noncrossing Graphs in Dependency Parsing</a>, arXiv:1706.03357 [cs.CL], 2017.
%F a(n) = (n+1)*C(3*n, n)/(2n+1) = (n+1)*[x^(n+1)]( Rev(x/c(x)) ) = (n+1)*A001764(n), c(x) the g.f. of A000108.
%F G.f.: A(x) = sin(arcsin((3^(3/2)*sqrt(x))/2)/3)/(sqrt(3)*sqrt(x)) + cos(arcsin((3^(3/2)* sqrt(x))/2)/3)/(2*sqrt(1-(27*x)/4)). - _Vladimir Kruchinin_, May 25 2012
%F 2*n*(2*n+1)*a(n) = 3*(13*n^2 -10*n +1)*a(n-1) -9*(3*n-4)*(3*n-5)*a(n-2). - _R. J. Mathar_, Nov 24 2012
%F a(n) = [x^n] ((1 - sqrt(1 - 4*x))/(2*x))^(n+1). - _Ilya Gutkovskiy_, Nov 01 2017
%p a:= n-> binomial(3*n,n)*(n+1)/(2*n+1):
%p seq(a(n), n=0..25); # _Alois P. Heinz_, Aug 15 2012
%t Table[Binomial[3n,n](n+1)/(2n+1), {n, 0, 25}] (* _Vincenzo Librandi_, Apr 08 2015 *)
%o (Magma) [(n+1)*Binomial(3*n,n)/(2*n+1): n in [0..25]]; // _Vincenzo Librandi_, Apr 08 2015
%o (SageMath) [(n+1)*binomial(3*n,n)/(2*n+1) for n in range(31)] # _G. C. Greubel_, Nov 09 2022
%o (PARI) a(n) = (n+1)*binomial(3*n,n)/(2*n+1); \\ _Michel Marcus_, Nov 12 2022
%Y Cf. A000108, A001764, A033184.
%Y Column k=2 of A214776.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Mar 27 2010