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Partial binomial sums of binomial(3n,n)/(2n+1) = A001764(n).
21

%I #19 Jul 25 2021 20:58:58

%S 1,2,6,25,126,704,4183,25897,165166,1077520,7156352,48222354,

%T 328859011,2265428728,15740837575,110187356134,776336572878,

%U 5501042194580,39177463572112,280277949384146,2013277273220064,14514764553512488,104993261648226446

%N Partial binomial sums of binomial(3n,n)/(2n+1) = A001764(n).

%H Nathaniel Johnston, <a href="/A188687/b188687.txt">Table of n, a(n) for n = 0..400</a>

%H Paul Barry, <a href="https://arxiv.org/abs/2104.01644">Centered polygon numbers, heptagons and nonagons, and the Robbins numbers</a>, arXiv:2104.01644 [math.CO], 2021.

%F a(n) = Sum_{k=0..n} binomial(n,k)*binomial(3k,k)/(2k+1).

%F G.f.: (2/sqrt(3x*(1-x)))*sin((1/3)*arcsin(3/2*sqrt(3*x/(1-x)))).

%F Recurrence: 2*n*(2*n+1)*a(n) = (39*n^2-35*n+8)*a(n-1) - 2*(n-1)*(33*n-32)*a(n-2) + 31*(n-2)*(n-1)*a(n-3). - _Vaclav Kotesovec_, Oct 20 2012

%F a(n) ~ 31^(n+3/2)/(3^4*2^(2*n+2)*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Oct 20 2012

%F G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x * (1 - x) * A(x)^3. - _Ilya Gutkovskiy_, Jul 25 2021

%t Table[Sum[Binomial[n,k]Binomial[3k,k]/(2k+1),{k,0,n}],{n,0,22}]

%o (Maxima) makelist(sum(binomial(n,k)*binomial(3*k,k)/(2*k+1),k,0,n),n,0,20);

%Y Cf. A005809, A001764, A188675, A188676, A104859, A188678, A188679, A188680, A188681, A188682, A188683, A188684, A188685, A188686.

%K nonn,easy

%O 0,2

%A _Emanuele Munarini_, Apr 08 2011