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%I #44 Feb 24 2023 02:04:02
%S 1,4,21,126,819,5616,40014,293436,2200770,16805880,130245570,
%T 1021926780,8102419470,64819355760,522606055815,4242331511910,
%U 34645707347265,284459491903860,2346790808206845,19444838125142430,161745698950048395,1350224965148230080
%N Expansion of 2F1( [1, 4/3]; [3]; 9*x).
%C Close to A003168.
%C Can be seen as a degree 3 analog of the Catalan numbers A000108 (which would be degree 2).
%H Vincenzo Librandi, <a href="/A185047/b185047.txt">Table of n, a(n) for n = 0..1000</a>
%F a(n) = -9^(n+1)*binomial(n+1/3, n+2). - _Karol A. Penson_, Nov 06 2015
%F a(n) = (1/(6*sqrt(3)*Pi))*Integral_{x = 0..9} x^n*x^(1/3)*(9 - x)^(2/3). Cf. A034164. - _Peter Bala_, Nov 17 2015
%F O.g.f.: (1 - (1-9*x)^(2/3) - 6*x)/(9*x^2).
%F D-finite with recurrence (n+2)*a(n) +3*(-3*n-1)*a(n-1)=0. - _R. J. Mathar_, Jul 27 2022
%F Sum_{n>=0} 1/a(n) = (3/512)*(92 + 15*Pi*sqrt(3) + 45*log(3)). - _Amiram Eldar_, Dec 18 2022
%F a(n) = ((n + 3)/3) * Product_{1 <= i <= j <= n} (2*i + j + 3)/(2*i + j - 1). - _Peter Bala_, Feb 22 2023
%p A185047:=n->-9^(n+1)*binomial(n+1/3, n+2): seq(A185047(n), n=0..30); # _Wesley Ivan Hurt_, Feb 16 2017
%t CoefficientList[Series[ HypergeometricPFQ[{1, 4/3}, {3}, 9 x], {x, 0, 20}], x]
%Y Cf. A000108, A003168, A034164.
%K nonn,easy
%O 0,2
%A _Olivier Gérard_, Feb 15 2011
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