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a(n) = (3^n/n!) * Product_{k=0..n-1} (3*k - 2).
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%I #32 Oct 15 2024 15:42:16

%S 1,-6,-9,-36,-189,-1134,-7371,-50544,-360126,-2640924,-19806930,

%T -151252920,-1172210130,-9197341020,-72921775230,-583374201840,

%U -4703454502335,-38180983607190,-311811366125385,-2560135427134740,-21121117273861605,-175003543126281870,-1455711290550435555

%N a(n) = (3^n/n!) * Product_{k=0..n-1} (3*k - 2).

%H G. C. Greubel, <a href="/A004989/b004989.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n) ~ -(2/3)*Gamma(1/3)^-1*n^(-5/3)*3^(2*n)*(1 + (5/9)*n^-1 + ...).

%F G.f.: (1-9*x)^(2/3).

%F D-finite with recurrence: n*a(n) +3*(-3*n+5)*a(n-1)=0. - _R. J. Mathar_, Jan 17 2020

%F Sum_{n>=0} 1/a(n) = 99/128 - 5*sqrt(3)*Pi/512 - 15*log(3)/512. - _Amiram Eldar_, Dec 02 2022

%F From _Peter Bala_, Oct 14 2024: (Start)

%F a(n) = -1/(sqrt(3)*Pi) * 9^n * Gamma(2/3)*Gamma(n-2/3)/Gamma(n+1).

%F For n >= 1, a(n) = - Integral_{x = 0..9} x^n * w(x) dx, where w(x) = sqrt(3)/(2*Pi) * x^(1/3)*(9 - x)^(2/3)/x^2. (End)

%p a:= n-> (3^n/n!)*product(3*k-2, k=0..n-1); seq(a(n), n=0..25); # _G. C. Greubel_, Aug 22 2019

%t Table[9^n*Pochhammer[-2/3, n]/n!, {n,0,25}] (* _G. C. Greubel_, Aug 22 2019 *)

%o (PARI) a(n)=if(n<0,0,prod(k=0,n-1,3*k-2)*3^n/n!)

%o (Magma) [1] cat [3^n*(&*[3*k-2: k in [0..n-1]])/Factorial(n): n in [1..25]]; // _G. C. Greubel_, Aug 22 2019

%o (Sage) [9^n*rising_factorial(-2/3, n)/factorial(n) for n in (0..25)] # _G. C. Greubel_, Aug 22 2019

%o (GAP) List([0..25], n-> 3^n*Product([0..n-1], k-> 3*k-2)/Factorial(n) ); # _G. C. Greubel_, Aug 22 2019

%Y Cf. A004988, A004990, A004991, A004992, A025748, A155579, A185047.

%K sign,easy

%O 0,2

%A Joe Keane (jgk(AT)jgk.org)