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%I #23 Dec 23 2022 07:50:20
%S 1,8,88,1232,20944,418880,9634240,250490240,7264216960,232454942720,
%T 8135922995200,309165073817600,12675768026521600,557733793166950400,
%U 26213488278846668800,1310674413942333440000,69465743938943672320000,3890081660580845649920000
%N a(n) = (3*n+5)!!!/5!!!.
%C Related to A008544(n+1) ((3*n+2)!!! triple factorials).
%C Row m=5 of the array A(4; m,n) := ((3*n+m)(!^3))/m(!^3), m >= 0, n >= 0.
%H G. C. Greubel, <a href="/A051605/b051605.txt">Table of n, a(n) for n = 0..378</a>
%F a(n) = ((3*n+5)(!^3))/5(!^3).
%F E.g.f.: 1/(1-3*x)^(8/3).
%F a(n) = 3^n*(n+5/3)!/(5/3)!. - _Paul Barry_, Sep 04 2005
%F a(n) = (3*n+5)*a(n-1). - _R. J. Mathar_, Nov 13 2012
%F Sum_{n>=0} 1/a(n) = 1 + 3*(9*e)^(1/3)*(Gamma(8/3) - Gamma(8/3, 1/3)). - _Amiram Eldar_, Dec 23 2022
%t RecurrenceTable[{a[0]==1,a[n]==(3n+5)a[n-1]},a,{n,20}] (* _Harvey P. Dale_, Oct 19 2013 *)
%t With[{nn = 30}, CoefficientList[Series[1/(1 - 3*x)^(8/3), {x, 0, nn}], x]*Range[0, nn]!] (* _G. C. Greubel_, Aug 15 2018 *)
%o (PARI) x='x+O('x^30); Vec(serlaplace(1/(1-3*x)^(8/3))) \\ _G. C. Greubel_, Aug 15 2018
%o (Magma) m:=30; R<x>:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!(1/(1-3*x)^(8/3))); [Factorial(n-1)*b[n]: n in [1..m]]; // _G. C. Greubel_, Aug 15 2018
%Y Cf. A032031, A007559(n+1), A034000(n+1), A034001(n+1), A051604 (rows m=0..4).
%K easy,nonn
%O 0,2
%A _Wolfdieter Lang_
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