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%I #21 Apr 02 2020 15:42:23
%S 1,5,34,314,3784,56792,1022320,21468848,515252608,13911820928,
%T 417354628864,13772702754560,495817299168256,19336874667570176,
%U 812148736037963776,36546693121708402688,1754241269842003394560
%N E.g.f. exp(2x)/(1-3x).
%C Second binomial transform of n!3^n.
%F a(n) = 3n*a(n-1)+2^n, n>0, a(0)=1.
%F a(n) ~ n! * exp(2/3) * 3^n. - _Vaclav Kotesovec_, Aug 04 2014
%F a(n) +(-3*n-2)*a(n-1) +6*(n-1)*a(n-2)=0. - _R. J. Mathar_, Dec 21 2014
%F From _Peter Bala_, Jan 30 2015: (Start)
%F a(n) = int {x = 0..inf} (3*x + 2)^n*exp(-x) dx.
%F The e.g.f. y = exp(2*x)/(1 - 3*x) satisfies the differential equation (1 - 3*x)*y' = (5 - 6*x)*y. Mathar's recurrence above follows easily from this.
%F The sequence b(n) = 3^n*n! also satisfies Mathar's recurrence with b(0) = 1, b(1) = 3. This leads to the continued fraction representation a(n) = 3^n*n!*( 1 + 2/(3 - 6/(8 - 12/(11 - ... - (6*n - 6)/(3*n + 2) )))) for n >= 2. Taking the limit gives the continued fraction representation exp(2/3) = 1 + 2/(3 - 6/(8 - 12/(11 - ... - (6*n - 6)/((3*n + 2) - ... )))). (End)
%F a(n) = 3^n*exp(2/3)*Gamma(n+1,2/3). - _Gerry Martens_, Jul 24 2015
%t With[{nn=20},CoefficientList[Series[Exp[2x]/(1-3x),{x,0,nn}],x] Range[ 0,nn]!] (* _Harvey P. Dale_, Apr 02 2020 *)
%Y Cf. A010844, A097819.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Aug 26 2004
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