A097817 E.g.f. exp(2x)/(1-3x).
1, 5, 34, 314, 3784, 56792, 1022320, 21468848, 515252608, 13911820928, 417354628864, 13772702754560, 495817299168256, 19336874667570176, 812148736037963776, 36546693121708402688, 1754241269842003394560
Offset: 0
Programs
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Mathematica
With[{nn=20},CoefficientList[Series[Exp[2x]/(1-3x),{x,0,nn}],x] Range[ 0,nn]!] (* Harvey P. Dale, Apr 02 2020 *)
Formula
a(n) = 3n*a(n-1)+2^n, n>0, a(0)=1.
a(n) ~ n! * exp(2/3) * 3^n. - Vaclav Kotesovec, Aug 04 2014
a(n) +(-3*n-2)*a(n-1) +6*(n-1)*a(n-2)=0. - R. J. Mathar, Dec 21 2014
From Peter Bala, Jan 30 2015: (Start)
a(n) = int {x = 0..inf} (3*x + 2)^n*exp(-x) dx.
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.
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)
a(n) = 3^n*exp(2/3)*Gamma(n+1,2/3). - Gerry Martens, Jul 24 2015
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