A097819 - OEIS

%I #18 Jun 16 2016 16:25:51

%S 1,7,65,807,12993,260103,6243201,174811815,5593984641,201383466759,

%T 8055338729409,354434904271143,17012875405546305,884669521090002183,

%U 49541493181044905217,2972489590862708661927,190239333815213397410049,12936274699434511153023495

%N E.g.f. exp(3x)/(1-4x).

%C Third binomial transform of n!4^n.

%H Harvey P. Dale, <a href="/A097819/b097819.txt">Table of n, a(n) for n = 0..365</a>

%F a(n) = 4*a(n-1)+3^n, n>0, a(0)=1.

%F a(n) + (-4*n-3)*a(n-1) + 12*(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} (4*x + 3)^n*exp(-x) dx.

%F a(n) ~ 4^n*n!*exp(3/4).

%F The e.g.f. y = exp(3*x)/(1 - 4*x) satisfies the differential equation (1 - 4*x)*y' = (7 - 12*x)*y. Mathar's recurrence above follows easily from this.

%F The sequence b(n) := 4^n*n! also satisfies Mathar's recurrence with b(0) = 1, b(1) = 4. This leads to the continued fraction representation a(n) = 4^n*n!*( 1 + 3/(4 - 12/(11 - 24/(15 - ... - (12*n - 12)/(4*n + 3) )))) for n >= 2. Taking the limit gives the continued fraction representation exp(3/4) = 1 + 3/(4 - 12/(11 - 24/(15 - ... - (12*n - 12)/((4*n + 3) - ... )))). (End)

%F a(n) = 4^n*exp(3/4)*Gamma(n+1,3/4). - _Gerry Martens_, Jul 24 2015

%t With[{nn=20},CoefficientList[Series[Exp[3x]/(1-4x),{x,0,nn}],x] Range[0,nn]!] (* _Harvey P. Dale_, Jun 16 2016 *)

%Y Cf. A010844, A097817.

%K easy,nonn

%O 0,2

%A _Paul Barry_, Aug 26 2004