A097819 E.g.f. exp(3x)/(1-4x).

1, 7, 65, 807, 12993, 260103, 6243201, 174811815, 5593984641, 201383466759, 8055338729409, 354434904271143, 17012875405546305, 884669521090002183, 49541493181044905217, 2972489590862708661927, 190239333815213397410049, 12936274699434511153023495

Offset: 0

Author

Paul Barry, Aug 26 2004

Keywords

• easy
• nonn

Comments

Third binomial transform of n!4^n.

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..365

Crossrefs

Cf. A010844, A097817.

Programs

Mathematica

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

Formula

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

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

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

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.

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)

a(n) = 4^n*exp(3/4)*Gamma(n+1,3/4). - Gerry Martens, Jul 24 2015