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A097819
E.g.f. exp(3x)/(1-4x).
3
1, 7, 65, 807, 12993, 260103, 6243201, 174811815, 5593984641, 201383466759, 8055338729409, 354434904271143, 17012875405546305, 884669521090002183, 49541493181044905217, 2972489590862708661927, 190239333815213397410049, 12936274699434511153023495
OFFSET
0,2
COMMENTS
Third binomial transform of n!4^n.
LINKS
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
MATHEMATICA
With[{nn=20}, CoefficientList[Series[Exp[3x]/(1-4x), {x, 0, nn}], x] Range[0, nn]!] (* Harvey P. Dale, Jun 16 2016 *)
CROSSREFS
Sequence in context: A300488 A085283 A069015 * A152525 A272646 A220067
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Aug 26 2004
STATUS
approved