OFFSET
0,3
COMMENTS
REFERENCES
S. Falcon, Iterated Binomial Transforms of the k-Fibonacci Sequence, British Journal of Mathematics & Computer Science, 4 (22): 2014.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..200
S. Falcon, Iterated Binomial Transforms of the k-Fibonacci Sequence, British Journal of Mathematics & Computer Science, 4 (22): 2014.
Index entries for linear recurrences with constant coefficients, signature (11,-29).
FORMULA
a(n) = 11*a(n-1) - 29*a(n-2), a(0)=0, a(1)=1.
a(n) = ((sqrt(5)/2 + 11/2)^n - (11/2 - sqrt(5)/2)^n)/sqrt(5).
G.f.: x/(1 - 11*x + 29*x^2). - adapted by Vincenzo Librandi, Aug 09 2013
E.g.f.: 2*exp(11*x/2)*sinh(sqrt(5)*x/2)/sqrt(5). - Ilya Gutkovskiy, Aug 12 2017
MAPLE
seq(coeff(series(x/(1-11*x+29*x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Aug 13 2019
MATHEMATICA
LinearRecurrence[{11, -29}, {0, 1}, 30] (* Vladimir Joseph Stephan Orlovsky, Jan 31 2011; modified by G. C. Greubel, Aug 13 2019 *)
CoefficientList[Series[x/(1 -11x +29x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 09 2013 *)
PROG
(Sage) [lucas_number1(n, 11, 29) for n in range(0, 21)] # Zerinvary Lajos, Apr 27 2009
(Magma) [n le 2 select (n-1) else 11*Self(n-1)-29*Self(n-2): n in [1..25]]; // Vincenzo Librandi, Aug 09 2013
(PARI) my(x='x+O('x^30)); Vec(x/(1-11*x+29*x^2)) \\ G. C. Greubel, Aug 13 2019
(GAP) a:=[0, 1];; for n in [3..30] do a[n]:=11*a[n-1]-29*a[n-2]; od; a; # G. C. Greubel, Aug 13 2019
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Mar 24 2003
STATUS
approved