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A081575
Fifth binomial transform of Fibonacci numbers F(n).
6
0, 1, 11, 92, 693, 4955, 34408, 234793, 1584891, 10624804, 70911005, 471901739, 3134499984, 20794349393, 137837343787, 913174649260, 6047638172037, 40041955063867, 265079998713464, 1754663288995961, 11613976216265115
OFFSET
0,3
COMMENTS
Binomial transform of A081574.
Case k=5 of family of recurrences a(n) = (2k+1)*a(n-1) - A028387(k-1)*a(n-2), a(0)=0, a(1)=1.
REFERENCES
S. Falcon, Iterated Binomial Transforms of the k-Fibonacci Sequence, British Journal of Mathematics & Computer Science, 4 (22): 2014.
LINKS
S. Falcon, Iterated Binomial Transforms of the k-Fibonacci Sequence, British Journal of Mathematics & Computer Science, 4 (22): 2014.
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
Sequence in context: A044262 A044643 A025507 * A016150 A115203 A164547
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Mar 24 2003
STATUS
approved