%I #33 May 04 2023 20:37:34
%S 0,1,11,92,693,4955,34408,234793,1584891,10624804,70911005,471901739,
%T 3134499984,20794349393,137837343787,913174649260,6047638172037,
%U 40041955063867,265079998713464,1754663288995961,11613976216265115
%N Fifth binomial transform of Fibonacci numbers F(n).
%C Binomial transform of A081574.
%C 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.
%D S. Falcon, Iterated Binomial Transforms of the k-Fibonacci Sequence, British Journal of Mathematics & Computer Science, 4 (22): 2014.
%H Vincenzo Librandi, <a href="/A081575/b081575.txt">Table of n, a(n) for n = 0..200</a>
%H S. Falcon, <a href="https://doi.org/10.9734/BJMCS/2014/11783">Iterated Binomial Transforms of the k-Fibonacci Sequence</a>, British Journal of Mathematics & Computer Science, 4 (22): 2014.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (11,-29).
%F a(n) = 11*a(n-1) - 29*a(n-2), a(0)=0, a(1)=1.
%F a(n) = ((sqrt(5)/2 + 11/2)^n - (11/2 - sqrt(5)/2)^n)/sqrt(5).
%F G.f.: x/(1 - 11*x + 29*x^2). - adapted by _Vincenzo Librandi_, Aug 09 2013
%F E.g.f.: 2*exp(11*x/2)*sinh(sqrt(5)*x/2)/sqrt(5). - _Ilya Gutkovskiy_, Aug 12 2017
%p seq(coeff(series(x/(1-11*x+29*x^2), x, n+1), x, n), n = 0..30); # _G. C. Greubel_, Aug 13 2019
%t LinearRecurrence[{11,-29}, {0,1}, 30] (* _Vladimir Joseph Stephan Orlovsky_, Jan 31 2011; modified by _G. C. Greubel_, Aug 13 2019 *)
%t CoefficientList[Series[x/(1 -11x +29x^2), {x, 0, 30}], x] (* _Vincenzo Librandi_, Aug 09 2013 *)
%o (Sage) [lucas_number1(n,11,29) for n in range(0, 21)] # _Zerinvary Lajos_, Apr 27 2009
%o (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
%o (PARI) my(x='x+O('x^30)); Vec(x/(1-11*x+29*x^2)) \\ _G. C. Greubel_, Aug 13 2019
%o (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
%Y Cf. A000045, A081570.
%K easy,nonn
%O 0,3
%A _Paul Barry_, Mar 24 2003