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A081575
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Fifth binomial transform of Fibonacci numbers F(n).
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6
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0, 1, 11, 92, 693, 4955, 34408, 234793, 1584891, 10624804, 70911005, 471901739, 3134499984, 20794349393, 137837343787, 913174649260, 6047638172037, 40041955063867, 265079998713464, 1754663288995961, 11613976216265115
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OFFSET
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0,3
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COMMENTS
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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.
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REFERENCES
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S. Falcon, Iterated Binomial Transforms of the k-Fibonacci Sequence, British Journal of Mathematics & Computer Science, 4 (22): 2014.
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LINKS
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FORMULA
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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).
E.g.f.: 2*exp(11*x/2)*sinh(sqrt(5)*x/2)/sqrt(5). - Ilya Gutkovskiy, Aug 12 2017
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MAPLE
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seq(coeff(series(x/(1-11*x+29*x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Aug 13 2019
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MATHEMATICA
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CoefficientList[Series[x/(1 -11x +29x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 09 2013 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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