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%I #56 Feb 11 2023 10:37:35
%S 0,5,5,10,15,25,40,65,105,170,275,445,720,1165,1885,3050,4935,7985,
%T 12920,20905,33825,54730,88555,143285,231840,375125,606965,982090,
%U 1589055,2571145,4160200,6731345,10891545,17622890,28514435,46137325,74651760,120789085
%N Fibonacci sequence beginning 0, 5.
%D A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, pp. 15, 34, 52.
%H G. C. Greubel, <a href="/A022088/b022088.txt">Table of n, a(n) for n = 0..1000</a>
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H Kristina Lund, Steven Schlicker and Patrick Sigmon, <a href="http://dx.doi.org/10.2140/involve.2008.1.197">Fibonacci sequences and the space of compact sets</a>, Involve, 1:2 (2008), pp. 159-165.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,1).
%F a(n) = round( (2*phi-1)*phi^n ) for n>3. - _Thomas Baruchel_, Sep 08 2004
%F a(n) = 5*Fibonacci(n).
%F a(n) = A119457(n+3,n-1) for n>1. - _Reinhard Zumkeller_, May 20 2006
%F G.f.: 5*x/(1-x-x^2). - _Philippe Deléham_, Nov 20 2008
%F a(n+2) = A014217(n+4) - A014217(n). - _Paul Curtz_, Dec 22 2008
%F a(n) = sqrt(5*(A000032(n)^2 - 4*(-1)^n)). - _Alexander Samokrutov_, Sep 02 2015
%F From _Tom Copeland_, Jan 25 2016: (Start)
%F The o.g.f. for the shifted series b(0)=0 and b(n) = a(n+1) is G(x) = 5*x*(1+x)/(1-x*(1+x)) = 5 L(-Cinv(-x)), where L(x) = x/(1-x) with inverse Linv(x) = x/(1+x) and Cinv(x) = x*(1-x), the inverse of the o.g.f. for the shifted Catalan numbers of A000108, C(x) = (1-sqrt(1-4*x))/2. Then Ginv(x) = -C(-Linv(x/5)) = (-1 + sqrt(1+4*x/(5+x)))/2.
%F a(n+1) = 5*Sum_{k=0..n} binomial(n-k,k) = 5 * A000045(n+1), from A267633, with the convention for zeros of the binomial assumed there.
%F (End)
%F For n > 0, 1/a(n) = Sum_{k>=1} F(n*k)/(L(n+1)^(k+1)), where F(n) = A000045(n) and L(n) = A000032(n). - _Diego Rattaggi_, Oct 26 2022
%t LinearRecurrence[{1,1},{0,5},40] (* _Harvey P. Dale_, Jan 13 2012 *)
%t 5*Fibonacci[Range[0, 50]] (* _G. C. Greubel_, Feb 10 2023 *)
%o (Magma) [5*Fibonacci(n): n in [1..40]]; // _Vincenzo Librandi_, Sep 03 2015
%o (PARI) a(n) = 5*fibonacci(n); \\ _Michel Marcus_, Sep 03 2015
%o (SageMath) [5*fibonacci(n) for n in range(51)] # _G. C. Greubel_, Feb 10 2023
%Y Cf. A000032, A000045, A000108, A014217, A267633.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_
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