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a(n) = Lucas(n+2) - 3.
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%I #59 Sep 08 2022 08:44:49

%S 1,4,8,15,26,44,73,120,196,319,518,840,1361,2204,3568,5775,9346,15124,

%T 24473,39600,64076,103679,167758,271440,439201,710644,1149848,1860495,

%U 3010346,4870844,7881193,12752040,20633236,33385279,54018518

%N a(n) = Lucas(n+2) - 3.

%C Sum of the first n Lucas numbers, that is, A000204(1) to A000204(n). - _T. D. Noe_, Oct 10 2005

%H G. C. Greubel, <a href="/A027961/b027961.txt">Table of n, a(n) for n = 1..2500</a>

%H E. S. Egge and T. Mansour, <a href="https://arxiv.org/abs/math/0203226">Restricted permutations, Fibonacci numbers and k-generalized Fibonacci numbers</a>, arXiv:math/0203226 [math.CO], 2002.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-1).

%F a(0) = 0, a(1) = 1, a(n) = a(n-1) + a(n-2) + 3.

%F a(n) = A000204(n+2) - 3 = A000045(2n+4)/A000045(n+2) - 3. - _Benoit Cloitre_, Jan 05 2003

%F G.f.: x*(1+2*x)/((1-x)*(1-x-x^2)). Differences of A023537. - _Ralf Stephan_, Feb 07 2004

%F a(n) = A101220(3, 1, n). - _Ross La Haye_, Jan 28 2005

%F a(n) = F(n) + F(n+2) - 3, n >= 2, where F(n) is the n-th Fibonacci number. - _Zerinvary Lajos_, Jan 31 2008

%F a(n) = Sum_{k=1..n} ((-1/phi)^k + (phi)^k) where phi = 1/2+1/2*sqrt(5). - _Dimitri Papadopoulos_, Jan 07 2016

%F a(n) = 2*a(n-1)-a(n-3) for n>3. - _Wesley Ivan Hurt_, Jan 07 2016

%p a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]+a[n-2]+3 od: seq(a[n],n=1..40); # _Miklos Kristof_, Mar 09 2005

%p with(combinat): seq(fibonacci(n)+fibonacci(n+2)-3, n=2..40); # _Zerinvary Lajos_, Jan 31 2008

%p g:=(1+z^2)/(1-z-z^2): gser:=series(g, z=0, 43): seq(coeff(gser, z, n)-3, n=3..40); # _Zerinvary Lajos_, Jan 09 2009

%t LucasL[Range[3, 40]] - 3 (* _Alonso del Arte_, Sep 26 2013 *)

%o (Magma) [Lucas(n+2)-3: n in [1..40]]; // _Vincenzo Librandi_, Apr 16 2011

%o (PARI) vector(40, n, fibonacci(n+3) +fibonacci(n+1) -3) \\ _G. C. Greubel_, Dec 18 2017

%o (PARI) first(n) = Vec(x*(1+2*x)/((1-x)*(1-x-x^2)) + O(x^(n+1))) \\ _Iain Fox_, Dec 19 2017

%o (Sage) [lucas_number2(n+2, 1, -1) -3 for n in (1..40)] # _G. C. Greubel_, Jun 01 2019

%o (GAP) List([1..40], n-> Lucas(1, -1, n+2)[2] -3 ) # _G. C. Greubel_, Jun 01 2019

%Y T(n, n+1), T given by A027960.

%Y Cf. A000045, A000204, A023537, A101220.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_