A027961 a(n) = Lucas(n+2) - 3.

1, 4, 8, 15, 26, 44, 73, 120, 196, 319, 518, 840, 1361, 2204, 3568, 5775, 9346, 15124, 24473, 39600, 64076, 103679, 167758, 271440, 439201, 710644, 1149848, 1860495, 3010346, 4870844, 7881193, 12752040, 20633236, 33385279, 54018518

Offset: 1

Author

Clark Kimberling

Keywords

• nonn
• easy

Comments

Sum of the first n Lucas numbers, that is, A000204(1) to A000204(n). - T. D. Noe, Oct 10 2005

Links

• G. C. Greubel, Table of n, a(n) for n = 1..2500
• E. S. Egge and T. Mansour, Restricted permutations, Fibonacci numbers and k-generalized Fibonacci numbers, arXiv:math/0203226 [math.CO], 2002.
• Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

Crossrefs

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

Cf. A000045, A000204, A023537, A101220.

Programs

GAP

List([1..40], n-> Lucas(1, -1, n+2)[2] -3 ); # G. C. Greubel, Jun 01 2019

Magma

[Lucas(n+2)-3: n in [1..40]]; // Vincenzo Librandi, Apr 16 2011

Maple

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
with(combinat): seq(fibonacci(n)+fibonacci(n+2)-3, n=2..40); # Zerinvary Lajos, Jan 31 2008
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

Mathematica

LucasL[Range[3, 40]] - 3 (* Alonso del Arte, Sep 26 2013 *)

PARI

vector(40, n, fibonacci(n+3) +fibonacci(n+1) -3) \\ G. C. Greubel, Dec 18 2017

PARI

first(n) = Vec(x*(1+2*x)/((1-x)*(1-x-x^2)) + O(x^(n+1))) \\ Iain Fox, Dec 19 2017

Sage

[lucas_number2(n+2, 1, -1) -3 for n in (1..40)] # G. C. Greubel, Jun 01 2019

Formula

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

a(n) = A000204(n+2) - 3 = A000045(2n+4)/A000045(n+2) - 3. - Benoit Cloitre, Jan 05 2003

G.f.: x*(1+2*x)/((1-x)*(1-x-x^2)). Differences of A023537. - Ralf Stephan, Feb 07 2004

a(n) = A101220(3, 1, n). - Ross La Haye, Jan 28 2005

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

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

a(n) = 2*a(n-1)-a(n-3) for n>3. - Wesley Ivan Hurt, Jan 07 2016