OFFSET
0,3
COMMENTS
Create a triangle with T(n,1) = L(n-1) for L a Lucas number and the other side T(n,n) = L(2*(n-1)). Interior elements are defined as T(r,c) = T(r-1,c-1) + T(r-1,c). Half the sum of the terms in row(n)=a(n) for n=1,2,3... - J. M. Bergot, Dec 15 2012
LINKS
Robert Israel, Table of n, a(n) for n = 0..2380
Index entries for linear recurrences with constant coefficients, signature (4,-3,-2,1).
FORMULA
G.f.: x*(1-2*x+2*x^2)/( (1-x-x^2)*(1-3*x+x^2) ). - R. J. Mathar, Dec 17 2012
a(n) = Lucas(n)*(Lucas(n) - 1)/2 - (-1)^n = binomial(Lucas(n), 2) - (-1)^n. - Vladimir Reshetnikov, Sep 27 2016
E.g.f.: (1/2)*exp(-2*x/(1+sqrt(5)))*(-1 + exp(x))*(1 + exp(sqrt(5)*x)). - Stefano Spezia, Dec 15 2019
MAPLE
Lucas:= n -> combinat:-fibonacci(n+1)+combinat:-fibonacci(n-1):
seq((Lucas(2*n)-Lucas(n))/2, n=0..100); # Robert Israel, Sep 15 2016
MATHEMATICA
Table[(LucasL[2n] - LucasL[n])/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)
PROG
(PARI) x='x+O('x^30); concat([0], Vec(x*(1-2*x+2*x^2)/((1-x-x^2)*(1-3*x+x^2)) )) \\ G. C. Greubel, Dec 02 2017
(Magma) [(Lucas(2*n) - Lucas(n))/2: n in [0..30]]; // G. C. Greubel, Dec 02 2017
(Sage) [(lucas_number2(2*n, 1, -1) - lucas_number2(n, 1, -1))/2 for n in (0..30)] # G. C. Greubel, Dec 15 2019
(GAP) List([0..30], n-> (Lucas(1, -1, 2*n)[2] - Lucas(1, -1, n)[2])/2 ); # G. C. Greubel, Dec 15 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
Corrected by Franklin T. Adams-Watters, Oct 25 2006
Corrected by T. D. Noe, Nov 01 2006
STATUS
approved