OFFSET
0,2
COMMENTS
Generally, F(n)*L(n+k) = F(2*n + k) + F(k)*(-1)^(n+1). If k=0 the sequence is A001906; if k=1 it is A081714.
a(n) is the maximum area of a quadrilateral with lengths of sides in order F(n), F(n), L(n-1), L(n-1) for n>1. - J. M. Bergot, Jan 28 2016
Can be obtained (up to signs) by setting x = F(n)/F(n+1) in g.f. for Lucas numbers - see Pongsriiam. - N. J. A. Sloane, Mar 23 2017
LINKS
Robert Israel, Table of n, a(n) for n = 0..2370
Prapanpong Pongsriiam, Integral Values of the Generating Functions of Fibonacci and Lucas Numbers, College Math. J., 48 (No. 2 2017), pp 97ff.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).
FORMULA
a(n) = F(2*n - 1) + (-1)^(n+1), assuming F(0)=0 and L(0)=2.
From R. J. Mathar, Apr 16 2009: (Start)
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: x*(2-3*x)/((1+x)*(x^2-3*x+1)). (End)
a(n) = (2^(-1-n)*(-5*(-1)^n*2^(1+n) - (-5+sqrt(5))*(3+sqrt(5))^n + (3-sqrt(5))^n*(5+sqrt(5))))/5. - Colin Barker, Apr 05 2016
a(n+1) = A081714(n) + 2*(-1)^n. - A.H.M. Smeets, Feb 26 2022
EXAMPLE
a(5) = 35 because F(5)*L(4) = 5*7.
MAPLE
seq(combinat:-fibonacci(2*n-1)+(-1)^(n+1), n=0..50); # Robert Israel, Jan 28 2016
MATHEMATICA
Table[Fibonacci[n] LucasL[n - 1], {n, 0, 31}] (* Michael De Vlieger, Jan 29 2016 *)
PROG
(PARI) concat( 0, Vec(-x*(-2+3*x)/((1+x)*(x^2-3*x+1)) + O(x^40))) \\ Michel Marcus, Jan 28 2016
(Magma) [Fibonacci(n)*Lucas(n-1): n in [0..30]]; // G. C. Greubel, Dec 21 2017
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Axel Harvey, Mar 08 2007
EXTENSIONS
More terms from Michel Marcus, Jan 28 2016
STATUS
approved