%I #43 Sep 08 2022 08:45:30
%S 0,2,1,6,12,35,88,234,609,1598,4180,10947,28656,75026,196417,514230,
%T 1346268,3524579,9227464,24157818,63245985,165580142,433494436,
%U 1134903171,2971215072,7778742050,20365011073,53316291174,139583862444,365435296163,956722026040,2504730781962
%N a(n) = Fibonacci(n)*Lucas(n-1).
%C 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.
%C 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
%C 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
%H Robert Israel, <a href="/A128534/b128534.txt">Table of n, a(n) for n = 0..2370</a>
%H Prapanpong Pongsriiam, <a href="http://www.jstor.org/stable/10.4169/college.math.j.48.2.97">Integral Values of the Generating Functions of Fibonacci and Lucas Numbers</a>, College Math. J., 48 (No. 2 2017), pp 97ff.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).
%F a(n) = F(2*n - 1) + (-1)^(n+1), assuming F(0)=0 and L(0)=2.
%F From _R. J. Mathar_, Apr 16 2009: (Start)
%F a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
%F G.f.: x*(2-3*x)/((1+x)*(x^2-3*x+1)). (End)
%F 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
%F a(n+1) = A081714(n) + 2*(-1)^n. - _A.H.M. Smeets_, Feb 26 2022
%e a(5) = 35 because F(5)*L(4) = 5*7.
%p seq(combinat:-fibonacci(2*n-1)+(-1)^(n+1),n=0..50); # _Robert Israel_, Jan 28 2016
%t Table[Fibonacci[n] LucasL[n - 1], {n, 0, 31}] (* _Michael De Vlieger_, Jan 29 2016 *)
%o (PARI) concat( 0, Vec(-x*(-2+3*x)/((1+x)*(x^2-3*x+1)) + O(x^40))) \\ _Michel Marcus_, Jan 28 2016
%o (Magma) [Fibonacci(n)*Lucas(n-1): n in [0..30]]; // _G. C. Greubel_, Dec 21 2017
%Y Cf. A000032, A000045, A001906, A081714, A128533, A128535.
%K easy,nonn
%O 0,2
%A _Axel Harvey_, Mar 08 2007
%E More terms from _Michel Marcus_, Jan 28 2016