A128535 a(n) = F(n)*L(n-2) where F = Fibonacci and L = Lucas numbers.

0, -1, 2, 2, 9, 20, 56, 143, 378, 986, 2585, 6764, 17712, 46367, 121394, 317810, 832041, 2178308, 5702888, 14930351, 39088170, 102334154, 267914297, 701408732, 1836311904, 4807526975, 12586269026, 32951280098, 86267571273, 225851433716, 591286729880

Offset: 0

Author

Axel Harvey, Mar 09 2007

Keywords

• sign
• easy

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.

For n>2, a(n) is twice the area of the triangle with vertices at (F(n-3), F(n-2)), (F(n-1), F(n)), and (L(n), L(n-1)), where F(n)=A000045(n) and L(n)=A000032(n). - J. M. Bergot, May 22 2014

a(n) is the maximum area of a quadrilateral with lengths of sides in order L(n-2), L(n-2), F(n), F(n) for n>2. - J. M. Bergot, Jan 28 2016

Examples

			a(7) = 143 because F(7)*L(5) = 13*11.
G.f. = -x + 2*x^2 + 2*x^3 + 9*x^4 + 20*x^5 + 56*x^6 + 143*x^7 + ...
		

References

• Mohammad K. Azarian, Identities Involving Lucas or Fibonacci and Lucas Numbers as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 45, 2012, pp. 2221-2227.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Crossrefs

Cf. A001906, A081714, A128533, A128534.

Programs

Magma

[Fibonacci(n)*Lucas(n-2): n in [0..30]]; // Vincenzo Librandi, Feb 20 2013

Maple

a:= n-> (<<0|1|0>, <0|0|1>, <-1|2|2>>^n. <<0,-1,2>>)[1, 1]:
seq(a(n), n=0..30);  # Alois P. Heinz, Sep 28 2016

Mathematica

Table[Fibonacci[i]LucasL[i-2], {i,0,30}] (* Harvey P. Dale, Feb 16 2011 *)
LinearRecurrence[{2, 2, -1}, {0, -1, 2}, 40] (* Vincenzo Librandi, Feb 20 2013 *)
a[ n_] := Fibonacci[2 n - 2] + (-1)^n; (* Michael Somos, May 26 2014 *)

PARI

a(n)=([0,1,0; 0,0,1; -1,2,2]^n*[0;-1;2])[1,1] \\ Charles R Greathouse IV, Feb 01 2016

PARI

a(n) = round(((-1)^n+(2^(-1-n)*(-(3-sqrt(5))^n*(3+sqrt(5))-(-3+sqrt(5))*(3+sqrt(5))^n))/sqrt(5))) \\ Colin Barker, Sep 28 2016

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*(-1+4*x)/((1+x)*(x^2-3*x+1)). (End)

a(n+1) = - A116697(2*n). - Reinhard Zumkeller, Feb 25 2011

a(-n) = - A128533(n). - Michael Somos, May 26 2014

0 = a(n)*(+4*a(n) + a(n+1) - 17*a(n+2)) + a(n+1)*(-14*a(n+1) + a(n+2)) + a(n+2)*(+4*a(n+2)) for all n in Z. - Michael Somos, May 26 2014

a(n) = ((-1)^n+(2^(-1-n)*(-(3-sqrt(5))^n*(3+sqrt(5))-(-3+sqrt(5))*(3+sqrt(5))^n)) / sqrt(5)). - Colin Barker, Sep 28 2016

Extensions

More terms from Harvey P. Dale, Feb 16 2011