A163196 a(n) = L(n)^2 * F(n+1)^2 * L(n-1) * F(n+2), where F(n) and L(n) are the Fibonacci and Lucas numbers, respectively.

-4, 4, 108, 2160, 39200, 704704, 12648636, 226979172, 4072998380, 73087049200, 1311494037696, 23533806023424, 422297015415548, 7577812474157380, 135978327526488300, 2440032083021144304, 43784599166902574816, 785682752921352087232, 14098504953417767184060, 252987406408599326907300

Offset

0,1

Comments

Natural bilateral extension (brackets mark index 0): ..., 39204, 2156, 112, 0, 0, [-4], 4, 108, 2160, 39200, 704704, ... This is A163194-reversed followed by A163196. That is, A163196(-n) = A163194(n-1).

Links

G. C. Greubel, Table of n, a(n) for n = 0..500

Stuart Clary and Paul D. Hemenway, On sums of cubes of Fibonacci numbers, Applications of Fibonacci Numbers, Vol. 5 (St. Andrews, 1992), 123-136, Kluwer Acad. Publ., 1993. See equation (3).

Index entries for linear recurrences with constant coefficients, signature (20,-35,-35,20,-1).

Formula

Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).

a(n) = L(n)^2 * F(n+1)^2 * L(n-1) * F(n+2).

a(n) = (1/5)*(F(6n+3) - 12*F(2n+1) - 10*(-1)^n).

a(n) = F(2n+1)^3 - 3*F(2n+1) - 2*(-1)^n.

a(n) = 4*Sum_{k=2..n} F(2k)^3 = 4*A163199(n) if n is even.

a(n) = 4*Sum_{k=1..n} F(2k)^3 = 4*A163198(n) if n is odd.

a(n) - 21 a(n-1) + 56 a(n-2) - 21 a(n-3) + a(n-4) = - 200*(-1)^n.

a(n) - 20 a(n-1) + 35 a(n-2) + 35 a(n-3) - 20 a(n-4) + a(n-5) = 0.

G.f.: (-4 + 84*x - 112*x^2)/(1 - 20*x + 35*x^2 + 35*x^3 - 20*x^4 + x^5) = -4*(1 - 21*x + 28*x^2)/((1 + x)*(1 - 3*x + x^2)*(1 - 18*x + x^2)).

A163194(n) - a(n) = 4*(-1)^n.

Mathematica

a[n_Integer] := LucasL[n]^2*Fibonacci[n+1]^2*LucasL[n-1] *Fibonacci[n+2]
LinearRecurrence[{20, -35, -35, 20, -1}, {-4, 4, 108, 2160, 39200}, 50] (* or *) Table[(1/5)*(Fibonacci[6*n+3] - 12*Fibonacci[2*n+1] - 10*(-1)^n), {n, 0, 25}] (* G. C. Greubel, Dec 09 2016 *)

Prog

(PARI) Vec( -4*(1 - 21*x + 28*x^2)/((1 + x)*(1 - 3*x + x^2)*(1 - 18*x + x^2)) + O(x^50)) \\ G. C. Greubel, Dec 09 2016
(PARI) for(n=0, 30, print1((1/5)*(fibonacci(6*n+3) - 12*fibonacci(2*n+1) - 10*(-1)^n), ", ")) \\ G. C. Greubel, Dec 21 2017
(Magma) [(Lucas(n)*Fibonacci(n+1))^2*(Lucas(n-1)*Fibonacci(n+2)): n in [0..30]]; // G. C. Greubel, Dec 21 2017

Crossrefs

Cf. A163194, A163195, A163197, A163198, A163199.

Sequence in context: A068337 A009534 A009559 * A219453 A283453 A055071

Adjacent sequences: A163193 A163194 A163195 * A163197 A163198 A163199

Keyword

sign,easy

Author

Stuart Clary, Jul 24 2009

Status

approved