A163197 a(n) = (1/4)* 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.

-1, 1, 27, 540, 9800, 176176, 3162159, 56744793, 1018249595, 18271762300, 327873509424, 5883451505856, 105574253853887, 1894453118539345, 33994581881622075, 610008020755286076, 10946149791725643704, 196420688230338021808, 3524626238354441796015, 63246851602149831726825

Offset: 0

Author

Stuart Clary, Jul 24 2009

Keywords

• sign
• easy

Comments

Natural bilateral extension (brackets mark index 0): ..., 9801, 539, 28, 0, 0, [-1], 1, 27, 540, 9800, 176176, ... This is A163195-reversed followed by A163197. That is, A163197(-n) = A163195(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).

Crossrefs

Cf. A163194, A163195, A163196, A163198, A163199.

Programs

Magma

[(1/4)*(Lucas(n)*Fibonacci(n+1))^2*Lucas(n-1)*Fibonacci(n+2): n in [0..30]]; // G. C. Greubel, Dec 21 2017

Mathematica

a[n_Integer] := (1/4)*LucasL[n]^2*Fibonacci[n+1]^2*LucasL[n-1]*Fibonacci[n+2]
LinearRecurrence[{20,-35,-35,20,-1},{-1,1,27,540,9800}, 50] (* or *) Table[(1/20)*(Fibonacci[6*n+3] - 12*Fibonacci[2*n+1] - 10*(-1)^n),{n,0,25}] (* G. C. Greubel, Dec 09 2016 *)

PARI

Vec(-(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, print((1/20)*(fibonacci(6*n+3) - 12*fibonacci(2*n+1) - 10*(-1)^n), ", ")) \\ G. C. Greubel, Dec 21 2017

Formula

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

a(n) = (1/4)*L(n)^2*F(n+1)^2*L(n-1)*F(n+2).

a(n) = (1/20)*(F(6n+3) - 12*F(2n+1) - 10*(-1)^n).

a(n) = (1/4)*(F(2n+1)^3 - 3*F(2n+1) - 2*(-1)^n).

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

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

a(n) - 21 a(n-1) + 56 a(n-2) - 21 a(n-3) + a(n-4) = - 50*(-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.: (-1 + 21*x - 28*x^2)/(1 - 20*x + 35*x^2 + 35*x^3 - 20*x^4 + x^5) = -(1 - 21*x + 28*x^2)/((1 + x)*(1 - 3*x + x^2)*(1 - 18*x + x^2)).

A163195(n) - a(n) = (-1)^n.