A163195 a(n) = (1/4)*F(n)^2 * L(n+1)^2 * F(n-1) * L(n+2), where F(n) and L(n) are the Fibonacci and Lucas numbers, respectively.

0, 0, 28, 539, 9801, 176175, 3162160, 56744792, 1018249596, 18271762299, 327873509425, 5883451505855, 105574253853888, 1894453118539344, 33994581881622076, 610008020755286075, 10946149791725643705, 196420688230338021807, 3524626238354441796016, 63246851602149831726824

Offset

0,3

Comments

Natural bilateral extension (brackets mark index 0): ..., 9800, 540, 27, 1, -1, [0], 0, 28, 539, 9801, 176175, ... This is A163197-reversed followed by A163195. That is, A163195(-n) = A163197(n-1).

References

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).

Links

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

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) = (1/4)*F(n)^2 * L(n+1)^2 * F(n-1) * L(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=1..n} F(2k)^3 = A163198(n) if n is even.

a(n) = Sum_{k=2..n} F(2k)^3 = A163199(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.: (28*x^2 - 21*x^3 + x^4)/(1 - 20*x + 35*x^2 + 35*x^3 - 20*x^4 + x^5) = x^2*(28 - 21*x + x^2)/((1 + x)*(1 - 3*x + x^2)*(1 - 18*x + x^2)).

a(n) - A163197(n) = (-1)^n.

Mathematica

a[n_Integer] := (1/4)*Fibonacci[n]^2 * LucasL[n+1]^2 * Fibonacci[n-1] * LucasL[n+2]
LinearRecurrence[{20, -35, -35, 20, -1}, {0, 0, 28, 539, 9801}, 50] (* or *) Table[(Fibonacci[6n+3] - 12*Fibonacci[2n+1] + 10*(-1)^n)/20, {n, 1, 25}] (* G. C. Greubel, Dec 09 2016 *)

Prog

(PARI) for(n=0, 30, print1((fibonacci(6*n+3) - 12*fibonacci(2*n+1) + 10*(-1)^n)/20, ", ")) \\ G. C. Greubel, Dec 21 2017
(Magma) [(1/4)*(Fibonacci(n)*Lucas(n+1))^2*(Fibonacci(n-1)*Lucas(n+2)): n in [0..30]]; // G. C. Greubel, Dec 21 2017

Crossrefs

Cf. A163194, A163196, A163197, A163198, A163199.

Sequence in context: A020758 A022000 A020569 * A092708 A163198 A278190

Adjacent sequences: A163192 A163193 A163194 * A163196 A163197 A163198

Keyword

nonn,easy

Author

Stuart Clary, Jul 24 2009

Status

approved