OFFSET
0,3
COMMENTS
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
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
KEYWORD
nonn,easy
AUTHOR
Stuart Clary, Jul 24 2009
STATUS
approved