A163201 Alternating sum of the cubes of the first n even-indexed Fibonacci numbers.

0, -1, 26, -486, 8775, -157600, 2828384, -50754249, 910750554, -16342762150, 293258984975, -5262319011456, 94428483336576, -1694450381348881, 30405678381733850, -545607760491930150, 9790534010478427479, -175684004428133950624, 3152521545695969823584, -56569703818099420107225

Offset

0,3

Comments

Natural bilateral extension (brackets mark index 0): ..., 8775, -486, 26, -1, 0, [0], -1, 26, -486, 8775, -157600, ... This is A163201-reversed followed by A163201. That is, a(-n) = a(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. For the factored closed form, let alpha equal the imaginary unit in Equation (21).

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) = Sum_{k=1..n} (-1)^k F(2k)^3.

a(n) = (1/50)*(L(6n+3) - 6 L(2n+1) + 2) if n is even.

a(n) = -(1/50)*(L(6n+3) - 6 L(2n+1) - 2) if n is odd.

a(n) = (1/2) * F(n)^2 * F(n+1)^2 * (L(2n+1) + 2) if n is even.

a(n) = -(1/2) * F(n)^2 * F(n+1)^2 * (L(2n+1) - 2) if n is odd.

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

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

G.f.: (-x + 6*x^2 - x^3)/(1 + 20*x + 35*x^2 - 35*x^3 - 20*x^4 - x^5) = -x*(1 - 6*x + x^2)/((1 - x)*(1 + 3*x + x^2)*(1 + 18*x + x^2)).

Mathematica

a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[2k]^3, {k, 1, n} ], -Sum[ (-1)^k Fibonacci[-2k]^3, {k, 1, -n - 1} ] ]
LinearRecurrence[{-20, -35, 35, 20, 1}, {0, -1, 26, -486, 8775}, 50] (* or *) Table[(-1)^n*(1/50)*(LucasL[6 n + 3] - 6 LucasL[2 n + 1] + 2*(-1)^n), {n, 0, 25}] (* G. C. Greubel, Dec 10 2016 *)

Prog

(PARI) concat([0], Vec(-x*(1 - 6*x + x^2)/((1 - x)*(1 + 3*x + x^2)*(1 + 18*x + x^2)) + O(x^50))) \\ G. C. Greubel, Dec 10 2016
(Magma) [(-1)^n*(1/50)*(Lucas(6*n+3)-6*Lucas(2*n+1)+2*(-1)^n): n in [0..20]]; // Vincenzo Librandi, Dec 10 2016

Crossrefs

Cf. A119284, A163198, A163200, A163202.

Sequence in context: A018208 A240190 A001723 * A205990 A230247 A058461

Adjacent sequences: A163198 A163199 A163200 * A163202 A163203 A163204

Keyword

sign,easy

Author

Stuart Clary, Jul 24 2009

Status

approved