A163198 Sum of the cubes of the first n even-indexed Fibonacci numbers.

0, 1, 28, 540, 9801, 176176, 3162160, 56744793, 1018249596, 18271762300, 327873509425, 5883451505856, 105574253853888, 1894453118539345, 33994581881622076, 610008020755286076, 10946149791725643705, 196420688230338021808, 3524626238354441796016, 63246851602149831726825

Offset: 0

Author

Stuart Clary, Jul 24 2009

Keywords

• nonn
• easy

Comments

Natural bilateral extension (brackets mark index 0): ..., 9801, 540, 28, 1, 0, [0], 1, 28, 540, 9801, 176176, ... This is A163198-reversed followed by A163198. That is, A163198(-n) = A163198(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 equations (3), (46), (47), and (49).
• R. S. Melham, Some conjectures concerning sums of odd powers of Fibonacci and Lucas numbers, The Fibonacci Quart. 46/47 (2008/2009), no. 4, 312-315.
• K. Ozeki, On Melham's sum, The Fibonacci Quart. 46/47 (2008/2009), no. 2, 107-110.
• H. Prodinger, On a sum of Melham and its variants, The Fibonacci Quart. 46/47 (2008/2009), no. 3, 207-215.
• K. Subba Rao, Some properties of Fibonacci numbers, Amer. Math. Monthly, 60(10):680-684, Dec. 1953. See page 682.
• Index entries for linear recurrences with constant coefficients, signature (22,-77,77,-22,1).

Crossrefs

Cf. A163194, A163195, A163196, A163197, A163199, A163200, A163201, A163202, A005968, A215039 (first differences).

Programs

Mathematica

a[n_Integer] := If[ n >= 0, Sum[ Fibonacci[2k]^3, {k, 1, n} ], -Sum[ Fibonacci[-2k]^3, {k, 1, -n - 1} ] ]
LinearRecurrence[{22, -77, 77, -22, 1}, {0, 1, 28, 540, 9801}, 50] (* G. C. Greubel, Dec 09 2016 *)
Accumulate[Fibonacci[Range[0,40,2]]^3] (* Harvey P. Dale, Nov 15 2023 *)

PARI

a(n) = sum(k=1, n, fibonacci(2*k)^3); \\ Michel Marcus, Feb 29 2016

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 09 2016

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} F(2k)^3.

a(n) = A163199(n) + 1.

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

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

a(n) = (1/4)*F(n)^2*L(n+1)^2*F(n-1)*L(n+2) = A163195(n) if n is even.

a(n) = (1/4)*L(n)^2*F(n+1)^2*L(n-1)*F(n+2) = A163197(n) if n is odd.

a(n) - 21 a(n-1) + 56 a(n-2) - 21 a(n-3) + a(n-4) = 8.

a(n) - 22 a(n-1) + 77 a(n-2) - 77 a(n-3) + 22 a(n-4) - a(n-5) = 0.

G.f.: (x + 6*x^2 + x^3)/(1 - 22*x + 77*x^2 - 77*x^3 + 22*x^4 - x^5) = x*(1 + 6*x + x^2)/((1 - x)*(1 - 3*x + x^2 )*(1 - 18*x + x^2)).

a(n) = (F(2*n+1)-1)^2*(F(2*n+1) + 2)/4, n>=0. See the Melham reference for a general conjecture. - Wolfdieter Lang, Aug 10 2012

Extensions

Melham and Ozeki references from Wolfdieter Lang, Aug 10 2012. Also Prodinger reference added, Oct 11 2012.