A163200 Sum of the cubes of the first n odd-indexed Fibonacci numbers.

0, 1, 9, 134, 2331, 41635, 746604, 13395941, 240376941, 4313380114, 77400441855, 1388894512391, 24922700621784, 447219716262409, 8025032191009041, 144003359719040030, 2584035442744223139, 46368634609657371691, 832051387531037141316, 14930556340948876798829

Offset: 0

Author

Stuart Clary, Jul 24 2009

Keywords

• nonn
• easy

Comments

Natural bilateral extension (brackets mark index 0): ..., -41635, -2331, -134, -9, -1, [0], 1, 9, 134, 2331, 41635, ... This is (-A163200)-reversed followed by A163200, without repeating the 0. That is, a(-n) = -a(n). Thus a(n) is an odd function of n.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..500
• 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 (21,-56,21,-1).

Crossrefs

Cf. A005968, A163198, A163201, A163202.

Programs

Magma

[(1/4)*Fibonacci(2*n)*(Fibonacci(2*n)^2+3): n in [0..20]]; // Vincenzo Librandi, Dec 10 2016

Mathematica

a[n_Integer] := If[ n >= 0, Sum[ Fibonacci[2k-1]^3, {k, 1, n} ], -Sum[ Fibonacci[-2k+1]^3, {k, 1, -n} ] ]
LinearRecurrence[{21,-56,21,-1}, {0,1,9,134}, 50] (* or *) Table[(1/20)*(Fibonacci[6*n] + 12*Fibonacci[2*n]),{n,0,25}] (* G. C. Greubel, Dec 09 2016 *)
Join[{0},Accumulate[Fibonacci[Range[1,41,2]]^3]] (* Harvey P. Dale, Jul 20 2021 *)

PARI

concat([0],Vec(x*(1 - 12*x + x^2)/((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-1)^3.

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

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

a(n) = (1/20)*F(2n)*(L(4n) + 13).

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

a(n) - 21*a(n-1) + 56*a(n-2) - 21*a(n-3) + a(n-4) = 0.

G.f.: (x - 12*x^2 + x^3)/(1 - 21*x + 56*x^2 - 21*x^3 + x^4) = x*(1 - 12*x + x^2)/((1 - 3*x + x^2 )*(1 - 18*x + x^2)).