A119284 Alternating sum of the cubes of the first n Fibonacci numbers.

0, -1, 0, -8, 19, -106, 406, -1791, 7470, -31834, 134541, -570428, 2415556, -10233781, 43348852, -183632148, 777872655, -3295130518, 13958382186, -59128679555, 250473067570, -1061021002966, 4494556993465, -19039249115928, 80651553232104, -341645462408521, 1447233402276936, -6130579072469696, 25969549690613035, -110008777837417954, 466004661036246046

Offset: 0

Author

Stuart Clary, May 13 2006

Keywords

• sign
• easy

Comments

Natural bilateral extension (brackets mark index 0): ..., 674, 162, 37, 10, 2, 1, 0, [0], -1, 0, -8, 19, -106, 406, -1791, ... This is A005968-reversed followed by A119284.

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (-2,9,-3,-4,1).

Crossrefs

Cf. A005968, A119282, A119283, A119285, A119286, A119287, A128696, A128698.

Programs

Mathematica

a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[k]^3, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k]^3, {k, 1, -n - 1} ] ]
Accumulate[Times@@@Partition[Riffle[Fibonacci[Range[0,30]]^3,{1,-1},{2,-1,2}],2]] (* or *) LinearRecurrence[{-2,9,-3,-4,1},{0,-1,0,-8,19},40] (* Harvey P. Dale, Aug 23 2020 *)

Formula

Let F(n) be the Fibonacci number A000045(n).

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

Closed form: a(n) = (-1)^n F(3n+1)/10 - 3 F(n+2)/5 + 1/2.

Recurrence: a(n) + 2 a(n-1) - 9 a(n-2) + 3 a(n-3) + 4 a(n-4) - a(n-5) = 0.

G.f.: A(x) = (-x - 2 x^2 + x^3)/(1 + 2 x - 9 x^2 + 3 x^3 + 4 x^4 - x^5) = x(-1 - 2 x + x^2)/((1 - x)(1 - x - x^2 )(1 + 4 x - x^2)).