cp's OEIS Frontend

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

%I A119284 #20 Aug 23 2020 12:35:59
%S A119284 0,-1,0,-8,19,-106,406,-1791,7470,-31834,134541,-570428,2415556,
%T A119284 -10233781,43348852,-183632148,777872655,-3295130518,13958382186,
%U A119284 -59128679555,250473067570,-1061021002966,4494556993465,-19039249115928,80651553232104,-341645462408521,1447233402276936,-6130579072469696,25969549690613035,-110008777837417954,466004661036246046
%N A119284 Alternating sum of the cubes of the first n Fibonacci numbers.
%C A119284 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.
%H A119284 Harvey P. Dale, <a href="/A119284/b119284.txt">Table of n, a(n) for n = 0..1000</a>
%H A119284 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (-2,9,-3,-4,1).
%F A119284 Let F(n) be the Fibonacci number A000045(n).
%F A119284 a(n) = Sum_{k=1..n} (-1)^k F(k)^3.
%F A119284 Closed form: a(n) = (-1)^n F(3n+1)/10 - 3 F(n+2)/5 + 1/2.
%F A119284 Recurrence: a(n) + 2 a(n-1) - 9 a(n-2) + 3 a(n-3) + 4 a(n-4) - a(n-5) = 0.
%F A119284 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)).
%t A119284 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} ] ]
%t A119284 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 *)
%Y A119284 Cf. A005968, A119282, A119283, A119285, A119286, A119287, A128696, A128698.
%K A119284 sign,easy
%O A119284 0,4
%A A119284 _Stuart Clary_, May 13 2006