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%I A119285 #18 Apr 02 2018 15:24:06
%S A119285 0,-1,0,-16,65,-560,3536,-25025,169456,-1166880,7983745,-54758496,
%T A119285 375223200,-2572072321,17628580320,-120829829680,828175410881,
%U A119285 -5676410656400,38906666170736,-266670338968385,1827785480332240,-12527828615754816,85867013279034625,-588541268397840576,4033921854875707200,-27648911743562183425
%N A119285 Alternating sum of the fourth powers of the first n Fibonacci numbers.
%C A119285 Natural bilateral extension (brackets mark index 0): ..., -3536, 560, -65, 16, 0, 1, 0, [0], -1, 0, -16, 65, -560, 3536, -25025, ... This is (-A119285)-reversed followed by A119285.
%H A119285 Kunle Adegoke, <a href="https://arxiv.org/abs/1706.00407">Sums of fourth powers of Fibonacci and Lucas numbers</a>, arXiv:1706.00407 [math.NT], 2017.
%H A119285 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (-5,15,15,-5,-1).
%F A119285 Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).
%F A119285 a(n) = Sum_{k=1..n} (-1)^k F(k)^4.
%F A119285 Closed form: a(n) = (-1)^n L(4n+2)/75 - (4/25) L(2n+1) + (-1)^n 3/25.
%F A119285 Factored closed form: a(n) = (-1)^n (1/3) F(n-2) F(n) F(n+1) F(n+3).
%F A119285 Recurrence: a(n) + 5 a(n-1) - 15 a(n-2) - 15 a(n-3) + 5 a(n-4) + a(n-5) = 0.
%F A119285 G.f.: A(x) = (-x - 5 x^2 - x^3)/(1 + 5 x - 15 x^2 - 15 x^3 + 5 x^4 + x^5) = -x(1 + 5 x + x^2)/((1 + x)(1 - 3 x + x^2)(1 + 7 x + x^2)).
%t A119285 a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[k]^4, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k]^4, {k, 1, -n - 1} ] ]
%t A119285 LinearRecurrence[{-5,15,15,-5,-1},{0,-1,0,-16,65},30] (* _Harvey P. Dale_, Apr 02 2018 *)
%Y A119285 Cf. A005969, A119282, A119283, A119284, A119286, A119287, A128696, A128698.
%K A119285 sign,easy
%O A119285 0,4
%A A119285 _Stuart Clary_, May 13 2006

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