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%I A119286 #15 Jun 24 2018 12:47:53
%S A119286 0,-1,0,-32,211,-2914,29854,-341439,3742662,-41692762,461591613,
%T A119286 -5122467836,56794896388,-629924960005,6985721085652,-77473909014348,
%U A119286 859194263419359,-9528629686028398,105674040835291026,-1171943417651373875,12997050199917354250,-144139501695851560726,1598531543102764228825,-17727986584911448406232,196606383515036414871336,-2180398207207766329269289
%N A119286 Alternating sum of the fifth powers of the first n Fibonacci numbers.
%C A119286 Natural bilateral extension (brackets mark index 0): ..., 3402, 277, 34, 2, 1, 0, [0], -1, 0, -32, 211, -2914, 29854, ... This is A098531-reversed followed by A119286.
%H A119286 <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (-7,48,20,-100,32,9,-1)
%F A119286 Let F(n) be the Fibonacci number A000045(n).
%F A119286 a(n) = Sum_{k=1..n} (-1)^k F(k)^5.
%F A119286 Closed form: a(n) = (-1)^n (1/275)(F(5n+1) + 2 F(5n+3)) - (1/10) F(3n+2) + (-1)^n (2/5) F(n-1) - 7/22; here F(5n+1) + 2 F(5n+3) = A001060(5n+1) = A013655(5n+2).
%F A119286 Recurrence: a(n) + 7 a(n-1) - 48 a(n-2) - 20 a(n-3) + 100 a(n-4) - 32 a(n-5) - 9 a(n-6) + a(n-7) = 0.
%F A119286 G.f.: A(x) = (-x - 7 x^2 + 16 x^3 + 7 x^4 - x^5)/(1 + 7 x - 48 x^2 - 20 x^3 + 100 x^4 - 32 x^5 - 9 x^6 + x^7) = -x(1 + 7 x - 16 x^2 - 7 x^3 + x^4)/((1 - x)(1 + x - x^2)(1 - 4 x - x^2)(1 + 11 x - x^2)).
%t A119286 a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[k]^5, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k]^5, {k, 1, -n - 1} ] ]
%t A119286 LinearRecurrence[{-7,48,20,-100,32,9,-1},{0,-1,0,-32,211,-2914,29854},30] (* _Harvey P. Dale_, Jun 24 2018 *)
%Y A119286 Cf. A098531, A119282, A119283, A119284, A119285, A119287, A128696, A128698.
%K A119286 sign,easy
%O A119286 0,4
%A A119286 _Stuart Clary_, May 13 2006

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