%I #15 Dec 09 2016 15:47:36

%S 0,-1,0,-64,665,-14960,247184,-4579625,81186496,-1463617920,

%T 26217022705,-470764268256,8445336180000,-151560390359569,

%U 2719538168853120,-48800836192146880,875690649999921929,-15713664197268146000,281970036429821245616,-5059748557502924705465,90793493265349521060160,-1629223203785737022267136,29235223670642547226470625

%N Alternating sum of the sixth powers of the first n Fibonacci numbers.

%C Natural bilateral extension (brackets mark index 0): ..., 14960, -665, 64, 0, 1, 0, [0], -1, 0, -64, 665, -14960, 247184, ... This is (-A119287)-reversed followed by A119287.

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (-12,117,156,-520,156,117,-12,-1).

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

%F a(n) = Sum_{k=1..n} (-1)^k F(k)^6.

%F a(n) = (-1)^n (1/250) F(6n+3) - (6/125) F(4n+2) + (-1)^n (3/25) F(2n+1) - (2/25)(2 n + 1).

%F Recurrence: a(n) + 12 a(n-1) - 117 a(n-2) - 156 a(n-3) + 520 a(n-4) - 156 a(n-5) - 117 a(n-6) + 12 a(n-7) + a(n-8) = 0.

%F G.f.: A(x) = (-x - 12 x^2 + 53 x^3 + 53 x^4 - 12 x^5 - x^6)/(1 + 12 x - 117 x^2 - 156 x^3 + 520 x^4 - 156 x^5 - 117 x^6 + 12 x^7 + x^8) = -x(1 + x)(1 + 11 x - 64 x^2 + 11 x^3 + x^4)/((1 - x)^2 (1 + 3 x + x^2)(1 - 7 x + x^2)(1 + 18 x + x^2)).

%t a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[k]^6, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k]^6, {k, 1, -n - 1} ] ]

%t Accumulate[Times@@@Partition[Riffle[Fibonacci[Range[0,30]]^6,{1,-1},{2,-1,2}],2]] (* _Harvey P. Dale_, Jul 23 2013 *)

%Y Cf. A098532, A119282, A119283, A119284, A119285, A119286, A128696, A128698.

%K sign,easy

%O 0,4

%A _Stuart Clary_, May 13 2006