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%I #9 Jun 13 2015 00:54:08
%S 0,-1,15,-610,27951,-1308385,61433856,-2885861665,135572548335,
%T -6369013518946,299207991620175,-14056406104466881,660351875572408320,
%U -31022481722865482305,1457396288941918481871,-68466603097469928960610
%N Alternating sum of the fourth powers of the first n odd-indexed Fibonacci numbers.
%C Natural bilateral extension (brackets mark index 0): ..., -1308385, 27951, -610, 15, -1, [0], -1, 15, -610, 27951, -1308385, ... That is, a(-n) = a(n).
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (-54,-330,0,330,54,1)
%F Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).
%F a(n) = sum_{k=1..n} (-1)^k F(2k-1)^4.
%F Closed form: a(n) = (-1)^n (1/525)(3 L(8n) + 28 L(4n) + 63 - (-1)^n 125).
%F Alternate closed form: a(n) = (1/21) F(2n)^2 (3 F(2n)^2 + 8) if n is even, a(n) = -(1/21)(3 F(2n)^4 + 8 F(2n)^2 + 10) if n is odd.
%F Recurrence: a(n) + 54 a(n-1) + 330 a(n-2) - 330 a(n-4) - 54 a(n-5) - a(n-6) = 0.
%F G.f.: A(x) = -(x + 39 x^2 + 130 x^3 + 39 x^4 + x^5)/(1 + 54 x + 330 x^2 - 330 x^4 - 54 x^5 - x^6) = -x(1 + 39 x + 130 x^2 + 39 x^3 + x^4)/((1 - x)(1 + x)(1 + 7 x + x^2)(1 + 47 x + x^2)).
%t a[n_Integer] := (-1)^n (1/525)(3*LucasL[8n] + 28*LucasL[4n] + 63 - (-1)^n 125); Table[a[n], {n, 0, 20}]
%Y Cf. A203169, A203170, A203171.
%Y Cf. A156089, A163202.
%K sign,easy
%O 0,3
%A _Stuart Clary_, Dec 30 2011
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