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%I #8 Feb 10 2014 01:31:47
%S 0,-1,3,-22,147,-1009,6912,-47377,324723,-2225686,15255075,-104559841,
%T 716663808,-4912086817,33667943907,-230763520534,1581676699827,
%U -10840973378257,74305136947968,-509294985257521,3490759759854675
%N Alternating sum of the squares of the first n odd-indexed Fibonacci numbers.
%C Natural bilateral extension (brackets mark index 0): ..., 6912, -1009, 147, -22, 3, -1, [0], -1, 3, -22, 147, -1009, 6912, ... This is A156089-reversed followed by A156089, without repeating the 0. That is, A156089(-n) = A156089(n).
%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)^2.
%F Closed form: a(n) = (-1)^n (L(4n) + 3)/15 - 1/3.
%F Factored closed form: a(n) = (1/3) F(2n)^2 if n is even.
%F Not-so-factored closed form: a(n) = -(F(2n)^2 + 2)/3 if n is odd.
%F Recurrence: a(n) + 7 a(n-1) - 7 a(n-3) - a(n-4) = 0.
%F G.f.: A(x) = (-x - 4 x^2 - x^3)/(1 + 7 x - 7 x^3 - x^4) = -x (1 + 4 x + x^2)/((1 - x)(1 + x)(1 + 7 x + x^2)).
%t a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[2k-1]^2, {k, 1, n} ], -Sum[ -(-1)^k Fibonacci[-2k+1]^2, {k, 1, -n} ] ]
%t Join[{0},Accumulate[Times@@@Partition[Riffle[Fibonacci[Range[1,43,2]]^2, {-1,1}],2]]] (* _Harvey P. Dale_, Aug 18 2011 *)
%Y Cf. A103434, A103433, A156088.
%K sign,easy
%O 0,3
%A _Stuart Clary_, Feb 04 2009
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