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%I #5 Feb 10 2014 01:31:32
%S 0,-9,432,-20304,953865,-44811360,2105180064,-98898651657,
%T 4646131447824,-218269279396080,10254010000167945,-481720200728497344,
%U 22630595424239207232,-1063156264738514242569,49945713847285930193520,-2346385394557700204852880
%N Alternating sum of the squares of the first n Fibonacci numbers with index divisible by 4.
%C Natural bilateral extension (brackets mark index 0): ..., -953865, 20304, -432, 9, 0, [0], -9, 432, -20304, 953865, -44811360, ... This is (-A156092)-reversed followed by A156092. That is, A156092(-n) = -A156092(n-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(4k)^2.
%F Closed form: a(n) = (-1)^n (L(8n+4) - 7)/35.
%F Factored closed form: a(n) = (-1)^n F(4n) F(4n+4)/7.
%F Recurrence: a(n) + 47 a(n-1) + a(n-2) = (-1)^n 9.
%F Recurrence: a(n) + 48 a(n-1) + 48 a(n-2) + a(n-3) = 0.
%F G.f.: A(x) = -9 x/(1 + 48 x + 48 x^2 + x^3) = -9 x/((1 + x)(1 + 47 x + x^2)).
%t a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[4k]^2, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[-4k]^2, {k, 1, -n - 1} ] ]
%Y Cf. A156086, A156087, A156093.
%K sign,easy
%O 0,2
%A _Stuart Clary_, Feb 04 2009
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