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 A156093 One ninth of the alternating sum of the squares of the first n Fibonacci numbers with index divisible by 4. 8

%I

%S 0,-1,48,-2256,105985,-4979040,233908896,-10988739073,516236827536,

%T -24252142155120,1139334444463105,-53524466747610816,

%U 2514510602693245248,-118128473859834915841,5549523760809547799280,-260709488284188911650320

%N One ninth of the alternating sum of the squares of the first n Fibonacci numbers with index divisible by 4.

%C Natural bilateral extension (brackets mark index 0): ..., -105985, 2256, -48, 1, 0, [0], -1, 48, -2256, 105985, -4979040, ... This is (-A156093)-reversed followed by A156093. That is, A156093(-n) = -A156093(n-1).

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (-48,-48,-1).

%F Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).

%F a(n) = (1/9) sum_{k=1..n} (-1)^k F(4k)^2.

%F Closed form: a(n) = (-1)^n (L(8n+4) - 7)/315.

%F Factored closed form: a(n) = (-1)^n F(4n) F(4n+4)/63.

%F Recurrence: a(n) + 47 a(n-1) + a(n-2) = (-1)^n.

%F Recurrence: a(n) + 48 a(n-1) + 48 a(n-2) + a(n-3) = 0.

%F G.f.: A(x) = -x/(1 + 48 x + 48 x^2 + x^3) = -x/((1 + x)(1 + 47 x + x^2)).

%t a[n_Integer] := If[ n >= 0, Sum[ (-1)^k (1/9) Fibonacci[4k]^2, {k, 1, n} ], Sum[ -(-1)^k (1/9) Fibonacci[-4k]^2, {k, 1, -n - 1} ] ]

%Y Cf. A156086, A156087, A156092.

%K sign,easy

%O 0,3

%A _Stuart Clary_, Feb 04 2009

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Last modified June 23 18:45 EDT 2021. Contains 345402 sequences. (Running on oeis4.)