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A156089
Alternating sum of the squares of the first n odd-indexed Fibonacci numbers.
3
0, -1, 3, -22, 147, -1009, 6912, -47377, 324723, -2225686, 15255075, -104559841, 716663808, -4912086817, 33667943907, -230763520534, 1581676699827, -10840973378257, 74305136947968, -509294985257521, 3490759759854675
OFFSET
0,3
COMMENTS
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).
FORMULA
Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).
a(n) = sum_{k=1..n} (-1)^k F(2k-1)^2.
Closed form: a(n) = (-1)^n (L(4n) + 3)/15 - 1/3.
Factored closed form: a(n) = (1/3) F(2n)^2 if n is even.
Not-so-factored closed form: a(n) = -(F(2n)^2 + 2)/3 if n is odd.
Recurrence: a(n) + 7 a(n-1) - 7 a(n-3) - a(n-4) = 0.
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)).
MATHEMATICA
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} ] ]
Join[{0}, Accumulate[Times@@@Partition[Riffle[Fibonacci[Range[1, 43, 2]]^2, {-1, 1}], 2]]] (* Harvey P. Dale, Aug 18 2011 *)
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Stuart Clary, Feb 04 2009
STATUS
approved