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Search: id:A119283
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| A119283 |
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Alternating sum of the squares of the first n Fibonacci numbers. |
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+0
10
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| 0, -1, 0, -4, 5, -20, 44, -125, 316, -840, 2185, -5736, 15000, -39289, 102840, -269260, 704909, -1845500, 4831556, -12649205, 33116020, -86698896, 226980625, -594243024, 1555748400, -4073002225, 10663258224, -27916772500, 73087059221, -191344405220, 500946156380, -1311494063981, 3433536035500, -8989114042584, 23533806092185, -61612304234040
(list; graph; listen)
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OFFSET
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0,4
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COMMENT
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Natural bilateral extension (brackets mark index 0): ..., 840, -316, 125, -44, 20, -5, 4, 0, 1, 0, [0], -1, 0, -4, 5, -20, 44, -125, 316, -840, 2185, ... This is (-A119283)-reversed followed by A119283.
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FORMULA
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Let F(n) be the Fibonacci number A000045(n).
a(n) = sum_{k=1..n} (-1)^k F(k)^2
Closed form: a(n) = (-1)^n F(2n+1)/5 - (2 n + 1)/5
Recurrence: a(n) + a(n-1) - 4 a(n-2) + a(n-3) + a(n-4) = 0
G.f.: A(x) = (-x - x^2)/(1 + x - 4 x^2 + x^3 + x^4) = -x(1 + x)/((1 - x)^2 (1 + 3 x + x^2))
a(n)= [ -1-2*n+(-1)^n*( A001906(n+1)-A001906(n) )]/5. - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 16 2007
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MATHEMATICA
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a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[k]^2, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k]^2, {k, 1, -n - 1} ] ]
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CROSSREFS
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Cf. A001654, A119282, A119284, A119285, A119286, A119287
Cf. A001654, A119282, A119284, A119285, A119286, A119287, A128696, A128698.
Adjacent sequences: A119280 A119281 A119282 this_sequence A119284 A119285 A119286
Sequence in context: A059182 A027958 A064670 this_sequence A057781 A081713 A120697
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KEYWORD
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sign,easy
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AUTHOR
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Stuart Clary (clary(AT)uakron.edu), May 13, 2006
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