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Search: id:A119284
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| A119284 |
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Alternating sum of the cubes of the first n Fibonacci numbers. |
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+0
11
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| 0, -1, 0, -8, 19, -106, 406, -1791, 7470, -31834, 134541, -570428, 2415556, -10233781, 43348852, -183632148, 777872655, -3295130518, 13958382186, -59128679555, 250473067570, -1061021002966, 4494556993465, -19039249115928, 80651553232104, -341645462408521, 1447233402276936, -6130579072469696, 25969549690613035, -110008777837417954, 466004661036246046
(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): ..., 674, 162, 37, 10, 2, 1, 0, [0], -1, 0, -8, 19, -106, 406, -1791, ... This is A005968-reversed followed by A119284.
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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)^3
Closed form: a(n) = (-1)^n F(3n+1)/10 - 3 F(n+2)/5 + 1/2
Recurrence: a(n) + 2 a(n-1) - 9 a(n-2) + 3 a(n-3) + 4 a(n-4) - a(n-5) = 0
G.f.: A(x) = (-x - 2 x^2 + x^3)/(1 + 2 x - 9 x^2 + 3 x^3 + 4 x^4 - x^5) = x(-1 - 2 x + x^2)/((1 - x)(1 - x - x^2 )(1 + 4 x - x^2))
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MATHEMATICA
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a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[k]^3, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k]^3, {k, 1, -n - 1} ] ]
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CROSSREFS
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Cf. A005968, A119282, A119283, A119285, A119286, A119287
Cf. A005968, A119282, A119283, A119285, A119286, A119287, A128696, A128698.
Sequence in context: A091560 A153703 A061877 this_sequence A153704 A029845 A124972
Adjacent sequences: A119281 A119282 A119283 this_sequence A119285 A119286 A119287
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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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