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Search: id:A119287
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| A119287 |
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Alternating sum of the sixth powers of the first n Fibonacci numbers. |
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+0
9
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| 0, -1, 0, -64, 665, -14960, 247184, -4579625, 81186496, -1463617920, 26217022705, -470764268256, 8445336180000, -151560390359569, 2719538168853120, -48800836192146880, 875690649999921929, -15713664197268146000, 281970036429821245616, -5059748557502924705465, 90793493265349521060160, -1629223203785737022267136, 29235223670642547226470625
(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): ..., 14960, -665, 64, 0, 1, 0, [0], -1, 0, -64, 665, -14960, 247184, ... This is (-A119287)-reversed followed by A119287.
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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)^6
Closed form: a(n) = (-1)^n (1/250) F(6n+3) - (6/125) F(4n+2) + (-1)^n (3/25) F(2n+1) - (2/25)(2 n + 1)
Recurrence: a(n) + 12 a(n-1) - 117 a(n-2) - 156 a(n-3) + 520 a(n-4) - 156 a(n-5) - 117 a(n-6) + 12 a(n-7) + a(n-8) = 0
G.f.: A(x) = (-x - 12 x^2 + 53 x^3 + 53 x^4 - 12 x^5 - x^6)/(1 + 12 x - 117 x^2 - 156 x^3 + 520 x^4 - 156 x^5 - 117 x^6 + 12 x^7 + x^8) = -x(1 + x)(1 + 11 x - 64 x^2 + 11 x^3 + x^4)/((1 - x)^2 (1 + 3 x + x^2)(1 - 7 x + x^2)(1 + 18 x + x^2))
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MATHEMATICA
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a[n_Integer] := If[ n >= 0, Sum[ (-1)^k Fibonacci[k]^6, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[ -k]^6, {k, 1, -n - 1} ] ]
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CROSSREFS
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Cf. A098532, A119282, A119283, A119284, A119285, A119286
Cf. A098532, A119282, A119283, A119284, A119285, A119286, A128696, A128698.
Sequence in context: A091083 A136950 A136957 this_sequence A008513 A048392 A074471
Adjacent sequences: A119284 A119285 A119286 this_sequence A119288 A119289 A119290
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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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