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Search: id:A156093
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| A156093 |
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One ninth of the alternating sum of the squares of the first n Fibonacci numbers with index divisible by 4. |
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+0
5
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| 0, -1, 48, -2256, 105985, -4979040, 233908896, -10988739073, 516236827536, -24252142155120, 1139334444463105, -53524466747610816, 2514510602693245248, -118128473859834915841, 5549523760809547799280, -260709488284188911650320
(list; graph; listen)
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OFFSET
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0,3
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COMMENT
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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).
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FORMULA
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Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).
a(n) = (1/9) sum_{k=1..n} (-1)^k F(4k)^2
Closed form: a(n) = (-1)^n (L(8n+4) - 7)/315
Factored closed form: a(n) = (-1)^n F(n) F(4n+4)/63
Recurrence: a(n) + 47 a(n-1) + a(n-2) = (-1)^n
Recurrence: a(n) + 48 a(n-1) + 48 a(n-2) + a(n-3) = 0
G.f.: A(x) = -x/(1 + 48 x + 48 x^2 + x^3) = -x/((1 + x)(1 + 47 x + x^2))
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MATHEMATICA
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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} ] ]
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CROSSREFS
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Cf. A156086, A156087, A156092
Sequence in context: A152068 A165043 A049678 this_sequence A063822 A158783 A009992
Adjacent sequences: A156090 A156091 A156092 this_sequence A156094 A156095 A156096
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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), Feb 4, 2009
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