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Search: id:A156092
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| A156092 |
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Alternating sum of the squares of the first n Fibonacci numbers with index divisible by 4. |
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+0
4
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| 0, -9, 432, -20304, 953865, -44811360, 2105180064, -98898651657, 4646131447824, -218269279396080, 10254010000167945, -481720200728497344, 22630595424239207232, -1063156264738514242569, 49945713847285930193520, -2346385394557700204852880
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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Natural bilateral extension (brackets mark index 0): ..., -953865, 20304, -432, 9, 0, [0], -9, 432, -20304, 953865, -44811360, ... This is (-A156092)-reversed followed by A156092. That is, A156092(-n) = -A156092(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) = sum_{k=1..n} (-1)^k F(4k)^2
Closed form: a(n) = (-1)^n (L(8n+4) - 7)/35
Factored closed form: a(n) = (-1)^n F(4n) F(4n+4)/7
Recurrence: a(n) + 47 a(n-1) + a(n-2) = (-1)^n 9
Recurrence: a(n) + 48 a(n-1) + 48 a(n-2) + a(n-3) = 0
G.f.: A(x) = -9 x/(1 + 48 x + 48 x^2 + x^3) = -9 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 Fibonacci[4k]^2, {k, 1, n} ], Sum[ -(-1)^k Fibonacci[-4k]^2, {k, 1, -n - 1} ] ]
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CROSSREFS
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Cf. A156086, A156087, A156093
Sequence in context: A024123 A081481 A160376 this_sequence A092813 A167720 A069073
Adjacent sequences: A156089 A156090 A156091 this_sequence A156093 A156094 A156095
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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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