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A156094
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5 F(2n) (F(2n) - 1) + 1 where F(n) denotes the n-th Fibonacci number.
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4
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1, 1, 31, 281, 2101, 14851, 102961, 708761, 4865911, 33372361, 228792301, 1568309051, 10749725281, 73680695281, 505017569551, 3461448647801, 23725139605861, 162614572159411, 1114576979567761, 7639424583421961, 52361395886149351
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| Natural bilateral extension (brackets mark index 0): ..., 15401, 2311, 361, 61, 11, [1], 1, 31, 281, 2101, 14851, ... This is A156095-reversed followed by A156094, without repeating the central 1. That is, A156094(-n) = A156095(n).
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FORMULA
| Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).
Alternate formula: a(n) = L(4n) - 5 F(2n) - 1
Recurrence: a(n) - 10 a(n-1) + 23 a(n-2) - 10 a(n-3) + a(n-4) = -5
Recurrence: a(n) - 11 a(n-1) + 33 a(n-2) - 33 a(n-3) + 11 a(n-4) - a(n-5) = 0
G.f.: A(x) = (1 - 10 x + 53 x^2 - 60 x^3 + 11 x^4)/(1 - 11 x + 33 x^2 - 33 x^3 + 11 x^4 - x^5) = (1 - 10 x + 53 x^2 - 60 x^3 + 11 x^4)/((1 - x) (1 - 7 x + x^2) (1 - 3 x + x^2))
a(n)=A056854(n)-5*A001906(n)-1. [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Feb 23 2009]
a(n)=((2*sqrt(5))/2)*(((3-sqrt(5))/2)^n-((3+sqrt(5))/2)^n)+((7+3*sqrt(5))/2)^n+((7-3*sqrt(5))/2)^n-1. [Tim Monahan Aug 15 2011]
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MATHEMATICA
| a[n_Integer] := 5 Fibonacci[2n] (Fibonacci[2n] - 1) + 1
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CROSSREFS
| Cf. A124296, A124297, A001603, A001604, A156095
Sequence in context: A126526 A008386 A161558 * A115151 A001298 A027841
Adjacent sequences: A156091 A156092 A156093 * A156095 A156096 A156097
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KEYWORD
| nonn,easy
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AUTHOR
| Stuart Clary (clary(AT)uakron.edu), Feb 4, 2009
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