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A156095
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5 F(2n) (F(2n) + 1) + 1 where F(n) denotes the n-th Fibonacci number.
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3
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1, 11, 61, 361, 2311, 15401, 104401, 712531, 4875781, 33398201, 228859951, 1568486161, 10750188961, 73681909211, 505020747661, 3461456968201, 23725161388951, 162614629188281, 1114577128871281, 7639424974303651, 52361396909490901
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| Natural bilateral extension (brackets mark index 0): ..., 14851, 2101, 281, 31, 1, [1], 11, 61, 361, 2311, 15401, ... This is A156095-reversed followed by A156095, without repeating the central 1. That is, A156095(-n) = A156094(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 - 27 x^2 + 20 x^3 + x^4)/(1 - 11 x + 33 x^2 - 33 x^3 + 11 x^4 - x^5) = (1 - 27 x^2 + 20 x^3 + x^4)/((1 - x) (1 - 7 x + x^2) (1 - 3 x + x^2))
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, A156094
Sequence in context: A115565 A137410 A199414 * A068846 A092164 A088545
Adjacent sequences: A156092 A156093 A156094 * A156096 A156097 A156098
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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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