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A156095 5 F(2n) (F(2n) + 1) + 1 where F(n) denotes the n-th Fibonacci number. 4
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; text; internal format)
OFFSET
0,2
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).
LINKS
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
MATHEMATICA
a[n_Integer] := 5 Fibonacci[2n] (Fibonacci[2n] + 1) + 1
CROSSREFS
Sequence in context: A137410 A268863 A199414 * A068846 A092164 A354336
KEYWORD
nonn,easy
AUTHOR
Stuart Clary, Feb 04 2009
STATUS
approved

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Last modified April 23 19:56 EDT 2024. Contains 371916 sequences. (Running on oeis4.)