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 A156094 5 F(2n) (F(2n) - 1) + 1 where F(n) denotes the n-th Fibonacci number. 5
 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; text; internal format)
 OFFSET 0,3 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). 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 - 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. - R. J. Mathar, 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 MATHEMATICA a[n_Integer] := 5 Fibonacci[2n] (Fibonacci[2n] - 1) + 1 5(#*(#-1))&/@Fibonacci[Range[0, 40, 2]]+1 (* Harvey P. Dale, Jan 06 2013 *) CROSSREFS Cf. A124296, A124297, A001603, A001604, A156095 Sequence in context: A008386 A161558 A336279 * A221430 A115151 A327561 Adjacent sequences:  A156091 A156092 A156093 * A156095 A156096 A156097 KEYWORD nonn,easy AUTHOR Stuart Clary, Feb 04 2009 STATUS approved

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Last modified May 14 16:48 EDT 2021. Contains 343898 sequences. (Running on oeis4.)