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 A133641 a(n) = 2*L(n) + L(n-1) - n, L(n) = n-th Lucas number of A000032 starting (1,3,4,...). =. 0
 1, 5, 8, 14, 24, 41, 69, 115, 190, 312, 510, 831, 1351, 2193, 3556, 5762, 9332, 15109, 24457, 39583, 64058, 103660, 167738, 271419, 439179, 710621, 1149824, 1860470, 3010320, 4870817, 7881165, 12752011, 20633206, 33385248, 54018486, 87403767, 141422287, 228826089, 370248412 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS a(n)/a(n-1) tends to phi. LINKS W. Kuszmaul, Fast Algorithms for Finding Pattern Avoiders and Counting Pattern Occurrences in Permutations, arXiv preprint arXiv:1509.08216 [cs.DM], 2015-2017. FORMULA Given n-th Lucas number of A000032 starting (1, 3, 4, 7,...), a(n) = 2*L(n) + L(n-1) - n. G.f.: -x*(1-5*x^2+x^3+2*x+2*x^4)/(-1+x+x^2)/(-1+x)^2. - R. J. Mathar, Nov 14 2007 a(n) = A000032(n+2)-n = Fibonacci(n + 3) + Fibonacci(n + 1) - n, n>1. [R. J. Mathar, Jul 20 2009, extended by David A. Corneth, Aug 08 2018] EXAMPLE a(5) = 24 = 2*L(5) + L(4) - n = 2*11 + 7 - 5. MATHEMATICA a[1] = 1; a[n_] := LucasL[n+2] - n; Array[a, 14] (* Jean-François Alcover, Aug 08 2018, after R. J. Mathar *) PROG (PARI) a(n) = {if(n==1, 1, fibonacci(n+3)+fibonacci(n+1)-n)} \\ David A. Corneth, Aug 08 2018 CROSSREFS Cf. A000032. Sequence in context: A124011 A101835 A192522 * A164094 A246319 A302649 Adjacent sequences:  A133638 A133639 A133640 * A133642 A133643 A133644 KEYWORD nonn AUTHOR Gary W. Adamson, Sep 19 2007 STATUS approved

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Last modified April 20 18:45 EDT 2021. Contains 343137 sequences. (Running on oeis4.)