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A024831 a(n) = least m such that if r and s in {F(h)/F(2*h): h = 1,2,...,n} satisfy r < s, then r < k/m < s for some integer k, where F = A000045 (Fibonacci numbers). 1

%I #34 Feb 25 2024 13:06:19

%S 2,7,10,10,15,23,37,59,95,153,247,399,645,1043,1687,2729,4415,7143,

%T 11557,18699,30255,48953,79207,128159,207365,335523,542887,878409,

%U 1421295,2299703,3720997,6020699,9741695,15762393,25504087,41266479,66770565,108037043,174807607,282844649

%N a(n) = least m such that if r and s in {F(h)/F(2*h): h = 1,2,...,n} satisfy r < s, then r < k/m < s for some integer k, where F = A000045 (Fibonacci numbers).

%C Note that F(2*h)/F(h) = Lucas(h) for h > 0. - Editors.

%C For a guide to related sequences, see A001000. - _Clark Kimberling_, Aug 07 2012

%H W. Kuszmaul, <a href="http://arxiv.org/abs/1509.08216">Fast Algorithms for Finding Pattern Avoiders and Counting Pattern Occurrences in Permutations</a>, arXiv preprint arXiv:1509.08216 [cs.DM], 2015-2017.

%F From _Philippe Deléham_, Feb 06 2024: (Start)

%F a(n) = a(n-1) + a(n-2) - 1 for n >= 8.

%F a(n) = 2*a(n-1) - a(n-3) for n >= 9.

%F a(n) = 1 + A022112(n-3) for n >= 6.

%F a(n) = floor(((1 + sqrt(5))/2)*a(n-1)) for n >= 8.

%F G.f.: x^2*(x^6+3*x^5+2*x^4-8*x^3-4*x^2+3*x+2)/((x-1)*(x^2+x-1)).

%F (End)

%t leastSeparator[seq_] := Module[{n = 1},

%t Table[While[Or @@ (Ceiling[n #1[[1]]] <

%t 2 + Floor[n #1[[2]]] &) /@ (Sort[#1, Greater] &) /@

%t Partition[Take[seq, k], 2, 1], n++]; n, {k, 2, Length[seq]}]];

%t t = Table[N[Fibonacci[h]/Fibonacci[2 h]], {h, 1, 30}]

%t t1 = leastSeparator[t]

%t (* _Peter J. C. Moses_, Aug 01 2012 *)

%Y Cf. A001000, A001622, A022112.

%K nonn

%O 2,1

%A _Clark Kimberling_

%E All the terms were corrected by _Clark Kimberling_, Aug 07 2012

%E More terms from _Sean A. Irvine_, Jul 25 2019

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Last modified September 7 01:14 EDT 2024. Contains 375728 sequences. (Running on oeis4.)