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).

2, 7, 10, 10, 15, 23, 37, 59, 95, 153, 247, 399, 645, 1043, 1687, 2729, 4415, 7143, 11557, 18699, 30255, 48953, 79207, 128159, 207365, 335523, 542887, 878409, 1421295, 2299703, 3720997, 6020699, 9741695, 15762393, 25504087, 41266479, 66770565, 108037043, 174807607, 282844649

Offset: 2

Author

Clark Kimberling

Keywords

• nonn

Comments

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

For a guide to related sequences, see A001000. - Clark Kimberling, Aug 07 2012

Links

• W. Kuszmaul, Fast Algorithms for Finding Pattern Avoiders and Counting Pattern Occurrences in Permutations, arXiv preprint arXiv:1509.08216 [cs.DM], 2015-2017.

Crossrefs

Cf. A001000, A001622, A022112.

Programs

Mathematica

leastSeparator[seq_] := Module[{n = 1},
Table[While[Or @@ (Ceiling[n #1[[1]]] <
2 + Floor[n #1[[2]]] &) /@ (Sort[#1, Greater] &) /@
Partition[Take[seq, k], 2, 1], n++]; n, {k, 2, Length[seq]}]];
t = Table[N[Fibonacci[h]/Fibonacci[2 h]], {h, 1, 30}]
t1 = leastSeparator[t]
(* Peter J. C. Moses, Aug 01 2012 *)

Formula

From Philippe Deléham, Feb 06 2024: (Start)

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

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

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

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

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)).

(End)

Extensions

All the terms were corrected by Clark Kimberling, Aug 07 2012

More terms from Sean A. Irvine, Jul 25 2019