A024851 Least m such that if r and s in {-F(2*h) + tau*F(2*h-1): h = 1,2,...,n} satisfy r < s, then r < k/m < s for some integer k, where F = A000045 (Fibonacci numbers) and tau = (1+sqrt(5))/2 (golden ratio).
2, 5, 12, 30, 77, 200, 522, 1365, 3572, 9350, 24477, 64080, 167762, 439205, 1149852, 3010350, 7881197, 20633240
Offset: 2
Examples
Referring to the terminology introduced at A001000, m=5 is the (1st) separator of the set S = {f(1),f(2),f(3)}, where f(h) = - F(2*h) + tau*F(2*h-1). That is, a(3) = 5, since 1/5 < f(3) < 2/5 < f(2) < 3/5 < f(1), whereas fractions k/m for m<5 do not separate the elements of S in this manner.
Crossrefs
Cf. A001000.
Programs
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Mathematica
f[n_] := f[n] = -Fibonacci[2 n] + GoldenRatio*Fibonacci[2 n - 1] 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[f[h], 40], {h, 1, 18}] (* A024851 *) t1 = leastSeparator[t] (* Peter J. C. Moses, Aug 01 2012 *)
Extensions
Extended, corrected, and edited by Clark Kimberling, Aug 09 2012
a(19) from Sean A. Irvine, Jul 26 2019
Comments