A024851 - OEIS

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

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OFFSET

2,1

COMMENTS

Possibly a duplicate of A188378. For a guide to related sequences, see A001000. - Clark Kimberling, Aug 09 2012

LINKS

Table of n, a(n) for n=2..19.

EXAMPLE

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.

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 *)

CROSSREFS

Cf. A001000.

Sequence in context: A026938 A086622 A253831 * A188378 A145267 A103287