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A024851
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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).
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3
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2, 5, 12, 30, 77, 200, 522, 1365, 3572, 9350, 24477, 64080, 167762, 439205, 1149852, 3010350, 7881197, 20633240
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OFFSET
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2,1
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COMMENTS
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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]
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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