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%I #41 Jan 09 2024 16:09:14
%S 2,5,12,30,77,200,522,1365,3572,9350,24477,64080,167762,439205,
%T 1149852,3010350,7881197,20633240
%N 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).
%C Possibly a duplicate of A188378. For a guide to related sequences, see A001000. - _Clark Kimberling_, Aug 09 2012
%e 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.
%t f[n_] := f[n] = -Fibonacci[2 n] + GoldenRatio*Fibonacci[2 n - 1]
%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[f[h], 40], {h, 1, 18}] (* A024851 *)
%t t1 = leastSeparator[t]
%t (* _Peter J. C. Moses_, Aug 01 2012 *)
%Y Cf. A001000.
%K nonn,more
%O 2,1
%A _Clark Kimberling_
%E Extended, corrected, and edited by _Clark Kimberling_, Aug 09 2012
%E a(19) from _Sean A. Irvine_, Jul 26 2019
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