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a(n) = least m such that if r and s in {1/1, 1/2, 1/3, ..., 1/n} satisfy r < s, then r < k/m < (k+1)/m < s for some integer k.
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%I #19 Dec 14 2021 10:30:54

%S 5,11,19,29,41,61,79,106,129,163,191,232,265,313,365,407,466,529,579,

%T 649,723,781,862,947,1013,1105,1201,1301,1379,1486,1597,1712,1801,

%U 1923,2049,2179,2279,2416,2557,2702,2813,2965,3121,3281,3445,3571,3742,3917,4096

%N a(n) = least m such that if r and s in {1/1, 1/2, 1/3, ..., 1/n} satisfy r < s, then r < k/m < (k+1)/m < s for some integer k.

%C For a guide to related sequences, see A001000. - _Peter J. C. Moses_, Aug 08 2012

%H Clark Kimberling, <a href="/A024833/b024833.txt">Table of n, a(n) for n = 2..300</a>

%e Using the terminology introduced at A001000, the 2nd separator of the set {1/3, 1/2, 1} is a(3) = 11, since 1/3 < 4/11 < 5/11 < 1/2 < 6/11 < 7/11 < 1 and 11 is the least m for which 1/3, 1/2, 1 are thus separated using numbers k/m. - _Clark Kimberling_, Aug 08 2012

%t leastSeparatorS[seq_, s_] := Module[{n = 1},

%t Table[While[Or @@ (Ceiling[n #1[[1]]] <

%t s + 1 + Floor[n #1[[2]]] &) /@ (Sort[#1, Greater] &) /@

%t Partition[Take[seq, k], 2, 1], n++]; n, {k, 2, Length[seq]}]];

%t t = Map[leastSeparatorS[1/Range[50], #] &, Range[5]];

%t TableForm[t]

%t t[[2]] (* _Clark Kimberling_, Aug 08 2012 *)

%Y Cf. A001000, A071111, A024843, A024846.

%K nonn

%O 2,1

%A _Clark Kimberling_