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a(n) = least m such that if r and s in {1/1, 1/3, 1/5, ..., 1/(2n-1)} satisfy r < s, then r < k/m < (k+1)/m < s for some integer k.
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%I #13 Jun 27 2022 11:08:35

%S 4,13,26,43,64,100,133,183,226,290,343,421,484,576,676,757,871,993,

%T 1090,1226,1370,1483,1641,1807,1936,2116,2304,2500,2653,2863,3081,

%U 3307,3482,3722,3970,4226,4423,4693,4971,5257,5476,5776,6084,6400,6724,6973,7311,7657,8011

%N a(n) = least m such that if r and s in {1/1, 1/3, 1/5, ..., 1/(2n-1)} satisfy r < s, then r < k/m < (k+1)/m < s for some integer k.

%C For a guide to related sequences, see A001000. - _Clark Kimberling_, Aug 12 2012

%H Clark Kimberling, <a href="/A024834/b024834.txt">Table of n, a(n) for n = 2..100</a>

%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/(2*Range[50] - 1), #] &, Range[5]];

%t t[[2]] (* A024834 *)

%t (* _Peter J. C. Moses_, Aug 06 2012 *)

%Y Cf. A001000, A024835.

%K nonn

%O 2,1

%A _Clark Kimberling_