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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+2)/m < s for some integer k.
3

%I #13 Jun 27 2022 18:52:26

%S 5,19,41,71,109,155,222,287,376,460,571,673,806,926,1081,1219,1396,

%T 1552,1751,1926,2147,2380,2584,2839,3106,3338,3627,3928,4188,4511,

%U 4846,5134,5491,5860,6176,6567,6970,7385,7740,8177,8626,9087,9481,9964,10459,10966,11398

%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+2)/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="/A024841/b024841.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[[3]] (* A024841 *)

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

%Y Cf. A001000, A024842.

%K nonn

%O 2,1

%A _Clark Kimberling_