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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+3)/m < s for some integer k.
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%I #16 Aug 31 2022 23:06:37

%S 7,28,61,106,163,232,313,406,511,647,780,946,1105,1301,1486,1712,1923,

%T 2179,2416,2702,2965,3281,3570,3916,4231,4607,4999,5356,5778,6216,

%U 6613,7081,7565,8002,8516,9046,9523,10083,10659,11176,11782,12404,12961,13613,14281,14878

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

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

%C From _Jianing Song_, Aug 31 2022: (Start)

%C Smallest m such that ceiling(m/(2*j-1)) - floor(m/(2*j+1)) = 5 for 1 <= j <= n-1.

%C Obviously we have a(n) > 3/(1/(2*n-3) - 1/(2*n-1)) => a(n) >= 6*n^2 - 12*n + 5. On the other hand, a(n) <= 4/(1/(2*n-3) - 1/(2*n-1)) + 1 = 2*(2*n-1)*(2*n-3) + 1: if m >= 2*(2*n-1)*(2*n-3) + 1, then m/(2*j-1) - m/(2*j+1) > 4 => ceiling(m/(2*j-1)) - floor(m/(2*j+1)) = ceiling(m/(2*j-1)-floor(m/(2*j+1))) >= ceiling(m/(2*j-1) - m/(2*j+1)) >= 5 for 1 <= j <= n-1. (End)

%H Clark Kimberling, <a href="/A024844/b024844.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[[4]] (* A024844 *)

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

%o (PARI) a(n) = for(m=6*n^2-12*n+5, 8*n^2-16*n+7, forstep(j=n-1, 1, -1, if(-((-m)\(2*j-1)) - m\(2*j+1) < 5, break(), if(j==1, return(m))))) \\ _Jianing Song_, Aug 31 2022

%Y Cf. A001000, A024845.

%K nonn

%O 2,1

%A _Clark Kimberling_