cp's OEIS Frontend

A024844 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.

%I A024844 #16 Aug 31 2022 23:06:37
%S A024844 7,28,61,106,163,232,313,406,511,647,780,946,1105,1301,1486,1712,1923,
%T A024844 2179,2416,2702,2965,3281,3570,3916,4231,4607,4999,5356,5778,6216,
%U A024844 6613,7081,7565,8002,8516,9046,9523,10083,10659,11176,11782,12404,12961,13613,14281,14878
%N A024844 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 A024844 For a guide to related sequences, see A001000. - _Clark Kimberling_, Aug 12 2012
%C A024844 From _Jianing Song_, Aug 31 2022: (Start)
%C A024844 Smallest m such that ceiling(m/(2*j-1)) - floor(m/(2*j+1)) = 5 for 1 <= j <= n-1.
%C A024844 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 A024844 Clark Kimberling, <a href="/A024844/b024844.txt">Table of n, a(n) for n = 2..100</a>
%t A024844 leastSeparatorS[seq_, s_] := Module[{n = 1},
%t A024844 Table[While[Or @@ (Ceiling[n #1[[1]]] <
%t A024844 s + 1 + Floor[n #1[[2]]] &) /@ (Sort[#1, Greater] &) /@
%t A024844 Partition[Take[seq, k], 2, 1], n++]; n, {k, 2, Length[seq]}]];
%t A024844 t = Map[leastSeparatorS[1/(2*Range[50]-1), #] &, Range[5]];
%t A024844 t[[4]] (* A024844 *)
%t A024844 (* _Peter J. C. Moses_, Aug 06 2012 *)
%o A024844 (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 A024844 Cf. A001000, A024845.
%K A024844 nonn
%O A024844 2,1
%A A024844 _Clark Kimberling_