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.

7, 28, 61, 106, 163, 232, 313, 406, 511, 647, 780, 946, 1105, 1301, 1486, 1712, 1923, 2179, 2416, 2702, 2965, 3281, 3570, 3916, 4231, 4607, 4999, 5356, 5778, 6216, 6613, 7081, 7565, 8002, 8516, 9046, 9523, 10083, 10659, 11176, 11782, 12404, 12961, 13613, 14281, 14878

Offset: 2

Author

Clark Kimberling

Keywords

• nonn

Comments

For a guide to related sequences, see A001000. - Clark Kimberling, Aug 12 2012

From Jianing Song, Aug 31 2022: (Start)

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

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)

Links

• Clark Kimberling, Table of n, a(n) for n = 2..100

Crossrefs

Cf. A001000, A024845.

Programs

Mathematica

leastSeparatorS[seq_, s_] := Module[{n = 1},
Table[While[Or @@ (Ceiling[n #1[[1]]] <
s + 1 + Floor[n #1[[2]]] &) /@ (Sort[#1, Greater] &) /@
Partition[Take[seq, k], 2, 1], n++]; n, {k, 2, Length[seq]}]];
t = Map[leastSeparatorS[1/(2*Range[50]-1), #] &, Range[5]];
t[[4]] (* A024844 *)
(* Peter J. C. Moses, Aug 06 2012 *)

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