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

2,1

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

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 *)

Prog

(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

Crossrefs

Cf. A001000, A024845.

Sequence in context: A118120 A078307 A045551 * A230285 A033582 A176362

Adjacent sequences: A024841 A024842 A024843 * A024845 A024846 A024847

Keyword

nonn

Author

Clark Kimberling

Status

approved