A024845 a(n) = least m such that if r and s in {1/2, 1/4, 1/6, ..., 1/(2*n)} satisfy r < s, then r < k/m < (k+3)/m < s for some integer k.

15, 41, 79, 129, 191, 265, 351, 449, 577, 703, 861, 1013, 1201, 1379, 1597, 1801, 2049, 2279, 2557, 2813, 3121, 3403, 3741, 4049, 4417, 4801, 5151, 5565, 5995, 6385, 6845, 7321, 7751, 8257, 8779, 9249, 9801, 10369, 10879, 11477, 12091, 12641, 13285, 13945

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)) - floor(m/(2*j+2)) >= 5 for 1 <= j <= n-1.

Obviously we have a(n) > 3/(1/(2*n-2) - 1/(2*n)) = 6*n*(n-1). On the other hand, a(n) <= 4/(1/(2*n-2) - 1/(2*n)) + 1 = 8*n*(n-1) + 1: if m >= 8*n*(n-1) + 1, then m/(2*j) - m/(2*j+2) > 4 => ceiling(m/(2*j)) - floor(m/(2*j+2)) = ceiling(m/(2*j)-floor(m/(2*j+2))) >= ceiling(m/(2*j) - m/(2*j+2)) >= 5 for 1 <= j <= n-1.

a(n) is odd for all n: for even m, we have ceiling(m/(2*j)) = ceiling((m-1)/(2*j)) (otherwise (m-1)/(2*j) would be an integer, which is impossible), so ceiling(m/(2*j)) - floor(m/(2*j+2)) >= 5 implies ceiling((m-1)/(2*j)) - floor((m-1)/(2*j+2)) >= 5. (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]), #] &, Range[5]];
t[[4]] (* A024845 *)
(* Peter J. C. Moses, Aug 06 2012 *)

Prog

(PARI) a(n) = for(m=6*n^2-6*n+1, 8*n^2-8*n+1, forstep(j=n-1, 1, -1, if(-((-m)\(2*j)) - m\(2*j+2) < 5, break(), if(j==1, return(m))))) \\ Jianing Song, Aug 31 2022

Crossrefs

Cf. A001000, A024844.

Sequence in context: A103003 A350133 A239325 * A056698 A371108 A371050

Adjacent sequences: A024842 A024843 A024844 * A024846 A024847 A024848

Keyword

nonn

Author

Clark Kimberling

Status

approved