A214965 Least m > 0 such that for every r and s in the set S = {{h*e} : h = 1,..,n} of fractional parts, if r < s, then r < k/m < s for some integer k; m is the least separator of S as defined at A001000.
2, 3, 4, 6, 6, 7, 11, 11, 11, 18, 18, 18, 18, 25, 25, 25, 25, 25, 25, 25, 25, 25, 32, 32, 32, 32, 32, 32, 32, 32, 32, 35, 35, 35, 39, 39, 39, 39, 55, 61, 61, 66, 68, 69, 69, 69, 70, 70, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71
Offset: 2
Keywords
Examples
Write the sorted fractional parts {h*e}, for h=1..5, as f1,f2,f3,f4,f5. Then f1 < 2/6 < f2 < 3/6 < f3 < 4/6 < f5 < 5/6 < f6, and no such separation occurs using fractions k/m having m < 6; so a(5) = 6.
Links
- Clark Kimberling, Table of n, a(n) for n = 2..300
Programs
-
Mathematica
leastSeparatorShort[seq_, s_] := Module[{n = 1}, While[Or @@ (n #1[[1]] <= s + Floor[n #1[[2]]] &) /@ (Sort[#1, Greater] &) /@ Partition[seq, 2, 1], n++]; n]; Table[leastSeparatorShort[Sort[N[FractionalPart[E*Range[n]], 50]], 1], {n, 2, 100}] (* Peter J. C. Moses, Aug 01 2012 *)
Comments