A214921 Least m > 0 such that for every r and s in the set S = {{h*sqrt(2)} : h = 1,..,n} of fractional parts, if r < s, then r < k/m < s for some integer k.
2, 3, 4, 5, 7, 7, 12, 12, 12, 12, 12, 15, 15, 17, 17, 17, 23, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 33, 36, 36, 36, 41, 41, 41, 41, 41, 41, 41, 41, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70, 70
Offset: 2
Keywords
Examples
Write the fractional parts of h*sqrt(2) for h=1,2,...,6, sorted, as f1, f2, f3, f4, f5, f6. Then f1 < 1/7 < f2 < 2/7 < f3 < 3/7 < f4 < 4/7 < f5 < 5/7 < f6, and 7 is the least m for which such a separation by fractions k/m occurs, so that a(6)=7.
Links
- Clark Kimberling, Table of n, a(n) for n = 2..300
Programs
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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[Sqrt[2] Range[n]], 50]], 1], {n, 2, 100}] (* Peter J. C. Moses, Aug 01 2012 *)
Comments