A226763 Greatest k such that 1/k >= mean{1, 1/2, 1/3,..., 1/n}.

1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 16

Offset

1,2

Links

Clark Kimberling, Table of n, a(n) for n = 1..1000

Formula

a(n) = ceiling(n/{sum{1/k, k = 1..n}).

Example

1/3 < mean{1,1/2,1/3,...,1/9} < 1/4, so that a(9) = 4.

Mathematica

f[n_] := Mean[Table[1/k, {k, 1, n}]]
Table[Floor[1/f[n]], {n, 1, 120}] (* A226762  *)
Table[Ceiling[1/f[n]], {n, 1, 120}] (* A226763 *)

Crossrefs

Cf. A226762.

Sequence in context: A172263 A337635 A140437 * A050500 A076885 A274024

Adjacent sequences: A226760 A226761 A226762 * A226764 A226765 A226766

Keyword

nonn,easy

Author

Clark Kimberling, Jun 19 2013

Status

approved