A226763 Least k such that 1/k <= mean of {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

Author

Clark Kimberling, Jun 19 2013

Keywords

• nonn
• easy

Examples

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

Links

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

Crossrefs

Cf. A226762.

Programs

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}] (* this sequence *)

Formula

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

Extensions

Name corrected by Jason Yuen, Nov 02 2024