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%I #6 Jun 20 2013 12:32:33
%S 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,
%T 9,9,9,9,10,10,10,10,10,11,11,11,11,11,11,12,12,12,12,12,12,13,13,13,
%U 13,13,13,14,14,14,14,14,14,15,15,15,15,15,15,16
%N Greatest k such that 1/k >= mean{1, 1/2, 1/3,..., 1/n}.
%H Clark Kimberling, <a href="/A226763/b226763.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = ceiling(n/{sum{1/k, k = 1..n}).
%e 1/3 < mean{1,1/2,1/3,...,1/9} < 1/4, so that a(9) = 4.
%t f[n_] := Mean[Table[1/k, {k, 1, n}]]
%t Table[Floor[1/f[n]], {n, 1, 120}] (* A226762 *)
%t Table[Ceiling[1/f[n]], {n, 1, 120}] (* A226763 *)
%Y Cf. A226762.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Jun 19 2013
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