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Least m>0 such that n! <= m^m.
1

%I #7 Jul 25 2012 17:57:32

%S 1,2,3,3,4,5,6,6,7,8,9,10,10,11,12,13,14,14,15,16,17,18,18,19,20,21,

%T 22,22,23,24,25,26,27,27,28,29,30,31,32,32,33,34,35,36,37,37,38,39,40,

%U 41,42,42,43,44,45,46,47,47,48,49,50,51,52

%N Least m>0 such that n! <= m^m.

%C Conjecture: this sequence results from A118168 by deleting the first two 1s.

%H Clark Kimberling, <a href="/A214449/b214449.txt">Table of n, a(n) for n = 1..1000</a>

%e a(4) = 3 because 2^2 < 4! < 3^3.

%t Table[m = 1; While[n! > m^m, m++]; m, {n, 1, 100}]

%Y Cf. A214046.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Jul 18 2012