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%I #17 Aug 15 2012 18:00:06
%S 1,2,6,1,1,3,35,1,2,5,1,1,2,17,1,2,4,1,1,2,11,1,1,4,1,1,2,8,1,1,3,1,1,
%T 2,7,1,1,3,76,1,2,5,1,1,3,24,1,2,5,1,1,2,14,1,1,4,1,1,2,10,1,1,3,1,1,
%U 2,8,1,1,3,1,1,2,6,1,1,3,38,1,2,5,1,1,2,18,1,2,4,1,1,2,12,1,1,4,1,1,2,9,1
%N a(n) = floor(1/frac(n*e)).
%C The sequence is well defined, since frac(n*e)>0 for n>0.
%C Let b(n,m) = |{a(k)| 1<=k<=n, a(k)>=m}| be the number of the first n terms which are >= m >= 1.
%C Then, lim b(n,m)/n = 1/m for n-->oo since frac(n*e) is uniformly distributed. - _Hieronymus Fischer_, Apr 15 2012
%H Clark Kimberling, <a href="/A024573/b024573.txt">Table of n, a(n) for n = 1..1000</a>
%t f[n_] := Floor[1/FractionalPart[n*E]]; Array[f, 100] (* _Robert G. Wilson v_, Apr 17 2012 *)
%Y Cf. A024584, A022843.
%K nonn
%O 1,2
%A _Clark Kimberling_
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