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%I #21 Nov 16 2022 14:12:10
%S 7,3,2,1,1,1,1,7,3,2,1,1,1,1,8,3,2,1,1,1,1,8,3,2,1,1,1,1,9,4,2,1,1,1,
%T 1,10,4,2,1,1,1,1,11,4,2,1,1,1,1,12,4,2,1,1,1,1,14,4,2,2,1,1,1,16,4,2,
%U 2,1,1,1,18,5,2,2,1,1,1,22,5,3,2,1,1,1,28,5,3,2,1,1,1,37,5,3,2,1,1,1,56,6
%N a(n) = floor(1/frac(n*Pi)).
%C From _Hieronymus Fischer_, Apr 15 2012: (Start)
%C The sequence is well defined, since frac(n*Pi)>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*pi) is uniformly distributed. (End)
%H Clark Kimberling, <a href="/A024584/b024584.txt">Table of n, a(n) for n = 1..1000</a>
%t Table[Floor[1/FractionalPart[n Pi]], {n, 100}] (* _Bruno Berselli_, Apr 15 2012 *)
%o (PARI) a(n) = floor(1/frac(n*Pi)); \\ _Michel Marcus_, Nov 16 2022
%Y Cf. A024753, A022844.
%K nonn
%O 1,1
%A _Clark Kimberling_
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