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A024584
a(n) = floor(1/frac(n*Pi)).
4
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, 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, 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
OFFSET
1,1
COMMENTS
From Hieronymus Fischer, Apr 15 2012: (Start)
The sequence is well defined, since frac(n*Pi)>0 for n>0.
Let b(n,m)=|{a(k)| 1<=k<=n, a(k)>=m}| be the number of the first n terms which are >= m >= 1.
Then, lim b(n,m)/n = 1/m for n-->oo since frac(n*pi) is uniformly distributed. (End)
LINKS
MATHEMATICA
Table[Floor[1/FractionalPart[n Pi]], {n, 100}] (* Bruno Berselli, Apr 15 2012 *)
PROG
(PARI) a(n) = floor(1/frac(n*Pi)); \\ Michel Marcus, Nov 16 2022
CROSSREFS
Sequence in context: A355674 A256843 A033327 * A132713 A262312 A135296
KEYWORD
nonn
STATUS
approved