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Beatty sequence for cube root of Pi: a(n) = floor(n*Pi^(1/3)).
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%I #25 Feb 14 2017 17:44:21

%S 0,1,2,4,5,7,8,10,11,13,14,16,17,19,20,21,23,24,26,27,29,30,32,33,35,

%T 36,38,39,41,42,43,45,46,48,49,51,52,54,55,57,58,60,61,62,64,65,67,68,

%U 70,71,73,74,76,77,79,80,82,83,84

%N Beatty sequence for cube root of Pi: a(n) = floor(n*Pi^(1/3)).

%C Beatty complement of A248524. - _M. F. Hasler_, Oct 07 2014

%H G. C. Greubel, <a href="/A240977/b240977.txt">Table of n, a(n) for n = 0..5000</a>

%H <a href="/index/Be#Beatty">Index entries for sequences related to Beatty sequences</a>

%F a(n) = floor(n*(Pi^1/3)).

%t Table[Floor[n*(Pi^(1/3))], {n, 0, 50}] (* _G. C. Greubel_, Feb 14 2017 *)

%o (Java) static int a(int n) {return (int) (n*Math.pow(Math.PI,(1.0/3)));}

%o (PARI) a(n)=n\Pi^(-1/3) \\ _M. F. Hasler_, Oct 07 2014

%Y Cf. A092039 (Pi^(1/3)), A022844 (similar for Pi), A037086 (similar for sqrt(Pi)), A248524.

%K nonn

%O 0,3

%A _Sarah Nathanson_, Sep 30 2014

%E a(0)=0 prepended by _Eric M. Schmidt_, Oct 06 2014

%E Name edited by _M. F. Hasler_, Oct 07 2014