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%I #10 Oct 22 2016 02:58:39
%S 0,0,0,0,1,1,2,3,5,6,7,9,11,13,15,17,20,22,25,28,31,35,38,42,45,49,53,
%T 58,62,66,71,76,81,86,91,97,103,108,114,121,127,133,140,147,154,161,
%U 168,175,183,191,198,206,215,223,232,240,249,258,267,277,286,296,305,315
%N Floor of the area of a circle in terms of its circumference n.
%H G. C. Greubel, <a href="/A135607/b135607.txt">Table of n, a(n) for n = 0..1000</a>
%F Area of a circle of radius r is A = Pi*r^2. Circumference of a circle of radius r is n = 2*Pi*r. Then area in terms of the circumference n is A = n^2/(4*Pi).
%e For a circle of circumference 10, the floor of the area A = floor(100/4/Pi) = 7.
%t Table[Floor[n^2/(4*Pi)], {n,0,25}] (* _G. C. Greubel_, Oct 21 2016 *)
%o (PARI) g(n) = for(c=0,n,a=c^2/4/Pi;print1(floor(a)","))
%o (PARI) a(n) = n^2\(4*Pi); \\ _Michel Marcus_, Oct 22 2016
%Y Cf. A038130, A066643, A135039.
%K nonn
%O 0,7
%A _Cino Hilliard_, Feb 27 2008