A114378 - OEIS

%I #10 Nov 24 2014 16:10:04

%S 3,25,84,201,392,678,1077,1608,2290,3141,4181,5428,6902,8620,10602,

%T 12867,15434,18321,21548,25132,29094,33451,38223,43429,49087,55216,

%U 61835,68964,76620,84823,93591,102943,112899,123477,134695,146574

%N Area of annuli of consecutive integer thickness.

%C The annulus is the region between two concentric circles of radius r(i) and r(i+1). The area of the annulus is the area of the bigger circle minus the area of the smaller one or Pi(r(i+1)^2 - r(i)^2). Then for this sequence which defines the thickness or the annuli as the consecutive integers, we determine the area using the formula above and the summation formula for an arithmetic progression. Area of annulus(i+1) = Pi(r(i+1)*(r(i+1)+1)/2 - r(i)*(r(i)+1)/2). In other words, the annuli form concentric circles whose successive radii are the sum of the successive annuli up to that point.

%F a(n) = floor(Pi*n^3). - _Robert Israel_, Nov 24 2014

%e Any circle is an annulus formed by a circle of radius r and a circle of radius 0. So the integer area of the annulus of the unit circle is Pi(1^2 - 0^2) = 3, the first term in the sequence.

%p seq(floor(Pi*n^3),n=1..100); # _Robert Israel_, Nov 24 2014

%o (PARI) g(n) = for(j=1,n,x=j*(j+1)/2;y=(j-1)*(j)/2;print1(floor(Pi*(x^2-y^2))","))

%K easy,nonn

%O 1,1

%A _Cino Hilliard_, Feb 10 2006