

A114378


Area of annuli of consecutive integer thickness.


0



3, 25, 84, 201, 392, 678, 1077, 1608, 2290, 3141, 4181, 5428, 6902, 8620, 10602, 12867, 15434, 18321, 21548, 25132, 29094, 33451, 38223, 43429, 49087, 55216, 61835, 68964, 76620, 84823, 93591, 102943, 112899, 123477, 134695, 146574
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OFFSET

1,1


COMMENTS

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.


LINKS

Table of n, a(n) for n=1..36.


FORMULA

a(n) = floor(Pi*n^3).  Robert Israel, Nov 24 2014


EXAMPLE

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.


MAPLE

seq(floor(Pi*n^3), n=1..100); # Robert Israel, Nov 24 2014


PROG

(PARI) g(n) = for(j=1, n, x=j*(j+1)/2; y=(j1)*(j)/2; print1(floor(Pi*(x^2y^2))", "))


CROSSREFS

Sequence in context: A083298 A083222 A041565 * A075306 A183761 A212054
Adjacent sequences: A114375 A114376 A114377 * A114379 A114380 A114381


KEYWORD

easy,nonn


AUTHOR

Cino Hilliard, Feb 10 2006


STATUS

approved



