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A114378
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Area of annuli of consecutive integer thickness.
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0
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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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.
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FORMULA
| The annulus is the region formed by two concentric circles.
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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.
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PROG
| (PARI) g(n) = for(j=1, n, x=j*(j+1)/2; y=(j-1)*(j)/2; print1(floor(Pi*(x^2-y^2))", "))
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CROSSREFS
| Sequence in context: A083298 A083222 A041565 * A075306 A183761 A180324
Adjacent sequences: A114375 A114376 A114377 * A114379 A114380 A114381
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KEYWORD
| easy,nonn
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AUTHOR
| Cino Hilliard (hillcino368(AT)gmail.com), Feb 10 2006
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