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A069072
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a(n) = (2n+1)*(2n+2)*(2n+3).
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8
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6, 60, 210, 504, 990, 1716, 2730, 4080, 5814, 7980, 10626, 13800, 17550, 21924, 26970, 32736, 39270, 46620, 54834, 63960, 74046, 85140, 97290, 110544, 124950, 140556, 157410, 175560, 195054, 215940, 238266, 262080, 287430, 314364, 342930
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OFFSET
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0,1
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COMMENTS
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Terms are areas of primitive Pythagorean triangles whose odd sides differ by 2; e.g., the triangle with sides 8,15,17 has area 60. - Lekraj Beedassy, Apr 18 2003
Using (n, n+1), (n, n+2), and (n+1, n+2) to generate three unreduced Pythagorean triangles gives a sum of the areas for all three to be (2*n+1)*(2*n+2)*(2*n+3), which are three consecutive numbers. - J. M. Bergot, Aug 22 2011
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REFERENCES
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T. J. I'a. Bromwich, Introduction to the Theory of Infinite Series, Macmillan, 2nd. ed. 1949, p. 190.
Jolley, Summation of Series, Oxford (1961).
Konrad Knopp, Theory and application of infinite series, Dover, p. 269.
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LINKS
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FORMULA
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log(2) - 1/2 = Sum_{n>=0} 1/a(n); (1/2)*(1-log(2)) = Sum_{n>=0} (-1)^n/a(n). [Jolley eq 236 and 237]
Sum_{n>=0} x^n/a(n) = ((1+x)/sqrt(x)*log((1+sqrt x)/(1-sqrt x)) + 2*log(1-x)-2)/(4x). [Jolley eq 280 for 0<x<1]
Sum_{n>=0} (-x)^n/a(n) = (1-log(1+x) -(1-x)/sqrt(x)*arctan(x))/(2x). [Jolley eq 281 for 0<x<=1]
G.f.: 6*(1 + 6*x + x^2) / (x-1)^4 . - R. J. Mathar, Jun 09 2013
a(0)=6, a(1)=60, a(2)=210, a(3)=504, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Dec 08 2013
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MATHEMATICA
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Array[Times@@(2#+{1, 2, 3})&, 40, 0] (* or *) LinearRecurrence[{4, -6, 4, -1}, {6, 60, 210, 504}, 40] (* Harvey P. Dale, Dec 08 2013 *)
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PROG
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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