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A069072
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(2n+1)*(2n+2)*(2n+3).
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4
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,1
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COMMENTS
| a(n)=6*A000447(n+1). 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 (blekraj(AT)yahoo.com), 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
| 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
| Konrad Knopp, Theorie und Anwendung der unendlichen Reihen, Berlin, J. Springer, 1922. (Original german edition of "Theory and Application of Infinite Series")
S. Ramanujan, Notebook entry
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FORMULA
| ln(2)-1/2=sum(n=0, inf, 1/a(n)); (1/2)*(1-ln(2))=sum(n=0, inf, (-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]
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CROSSREFS
| Cf. A097321, A069140.
Sequence in context: A036283 A126576 A121287 * A074441 A006741 A120573
Adjacent sequences: A069069 A069070 A069071 * A069073 A069074 A069075
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KEYWORD
| easy,nonn
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AUTHOR
| Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 05 2002
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