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%I #51 Jan 15 2024 12:02:28
%S 6,60,210,504,990,1716,2730,4080,5814,7980,10626,13800,17550,21924,
%T 26970,32736,39270,46620,54834,63960,74046,85140,97290,110544,124950,
%U 140556,157410,175560,195054,215940,238266,262080,287430,314364,342930
%N a(n) = (2n+1)*(2n+2)*(2n+3).
%C 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
%C 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
%D T. J. I'a. Bromwich, Introduction to the Theory of Infinite Series, Macmillan, 2nd. ed. 1949, p. 190.
%D Jolley, Summation of Series, Oxford (1961).
%D Konrad Knopp, Theory and application of infinite series, Dover, p. 269.
%H M. Janjic and B. Petkovic, <a href="http://arxiv.org/abs/1301.4550">A Counting Function</a>, arXiv 1301.4550 [math.CO], 2013.
%H Konrad Knopp, <a href="http://name.umdl.umich.edu/ACM1954.0001.001">Theorie und Anwendung der unendlichen Reihen</a>, Berlin, J. Springer, 1922. (Original german edition of "Theory and Application of Infinite Series")
%H S. Ramanujan, <a href="http://www.imsc.res.in/~rao/ramanujan/NoteBooks/NoteBook2/chapterII/page1.htm">Notebook entry</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F 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]
%F 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]
%F 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]
%F a(n) = 6*A000447(n+1). - _Lekraj Beedassy_, Apr 18 2003
%F G.f.: 6*(1 + 6*x + x^2) / (x-1)^4 . - _R. J. Mathar_, Jun 09 2013
%F 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
%F a(n) = 2*A035328(n+1). - _J. M. Bergot_, Jan 02 2015
%t 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 *)
%o (PARI) a(n)=(2*n+1)*(2*n+2)*(2*n+3) \\ _Charles R Greathouse IV_, Oct 07 2015
%Y Cf. A097321, A069140, A000447, A035328.
%K easy,nonn
%O 0,1
%A _Benoit Cloitre_, Apr 05 2002
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