A069072 a(n) = (2n+1)*(2n+2)*(2n+3).

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

Offset: 0

Author

Benoit Cloitre, Apr 05 2002

Keywords

• easy
• nonn

Comments

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

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.

Links

• M. Janjic and B. Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
• 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
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A097321, A069140, A000447, A035328.

Programs

Mathematica

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 *)

PARI

a(n)=(2*n+1)*(2*n+2)*(2*n+3) \\ Charles R Greathouse IV, Oct 07 2015

Formula

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

Sum_{n>=0} (-x)^n/a(n) = (1-log(1+x) -(1-x)/sqrt(x)*arctan(x))/(2x). [Jolley eq 281 for 0

a(n) = 6*A000447(n+1). - Lekraj Beedassy, Apr 18 2003

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

a(n) = 2*A035328(n+1). - J. M. Bergot, Jan 02 2015

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