OFFSET
0,1
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).
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<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]
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
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 *)
PROG
(PARI) a(n)=(2*n+1)*(2*n+2)*(2*n+3) \\ Charles R Greathouse IV, Oct 07 2015
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Benoit Cloitre, Apr 05 2002
STATUS
approved