OFFSET
0,2
COMMENTS
Define b(n) = A000217(n), the triangular numbers. Using six consecutive terms to create the vertices of a triangle at points (b(n-2), b(n-1)), (b(n), b(n+1)), and (b(n+2), b(n+3)), one fourth the area of these triangles = a(n). - J. M. Bergot, Jul 30 2013
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
Takao Komatsu, Ritika Goel, and Neha Gupta, The Frobenius number for the triple of the 2-step star numbers, arXiv:2409.14788 [math.CO], 2024. See p. 2.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
a(n) = 4*n^2 + 8*n + 1.
a(n) = a(n-1) + 8*n + 4, with a(0)=1. - Vincenzo Librandi, Aug 08 2010
G.f.: (1 + 10*x - 3*x^2)/(1-x)^3. - Bruno Berselli, Apr 18 2011
E.g.f.: (1 + 12*x + 4*x^2)*exp(x). - G. C. Greubel, Dec 22 2022
From Amiram Eldar, Jan 18 2023: (Start)
Sum_{n>=0} 1/a(n) = 1/6 - cot(sqrt(3)*Pi/2)*sqrt(3)*Pi/12.
Sum_{n>=0} (-1)^n/a(n) = cosec(sqrt(3)*Pi/2)*sqrt(3)*Pi/12 - 1/6. (End)
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {1, 13, 33}, 51] (* Vladimir Joseph Stephan Orlovsky, Oct 25 2008 *)
PROG
(PARI) a(n)=4*n^2+8*n+1 \\ Charles R Greathouse IV, Jun 17 2017
(Magma) [4*n^2+8*n+1: n in [0..60]]; // G. C. Greubel, Dec 22 2022
(SageMath) [4*n^2+8*n+1 for n in range(61)] # G. C. Greubel, Dec 22 2022
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Apr 03 2003
STATUS
approved