%I #68 Sep 27 2024 03:43:41

%S 1,13,33,61,97,141,193,253,321,397,481,573,673,781,897,1021,1153,1293,

%T 1441,1597,1761,1933,2113,2301,2497,2701,2913,3133,3361,3597,3841,

%U 4093,4353,4621,4897,5181,5473,5773,6081,6397,6721,7053,7393,7741,8097,8461

%N Third row of number array A082105.

%C 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

%H G. C. Greubel, <a href="/A082109/b082109.txt">Table of n, a(n) for n = 0..1000</a>

%H Takao Komatsu, Ritika Goel, and Neha Gupta, <a href="https://arxiv.org/abs/2409.14788">The Frobenius number for the triple of the 2-step star numbers</a>, arXiv:2409.14788 [math.CO], 2024. See p. 2.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F a(n) = 4*n^2 + 8*n + 1.

%F a(n) = a(n-1) + 8*n + 4, with a(0)=1. - _Vincenzo Librandi_, Aug 08 2010

%F G.f.: (1 + 10*x - 3*x^2)/(1-x)^3. - _Bruno Berselli_, Apr 18 2011

%F E.g.f.: (1 + 12*x + 4*x^2)*exp(x). - _G. C. Greubel_, Dec 22 2022

%F From _Amiram Eldar_, Jan 18 2023: (Start)

%F Sum_{n>=0} 1/a(n) = 1/6 - cot(sqrt(3)*Pi/2)*sqrt(3)*Pi/12.

%F Sum_{n>=0} (-1)^n/a(n) = cosec(sqrt(3)*Pi/2)*sqrt(3)*Pi/12 - 1/6. (End)

%t LinearRecurrence[{3,-3,1}, {1,13,33}, 51] (* _Vladimir Joseph Stephan Orlovsky_, Oct 25 2008 *)

%o (PARI) a(n)=4*n^2+8*n+1 \\ _Charles R Greathouse IV_, Jun 17 2017

%o (Magma) [4*n^2+8*n+1: n in [0..60]]; // _G. C. Greubel_, Dec 22 2022

%o (SageMath) [4*n^2+8*n+1 for n in range(61)] # _G. C. Greubel_, Dec 22 2022

%Y Cf. A000217, A082105, A082108, A000217.

%Y Column 2 of array A188646.

%K easy,nonn

%O 0,2

%A _Paul Barry_, Apr 03 2003