A082109 Third row of number array A082105.
1, 13, 33, 61, 97, 141, 193, 253, 321, 397, 481, 573, 673, 781, 897, 1021, 1153, 1293, 1441, 1597, 1761, 1933, 2113, 2301, 2497, 2701, 2913, 3133, 3361, 3597, 3841, 4093, 4353, 4621, 4897, 5181, 5473, 5773, 6081, 6397, 6721, 7053, 7393, 7741, 8097, 8461
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Russ Cox et al., incorrect A082109 comment about A000217, SeqFan, 2024.
- 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).
Programs
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Magma
[4*n^2+8*n+1: n in [0..60]]; // G. C. Greubel, Dec 22 2022
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Mathematica
LinearRecurrence[{3,-3,1}, {1,13,33}, 51] (* Vladimir Joseph Stephan Orlovsky, Oct 25 2008 *) -
PARI
a(n)=4*n^2+8*n+1 \\ Charles R Greathouse IV, Jun 17 2017
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SageMath
[4*n^2+8*n+1 for n in range(61)] # G. C. Greubel, Dec 22 2022
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)