A006137 a(n) = 1 + n/2 + 9*n^2/2.
1, 6, 20, 43, 75, 116, 166, 225, 293, 370, 456, 551, 655, 768, 890, 1021, 1161, 1310, 1468, 1635, 1811, 1996, 2190, 2393, 2605, 2826, 3056, 3295, 3543, 3800, 4066, 4341, 4625, 4918, 5220, 5531, 5851, 6180, 6518, 6865, 7221, 7586, 7960, 8343, 8735, 9136, 9546
Offset: 0
Links
- Harvey P. Dale, Table of n, a(n) for n = 0..1000
- Nickolas Arustamyan, Christopher Cox, Erik Lundberg, Sean Perry, and Zvi Rosen, On the Number of Equilibria Balancing Newtonian Point Masses with a Central Force, arXiv:2106.11416 [math-ph], 2021.
- Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Programs
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Mathematica
Table[1+n/2+9 n^2/2,{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{1,6,20},40] (* Harvey P. Dale, Oct 05 2012 *)
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PARI
a(n)=1+n/2+9*n^2/2 \\ Charles R Greathouse IV, Jun 17 2017
Formula
a(n) = a(n-1) + 9*n - 4 (with a(0)=1). - Vincenzo Librandi, Nov 18 2010
From Klaus Purath, Jan 14 2022: (Start)
a(n) = A276819(n) + n.
From Stefano Spezia, Dec 25 2022: (Start)
O.g.f.: (1 + 3*x+ 5*x^2)/(1 - x)^3.
E.g.f.: exp(x)*(2 + 10*x + 9*x^2)/2. (End)
Comments