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a(n) = 1 + n/2 + 9*n^2/2.
5

%I #46 Feb 19 2024 01:49:00

%S 1,6,20,43,75,116,166,225,293,370,456,551,655,768,890,1021,1161,1310,

%T 1468,1635,1811,1996,2190,2393,2605,2826,3056,3295,3543,3800,4066,

%U 4341,4625,4918,5220,5531,5851,6180,6518,6865,7221,7586,7960,8343,8735,9136,9546

%N a(n) = 1 + n/2 + 9*n^2/2.

%C 72*a(n) - 71 = (18*n+1)^2 = A161705(n)^2 is a perfect square. - _Klaus Purath_, Jan 14 2022

%H Harvey P. Dale, <a href="/A006137/b006137.txt">Table of n, a(n) for n = 0..1000</a>

%H Nickolas Arustamyan, Christopher Cox, Erik Lundberg, Sean Perry, and Zvi Rosen, <a href="https://arxiv.org/abs/2106.11416">On the Number of Equilibria Balancing Newtonian Point Masses with a Central Force</a>, arXiv:2106.11416 [math-ph], 2021.

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

%F a(n) = a(n-1) + 9*n - 4 (with a(0)=1). - _Vincenzo Librandi_, Nov 18 2010

%F From _Klaus Purath_, Jan 14 2022: (Start)

%F a(n) = A276819(n) + n.

%F A003215(a(n)) - A003215(a(n)-3) = A002378(9*n). (End)

%F From _Stefano Spezia_, Dec 25 2022: (Start)

%F O.g.f.: (1 + 3*x+ 5*x^2)/(1 - x)^3.

%F E.g.f.: exp(x)*(2 + 10*x + 9*x^2)/2. (End)

%t 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 *)

%o (PARI) a(n)=1+n/2+9*n^2/2 \\ _Charles R Greathouse IV_, Jun 17 2017

%Y Cf. A002378, A003215, A161705, A276819.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_