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%I #43 Sep 08 2022 08:45:54
%S 5,27,65,119,189,275,377,495,629,779,945,1127,1325,1539,1769,2015,
%T 2277,2555,2849,3159,3485,3827,4185,4559,4949,5355,5777,6215,6669,
%U 7139,7625,8127,8645,9179,9729,10295,10877
%N a(n) = 8*n^2 + 14*n + 5.
%C A160050(4*n+1) = A033954(n); A160050(4*n+2) = A001107(n); the third quadrisection is a(n).
%C First 16 terms of clockwise spiral for odd numbers are as follows:
%C .
%C 13--15--17--19
%C | |
%C 11 1---3 21
%C | | |
%C 9---7---5 23
%C |
%C 31--29--27--25
%C .
%C a(n) comes from the third vertical.
%C Sequence found by reading the line from 5, in the direction 5, 27, in the square spiral whose vertices are the triangular numbers A000217. - _Omar E. Pol_, Dec 25 2011
%H Vincenzo Librandi, <a href="/A181890/b181890.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = A160050(4*n+3).
%F a(n) = (2*n+1)*(4*n+5).
%F a(n) = a(n-1) + 16*n + 6;
%F a(n) = 2*a(n-1) - a(n-2) + 16.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F G.f.: (5 + 12*x - x^2)/(1 - x)^3. - _Arkadiusz Wesolowski_, Dec 25 2011
%F a(n) = A014635(n+1) - 1. - _Omar E. Pol_, Dec 25 2011
%F From _Vaclav Kotesovec_, Aug 18 2018: (Start)
%F Sum_{n>=0} 1/a(n) = 2/3 - Pi/12 - log(2)/6 = 0.289342748774193011891907697817...
%F Sum_{n>=0} (-1)^n / a(n) = (1 + sqrt(2))*Pi/12 - 2/3 - sqrt(2)*log(tan(Pi/8))/6 = 0.173114712692423461587883724528539... (End)
%F a(n) = A014106(2*n+1). - _Rick L. Shepherd_, Aug 06 2019
%t Table[(2n+1)(4n+5),{n,0,40}] (* _Harvey P. Dale_, Feb 06 2011 *)
%t CoefficientList[Series[(5 + 12 x - x^2)/(1 - x)^3, {x, 0, 40}], x] (* _Vincenzo Librandi_, May 23 2014 *)
%o (Magma) [8*n^2+14*n+5: n in [0..700]]; // _Vincenzo Librandi_, Feb 01 2011
%o (PARI) a(n)=8*n^2+14*n+5 \\ _Charles R Greathouse IV_, Dec 21 2011
%Y Cf. A000217, A014106, A014635, A064038, A160050.
%K nonn,easy
%O 0,1
%A _Paul Curtz_, Feb 01 2011
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