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A016826 a(n) = (4n + 2)^2.

%I A016826 #45 Jan 29 2021 04:48:57
%S A016826 4,36,100,196,324,484,676,900,1156,1444,1764,2116,2500,2916,3364,3844,
%T A016826 4356,4900,5476,6084,6724,7396,8100,8836,9604,10404,11236,12100,12996,
%U A016826 13924,14884,15876,16900,17956
%N A016826 a(n) = (4n + 2)^2.
%C A016826 A bisection of A016742. Sequence arises from reading the line from 4, in the direction 4, 36, ... in the square spiral whose vertices are the squares A000290. - _Omar E. Pol_, May 24 2008
%H A016826 Harvey P. Dale, <a href="/A016826/b016826.txt">Table of n, a(n) for n = 0..1000</a>
%H A016826 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A016826 a(n) = a(n-1) + 32*n (with a(0)=4). - _Vincenzo Librandi_, Dec 15 2010
%F A016826 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), with a(0)=4, a(1)=36, a(2)=100. - _Harvey P. Dale_, Nov 24 2011
%F A016826 G.f.: -((4*(x^2+6*x+1))/(x-1)^3). - _Harvey P. Dale_, Nov 24 2011
%F A016826 a(n) = A000290(A016825(n)). - _Wesley Ivan Hurt_, Feb 24 2014
%F A016826 From _Amiram Eldar_, Jun 28 2020: (Start)
%F A016826 Sum_{n>=0} 1/a(n) = Pi^2/32.
%F A016826 Sum_{n>=0} (-1)^n/a(n) = G/4, where G is the Catalan constant (A006752). (End)
%F A016826 From _Amiram Eldar_, Jan 29 2021: (Start)
%F A016826 Product_{n>=0} (1 + 1/a(n)) = cosh(Pi/4).
%F A016826 Product_{n>=0} (1 - 1/a(n)) = 1/sqrt(2) (A010503). (End)
%p A016826 A016826:=n->(4*n + 2)^2; seq(A016826(n), n=0..40); # _Wesley Ivan Hurt_, Feb 24 2014
%t A016826 (4*Range[0,40]+2)^2 (* or *) LinearRecurrence[{3,-3,1},{4,36,100},40] (* _Harvey P. Dale_, Nov 24 2011 *)
%o A016826 (PARI) a(n)=(4*n+2)^2 \\ _Charles R Greathouse IV_, Oct 07 2015
%Y A016826 Equals A001539 + 1.
%Y A016826 Cf. A000290, A010503, A016742, A016754, A016802, A016814, A016838.
%K A016826 nonn,easy
%O A016826 0,1
%A A016826 _N. J. A. Sloane_