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A157913 a(n) = 64*n^2 - 16.

%I A157913 #28 Jan 16 2025 07:57:01
%S A157913 48,240,560,1008,1584,2288,3120,4080,5168,6384,7728,9200,10800,12528,
%T A157913 14384,16368,18480,20720,23088,25584,28208,30960,33840,36848,39984,
%U A157913 43248,46640,50160,53808,57584,61488,65520,69680,73968,78384,82928,87600,92400,97328,102384
%N A157913 a(n) = 64*n^2 - 16.
%C A157913 The identity (8*n^2 - 1)^2 - (64*n^2 - 16)*n^2 = 1 can be written as A157914(n)^2 - a(n)*n^2 = 1. - _Vincenzo Librandi_, Feb 09 2012
%H A157913 Vincenzo Librandi, <a href="/A157913/b157913.txt">Table of n, a(n) for n = 1..10000</a>
%H A157913 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A157913 From  _Vincenzo Librandi_, Feb 09 2012: (Start)
%F A157913 G.f.: -16*x*(3 + 6*x - x^2)/(x - 1)^3.
%F A157913 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
%F A157913 From _Amiram Eldar_, Mar 07 2023: (Start)
%F A157913 Sum_{n>=1} 1/a(n) = 1/32.
%F A157913 Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi-2)/64. (End)
%F A157913 From _Elmo R. Oliveira_, Jan 16 2025: (Start)
%F A157913 E.g.f.: 16*(exp(x)*(4*x^2 + 4*x - 1) + 1).
%F A157913 a(n) = 16*A000466(n). (End)
%t A157913 LinearRecurrence[{3, -3, 1}, {48, 240, 560}, 50] (* _Vincenzo Librandi_, Feb 09 2012 *)
%t A157913 64*Range[40]^2-16 (* _Harvey P. Dale_, Jul 27 2012 *)
%o A157913 (Magma) I:=[48, 240, 560]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // _Vincenzo Librandi_, Feb 09 2012
%o A157913 (PARI) for(n=1, 40, print1(64*n^2 - 16", ")); \\ _Vincenzo Librandi_, Feb 09 2012
%Y A157913 Cf. A000466, A157914.
%K A157913 nonn,easy
%O A157913 1,1
%A A157913 _Vincenzo Librandi_, Mar 09 2009