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%I #25 Mar 05 2023 03:06:17
%S 56,248,568,1016,1592,2296,3128,4088,5176,6392,7736,9208,10808,12536,
%T 14392,16376,18488,20728,23096,25592,28216,30968,33848,36856,39992,
%U 43256,46648,50168,53816,57592,61496,65528,69688,73976,78392,82936,87608,92408,97336,102392
%N a(n) = 64*n^2 - 8.
%C The identity (16*n^2 - 1)^2 - (64*n^2 - 8)*(2*n)^2 = 1 can be written as A141759(n)^2 - a(n)*A005843(n)^2 = 1. - _Vincenzo Librandi_, Feb 09 2012
%H Vincenzo Librandi, <a href="/A158487/b158487.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F From _Vincenzo Librandi_, Feb 09 2012: (Start)
%F G.f.: -8*x*(7 + 10*x - x^2)/(x - 1)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
%F From _Amiram Eldar_, Mar 05 2023: (Start)
%F Sum_{n>=1} 1/a(n) = (1 - cot(Pi/(2*sqrt(2)))*Pi/(2*sqrt(2)))/16.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/(2*sqrt(2)))*Pi/(2*sqrt(2)) - 1)/16. (End)
%t LinearRecurrence[{3, -3, 1}, {56, 248, 568}, 50] (* _Vincenzo Librandi_, Feb 09 2012 *)
%o (Magma) I:=[56, 248, 568]; [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 (PARI) for(n=1, 40, print1(64*n^2 - 8", ")); \\ _Vincenzo Librandi_, Feb 09 2012
%Y Cf. A005843, A141759.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 20 2009
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