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A157912 a(n) = 64*n^2 + 16.

%I A157912 #37 Jan 16 2025 07:33:14
%S A157912 80,272,592,1040,1616,2320,3152,4112,5200,6416,7760,9232,10832,12560,
%T A157912 14416,16400,18512,20752,23120,25616,28240,30992,33872,36880,40016,
%U A157912 43280,46672,50192,53840,57616,61520,65552,69712,74000,78416,82960,87632,92432,97360,102416
%N A157912 a(n) = 64*n^2 + 16.
%C A157912 The identity (8*n^2 + 1)^2 - (64*n^2 + 16)*n^2 = 1 can be written as A081585(n)^2 - a(n)*n^2 = 1. - _Vincenzo Librandi_, Feb 09 2012
%H A157912 Vincenzo Librandi, <a href="/A157912/b157912.txt">Table of n, a(n) for n = 1..10000</a>
%H A157912 Vincenzo Librandi, <a href="https://web.archive.org/web/20090309225914/http://mathforum.org/kb/message.jspa?messageID=5785989&amp;tstart=0">X^2-AY^2=1</a>, Math Forum, 2007. [Wayback Machine link]
%H A157912 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A157912 From _Vincenzo Librandi_, Feb 09 2012: (Start)
%F A157912 G.f.: x*(80+32*x+16*x^2)/(1-x)^3.
%F A157912 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
%F A157912 From _Amiram Eldar_, Mar 07 2023: (Start)
%F A157912 Sum_{n>=1} 1/a(n) = (coth(Pi/2)*Pi/2 - 1)/32.
%F A157912 Sum_{n>=1} (-1)^(n+1)/a(n) = (1 - cosech(Pi/2)*Pi/2)/32. (End)
%F A157912 From _Elmo R. Oliveira_, Jan 16 2025: (Start)
%F A157912 E.g.f.: 16*(exp(x)*(4*x^2 + 4*x + 1) - 1).
%F A157912 a(n) = 16*A053755(n). (End)
%p A157912 A157912:=n->64*n^2 + 16: seq(A157912(n), n=1..50); # _Wesley Ivan Hurt_, Oct 27 2014
%t A157912 64Range[50]^2+16  (* _Harvey P. Dale_, Mar 24 2011 *)
%t A157912 LinearRecurrence[{3, -3, 1}, {80, 272, 592}, 40] (* _Vincenzo Librandi_, Feb 09 2012 *)
%o A157912 (Magma) I:=[80, 272, 592]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]]; // _Vincenzo Librandi_, Feb 09 2012
%o A157912 (PARI) for(n=1, 40, print1(64*n^2 + 16", ")); \\ _Vincenzo Librandi_, Feb 09 2012
%Y A157912 Cf. A053755, A081585.
%K A157912 nonn,easy
%O A157912 1,1
%A A157912 _Vincenzo Librandi_, Mar 09 2009