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A158683 a(n) = 1024*n^2 - 32.

%I A158683 #28 Jan 16 2025 12:29:13
%S A158683 992,4064,9184,16352,25568,36832,50144,65504,82912,102368,123872,
%T A158683 147424,173024,200672,230368,262112,295904,331744,369632,409568,
%U A158683 451552,495584,541664,589792,639968,692192,746464,802784,861152,921568,984032,1048544,1115104,1183712
%N A158683 a(n) = 1024*n^2 - 32.
%C A158683 The identity (64*n^2 - 1)^2 - (1024*n^2 - 32)*(2*n)^2 = 1 can be written as A158684(n)^2 - a(n)*A005843(n)^2 = 1.
%H A158683 Vincenzo Librandi, <a href="/A158683/b158683.txt">Table of n, a(n) for n = 1..10000</a>
%H A158683 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 A158683 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A158683 G.f.: 32*x*(-31-34*x+x^2)/(x-1)^3.
%F A158683 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F A158683 From _Amiram Eldar_, Mar 21 2023: (Start)
%F A158683 Sum_{n>=1} 1/a(n) = (1 - cot(Pi/(4*sqrt(2)))*Pi/(4*sqrt(2)))/64.
%F A158683 Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/(4*sqrt(2)))*Pi/(4*sqrt(2)) - 1)/64. (End)
%F A158683 From _Elmo R. Oliveira_, Jan 16 2025: (Start)
%F A158683 E.g.f.: 32*(exp(x)*(32*x^2 + 32*x - 1) + 1).
%F A158683 a(n) = 32*A158563(n). (End)
%p A158683 A158683:=n->1024*n^2-32: seq(A158683(n), n=1..50); # _Wesley Ivan Hurt_, Nov 20 2014
%t A158683 LinearRecurrence[{3, -3, 1}, {992, 4064, 9184}, 50] (* _Vincenzo Librandi_, Feb 19 2012 *)
%o A158683 (Magma) I:=[992, 4064, 9184]; [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 19 2012
%o A158683 (PARI) for(n=1, 40, print1(1024*n^2 - 32", ")); \\ _Vincenzo Librandi_, Feb 19 2012
%Y A158683 Cf. A005843, A158563, A158684.
%K A158683 nonn,easy
%O A158683 1,1
%A A158683 _Vincenzo Librandi_, Mar 24 2009
%E A158683 Comment rewritten and formula replaced by _R. J. Mathar_, Oct 22 2009