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A158684 a(n) = 64*n^2 - 1.

%I A158684 #28 Jan 17 2025 06:54:29
%S A158684 63,255,575,1023,1599,2303,3135,4095,5183,6399,7743,9215,10815,12543,
%T A158684 14399,16383,18495,20735,23103,25599,28223,30975,33855,36863,39999,
%U A158684 43263,46655,50175,53823,57599,61503,65535,69695,73983,78399,82943,87615,92415,97343,102399
%N A158684 a(n) = 64*n^2 - 1.
%C A158684 The identity (64*n^2 - 1)^2 - (1024*n^2 - 32)*(2*n)^2 = 1 can be written as a(n)^2 - A158683(n)*A005843(n)^2 = 1.
%H A158684 Vincenzo Librandi, <a href="/A158684/b158684.txt">Table of n, a(n) for n = 1..10000</a>
%H A158684 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 A158684 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A158684 G.f.: x*(-63 - 66*x + x^2)/(x-1)^3.
%F A158684 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F A158684 From _Amiram Eldar_, Mar 21 2023: (Start)
%F A158684 Sum_{n>=1} 1/a(n) = (1 - cot(Pi/8)*Pi/8)/2 = (1 - (sqrt(2)+1)*Pi/8)/2.
%F A158684 Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/8)*Pi/8 - 1)/2. (End)
%F A158684 E.g.f.: 1 + exp(x)*(64*x^2 + 64*x - 1). - _Elmo R. Oliveira_, Jan 16 2025
%t A158684 LinearRecurrence[{3, -3, 1}, {63, 255, 575}, 50] (* _Vincenzo Librandi_, Feb 20 2012 *)
%t A158684 64*Range[40]^2-1 (* _Harvey P. Dale_, Jul 09 2017 *)
%o A158684 (Magma) I:=[63, 255, 575]; [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 20 2012
%o A158684 (PARI) for(n=1, 40, print1(64*n^2 - 1", ")); \\ _Vincenzo Librandi_, Feb 20 2012
%Y A158684 Cf. A005843, A158683.
%K A158684 nonn,easy
%O A158684 1,1
%A A158684 _Vincenzo Librandi_, Mar 24 2009
%E A158684 Comment rewritten and formula replaced by _R. J. Mathar_, Oct 22 2009