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A158686 a(n) = 64*n^2 + 1.

%I A158686 #32 Jan 17 2025 07:45:48
%S A158686 1,65,257,577,1025,1601,2305,3137,4097,5185,6401,7745,9217,10817,
%T A158686 12545,14401,16385,18497,20737,23105,25601,28225,30977,33857,36865,
%U A158686 40001,43265,46657,50177,53825,57601,61505,65537,69697,73985,78401,82945,87617,92417,97345,102401
%N A158686 a(n) = 64*n^2 + 1.
%C A158686 The identity (64*n^2 + 1)^2 - (1024*n^2 + 32)*(2*n)^2 = 1 can be written as a(n)^2 - A158685(n)*(A005843(n))^2 = 1.
%H A158686 Vincenzo Librandi, <a href="/A158686/b158686.txt">Table of n, a(n) for n = 0..1000</a>
%H A158686 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 A158686 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A158686 G.f.: -(1+62*x+65*x^2)/(x-1)^3.
%F A158686 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F A158686 From _Amiram Eldar_, Mar 21 2023: (Start)
%F A158686 Sum_{n>=0} 1/a(n) = (coth(Pi/8)*Pi/8 + 1)/2.
%F A158686 Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/8)*Pi/8 + 1)/2. (End)
%F A158686 From _Elmo R. Oliveira_, Jan 17 2025: (Start)
%F A158686 E.g.f.: exp(x)*(1 + 64*x + 64*x^2).
%F A158686 a(n) = A108211(2*n) for n > 0. (End)
%t A158686 64 Range[0, 40]^2 + 1 (* or *) LinearRecurrence[{3, -3, 1}, {1, 65, 257}, 40] (* _Harvey P. Dale_, Jan 24 2012 *)
%t A158686 CoefficientList[Series[- (1 + 62 x + 65 x^2) / (x - 1)^3, {x, 0, 40}], x] (* _Vincenzo Librandi_, Sep 11 2013 *)
%o A158686 (Magma) [64*n^2+1: n in [0..40]]; // _Vincenzo Librandi_, Sep 11 2013
%o A158686 (PARI) a(n)=64*n^2+1 \\ _Charles R Greathouse IV_, Jun 17 2017
%Y A158686 Cf. A005843, A108211, A158685.
%K A158686 nonn,easy
%O A158686 0,2
%A A158686 _Vincenzo Librandi_, Mar 24 2009
%E A158686 Comment rewritten, a(0) added and formula replaced by _R. J. Mathar_, Oct 22 2009