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a(n) = 512*n - 16.
3

%I #21 Sep 08 2022 08:45:42

%S 496,1008,1520,2032,2544,3056,3568,4080,4592,5104,5616,6128,6640,7152,

%T 7664,8176,8688,9200,9712,10224,10736,11248,11760,12272,12784,13296,

%U 13808,14320,14832,15344,15856,16368,16880,17392,17904,18416,18928

%N a(n) = 512*n - 16.

%C The identity (2048*n^2 - 128*n + 1)^2 - (16*n^2 - n)*(512*n - 16)^2 = 1 can be written as A157448(n)^2 - A157446(n)*a(n)^2 = 1 (see also second comment at A157446). - _Vincenzo Librandi_, Jan 26 2012

%H Vincenzo Librandi, <a href="/A157447/b157447.txt">Table of n, a(n) for n = 1..10000</a>

%H Vincenzo Librandi, <a href="http://mathforum.org/kb/message.jspa?messageID=5785989&amp;tstart=0">X^2-AY^2=1</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1).

%F a(n) = 2*a(n-1) - a(n-2). - _Vincenzo Librandi_, Jan 26 2012

%F G.f.: x*(16*x + 496)/(x-1)^2. - _Vincenzo Librandi_, Jan 26 2012

%t LinearRecurrence[{2,-1},{496,1008},40] (* _Vincenzo Librandi_, Jan 26 2012 *)

%t 512*Range[50]-16 (* _Harvey P. Dale_, Apr 09 2019 *)

%o (Magma) I:=[496, 1008]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; // _Vincenzo Librandi_, Jan 26 2012

%o (PARI) for(n=1, 22, print1(512*n - 16", ")); \\ _Vincenzo Librandi_, Jan 26 2012

%Y Cf. A157446, A157448.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Mar 01 2009