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a(n) = 128*n^2 - 32*n + 1.
2

%I #22 May 02 2024 16:12:14

%S 97,449,1057,1921,3041,4417,6049,7937,10081,12481,15137,18049,21217,

%T 24641,28321,32257,36449,40897,45601,50561,55777,61249,66977,72961,

%U 79201,85697,92449,99457,106721,114241,122017,130049,138337,146881,155681,164737,174049,183617

%N a(n) = 128*n^2 - 32*n + 1.

%C The identity (128*n^2 - 32*n + 1)^2 - (4*n^2 - n)*(64*n - 8)^2 = 1 can be written as a(n)^2 - A033991(n)*A157330(n)^2 = 1 (see also the second part of the comment at A157330). - _Vincenzo Librandi_, Jan 29 2012

%H Vincenzo Librandi, <a href="/A157331/b157331.txt">Table of n, a(n) for n = 1..10000</a>

%H Vincenzo Librandi, <a href="http://mathforum.org/kb/message.jspa?messageID=5773864&amp;tstart=0">X^2-AY^2=1</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F From _Vincenzo Librandi_, Jan 29 2012: (Start)

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

%F G.f.: x*(-97 - 158*x - x^2)/(x-1)^3. (End)

%F E.g.f.: exp(x)*(128*x^2 + 96*x + 1) - 1. - _Stefano Spezia_, May 02 2024

%t LinearRecurrence[{3,-3,1},{97,449,1057},40] (* _Vincenzo Librandi_, Jan 29 2012 *)

%o (Magma) I:=[97, 449, 1057]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]]; // _Vincenzo Librandi_, Jan 29 2012

%o (PARI) for(n=1, 40, print1(128*n^2 - 32*n + 1", ")); \\ _Vincenzo Librandi_, Jan 29 2012

%Y Cf. A033991, A157330.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Feb 27 2009