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%I #18 Sep 08 2022 08:45:42
%S 103651201,414662401,933033601,1658764801,2591856001,3732307201,
%T 5080118401,6635289601,8397820801,10367712001,12544963201,14929574401,
%U 17521545601,20320876801,23327568001,26541619201,29963030401,33591801601
%N a(n) = 103680000*n^2 - 28800*n + 1.
%C The identity (103680000*n^2 - 28800*n + 1)^2 - (3600*n^2 - n)*(1728000*n - 240)^2 = 1 can be written as a(n)^2 - A157857(n)*A157858(n)^2 = 1 (see second comment at A157858). - _Vincenzo Librandi_, Jan 25 2012
%H Vincenzo Librandi, <a href="/A157859/b157859.txt">Table of n, a(n) for n = 1..10000</a>
%H Vincenzo Librandi, <a href="http://mathforum.org/kb/message.jspa?messageID=5785989&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 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - _Vincenzo Librandi_, Jan 25 2012
%F G.f.: x*(-103651201 - 103708798*x - x^2)/(x-1)^3. - _Vincenzo Librandi_, Jan 25 2012
%t LinearRecurrence[{3,-3,1},{103651201,414662401,933033601},40] (* _Vincenzo Librandi_, Jan 25 2012 *)
%o (Magma) I:=[103651201, 414662401, 933033601]; [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_, Jan 25 2012
%o (PARI) for(n=1, 22, print1(103680000*n^2 - 28800*n + 1", ")); \\ _Vincenzo Librandi_, Jan 25 2012
%Y Cf. A157857, A157858.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 08 2009
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