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%I #24 Sep 08 2022 08:45:42
%S 5201,20401,45601,80801,126001,181201,246401,321601,406801,502001,
%T 607201,722401,847601,982801,1128001,1283201,1448401,1623601,1808801,
%U 2004001,2209201,2424401,2649601,2884801,3130001,3385201,3650401
%N a(n) = 5000*n^2 + 200*n + 1.
%C The identity (5000*n^2 + 200*n + 1)^2 - (25*n^2 + n)*(1000*n + 20)^2 = 1 can be written as a(n)^2 - A173089(n)*A157510(n)^2 = 1 (see also second part of the comment at A173089). - _Vincenzo Librandi_, Feb 04 2012
%H Vincenzo Librandi, <a href="/A157511/b157511.txt">Table of n, a(n) for n = 1..10000</a>
%H Vincenzo Librandi, <a href="http://mathforum.org/kb/message.jspa?messageID=5771301&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 _Harvey P. Dale_, May 24 2011: (Start)
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(1)=5201, a(2)=20401, a(3)=45601.
%F G.f.: -x*((5201 + x*(4798+x))/(x-1)^3). (End)
%t Table[5000n^2+200n+1,{n,40}] (* or *) LinearRecurrence[{3,-3,1},{5201,20401,45601},40] (* _Harvey P. Dale_, May 24 2011 *)
%o (Magma) I:=[5201, 20401, 45601]; [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 04 2012
%o (PARI) for(n=1, 40, print1(5000*n^2 + 200*n + 1", ")); \\ _Vincenzo Librandi_, Feb 04 2012
%Y Cf. A157510, A173089.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 02 2009
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