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%I #24 Nov 11 2024 20:30:40
%S 223,896,2019,3592,5615,8088,11011,14384,18207,22480,27203,32376,
%T 37999,44072,50595,57568,64991,72864,81187,89960,99183,108856,118979,
%U 129552,140575,152048,163971,176344,189167,202440,216163,230336,244959,260032
%N a(n) = 225*n^2 - 2*n.
%C The identity (225*n-1)^2-(225*n^2-2*n)*(15)^2=1 can be written as A158227(n)^2-a(n)*(15)^2=1.
%H Vincenzo Librandi, <a href="/A158226/b158226.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 E. J. Barbeau, <a href="http://www.math.toronto.edu/barbeau/home.html">Polynomial Excursions</a>, Chapter 10: <a href="http://www.math.toronto.edu/barbeau/hxpol10.pdf">Diophantine equations</a> (2010), pages 84-85 (row 15 in the first table at p. 85, case d(t) = t*(15^2*t-2)).
%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).
%F G.f.: x*(-223-227*x)/(x-1)^3.
%t LinearRecurrence[{3,-3,1},{223,896,2019},50]
%t Table[225n^2-2n,{n,40}] (* _Harvey P. Dale_, Feb 25 2021 *)
%o (Magma) I:=[223, 896, 2019]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]];
%o (PARI) a(n) = 225*n^2 - 2*n
%Y Cf. A158227.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 14 2009