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%I #23 Jan 31 2023 17:49:03
%S 839,3360,7563,13448,21015,30264,41195,53808,68103,84080,101739,
%T 121080,142103,164808,189195,215264,243015,272448,303563,336360,
%U 370839,407000,444843,484368,525575,568464,613035,659288,707223,756840,808139,861120
%N a(n) = 841*n^2 - 2*n.
%C The identity (841*n-1)^2-(841*n^2-2*n)*(29)^2 = 1 can be written as A158402(n)^2-a(n)*(29)^2 = 1.
%H Vincenzo Librandi, <a href="/A158401/b158401.txt">Table of n, a(n) for n = 1..10000</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*(29^2*t-2)).
%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) for n > 3.
%F G.f.: x*(-839-843*x)/(x-1)^3.
%p A158401:=n->841*n^2 - 2*n: seq(A158401(n), n=1..50); # _Wesley Ivan Hurt_, Oct 15 2017
%t LinearRecurrence[{3,-3,1},{839,3360,7563},50]
%t Table[841n^2-2n,{n,40}] (* _Harvey P. Dale_, Jan 31 2023 *)
%o (Magma) I:=[839, 3360, 7563]; [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) = 841*n^2 - 2*n.
%Y Cf. A158402.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 18 2009