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a(n) = 121*n^2 - 38*n + 3.
3

%I #23 Sep 15 2022 02:37:01

%S 86,411,978,1787,2838,4131,5666,7443,9462,11723,14226,16971,19958,

%T 23187,26658,30371,34326,38523,42962,47643,52566,57731,63138,68787,

%U 74678,80811,87186,93803,100662,107763,115106,122691,130518,138587,146898

%N a(n) = 121*n^2 - 38*n + 3.

%C The identity (14641*n^2 - 4598*n + 362)^2 - (121*n^2 - 38*n + 3)*(1331*n - 209)^2 = 1 can be written as A157445(n)^2 - a(n)*A157444(n)^2 = 1. - _Vincenzo Librandi_, Jan 26 2012

%C The continued fraction expansion of sqrt(a(n)) is [11n-2; {3, 1, 1, 1, 11n-3, 1, 1, 1, 3, 22n-4}]. - _Magus K. Chu_, Sep 13 2022

%H Vincenzo Librandi, <a href="/A157443/b157443.txt">Table of n, a(n) for n = 1..10000</a>

%H Vincenzo Librandi, <a href="http://mathforum.org/kb/message.jspa?messageID=5785989&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 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - _Vincenzo Librandi_, Jan 26 2012

%F G.f.: x*(-86 - 153*x - 3*x^2)/(x-1)^3. - _Vincenzo Librandi_, Jan 26 2012

%t LinearRecurrence[{3,-3,1},{86,411,978},40] (* _Vincenzo Librandi_, Jan 26 2012 *)

%o (Magma) I:=[86, 411, 978]; [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 26 2012

%o (PARI) for(n=1, 22, print1(121*n^2 - 38*n + 3", ")); \\ _Vincenzo Librandi_, Jan 26 2012

%Y Cf. A157444, A157445.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Mar 01 2009