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%I #34 Mar 16 2023 07:14:48
%S 2177,8449,18817,33281,51841,74497,101249,132097,167041,206081,249217,
%T 296449,347777,403201,462721,526337,594049,665857,741761,821761,
%U 905857,994049,1086337,1182721,1283201,1387777,1496449,1609217,1726081,1847041
%N a(n) = 2048n^2 + 128n + 1.
%C The identity (2048*n^2+128*n+1)^2-(16*n^2+n)*(512*n+16)^2=1 can be written as a(n)^2-A157474(n)*A157475(n)^2=1. [rewritten by _Bruno Berselli_, Aug 22 2011]
%C This is the case s=4 of the identity (8*n^2*s^4+8*n*s^2+1)^2 - (n^2*s^2+n)*(8*n*s^3+4*s)^2 = 1. - _Bruno Berselli_, Jan 25 2012
%H Vincenzo Librandi, <a href="/A157476/b157476.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 From _Harvey P. Dale_, Aug 15 2011: (Start)
%F G.f.: x*(-x^2-1918*x-2177)/(x-1)^3.
%F a(1)=2177, a(2)=8449, a(3)=18817, a(n)=3*a(n-1)-3*a(n-2)+a(n-3). (End)
%t Table[2048n^2+128n+1,{n,30}] (* or *) LinearRecurrence[{3,-3,1},{2177,8449,18817},30] (* _Harvey P. Dale_, Aug 15 2011 *)
%o (PARI) a(n)=2048*n^2+128*n+1 \\ _Charles R Greathouse IV_, Jun 17 2017
%Y Cf. A157474, A157475.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 01 2009
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