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%I #22 Mar 21 2023 05:20:58
%S 33,1122,4389,9834,17457,27258,39237,53394,69729,88242,108933,131802,
%T 156849,184074,213477,245058,278817,314754,352869,393162,435633,
%U 480282,527109,576114,627297,680658,736197,793914,853809,915882,980133,1046562,1115169,1185954,1258917
%N a(n) = 1089*n^2 + 33.
%C The identity (66*n^2 + 1)^2 - (1089*n^2 + 33)*(2*n)^2 = 1 can be written as A158689(n)^2 - a(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158688/b158688.txt">Table of n, a(n) for n = 0..10000</a>
%H Vincenzo Librandi, <a href="https://web.archive.org/web/20090309225914/http://mathforum.org/kb/message.jspa?messageID=5785989&tstart=0">X^2-AY^2=1</a>, Math Forum, 2007. [Wayback Machine link]
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F G.f.: -33*(1 + 31*x + 34*x^2)/(x-1)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F From _Amiram Eldar_, Mar 21 2023: (Start)
%F Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(33))*Pi/sqrt(33) + 1)/66.
%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(33))*Pi/sqrt(33) + 1)/66. (End)
%t LinearRecurrence[{3, -3, 1}, {33, 1122, 4389}, 50] (* _Vincenzo Librandi_, Feb 20 2012 *)
%o (Magma) I:=[33, 1122, 4389]; [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 20 2012
%o (PARI) for(n=0, 40, print1(1089*n^2 + 33", ")); \\ _Vincenzo Librandi_, Feb 20 2012
%Y Cf. A158689, A005843.
%K nonn,easy
%O 0,1
%A _Vincenzo Librandi_, Mar 24 2009
%E Comment rewritten, a(0) added and formula replaced by _R. J. Mathar_, Oct 22 2009
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