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a(n) = 121*n^2 - 11.
2

%I #32 Mar 06 2023 02:20:33

%S 110,473,1078,1925,3014,4345,5918,7733,9790,12089,14630,17413,20438,

%T 23705,27214,30965,34958,39193,43670,48389,53350,58553,63998,69685,

%U 75614,81785,88198,94853,101750,108889,116270,123893,131758,139865,148214,156805,165638,174713

%N a(n) = 121*n^2 - 11.

%C The identity (22*n^2 - 1)^2 - (121*n^2 - 11)*(2*n)^2 = 1 can be written as A158540(n)^2 - a(n)*A005843(n)^2 = 1. - _Vincenzo Librandi_, Feb 21 2012

%H Vincenzo Librandi, <a href="/A158539/b158539.txt">Table of n, a(n) for n = 1..10000</a>

%H Vincenzo Librandi, <a href="https://web.archive.org/web/20090309225914/http://mathforum.org/kb/message.jspa?messageID=5785989&amp;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 From _Vincenzo Librandi_, Feb 21 2012: (Start)

%F G.f.: 11*x*(-10 - 13*x + x^2)/(x-1)^3.

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)

%F From _Amiram Eldar_, Mar 06 2023: (Start)

%F Sum_{n>=1} 1/a(n) = (1 - cot(Pi/sqrt(11))*Pi/sqrt(11))/22.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(11))*Pi/sqrt(11) - 1)/22. (End)

%t LinearRecurrence[{3, -3, 1}, {110, 473, 1078}, 50] (* _Vincenzo Librandi_, Feb 21 2012 *)

%t 121 Range[40]^2-11 (* or *) CoefficientList[Series[(11(x^2-13x-10))/(x-1)^3,{x,0,40}],x] (* _Harvey P. Dale_, Aug 16 2021 *)

%o (Magma) I:=[110, 473, 1078]; [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 21 2012

%o (PARI) for(n=1, 40, print1(121*n^2 - 11", ")); \\ _Vincenzo Librandi_, Feb 21 2012

%Y Cf. A005843, A158540.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Mar 21 2009

%E Edited by _N. J. A. Sloane_, Oct 12 2009