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a(n) = 81*n^2 - 9.
3

%I #27 Oct 14 2023 16:46:42

%S 72,315,720,1287,2016,2907,3960,5175,6552,8091,9792,11655,13680,15867,

%T 18216,20727,23400,26235,29232,32391,35712,39195,42840,46647,50616,

%U 54747,59040,63495,68112,72891,77832,82935,88200,93627,99216,104967,110880,116955,123192

%N a(n) = 81*n^2 - 9.

%C The identity (18*n^2 - 1)^2 - (81*n^2 - 9)*(2*n)^2 = 1 can be written as A157910(n)^2 - a(n)*A005843(n)^2 = 1. - _Vincenzo Librandi_, Feb 08 2012

%H Vincenzo Librandi, <a href="/A157909/b157909.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 08 2012: (Start)

%F G.f.: -9*x*(8 + 11*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 07 2023: (Start)

%F Sum_{n>=1} 1/a(n) = 1/18 - Pi/(54*sqrt(3)).

%F Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(27*sqrt(3)) - 1/18. (End)

%t LinearRecurrence[{3, -3, 1}, {72, 315, 720}, 50] (* _Vincenzo Librandi_, Feb 08 2012 *)

%t 81*Range[40]^2-9 (* _Harvey P. Dale_, Oct 14 2023 *)

%o (Magma) I:=[72, 315, 720]; [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 08 2012

%o (PARI) for(n=1, 40, print1(81*n^2 - 9", ")); \\ _Vincenzo Librandi_, Feb 08 2012

%Y Cf. A005843, A157910.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Mar 09 2009