%I #23 Mar 05 2023 03:05:55
%S 20,95,220,395,620,895,1220,1595,2020,2495,3020,3595,4220,4895,5620,
%T 6395,7220,8095,9020,9995,11020,12095,13220,14395,15620,16895,18220,
%U 19595,21020,22495,24020,25595,27220,28895,30620,32395,34220,36095,38020,39995,42020,44095
%N a(n) = 25*n^2 - 5.
%C The identity (10*n^2-1)^2 - (25*n^2-5)*(2*n)^2 = 1 can be written as A158447(n)^2 - a(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158446/b158446.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&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 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F G.f: 5*x*(4+7*x-x^2)/(1-x)^3.
%F From _Amiram Eldar_, Mar 05 2023: (Start)
%F Sum_{n>=1} 1/a(n) = (1 - cot(Pi/sqrt(5))*Pi/sqrt(5))/10.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(5))*Pi/sqrt(5) - 1)/10. (End)
%t Table[25n^2-5,{n,50}]
%t LinearRecurrence[{3,-3,1},{20,95,220},40] (* _Harvey P. Dale_, May 05 2019 *)
%o (Magma) I:=[20, 95, 220]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]];
%o (PARI) a(n) = 25*n^2 - 5.
%Y Cf. A005843, A158447.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 19 2009
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