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%I #27 Mar 24 2023 06:53:38
%S 1640,6440,14440,25640,40040,57640,78440,102440,129640,160040,193640,
%T 230440,270440,313640,360040,409640,462440,518440,577640,640040,
%U 705640,774440,846440,921640,1000040,1081640,1166440,1254440,1345640,1440040,1537640,1638440,1742440
%N a(n) = 1600*n^2 + 40.
%C The identity (80*n^2 + 1)^2 - (1600*n^2 + 40)*(2*n)^2 = 1 can be written as A158776(n)^2 - a(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158775/b158775.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 From _R. J. Mathar_, Jul 26 2009: (Start)
%F G.f.: -40*x*(41 + 38*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 24 2023: (Start)
%F Sum_{n>=1} 1/a(n) = (coth(Pi/(2*sqrt(10)))*Pi/(2*sqrt(10)) - 1)/80.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (1 - cosech(Pi/(2*sqrt(10)))*Pi/(2*sqrt(10)))/80. (End)
%t LinearRecurrence[{3, -3, 1}, {1640, 6440, 14440}, 50] (* _Vincenzo Librandi_, Feb 20 2012 *)
%t 1600*Range[30]^2+ 40 (* _Harvey P. Dale_, Jun 02 2020 *)
%o (Magma) I:=[1640, 6440, 14440]; [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=1, 40, print1(1600*n^2 + 40", ")); \\ _Vincenzo Librandi_, Feb 20 2012
%Y Cf. A005843, A158776.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 26 2009
%E Edited by _R. J. Mathar_, Jul 26 2009
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