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%I #28 Jan 15 2025 17:30:40
%S 19,380,1463,3268,5795,9044,13015,17708,23123,29260,36119,43700,52003,
%T 61028,70775,81244,92435,104348,116983,130340,144419,159220,174743,
%U 190988,207955,225644,244055,263188,283043,303620,324919,346940,369683,393148,417335,442244
%N a(n) = 361*n^2 + 19.
%C The identity (38*n^2 + 1)^2 - (361*n^2 + 19)*(2*n)^2 = 1 can be written as A158593(n)^2 - a(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158592/b158592.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.: -19*(1 + 17*x + 20*x^2)/(x-1)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F From _Amiram Eldar_, Mar 14 2023: (Start)
%F Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(19))*Pi/sqrt(19) + 1)/38.
%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(19))*Pi/sqrt(19) + 1)/38. (End)
%F E.g.f.: 19*exp(x)*(1 + 19*x + 19*x^2). - _Elmo R. Oliveira_, Jan 15 2025
%t LinearRecurrence[{3, -3, 1}, {19, 380, 1463}, 40] (* _Vincenzo Librandi_, Feb 16 2012 *)
%o (Magma) I:=[19, 380, 1463]; [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 16 2012
%o (PARI) for(n=0, 40, print1(361*n^2 + 19", ")); \\ _Vincenzo Librandi_, Feb 16 2012
%Y Cf. A005843, A158593.
%K nonn,easy,changed
%O 0,1
%A _Vincenzo Librandi_, Mar 22 2009
%E Comment rewritten, formula replaced by _R. J. Mathar_, Oct 28 2009