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%I #24 Mar 14 2023 03:37:12
%S 342,1425,3230,5757,9006,12977,17670,23085,29222,36081,43662,51965,
%T 60990,70737,81206,92397,104310,116945,130302,144381,159182,174705,
%U 190950,207917,225606,244017,263150,283005,303582,324881,346902,369645,393110,417297,442206,467837
%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 in as A158596(n)^2 - a(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158595/b158595.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 G.f.: 19*x*(-18 - 21*x + 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>=1} 1/a(n) = (1 - cot(Pi/sqrt(19))*Pi/sqrt(19))/38.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(19))*Pi/sqrt(19) - 1)/38. (End)
%t LinearRecurrence[{3, -3, 1}, {342, 1425, 3230}, 50] (* _Vincenzo Librandi_, Feb 16 2012 *)
%o (Magma) I:=[342, 1425, 3230]; [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=1, 40, print1(361*n^2 - 19", ")); \\ _Vincenzo Librandi_, Feb 16 2012
%Y Cf. A005843, A158596.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 22 2009
%E Comment rewritten, formula replaced by _R. J. Mathar_, Oct 28 2009
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