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%I #24 Mar 14 2023 03:37:16
%S 37,151,341,607,949,1367,1861,2431,3077,3799,4597,5471,6421,7447,8549,
%T 9727,10981,12311,13717,15199,16757,18391,20101,21887,23749,25687,
%U 27701,29791,31957,34199,36517,38911,41381,43927,46549,49247,52021,54871,57797,60799,63877
%N a(n) = 38*n^2 - 1.
%C The identity (38*n^2 - 1)^2 - (361*n^2 - 19)*(2*n)^2 = 1 can be written as a(n)^2 - A158595(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158596/b158596.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.: x*(-37 - 40*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(38))*Pi/sqrt(38))/2.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(38))*Pi/sqrt(38) - 1)/2. (End)
%t LinearRecurrence[{3, -3, 1}, {37, 151, 341}, 50] (* _Vincenzo Librandi_, Feb 16 2012 *)
%o (Magma) I:=[37, 151, 341]; [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(38*n^2 - 1", ")); \\ _Vincenzo Librandi_, Feb 16 2012
%Y Cf. A005843, A158595.
%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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