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%I #27 Mar 16 2023 04:03:09
%S 41,167,377,671,1049,1511,2057,2687,3401,4199,5081,6047,7097,8231,
%T 9449,10751,12137,13607,15161,16799,18521,20327,22217,24191,26249,
%U 28391,30617,32927,35321,37799,40361,43007,45737,48551,51449,54431,57497,60647,63881,67199,70601
%N a(n) = 42*n^2 - 1.
%C The identity (42*n^2-1)^2 - (441*n^2-21)*(2*n)^2 = 1 can be written as a(n)^2-A145678(n)*A005843(n)^2 = 1. - _Vincenzo Librandi_, Feb 16 2012
%H Vincenzo Librandi, <a href="/A158626/b158626.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*(-41-44*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 16 2023: (Start)
%F Sum_{n>=1} 1/a(n) = (1 - cot(Pi/sqrt(42))*Pi/sqrt(42))/2.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(42))*Pi/sqrt(42) - 1)/2. (End)
%t LinearRecurrence[{3, -3, 1}, {41, 167, 377}, 50] (* _Vincenzo Librandi_, Feb 16 2012 *)
%o (Magma) I:=[41, 167, 377]; [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(42*n^2-1", ")); \\ _Vincenzo Librandi_, Feb 16 2012
%Y Cf. A158625, A145678, A005843.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 23 2009
%E Edited by _R. J. Mathar_, Jul 26 2009
%E A-number updated by _R. J. Mathar_, Oct 16 2009
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