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%I #22 Mar 20 2023 03:02:31
%S 1,59,233,523,929,1451,2089,2843,3713,4699,5801,7019,8353,9803,11369,
%T 13051,14849,16763,18793,20939,23201,25579,28073,30683,33409,36251,
%U 39209,42283,45473,48779,52201,55739,59393,63163,67049,71051,75169,79403,83753,88219,92801
%N a(n) = 58*n^2 + 1.
%C The identity (58*n^2 + 1)^2 - (841*n^2 + 29)*(2*n)^2 = 1 can be written as a(n)^2 - A158665(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158666/b158666.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.: -(1 + 56*x + 59*x^2)/(x-1)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F From _Amiram Eldar_, Mar 20 2023: (Start)
%F Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(58))*Pi/sqrt(58) + 1)/2.
%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(58))*Pi/sqrt(58) + 1)/2. (End)
%t LinearRecurrence[{3, -3, 1}, {1, 59, 233}, 50] (* _Vincenzo Librandi_, Feb 18 2012 *)
%o (Magma) I:=[1, 59, 233]; [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 18 2012
%o (PARI) for(n=0, 40, print1(58*n^2 + 1", ")); \\ _Vincenzo Librandi_, Feb 18 2012
%Y Cf. A005843, A158665.
%K nonn,easy
%O 0,2
%A _Vincenzo Librandi_, Mar 24 2009
%E Comment rephrased and redundant formula replaced by _R. J. Mathar_, Oct 19 2009
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