Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.
%I #28 Jan 15 2025 17:06:13
%S 23,552,2139,4784,8487,13248,19067,25944,33879,42872,52923,64032,
%T 76199,89424,103707,119048,135447,152904,171419,190992,211623,233312,
%U 256059,279864,304727,330648,357627,385664,414759,444912,476123,508392,541719,576104,611547,648048
%N a(n) = 529*n^2 + 23.
%C The identity (46*n^2 + 1)^2 - (529*n^2 + 23)*(2*n)^2 = 1 can be written as A158632(n)^2 - a(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158631/b158631.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 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F G.f.: -23*(1 + 21*x + 24*x^2)/(x-1)^3.
%F From _Amiram Eldar_, Mar 16 2023: (Start)
%F Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(23))*Pi/sqrt(23) + 1)/46.
%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(23))*Pi/sqrt(23) + 1)/46. (End)
%F E.g.f.: 23*exp(x)*(1 + 23*x + 23*x^2). - _Elmo R. Oliveira_, Jan 15 2025
%t LinearRecurrence[{3, -3, 1}, {23, 552, 2139}, 50] (* _Vincenzo Librandi_, Feb 17 2012 *)
%t 529*Range[0,40]^2+23 (* _Harvey P. Dale_, Jun 22 2014 *)
%o (Magma) I:=[23, 552, 2139]; [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 17 2012
%o (PARI) for(n=0, 40, print1(529*n^2 + 23", ")); \\ _Vincenzo Librandi_, Feb 17 2012
%Y Cf. A005843, A158632.
%K nonn,easy
%O 0,1
%A _Vincenzo Librandi_, Mar 23 2009
%E Comment rephrased and redundant formula replaced by _R. J. Mathar_, Oct 19 2009