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A158637 a(n) = 576*n^2 + 24.

%I A158637 #27 Jan 15 2025 22:08:40
%S A158637 24,600,2328,5208,9240,14424,20760,28248,36888,46680,57624,69720,
%T A158637 82968,97368,112920,129624,147480,166488,186648,207960,230424,254040,
%U A158637 278808,304728,331800,360024,389400,419928,451608,484440,518424,553560,589848,627288,665880,705624
%N A158637 a(n) = 576*n^2 + 24.
%C A158637 The identity (48*n^2 + 1)^2 - (576*n^2 + 24)*(2*n)^2 = 1 can be written as A158638(n)^2 - a(n)*A005843(n)^2 = 1.
%H A158637 Vincenzo Librandi, <a href="/A158637/b158637.txt">Table of n, a(n) for n = 0..10000</a>
%H A158637 Vincenzo Librandi, <a href="https://web.archive.org/web/20090309225914/http://mathforum.org/kb/message.jspa?messageID=5785989&amp;tstart=0">X^2-AY^2=1</a>, Math Forum, 2007. [Wayback Machine link]
%H A158637 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A158637 G.f.: -24*(1 + 22*x + 25*x^2)/(x-1)^3.
%F A158637 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F A158637 From _Amiram Eldar_, Mar 19 2023: (Start)
%F A158637 Sum_{n>=0} 1/a(n) = (coth(Pi/(2*sqrt(6)))*Pi/(2*sqrt(6)) + 1)/48.
%F A158637 Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(2*sqrt(6)))*Pi/(2*sqrt(6)) + 1)/48. (End)
%F A158637 From _Elmo R. Oliveira_, Jan 15 2025: (Start)
%F A158637 E.g.f.: 24*exp(x)*(1 + 24*x + 24*x^2).
%F A158637 a(n) = 24*A158547(n). (End)
%t A158637 24(24Range[0,40]^2+1) (* or *) LinearRecurrence[{3,-3,1},{24, 600, 2328}, 40] (* _Harvey P. Dale_, May 23 2011 *)
%o A158637 (Magma) I:=[24, 600, 2328]; [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 A158637 (PARI) for(n=0, 40, print1(576*n^2 + 24", ")); \\ _Vincenzo Librandi_, Feb 17 2012
%Y A158637 Cf. A005843, A158547, A158638.
%K A158637 nonn,easy
%O A158637 0,1
%A A158637 _Vincenzo Librandi_, Mar 23 2009
%E A158637 Comment rephrased and redundant formula replaced by _R. J. Mathar_, Oct 19 2009