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A158638 a(n) = 48*n^2 + 1.

%I A158638 #33 Jan 17 2025 07:38:51
%S A158638 1,49,193,433,769,1201,1729,2353,3073,3889,4801,5809,6913,8113,9409,
%T A158638 10801,12289,13873,15553,17329,19201,21169,23233,25393,27649,30001,
%U A158638 32449,34993,37633,40369,43201,46129,49153,52273,55489,58801,62209,65713,69313,73009,76801
%N A158638 a(n) = 48*n^2 + 1.
%C A158638 The identity (48*n^2 + 1)^2 - (576*n^2 + 24)*(2*n)^2 = 1 can be written as a(n)^2 - A158637(n)*A005843(n)^2 = 1.
%H A158638 Vincenzo Librandi, <a href="/A158638/b158638.txt">Table of n, a(n) for n = 0..10000</a>
%H A158638 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 A158638 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A158638 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F A158638 G.f.: -(1 + 46*x + 49*x^2)/(x-1)^3.
%F A158638 a(n) = 48*A000290(n) + 1. - _Wesley Ivan Hurt_, Dec 06 2013
%F A158638 From _Amiram Eldar_, Mar 19 2023: (Start)
%F A158638 Sum_{n>=0} 1/a(n) = (coth(Pi/(4*sqrt(3)))*Pi/(4*sqrt(3)) + 1)/2.
%F A158638 Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(4*sqrt(3)))*Pi/(4*sqrt(3)) + 1)/2. (End)
%F A158638 From _Elmo R. Oliveira_, Jan 17 2025: (Start)
%F A158638 E.g.f.: exp(x)*(1 + 48*x + 48*x^2).
%F A158638 a(n) = A158480(2*n). (End)
%p A158638 A158638:=n->48*n^2 + 1; seq(A158638(n), n=0..100); # _Wesley Ivan Hurt_, Dec 06 2013
%t A158638 LinearRecurrence[{3, -3, 1}, {1, 49, 193}, 50] (* _Vincenzo Librandi_, Feb 17 2012 *)
%t A158638 48*Range[0,40]^2+1 (* _Harvey P. Dale_, Mar 19 2013 *)
%o A158638 (Magma) I:=[1, 49, 193]; [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 A158638 (PARI) for(n=0, 40, print1(48*n^2 + 1", ")); \\ _Vincenzo Librandi_, Feb 17 2012
%Y A158638 Cf. A000290, A005843, A158480, A158637.
%K A158638 nonn,easy
%O A158638 0,2
%A A158638 _Vincenzo Librandi_, Mar 23 2009
%E A158638 Comment rephrased and redundant formula replaced by _R. J. Mathar_, Oct 19 2009