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A158658 a(n) = 56*n^2 - 1.

%I A158658 #28 Jan 17 2025 07:35:02
%S A158658 55,223,503,895,1399,2015,2743,3583,4535,5599,6775,8063,9463,10975,
%T A158658 12599,14335,16183,18143,20215,22399,24695,27103,29623,32255,34999,
%U A158658 37855,40823,43903,47095,50399,53815,57343,60983,64735,68599,72575,76663,80863,85175,89599
%N A158658 a(n) = 56*n^2 - 1.
%C A158658 The identity (56*n^2 - 1)^2 - (784*n^2 - 28)*(2*n)^2 = 1 can be written as a(n)^2 - A158657(n)*A005843(n)^2 = 1.
%H A158658 Vincenzo Librandi, <a href="/A158658/b158658.txt">Table of n, a(n) for n = 1..10000</a>
%H A158658 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 A158658 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A158658 G.f.: x*(-55 - 58*x + x^2)/(x-1)^3.
%F A158658 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F A158658 From _Amiram Eldar_, Mar 20 2023: (Start)
%F A158658 Sum_{n>=1} 1/a(n) = (1 - cot(Pi/(2*sqrt(14)))*Pi/(2*sqrt(14)))/2.
%F A158658 Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/(2*sqrt(14)))*Pi/(2*sqrt(14)) - 1)/2. (End)
%F A158658 From _Elmo R. Oliveira_, Jan 17 2025: (Start)
%F A158658 E.g.f.: exp(x)*(56*x^2 + 56*x - 1) + 1.
%F A158658 a(n) = A158485(2*n). (End)
%t A158658 LinearRecurrence[{3, -3, 1}, {55, 223, 503}, 50] (* _Vincenzo Librandi_, Feb 17 2012 *)
%t A158658 56 Range[40]^2-1 (* _Harvey P. Dale_, Oct 23 2022 *)
%o A158658 (Magma) I:=[55, 223, 503]; [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 A158658 (PARI) for(n=1, 40, print1(56*n^2 - 1", ")); \\ _Vincenzo Librandi_, Feb 17 2012
%Y A158658 Cf. A005843, A158485, A158657.
%K A158658 nonn,easy
%O A158658 1,1
%A A158658 _Vincenzo Librandi_, Mar 23 2009
%E A158658 Comment rephrased and redundant formula replaced by _R. J. Mathar_, Oct 19 2009