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A158776 a(n) = 80*n^2 + 1.

%I A158776 #31 Jan 25 2025 17:16:58
%S A158776 1,81,321,721,1281,2001,2881,3921,5121,6481,8001,9681,11521,13521,
%T A158776 15681,18001,20481,23121,25921,28881,32001,35281,38721,42321,46081,
%U A158776 50001,54081,58321,62721,67281,72001,76881,81921,87121,92481,98001,103681,109521,115521,121681
%N A158776 a(n) = 80*n^2 + 1.
%C A158776 The identity (80*n^2 + 1)^2 - (1600*n^2 + 40)*(2*n)^2 = 1 can be written as a(n)^2 - A158775(n)*A005843(n)^2 = 1.
%H A158776 Vincenzo Librandi, <a href="/A158776/b158776.txt">Table of n, a(n) for n = 0..10000</a>
%H A158776 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 A158776 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A158776 G.f.: -(1 + 78*x + 81*x^2)/(x-1)^3.
%F A158776 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F A158776 From _Amiram Eldar_, Mar 24 2023: (Start)
%F A158776 Sum_{n>=0} 1/a(n) = (coth(Pi/(4*sqrt(5)))*Pi/(4*sqrt(5)) + 1)/2.
%F A158776 Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(4*sqrt(5)))*Pi/(4*sqrt(5)) + 1)/2. (End)
%F A158776 From _Elmo R. Oliveira_, Jan 25 2025: (Start)
%F A158776 E.g.f.: exp(x)*(1 + 80*x + 80*x^2).
%F A158776 a(n) = A158493(2*n). (End)
%t A158776 LinearRecurrence[{3, -3, 1}, {1, 81, 321}, 50] (* _Vincenzo Librandi_, Feb 20 2012 *)
%t A158776 80 Range[0,50]^2+1 (* _Harvey P. Dale_, Jan 17 2021 *)
%o A158776 (PARI) a(n)=80*n^2+1 \\ _Charles R Greathouse IV_, Dec 23 2011
%o A158776 (Magma) I:=[1, 81, 321]; [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 20 2012
%Y A158776 Cf. A005843, A158493, A158775.
%K A158776 nonn,easy
%O A158776 0,2
%A A158776 _Vincenzo Librandi_, Mar 26 2009
%E A158776 Comment rewritten, a(0) added and formula replaced by _R. J. Mathar_, Oct 22 2009