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A157664 a(n) = 80000*n^2 + 800*n + 1.

%I A157664 #23 Sep 08 2022 08:45:42
%S A157664 80801,321601,722401,1283201,2004001,2884801,3925601,5126401,6487201,
%T A157664 8008001,9688801,11529601,13530401,15691201,18012001,20492801,
%U A157664 23133601,25934401,28895201,32016001,35296801,38737601,42338401,46099201
%N A157664 a(n) = 80000*n^2 + 800*n + 1.
%C A157664 The identity (80000n^2 + 800n + 1)^2 - (100n^2 + n)*(8000n + 40)^2 = 1 can be written as a(n)^2 - A055438(n)*A157663(n)^2 = 1. - _Vincenzo Librandi_, Feb 04 2012
%C A157664 This is the case s=10 of the identity (8*n^2*s^4 + 8*n*s^2 + 1)^2 - (n^2*s^2 + n)*(8*n*s^3 + 4*s)^2 = 1. - _Bruno Berselli_, Feb 04 2012
%H A157664 Vincenzo Librandi, <a href="/A157664/b157664.txt">Table of n, a(n) for n = 1..10000</a>
%H A157664 Vincenzo Librandi, <a href="http://mathforum.org/kb/message.jspa?messageID=5773864&amp;tstart=0">X^2-AY^2=1</a>
%H A157664 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A157664 G.f.: x*(80801 + 79198*x + x^2)/(1-x)^3. - _Vincenzo Librandi_, Feb 04 2012
%F A157664 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - _Vincenzo Librandi_, Feb 04 2012
%F A157664 E.g.f.: (1 + 80800*x + 80000*x^2)*exp(x) - 1. - _G. C. Greubel_, Nov 17 2018
%t A157664 LinearRecurrence[{3, -3, 1}, {80801, 321601, 722401}, 50] (* _Vincenzo Librandi_, Feb 04 2012 *)
%o A157664 (Magma) I:=[80801, 321601, 722401]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]]; // _Vincenzo Librandi_, Feb 04 2012
%o A157664 (PARI) for(n=1, 40, print1(80000*n^2 + 800*n + 1", ")); \\ _Vincenzo Librandi_, Feb 04 2012
%o A157664 (Sage) [80000*n^2+800*n+1 for n in (1..40)] # _G. C. Greubel_, Nov 17 2018
%o A157664 (GAP) List([1..40], n -> 80000*n^2+800*n+1); # _G. C. Greubel_, Nov 17 2018
%Y A157664 Cf. A055438, A157663.
%K A157664 nonn,easy
%O A157664 1,1
%A A157664 _Vincenzo Librandi_, Mar 04 2009