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A157663 a(n) = 8000*n + 40.

%I A157663 #29 Dec 31 2024 16:15:50
%S A157663 8040,16040,24040,32040,40040,48040,56040,64040,72040,80040,88040,
%T A157663 96040,104040,112040,120040,128040,136040,144040,152040,160040,168040,
%U A157663 176040,184040,192040,200040,208040,216040,224040,232040,240040,248040,256040
%N A157663 a(n) = 8000*n + 40.
%C A157663 The identity (80000*n^2 + 800*n + 1)^2 - (100*n^2 + n)*(8000*n + 40)^2 = 1 can be written as A157664(n)^2 - A055438(n)*a(n)^2 = 1 (see _Bruno Berselli_'s comment at A157664). - _Vincenzo Librandi_, Feb 04 2012
%H A157663 Vincenzo Librandi, <a href="/A157663/b157663.txt">Table of n, a(n) for n = 1..10000</a>
%H A157663 Vincenzo Librandi, <a href="http://mathforum.org/kb/message.jspa?messageID=5773864&amp;tstart=0">X^2-AY^2=1</a>
%H A157663 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1).
%F A157663 G.f.: x*(8040 - 40*x)/(1-x)^2. - _Vincenzo Librandi_, Feb 04 2012
%F A157663 a(n) = 2*a(n-1) - a(n-2). - _Vincenzo Librandi_, Feb 04 2012
%F A157663 E.g.f.: 40*(-1 + (1 + 200*x)*exp(x)). - _G. C. Greubel_, Nov 17 2018
%t A157663 LinearRecurrence[{2, -1}, {8040, 16040}, 50] (* _Vincenzo Librandi_, Feb 04 2012 *)
%t A157663 8000*Range[40]+40 (* _Harvey P. Dale_, Dec 31 2024 *)
%o A157663 (Magma) I:=[8040, 16040]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; // _Vincenzo Librandi_, Feb 04 2012
%o A157663 (PARI) for(n=1, 50, print1(8000*n + 40", ")); \\ _Vincenzo Librandi_, Feb 04 2012
%o A157663 (Sage) [40*(200*n + 1) for n in (1..40)] # _G. C. Greubel_, Nov 17 2018
%o A157663 (GAP) List([1..40], n -> 40*(200*n + 1)); # _G. C. Greubel_, Nov 17 2018
%Y A157663 Cf. A055438, A157664.
%K A157663 nonn,easy
%O A157663 1,1
%A A157663 _Vincenzo Librandi_, Mar 04 2009