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A158601 a(n) = 400*n^2 + 20.

%I A158601 #30 Jan 15 2025 22:18:56
%S A158601 20,420,1620,3620,6420,10020,14420,19620,25620,32420,40020,48420,
%T A158601 57620,67620,78420,90020,102420,115620,129620,144420,160020,176420,
%U A158601 193620,211620,230420,250020,270420,291620,313620,336420,360020,384420,409620,435620,462420,490020
%N A158601 a(n) = 400*n^2 + 20.
%C A158601 The identity (40*n^2 + 1)^2 - (400*n^2 + 20)*(2*n)^2 = 1 can be written as A158602(n)^2 - a(n)*A005843(n)^2 = 1.
%H A158601 Vincenzo Librandi, <a href="/A158601/b158601.txt">Table of n, a(n) for n = 0..10000</a>
%H A158601 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 A158601 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A158601 G.f.: -20*(1 + 18*x + 21*x^2)/(x-1)^3.
%F A158601 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F A158601 From _Amiram Eldar_, Mar 16 2023: (Start)
%F A158601 Sum_{n>=0} 1/a(n) = (coth(Pi/(2*sqrt(5)))*Pi/(2*sqrt(5)) + 1)/40.
%F A158601 Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(2*sqrt(5)))*Pi/(2*sqrt(5)) + 1)/40. (End)
%F A158601 From _Elmo R. Oliveira_, Jan 15 2025: (Start)
%F A158601 E.g.f.: 20*exp(x)*(1 + 20*x + 20*x^2).
%F A158601 a(n) = 20*A158493(n). (End)
%t A158601 400 Range[0,40]^2+20  (* _Harvey P. Dale_, Feb 05 2011 *)
%t A158601 LinearRecurrence[{3, -3, 1}, {20, 420, 1620}, 50] (* _Vincenzo Librandi_, Feb 16 2012 *)
%o A158601 (Magma) I:=[20, 420, 1620]; [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 16 2012
%o A158601 (PARI) for(n=0, 40, print1(400*n^2 + 20", ")); \\ _Vincenzo Librandi_, Feb 16 2012
%Y A158601 Cf. A005843, A158493, A158602.
%K A158601 nonn,easy
%O A158601 0,1
%A A158601 _Vincenzo Librandi_, Mar 22 2009
%E A158601 Comment rewritten, formula replaced by _R. J. Mathar_, Oct 28 2009