Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #27 Mar 16 2023 04:02:24
%S 20,420,1620,3620,6420,10020,14420,19620,25620,32420,40020,48420,
%T 57620,67620,78420,90020,102420,115620,129620,144420,160020,176420,
%U 193620,211620,230420,250020,270420,291620,313620,336420,360020,384420,409620,435620,462420,490020
%N a(n) = 400*n^2 + 20.
%C 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 Vincenzo Librandi, <a href="/A158601/b158601.txt">Table of n, a(n) for n = 0..10000</a>
%H Vincenzo Librandi, <a href="https://web.archive.org/web/20090309225914/http://mathforum.org/kb/message.jspa?messageID=5785989&tstart=0">X^2-AY^2=1</a>, Math Forum, 2007. [Wayback Machine link]
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F G.f.: -20*(1 + 18*x + 21*x^2)/(x-1)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F From _Amiram Eldar_, Mar 16 2023: (Start)
%F Sum_{n>=0} 1/a(n) = (coth(Pi/(2*sqrt(5)))*Pi/(2*sqrt(5)) + 1)/40.
%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(2*sqrt(5)))*Pi/(2*sqrt(5)) + 1)/40. (End)
%t 400 Range[0,40]^2+20 (* _Harvey P. Dale_, Feb 05 2011 *)
%t LinearRecurrence[{3, -3, 1}, {20, 420, 1620}, 50] (* _Vincenzo Librandi_, Feb 16 2012 *)
%o (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 (PARI) for(n=0, 40, print1(400*n^2 + 20", ")); \\ _Vincenzo Librandi_, Feb 16 2012
%Y Cf. A005843, A158602.
%K nonn,easy
%O 0,1
%A _Vincenzo Librandi_, Mar 22 2009
%E Comment rewritten, formula replaced by _R. J. Mathar_, Oct 28 2009