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A158602 a(n) = 40*n^2 + 1.

%I A158602 #33 Jan 17 2025 06:54:14
%S A158602 1,41,161,361,641,1001,1441,1961,2561,3241,4001,4841,5761,6761,7841,
%T A158602 9001,10241,11561,12961,14441,16001,17641,19361,21161,23041,25001,
%U A158602 27041,29161,31361,33641,36001,38441,40961,43561,46241,49001,51841,54761,57761,60841,64001
%N A158602 a(n) = 40*n^2 + 1.
%C A158602 The identity (40*n^2 + 1)^2 - (400*n^2 + 20)*(2*n)^2 = 1 can be written as a(n)^2 - A158601(n)*A005843(n)^2 = 1.
%H A158602 Vincenzo Librandi, <a href="/A158602/b158602.txt">Table of n, a(n) for n = 0..10000</a>
%H A158602 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 A158602 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A158602 G.f.: -(1 + 38*x + 41*x^2)/(x-1)^3.
%F A158602 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F A158602 From _Amiram Eldar_, Mar 16 2023: (Start)
%F A158602 Sum_{n>=0} 1/a(n) = (coth(Pi/(2*sqrt(10)))*Pi/(2*sqrt(10)) + 1)/2.
%F A158602 Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(2*sqrt(10)))*Pi/(2*sqrt(10)) + 1)/2. (End)
%F A158602 From _Elmo R. Oliveira_, Jan 16 2025: (Start)
%F A158602 E.g.f.: exp(x)*(1 + 40*x + 40*x^2).
%F A158602 a(n) = A158187(2*n). (End)
%p A158602 A158602:=n->40*n^2; seq(A158602(k), k=0..100); # _Wesley Ivan Hurt_, Sep 27 2013
%t A158602 40*Range[0,40]^2+1 (* or *) LinearRecurrence[{3,-3,1},{1,41,161},40] (* _Harvey P. Dale_, Jul 25 2011 *)
%t A158602 Table[40n^2+1, {n,0,100}] (* _Wesley Ivan Hurt_, Sep 27 2013 *)
%o A158602 (Magma) I:=[1,41,161]; [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 A158602 (PARI) for(n=0, 40, print1(40*n^2 + 1", ")); \\ _Vincenzo Librandi_, Feb 16 2012
%Y A158602 Cf. A005843, A158187, A158601.
%K A158602 nonn,easy
%O A158602 0,2
%A A158602 _Vincenzo Librandi_, Mar 22 2009
%E A158602 Comment rewritten, formula replaced by _R. J. Mathar_, Oct 28 2009