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A158603 a(n) = 441*n^2 + 21.

%I A158603 #27 Jan 15 2025 17:26:04
%S A158603 21,462,1785,3990,7077,11046,15897,21630,28245,35742,44121,53382,
%T A158603 63525,74550,86457,99246,112917,127470,142905,159222,176421,194502,
%U A158603 213465,233310,254037,275646,298137,321510,345765,370902,396921,423822,451605,480270,509817,540246
%N A158603 a(n) = 441*n^2 + 21.
%C A158603 The identity (42*n^2 + 1)^2 - (441*n^2 + 21)*(2*n)^2 = 1 can be written as A158604(n)^2 - a(n)*A005843(n)^2 = 1.
%H A158603 Vincenzo Librandi, <a href="/A158603/b158603.txt">Table of n, a(n) for n = 0..10000</a>
%H A158603 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 A158603 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A158603 G.f.: -21*(1 + 19*x + 22*x^2)/(x-1)^3.
%F A158603 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F A158603 From _Amiram Eldar_, Mar 16 2023: (Start)
%F A158603 Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(21))*Pi/sqrt(21) + 1)/42.
%F A158603 Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(21))*Pi/sqrt(21) + 1)/42. (End)
%F A158603 E.g.f.: 21*exp(x)*(1 + 21*x + 21*x^2). - _Elmo R. Oliveira_, Jan 15 2025
%t A158603 LinearRecurrence[{3, -3, 1}, {21, 462, 1785}, 50] (* _Vincenzo Librandi_, Feb 16 2012 *)
%o A158603 (Magma) I:=[21, 462, 1785]; [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 A158603 (PARI) for(n=0, 40, print1(441*n^2 + 21", ")); \\ _Vincenzo Librandi_, Feb 16 2012
%Y A158603 Cf. A005843, A158604.
%K A158603 nonn,easy
%O A158603 0,1
%A A158603 _Vincenzo Librandi_, Mar 22 2009
%E A158603 Comment rewritten, formula replaced by _R. J. Mathar_, Oct 28 2009