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%I #33 Feb 27 2024 12:05:10
%S 1,21,81,181,321,501,721,981,1281,1621,2001,2421,2881,3381,3921,4501,
%T 5121,5781,6481,7221,8001,8821,9681,10581,11521,12501,13521,14581,
%U 15681,16821,18001,19221,20481,21781,23121,24501,25921,27381,28881,30421,32001,33621,35281
%N a(n) = 20*n^2 + 1.
%C The identity (20*n^2 + 1)^2 - (100*n^2 + 10)*(2*n)^2 = 1 can be written as a(n)^2 - A158492(n)*A005843(n)^2 = 1. - _Vincenzo Librandi_, Feb 21 2012
%C Sequence found by reading the segment (1, 21) together with the line from 21, in the direction 21, 81, ..., in the square spiral whose vertices are the generalized dodecagonal numbers A195162. - _Omar E. Pol_, Nov 05 2012
%H Vincenzo Librandi, <a href="/A158493/b158493.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 From _Vincenzo Librandi_, Feb 21 2012: (Start)
%F G.f.: -(1 + 18*x + 21*x^2)/(x-1)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
%F From _Amiram Eldar_, Mar 06 2023: (Start)
%F Sum_{n>=0} 1/a(n) = (coth(Pi/(2*sqrt(5)))*Pi/(2*sqrt(5)) + 1)/2.
%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(2*sqrt(5)))*Pi/(2*sqrt(5)) + 1)/2. (End)
%t LinearRecurrence[{3, -3, 1}, {1, 21, 81}, 50] (* _Vincenzo Librandi_, Feb 21 2012 *)
%o (Magma) I:=[1, 21, 81]; [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 21 2012
%o (PARI) for(n=0, 40, print1(20*n^2 + 1", ")); \\ _Vincenzo Librandi_, Feb 21 2012
%Y Cf. A005843, A158492, A195162.
%K nonn,easy
%O 0,2
%A _Vincenzo Librandi_, Mar 21 2009
%E Edited by _N. J. A. Sloane_, Oct 12 2009
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