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%I #29 Mar 14 2023 03:37:29
%S 18,342,1314,2934,5202,8118,11682,15894,20754,26262,32418,39222,46674,
%T 54774,63522,72918,82962,93654,104994,116982,129618,142902,156834,
%U 171414,186642,202518,219042,236214,254034,272502,291618,311382,331794,352854,374562,396918
%N a(n) = 324*n^2 + 18.
%C The identity (36*n^2 + 1)^2 - (324*n^2 + 18)*(2*n)^2 = 1 can be written as A158591(n)^2 - a(n)* A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158590/b158590.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.: -18*(1 + 16*x + 19*x^2)/(x-1)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F From _Amiram Eldar_, Mar 14 2023: (Start)
%F Sum_{n>=0} 1/a(n) = (coth(Pi/(3*sqrt(2)))*Pi/(3*sqrt(2)) + 1)/36.
%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/(3*sqrt(2)))*Pi/(3*sqrt(2)) + 1)/36. (End)
%t LinearRecurrence[{3, -3, 1}, {18, 342, 1314}, 50] (* _Vincenzo Librandi_, Feb 16 2012 *)
%t 324 Range[0,40]^2+18 (* _Harvey P. Dale_, Nov 22 2018 *)
%o (Magma) I:=[18, 342, 1314]; [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(324*n^2 + 18", ")); \\ _Vincenzo Librandi_, Feb 16 2012
%Y Cf. A005843, A158591.
%K nonn,easy
%O 0,1
%A _Vincenzo Librandi_, Mar 22 2009
%E Comment rewritten, formula replaced by _R. J. Mathar_, Oct 28 2009
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