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%I #22 Mar 16 2023 04:04:51
%S 22,506,1958,4378,7766,12122,17446,23738,30998,39226,48422,58586,
%T 69718,81818,94886,108922,123926,139898,156838,174746,193622,213466,
%U 234278,256058,278806,302522,327206,352858,379478,407066,435622,465146,495638,527098,559526,592922
%N a(n) = 484*n^2 + 22.
%C The identity (44*n^2 + 1)^2 - (484*n^2 + 22)*(2*n)^2 = 1 can be written as A158630(n)^2 - a(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158629/b158629.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.: -22*(1 + 20*x + 23*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/sqrt(22))*Pi/sqrt(22) + 1)/44.
%F Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(22))*Pi/sqrt(22) + 1)/44. (End)
%t LinearRecurrence[{3, -3, 1}, {22, 506, 1958}, 50] (* _Vincenzo Librandi_, Feb 17 2012 *)
%o (Magma) I:=[22, 506, 1958]; [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 17 2012
%o (PARI) for(n=0, 40, print1(484*n^2 + 22", ")); \\ _Vincenzo Librandi_, Feb 17 2012
%Y Cf. A005843, A158630.
%K nonn,easy
%O 0,1
%A _Vincenzo Librandi_, Mar 23 2009
%E Comment rewritten, formula replaced by _R. J. Mathar_, Oct 28 2009
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