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%I #28 Mar 09 2023 02:43:56
%S 210,885,2010,3585,5610,8085,11010,14385,18210,22485,27210,32385,
%T 38010,44085,50610,57585,65010,72885,81210,89985,99210,108885,119010,
%U 129585,140610,152085,164010,176385,189210,202485,216210,230385,245010,260085,275610,291585,308010
%N a(n) = 225*n^2 - 15.
%C The identity (30*n^2 - 1)^2 - (225*n^2 - 15) * (2*n)^2 = 1 can be written as A158560(n)^2 - a(n) * A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158559/b158559.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F G.f.: 15*x*(-14 - 17*x + x^2)/(x-1)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F From _Amiram Eldar_, Mar 09 2023: (Start)
%F Sum_{n>=1} 1/a(n) = (1 - cot(Pi/sqrt(15))*Pi/sqrt(15))/30.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(15))*Pi/sqrt(15) - 1)/30. (End)
%t 15(15Range[40]^2-1) (* or *) LinearRecurrence[{3,-3,1},{210,885,2010},40] (* _Harvey P. Dale_, Jan 24 2012 *)
%o (Magma) I:=[210, 885, 2010]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]]; // _Vincenzo Librandi_, Feb 14 2012
%o (PARI) for(n=1, 40, print1(225*n^2 - 15", ")); \\ _Vincenzo Librandi_, Feb 05 2012
%Y Cf. A005843, A158560.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 21 2009
%E Comment rewritten by _R. J. Mathar_, Oct 16 2009
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