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a(n) = 841*n^2 - 29.
2

%I #28 Mar 20 2023 03:02:28

%S 812,3335,7540,13427,20996,30247,41180,53795,68092,84071,101732,

%T 121075,142100,164807,189196,215267,243020,272455,303572,336371,

%U 370852,407015,444860,484387,525596,568487,613060,659315,707252,756871,808172,861155,915820,972167,1030196

%N a(n) = 841*n^2 - 29.

%C The identity (58*n^2-1)^2 - (841*n^2-29)*(2*n)^2 = 1 can be written as A158668(n)^2 - a(n)*A005843(n)^2 = 1.

%H Vincenzo Librandi, <a href="/A158667/b158667.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.: 29*x*(-28-31*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 20 2023: (Start)

%F Sum_{n>=1} 1/a(n) = (1 - cot(Pi/sqrt(29))*Pi/sqrt(29))/58.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(29))*Pi/sqrt(29) - 1)/58. (End)

%p A158667:=n->841*n^2 - 29; seq(A158667(k), k=1..50); # _Wesley Ivan Hurt_, Nov 01 2013

%t LinearRecurrence[{3, -3, 1}, {812, 3335, 7540}, 40] (* or *) 29 (29 Range[40]^2 - 1) (* _Harvey P. Dale_, Oct 31 2011 *)

%o (Magma) I:=[812,3335,7540]; [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 18 2012

%o (PARI) for(n=1, 40, print1(841*n^2-29", ")); \\ _Vincenzo Librandi_, Feb 18 2012

%Y Cf. A005843, A158668.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Mar 24 2009

%E Comment rephrased and redundant formula replaced by _R. J. Mathar_, Oct 19 2009