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a(n) = 529*n^2 - 23.
2

%I #23 Mar 16 2023 04:04:00

%S 506,2093,4738,8441,13202,19021,25898,33833,42826,52877,63986,76153,

%T 89378,103661,119002,135401,152858,171373,190946,211577,233266,256013,

%U 279818,304681,330602,357581,385618,414713,444866,476077,508346,541673,576058,611501,648002

%N a(n) = 529*n^2 - 23.

%C The identity (46*n^2-1)^2-(529*n^2-23)*(2*n)^2 = 1 can be written as A158634(n)^2-a(n)*A005843(n)^2 = 1.

%H Vincenzo Librandi, <a href="/A158633/b158633.txt">Table of n, a(n) for n = 1..10000</a>

%H Vincenzo Librandi, <a href="https://web.archive.org/web/20090309225914/http://mathforum.org/kb/message.jspa?messageID=5785989&amp;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.: 23*x*(-22-25*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 16 2023: (Start)

%F Sum_{n>=1} 1/a(n) = (1 - cot(Pi/sqrt(23))*Pi/sqrt(23))/46.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(23))*Pi/sqrt(23) - 1)/46. (End)

%p A158633:=n->529*n^2 - 23: seq(A158633(n), n=1..50); # _Wesley Ivan Hurt_, Jan 28 2017

%t a[n_] := 529*n^2 - 23; Array[a, 50] (* _Amiram Eldar_, Mar 16 2023 *)

%o (PARI) for(n=1, 40, print1(529*n^2 - 23", ")); \\ _Vincenzo Librandi_, Feb 17 2012

%Y Cf. A005843, A158634.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Mar 23 2009

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