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a(n) = 576*n^2 - 24.
1

%I #22 Mar 19 2023 02:47:41

%S 552,2280,5160,9192,14376,20712,28200,36840,46632,57576,69672,82920,

%T 97320,112872,129576,147432,166440,186600,207912,230376,253992,278760,

%U 304680,331752,359976,389352,419880,451560,484392,518376,553512,589800,627240,665832,705576

%N a(n) = 576*n^2 - 24.

%C The identity (48*n^2 - 1)^2 - (576*n^2 - 24)*(2*n)^2 = 1 can be written as A065532(n)^2 - a(n)*A005843(n)^2 = 1.

%H Vincenzo Librandi, <a href="/A158636/b158636.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.: 24*x*(-23 - 26*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 19 2023: (Start)

%F Sum_{n>=1} 1/a(n) = (1 - cot(Pi/(2*sqrt(6)))*Pi/(2*sqrt(6)))/48.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/(2*sqrt(6)))*Pi/(2*sqrt(6)) - 1)/48. (End)

%t LinearRecurrence[{3, -3, 1}, {552, 2280, 5160}, 50] (* _Vincenzo Librandi_, Feb 17 2012 *)

%o (Magma) I:=[552, 2280, 5160]; [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=1, 40, print1(576*n^2 - 24", ")); \\ _Vincenzo Librandi_, Feb 17 2012

%Y Cf. A005843, A065532.

%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