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

%I #29 Mar 06 2023 02:21:12

%S 23,95,215,383,599,863,1175,1535,1943,2399,2903,3455,4055,4703,5399,

%T 6143,6935,7775,8663,9599,10583,11615,12695,13823,14999,16223,17495,

%U 18815,20183,21599,23063,24575,26135,27743,29399,31103,32855,34655,36503,38399,40343

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

%C The identity (24*n^2 - 1)^2 - (144*n^2 - 12)*(2*n)^2 = 1 can be written as a(n)^2 - A158543(n)*A005843(n)^2 = 1.

%H Vincenzo Librandi, <a href="/A158544/b158544.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 From _Vincenzo Librandi_, Feb 14 2012: (Start)

%F G.f.: -x*(23 + 26*x - x^2)/(x - 1)^3.

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)

%F From _Amiram Eldar_, Mar 06 2023: (Start)

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

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

%t LinearRecurrence[{3, -3, 1}, {23, 95, 215}, 50] (* _Vincenzo Librandi_, Feb 14 2012 *)

%o (Magma) I:=[23, 95, 215]; [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 14 2012

%o (PARI) for(n=1, 40, print1(24*n^2 - 1", ")); \\ _Vincenzo Librandi_, Feb 14 2012

%Y Cf. A005843, A158543.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Mar 21 2009

%E Comment rewritten by _R. J. Mathar_, Oct 16 2009