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a(n) = 2025*n^2 - 3401*n + 1428.
3

%I #22 Sep 08 2022 08:45:41

%S 52,2726,9450,20224,35048,53922,76846,103820,134844,169918,209042,

%T 252216,299440,350714,406038,465412,528836,596310,667834,743408,

%U 823032,906706,994430,1086204,1182028,1281902,1385826,1493800,1605824,1721898

%N a(n) = 2025*n^2 - 3401*n + 1428.

%C The identity (32805000*n^2 - 10513800*n + 842401)^2 - (2025*n^2 - 3401*n + 1428)*(729000*n - 116820)^2 = 1 can be written as A157079(n)^2 - a(n)*A156866(n)^2 = 1.

%H Vincenzo Librandi, <a href="/A156854/b156854.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 a(n) = 3*a(n-1) -3*a(n-2) +a(n-3).

%F G.f.: x*(52 +2570*x +1428*x^2)/(1-x)^3.

%F E.g.f.: -1428 + (1428 - 1376*x + 2025*x^2)*exp(x). - _G. C. Greubel_, Jan 28 2022

%t LinearRecurrence[{3,-3,1},{52,2726,9450},40]

%o (Magma) I:=[52, 2726, 9450]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]];

%o (PARI) a(n)=2025*n^2-3401*n+1428 \\ _Charles R Greathouse IV_, Dec 23 2011

%o (Sage) [(81*n -68)*(25*n -21) for n in (1..40)] # _G. C. Greubel_, Jan 28 2022

%Y Cf. A156853, A156866, A157079.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Feb 17 2009