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a(n) = 2025*n^2 - 649*n + 52.
4

%I #23 Sep 08 2022 08:45:41

%S 1428,6854,16330,29856,47432,69058,94734,124460,158236,196062,237938,

%T 283864,333840,387866,445942,508068,574244,644470,718746,797072,

%U 879448,965874,1056350,1150876,1249452,1352078,1458754,1569480,1684256

%N a(n) = 2025*n^2 - 649*n + 52.

%C The identity (32805000*n^2 - 55096200*n + 23133601)^2 - (2025*n^2 - 649*n + 52)*(729000*n - 612180)^2 = 1 can be written as A157078(n)^2 - a(n)*A156865(n)^2 = 1.

%H Vincenzo Librandi, <a href="/A156853/b156853.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*(1428 + 2570*x + 52*x^2)/(1-x)^3.

%F E.g.f.: -52 + (52 + 1376*x + 2025*x^2)*exp(x). - _G. C. Greubel_, Jan 27 2022

%t LinearRecurrence[{3,-3,1},{1428,6854,16330},40]

%o (Magma) I:=[1428, 6854, 16330]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..30]];

%o (PARI) a(n)=2025*n^2-649*n+52 \\ _Charles R Greathouse IV_, Dec 23 2011

%o (Sage) [(25*n -4)*(81*n -13) for n in (1..30)] # _G. C. Greubel_, Jan 27 2022

%Y Cf. A156865, A157078.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Feb 17 2009