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a(n) = (2*n - 11)*n^2.
1

%I #13 Jul 30 2016 19:55:48

%S 0,-9,-28,-45,-48,-25,36,147,320,567,900,1331,1872,2535,3332,4275,

%T 5376,6647,8100,9747,11600,13671,15972,18515,21312,24375,27716,31347,

%U 35280,39527,44100,49011,54272,59895

%N a(n) = (2*n - 11)*n^2.

%H Ivan Panchenko, <a href="/A015245/b015245.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F G.f.: x*(-9 + 8*x + 13*x^2)/(1-x)^4. - _Ivan Panchenko_, Nov 09 2013

%F From _G. C. Greubel_, Jul 30 2016: (Start)

%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).

%F E.g.f.: x*(-9 - 5*x + 2*x^2)*exp(x). (End)

%t Table[(2*n - 11)*n^2,{n,0,25}] (* or *) LinearRecurrence[{4,-6,4,-1},{0, -9, -28, -45},25] (* _G. C. Greubel_, Jul 30 2016 *)

%o (PARI) a(n)=(2*n-11)*n^2 \\ _Charles R Greathouse IV_, Jul 30 2016

%K sign,easy

%O 0,2

%A _N. J. A. Sloane_, Dec 11 1999