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a(n) = n^2*(n+1)^2*(4*n^2 - 5*n + 4)/12.
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%I #26 Sep 24 2023 11:10:03

%S 0,1,30,300,1600,5925,17346,43120,95040,191025,356950,628716,1054560,

%T 1697605,2638650,3979200,5844736,8388225,11793870,16281100,22108800,

%U 29579781,39045490,50910960,65640000,83760625,105870726,132643980,164836000,203290725,248947050

%N a(n) = n^2*(n+1)^2*(4*n^2 - 5*n + 4)/12.

%D T. A. Gulliver, Sequences from Cubes of Integers, Int. Math. Journal, 4 (2003), 439-445.

%H Vincenzo Librandi, <a href="/A101381/b101381.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (7,-21,35,-35,21,-7,1)

%F G.f.: x*(1 + 23*x + 111*x^2 + 95*x^3 + 10*x^4) / (1-x)^7. - _R. J. Mathar_, Jun 15 2011

%F E.g.f.: x*(12 + 168*x + 426*x^2 + 288*x^3 + 63*x^4 + 4*x^5)*exp(x)/12. - _G. C. Greubel_, Mar 11 2021

%p A101381:= n-> n^2*(n+1)^2*(4*n^2-5*n+4)/12: seq(A101381(n), n=0..35); # _G. C. Greubel_, Mar 11 2021

%t Table[n^2 (n+1)^2 (4n^2-5n+4)/12,{n,0,30}] (* or *) LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,1,30,300,1600,5925,17346},40] (* _Harvey P. Dale_, Feb 03 2021 *)

%o (Magma) [n^2*(n+1)^2*(4*n^2-5*n+4)/12: n in [0..40]]; // _Vincenzo Librandi_, Jun 15 2011

%o (PARI) vector(35, n, my(m=n-1); m^2*(m+1)^2*(4*m^2-5*m+4)/12) \\ _G. C. Greubel_, Mar 11 2021

%o (Sage) [n^2*(n+1)^2*(4*n^2-5*n+4)/12 for n in (0..35)] # _G. C. Greubel_, Mar 11 2021

%K nonn,easy

%O 0,3

%A _N. J. A. Sloane_, Jan 15 2005