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a(n) = 8*n^3*((2*n-1)^2 - 4*n + 4).
2

%I #9 Oct 21 2022 21:38:18

%S 0,8,320,3672,18944,65000,174528,397880,806912,1498824,2600000,

%T 4269848,6704640,10141352,14861504,21195000,29523968,40286600,

%U 53980992,71168984,92480000,118614888,150349760,188539832,234123264,288125000,351660608,425940120

%N a(n) = 8*n^3*((2*n-1)^2 - 4*n + 4).

%D L. U. Uko, A census of prime-order uniform step magic squares, Abstracts Amer. Math. Soc., Vol. 24, No. 1, 2003, #983-05-194.

%H G. C. Greubel, <a href="/A079504/b079504.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).

%F From _G. C. Greubel_, Jan 18 2019: (Start)

%F G.f.: 8*x*(1 +34*x +234*x^2 +194*x^3 +17*x^4)/(1-x)^6.

%F E.g.f.: 8*x*(1 +19*x +57*x^2 +32*x^3 +4*x^4)*exp(x). (End)

%t Table[8*n^3*((2*n-1)^2-4*n+4), {n, 0, 30}] (* or *) LinearRecurrence[{6, -15,20,-15,6,-1, {0,8,320,3672,18944,65000}, 30] (* _G. C. Greubel_, Jan 18 2019 *)

%o (PARI) vector(30, n, n--; 8*n^3*((2*n-1)^2-4*n+4)) \\ _G. C. Greubel_, Jan 18 2019

%o (Magma) [8*n^3*((2*n-1)^2-4*n+4): n in [0..30]]; // _G. C. Greubel_, Jan 18 2019

%o (Sage) [8*n^3*((2*n-1)^2-4*n+4) for n in (0..30)] # _G. C. Greubel_, Jan 18 2019

%o (GAP) List([0..30], n -> 8*n^3*((2*n-1)^2-4*n+4)); # _G. C. Greubel_, Jan 18 2019

%Y Cf. A079503.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_, Jan 21 2003