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a(n) = n*(n+1)*(2*n^3 - n^2 + 2)^2/6.
2

%I #18 Sep 08 2022 08:45:16

%S 0,3,196,4418,43320,257645,1108828,3810996,11105328,28524615,66322740,

%T 142270678,285547496,541981713,980925260,1704069160,2856536928,

%U 4640618571,7332534948,11302649130,17039568280,25178606453,36535105596,52143138908,73300147600

%N a(n) = n*(n+1)*(2*n^3 - n^2 + 2)^2/6.

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

%H Vincenzo Librandi, <a href="/A101382/b101382.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (9,-36,84,-126,126,-84,36,-9,1).

%F G.f.: x*(3 + 169*x + 2762*x^2 + 10362*x^3 + 10727*x^4 + 2749*x^5 + 108*x^6)/(1 - x)^9. - _Ilya Gutkovskiy_, Feb 24 2017

%F E.g.f.: x*(18 + 570*x + 3839*x^2 + 6703*x^3 + 4164*x^4 + 1061*x^5 + 112*x^6 + 4*x^7)*exp(x)/6. - _G. C. Greubel_, Mar 11 2021

%p A101382:= n-> n*(n+1)*(2*n^3-n^2+2)^2/6: seq(A101382(n), n=0..35); # _G. C. Greubel_, Mar 11 2021

%t Table[n*(n+1)*(2*n^3-n^2+2)^2/6, {n, 0, 35}] (* _G. C. Greubel_, Mar 11 2021 *)

%o (Magma) [n*(n+1)*(2*n^3-n^2+2)^2/6: n in [0..40]]; // _Vincenzo Librandi_, May 26 2011

%o (PARI) a(n)=n*(n+1)*(2*n^3-n^2+2)^2/6 \\ _Charles R Greathouse IV_, Feb 24 2017

%o (Sage) [n*(n+1)*(2*n^3-n^2+2)^2/6 for n in (0..35)] # _G. C. Greubel_, Mar 11 2021

%Y Cf. A101383.

%K nonn,easy

%O 0,2

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