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

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

%S 0,1,14,94,380,1135,2786,5964,11544,20685,34870,55946,86164,128219,

%T 185290,261080,359856,486489,646494,846070,1092140,1392391,1755314,

%U 2190244,2707400,3317925,4033926,4868514,5835844,6951155,8230810,9692336,11354464,13237169

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

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

%H Vincenzo Librandi, <a href="/A101383/b101383.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 G.f.: x*(1 + 8*x + 25*x^2 + 6*x^3)/(1 - x)^6. - _Ilya Gutkovskiy_, Feb 24 2017

%F E.g.f.: x*(6 + 36*x + 55*x^2 + 21*x^3 + 2*x^4)*exp(x)/6. - _G. C. Greubel_, Mar 11 2021

%p A101383:= n-> n*(n+1)*(2*n^3-n^2+2)/6: seq(A101383(n), n=0..35); # _G. C. Greubel_, Mar 11 2021

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

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

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

%Y Cf. A101382.

%K nonn,easy

%O 0,3

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