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

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

%S 0,0,1,60,594,3040,10875,30996,75460,163584,324405,599500,1046166,

%T 1740960,2783599,4301220,6453000,9435136,13486185,18892764,25995610,

%U 35196000,46962531,61838260,80448204,103507200,131828125,166330476,208049310,258144544,317910615

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

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

%H Vincenzo Librandi, <a href="/A101384/b101384.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^2*(1 + 53*x + 195*x^2 + 107*x^3 + 4*x^4)/(1 - x)^7. - _Ilya Gutkovskiy_, Feb 24 2017

%F E.g.f.: x^2*(1 + 19*x + 30*x^2 + 11*x^3 + x^4)*exp(x)/2. - _G. C. Greubel_, Mar 11 2021

%p A101384:= n-> n*(n-1)^3*(n^2 -n -1)/2: seq(A101384(n), n=0..35); # _G. C. Greubel_, Mar 11 2021

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

%o (Magma) [n*(n-1)^3*(n^2-n-1)/2: n in [0..40]]; // _Vincenzo Librandi_, Jun 15 2011

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

%Y Cf. A062392.

%K nonn,easy

%O 0,4

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