login
a(n) = n^5 - n^2*(n^2 - 1)/2.
1

%I #17 Oct 21 2022 21:40:30

%S 0,1,26,207,904,2825,7146,15631,30752,55809,95050,153791,238536,

%T 357097,518714,734175,1015936,1378241,1837242,2411119,3120200,3987081,

%U 5036746,6296687,7797024,9570625,11653226,14083551,16903432,20157929,23895450,28167871

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

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

%H Vincenzo Librandi, <a href="/A100242/b100242.txt">Table of n, a(n) for n = 0..2000</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 a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6); a(0)=0, a(1)=1, a(2)=26, a(3)=207, a(4)=904, a(5)=2825. - _Harvey P. Dale_, Aug 15 2011

%F G.f.: (x^5 + 32*x^4 + 66*x^3 + 20*x^2 + x)/(x-1)^6. - _Harvey P. Dale_, Aug 15 2011

%t Table[n^5-(n^2 (n^2-1))/2,{n,0,40}] (* or *) LinearRecurrence[ {6,-15,20,-15,6,-1},{0,1,26,207,904,2825},40] (* _Harvey P. Dale_, Aug 15 2011 *)

%o (Magma) [n^5-n^2*(n^2-1)/2: n in [0..50]]; // _Vincenzo Librandi_, May 15 2011

%o (PARI) a(n)=n^5-n^2*(n^2-1)/2 \\ _Charles R Greathouse IV_, Oct 21 2022

%K nonn,easy

%O 0,3

%A _N. J. A. Sloane_, Jan 12 2005