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

%I #15 Sep 08 2022 08:45:49

%S 0,1,1152,91854,2129920,24609375,182238336,991545772,4303355904,

%T 15712053165,50050000000,142763014746,371719176192,896488062379,

%U 2025520479360,4326159375000,8798240505856,17139436032537

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

%H Vincenzo Librandi, <a href="/A170775/b170775.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_11">Index entries for linear recurrences with constant coefficients</a>, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).

%F G.f.: (x*(x*(x*(x*(x*(x*(x*(x*(x*(896*x+74541)+1100004)+4878360)+ 7862124)+ 4859754)+1103484)+78096)+1140)+1))/(x-1)^12. - _Harvey P. Dale_, Oct 25 2011

%t Table[(n^8 (n^3+1))/2,{n,0,20}] (* _Harvey P. Dale_, Oct 25 2011 *)

%o (Magma) [n^8*(n^3+1)/2: n in [0..30]]; // _Vincenzo Librandi_, Aug 26 2011

%o (PARI) for(n=0,30, print1(n^8*(n^3+1)/2, ", ")) \\ _G. C. Greubel_, Dec 05 2017

%K nonn,easy

%O 0,3

%A _N. J. A. Sloane_, Dec 11 2009