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a(n) = n^7*(n^3 + 1)/2.
1

%I #13 Sep 08 2022 08:45:49

%S 0,1,576,30618,532480,4921875,30373056,141649396,537919488,1745783685,

%T 5005000000,12978455886,30976598016,68960620183,144680034240,

%U 288410625000,549890031616,1008202119561,1785539723328,3065980064770

%N a(n) = n^7*(n^3 + 1)/2.

%H Vincenzo Librandi, <a href="/A168660/b168660.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 From _G. C. Greubel_, Jul 28 2016: (Start)

%F G.f.: x*(1 + 565*x + 24337*x^2 + 227197*x^3 + 653875*x^4 + 656479*x^5 + 227995*x^6 + 23503*x^7 + 448*x^8)/(1 - x)^11.

%F E.g.f.: (1/2)*x*(2 + 574*x + 9631*x^2 + 34455*x^3 + 42665*x^4 + 22848*x^5 + 5881*x^6 + 750*x^7 + 45*x^8 + x^9)*exp(x). (End)

%t Table[n^7*(n^3 + 1)/2, {n,0,50}] (* _G. C. Greubel_, Jul 28 2016 *)

%o (Magma) [n^7*(n^3+1)/2: n in [0..20]]; // _Vincenzo Librandi_, Aug 28 2011

%o (PARI) a(n)=n^7*(n^3+1)/2 \\ _Charles R Greathouse IV_, Jul 29 2016

%K nonn,easy

%O 0,3

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