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a(n) = n*(n^8+1)/2.
4

%I #17 Sep 08 2022 08:45:48

%S 0,1,257,9843,131074,976565,5038851,20176807,67108868,193710249,

%T 500000005,1178973851,2579890182,5302249693,10330523399,19221679695,

%U 34359738376,59293938257,99179645193,161343848899,256000000010,397140023301,603634608907

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

%H G. C. Greubel, <a href="/A168116/b168116.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

%F G.f.: x*(1 + 247*x + 7318*x^2 + 44089*x^3 + 78130*x^4 + 44089*x^5 + 7318*x^6 + 247*x^7 + x^8)/(1 - x)^10. - _R. J. Mathar_, Oct 13 2011

%F a(n) = 10*a(n-1) - 45*a(n-2) + 120*a(n-3) - 210*a(n-4) + 252*a(n-5) - 210*a(n-6) + 120*a(n-7) - 45*a(n-8) + 10*a(n-9) - a(n-10). - _Wesley Ivan Hurt_, Jun 26 2022

%t Table[n*(n^8 + 1)/2, {n, 0, 50}] (* _G. C. Greubel_, Jul 13 2016 *)

%o (Magma) [n*(n^8+1)/2: n in [0..25]]; // _Vincenzo Librandi_, Jul 14 2016

%K nonn

%O 0,3

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