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a(n) = 2^n-n^11.
2

%I #19 Jan 29 2026 20:17:05

%S 1,1,-2044,-177139,-4194288,-48828093,-362796992,-1977326615,

%T -8589934336,-31381059097,-99999998976,-285311668563,-743008366592,

%U -1792160385845,-4049565153280,-8649755826607,-17592185978880

%N a(n) = 2^n-n^11.

%H Vincenzo Librandi, <a href="/A024021/b024021.txt">Table of n, a(n) for n = 0..235</a>

%H <a href="/index/Rec#order_13">Index entries for linear recurrences with constant coefficients</a>, signature (14,-90,352,-935,1782,-2508,2640,-2079,1210,-506,144,-25,2).

%F G.f.: -(3*x^12 +4059*x^11 +303304*x^10 +4254119*x^9 +17273235*x^8 +21709590*x^7 +3752904*x^6 -5331930*x^5 -1897719*x^4 -148785*x^3 -1968*x^2 -13*x +1) / ((x -1)^12*(2*x -1)). - _Colin Barker_, Oct 04 2014

%t Table[2^n-n^11,{n,0,20}] (* _Harvey P. Dale_, Aug 17 2019 *)

%o (Magma) [2^n-n^11: n in [0..30] ]; // _Vincenzo Librandi_, Apr 30 2011

%K sign,easy

%O 0,3

%A _N. J. A. Sloane_