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a(n) = 11^n - n^12.
1

%I #22 Sep 08 2022 08:44:48

%S 1,10,-3975,-530110,-16762575,-243979574,-2175010775,-13821800030,

%T -68505117855,-280071588790,-974062575399,-2853116706110,

%U -5777672071535,11224627021450,323055921207945,4047501831525026,45668254886861505

%N a(n) = 11^n - n^12.

%H Vincenzo Librandi, <a href="/A024139/b024139.txt">Table of n, a(n) for n = 0..300</a>

%H <a href="/index/Rec#order_14">Index entries for linear recurrences with constant coefficients</a>, signature (24,-221,1144,-3861,9152,-15873,20592,-20163,14872,-8151,3224,-871,144,-11).

%p seq(11^k-k^12,k=0..20); # _Muniru A Asiru_, Jul 15 2018

%t Table[11^n-n^12,{n,0,60}] (* _Vladimir Joseph Stephan Orlovsky_, Feb 15 2011 *)

%o (Magma) [11^n-n^12: n in [0..20]]; // _Vincenzo Librandi_, Jul 01 2011

%o (PARI) a(n)=11^n-n^12 \\ _Charles R Greathouse IV_, Jul 01 2011

%o (GAP) List([0..20],n->11^n-n^12); # _Muniru A Asiru_, Jul 15 2018

%K sign,easy

%O 0,2

%A _N. J. A. Sloane_