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a(n) = 12^n - n^3.
2

%I #19 Sep 08 2022 08:44:48

%S 1,11,136,1701,20672,248707,2985768,35831465,429981184,5159779623,

%T 61917363224,743008369357,8916100446528,106993205376875,

%U 1283918464546120,15407021574582993,184884258895032320,2218611106740432079,26623333280885238072,319479999370622919989

%N a(n) = 12^n - n^3.

%H Vincenzo Librandi, <a href="/A024143/b024143.txt">Table of n, a(n) for n = 0..300</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (16,-54,76,-49,12).

%F From _Colin Barker_, Oct 11 2018: (Start)

%F G.f.: (1 - 5*x + 14*x^2 + 43*x^3 + 13*x^4) / ((1 - x)^4*(1 - 12*x)).

%F a(n) = 16*a(n-1) - 54*a(n-2) + 76*a(n-3) - 49*a(n-4) + 12*a(n-5) for n>4.

%F (End)

%o (Magma) [12^n-n^3: n in [0..20]]; // _Vincenzo Librandi_, Jul 01 2011

%o (PARI) a(n)=12^n-n^3 \\ _Charles R Greathouse IV_, Jul 01 2011

%o (PARI) Vec((1 - 5*x + 14*x^2 + 43*x^3 + 13*x^4) / ((1 - x)^4*(1 - 12*x)) + O(x^20)) \\ _Colin Barker_, Oct 11 2018

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_