%I #38 Sep 11 2024 10:57:34

%S 1,3,14,61,252,1019,4090,16377,65528,262135,1048566,4194293,16777204,

%T 67108851,268435442,1073741809,4294967280,17179869167,68719476718,

%U 274877906925,1099511627756,4398046511083,17592186044394,70368744177641,281474976710632,1125899906842599

%N a(n) = 4^n - n.

%H Vincenzo Librandi, <a href="/A024037/b024037.txt">Table of n, a(n) for n = 0..200</a>

%H Guo-Niu Han, <a href="/A196265/a196265.pdf">Enumeration of Standard Puzzles</a>, 2011. [Cached copy]

%H Guo-Niu Han, <a href="https://arxiv.org/abs/2006.14070">Enumeration of Standard Puzzles</a>, arXiv:2006.14070 [math.CO], 2020.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (6,-9,4).

%F From _Vincenzo Librandi_, Jun 16 2013: (Start)

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

%F a(n) = 6*a(n-1) - 9*a(n-2) + 4*a(n-3). (End)

%F E.g.f.: exp(x)*(exp(3*x) - x). - _Elmo R. Oliveira_, Sep 10 2024

%t Table[4^n - n, {n, 0, 30}] (* or *) CoefficientList[Series[(1 - 3 x + 5 x^2) / ((1 - 4 x) (1 - x)^2), {x, 0, 30}], x] (* _Vincenzo Librandi_, Jun 16 2013 *)

%o (Magma) [4^n - n: n in [0..35]]; // _Vincenzo Librandi_, May 13 2011

%o (Magma) I:=[1, 3, 14]; [n le 3 select I[n] else 6*Self(n-1)-9*Self(n-2)+4*Self(n-3): n in [1..30]]; // _Vincenzo Librandi_, Jun 16 2013

%o (PARI) a(n)=4^n-n \\ _Charles R Greathouse IV_, Sep 24 2015

%Y Cf. numbers of the form k^n - n: A000325 (k=2), A024024 (k=3), this sequence (k=4), A024050 (k=5), A024063 (k=6), A024076 (k=7), A024089 (k=8), A024102 (k=9), A024115 (k=10), A024128 (k=11), A024141 (k=12).

%Y Cf. A140660 (first differences).

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_