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a(n) = (n^3 - n)*4^n.
8

%I #27 Oct 02 2022 10:32:40

%S 0,96,1536,15360,122880,860160,5505024,33030144,188743680,1038090240,

%T 5536481280,28789702656,146565758976,732828794880,3607772528640,

%U 17523466567680,84112639524864,399535037743104,1880164883496960,8774102789652480,40637949762600960

%N a(n) = (n^3 - n)*4^n.

%H Vincenzo Librandi, <a href="/A128962/b128962.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (16,-96,256,-256).

%F G.f.: 96*x^2/(1-4*x)^4. - _Vincenzo Librandi_, Feb 09 2013

%F a(n) = 16*a(n-1) - 96*a(n-2) + 256*a(n-3) - 256*a(n-4). - _Vincenzo Librandi_, Feb 09 2013

%F a(n) = 96*A038846(n-2) for n>1. - _Bruno Berselli_, Feb 10 2013

%F From _Amiram Eldar_, Oct 02 2022: (Start)

%F a(n) = A007531(n+1)*A000302(n).

%F Sum_{n>=2} 1/a(n) = (9/8)*log(4/3) - 5/16.

%F Sum_{n>=2} (-1)^n/a(n) = (25/8)*log(5/4) - 11/16. (End)

%t CoefficientList[Series[96 x / (1-4 x)^4, {x, 0, 30}], x] (* _Vincenzo Librandi_, Feb 09 2013 *)

%t Table[(n^3-n)4^n,{n,20}] (* or *) LinearRecurrence[{16,-96,256,-256},{0,96,1536,15360},20] (* _Harvey P. Dale_, Dec 31 2018 *)

%o (Magma) [(n^3-n)*4^n: n in [1..20]]; // _Vincenzo Librandi_, Feb 09 2013

%Y Cf. A000302, A007531, A036289, A128796.

%Y Cf. A128960, A128961, A128963, A128964, A128965, A128967, A128969.

%K nonn,easy

%O 1,2

%A _Mohammad K. Azarian_, Apr 28 2007

%E Offset corrected by _Mohammad K. Azarian_, Nov 20 2008