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a(n) = 6^n * n*(n+1).
1

%I #27 Nov 18 2024 12:05:02

%S 0,12,216,2592,25920,233280,1959552,15676416,120932352,906992640,

%T 6651279360,47889211392,339578044416,2377046310912,16456474460160,

%U 112844396298240,767341894828032,5179557790089216,34733505180598272

%N a(n) = 6^n * n*(n+1).

%H Vincenzo Librandi, <a href="/A116164/b116164.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (18,-108,216).

%F G.f.: 12*x/(1-6*x)^3. - _Vincenzo Librandi_, Feb 28 2013

%F a(n) = 18*a(n-1) - 108*a(n-2) + 216*a(n-3). - _Vincenzo Librandi_, Feb 28 2013

%F a(n) = 12*A081136(n+1). - _Bruno Berselli_, Feb 28 2013

%F E.g.f.: 12*x*(1 + 3*x)*exp(6*x). - _G. C. Greubel_, May 10 2019

%F From _Amiram Eldar_, Jul 20 2020: (Start)

%F Sum_{n>=1} 1/a(n) = 1 - 5*log(6/5).

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 7*log(7/6) - 1. (End)

%t Table[(n^2 + n) 6^n, {n, 0, 30}] (* or *) CoefficientList[Series[12 x/(1 - 6 x)^3, {x, 0, 30}], x] (* _Vincenzo Librandi_, Feb 28 2013 *)

%o (Magma) [(n^2+n)*6^n: n in [0..30]]; // _Vincenzo Librandi_, Feb 28 2013

%o (Magma) I:=[0,12,216]; [n le 3 select I[n] else 18*Self(n-1)-108*Self(n-2)+216*Self(n-3): n in [1..30]]; // _Vincenzo Librandi_, Feb 28 2013

%o (PARI) a(n)=(n^2+n)*6^n \\ _Charles R Greathouse IV_, Feb 28 2013

%o (Sage) [6^n*n*(n+1) for n in (0..30)] # _G. C. Greubel_, May 10 2019

%o (GAP) List([0..30], n-> 6^n*n*(n+1) ); # _G. C. Greubel_, May 10 2019

%Y Cf. A007758, A036289, A081136, A128796.

%K nonn,easy

%O 0,2

%A _Mohammad K. Azarian_, Apr 08 2007