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A116144 a(n) = 4^n * n*(n+1).

%I A116144 #39 Dec 08 2024 04:00:06
%S A116144 0,8,96,768,5120,30720,172032,917504,4718592,23592960,115343360,
%T A116144 553648128,2617245696,12213813248,56371445760,257698037760,
%U A116144 1168231104512,5257039970304,23502061043712,104453604638720,461794883665920
%N A116144 a(n) = 4^n * n*(n+1).
%H A116144 Vincenzo Librandi, <a href="/A116144/b116144.txt">Table of n, a(n) for n = 0..1000</a>
%H A116144 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (12,-48,64).
%F A116144 From _R. J. Mathar_, Dec 19 2008: (Start)
%F A116144 G.f.: 8*x/(1-4*x)^3.
%F A116144 a(n) = 8*A038845(n-1). (End)
%F A116144 a(n) = 12*a(n-1) -48*a(n-2) +64*a(n-3). - _Vincenzo Librandi_, Feb 28 2013
%F A116144 E.g.f.: 8*x*(1 + 2*x)*exp(4*x). - _G. C. Greubel_, May 10 2019
%F A116144 From _Amiram Eldar_, Jul 20 2020: (Start)
%F A116144 Sum_{n>=1} 1/a(n) = 1 - 3*log(4/3).
%F A116144 Sum_{n>=1} (-1)^(n+1)/a(n) = 5*log(5/4) - 1. (End)
%t A116144 Table[(n^2 + n)*4^n, {n, 0, 30}] (* _Vincenzo Librandi_, Feb 28 2013 *)
%t A116144 LinearRecurrence[{12,-48,64},{0,8,96},30] (* _Harvey P. Dale_, Feb 27 2015 *)
%o A116144 (Magma) [(n^2+n)*4^n: n in [0..30]]; // _Vincenzo Librandi_, Feb 28 2013
%o A116144 (Magma) I:=[0,8,96]; [n le 3 select I[n] else 12*Self(n-1)-48*Self(n-2)+64*Self(n-3): n in [1..30]]; // _Vincenzo Librandi_, Feb 28 2013
%o A116144 (PARI) a(n)=(n^2+n)*4^n \\ _Charles R Greathouse IV_, Feb 28 2013
%o A116144 (Sage) [4^n*n*(n+1) for n in (0..30)] # _G. C. Greubel_, May 10 2019
%o A116144 (GAP) List([0..30], n-> 4^n*n*(n+1)); # _G. C. Greubel_, May 10 2019
%Y A116144 Cf. A007758, A036289, A038845, A128796.
%K A116144 nonn,easy
%O A116144 0,2
%A A116144 _Mohammad K. Azarian_, Apr 08 2007