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%I #29 Sep 08 2022 08:45:38
%S 0,4,20,76,260,844,2660,8236,25220,76684,232100,700396,2109380,
%T 6344524,19066340,57264556,171924740,516036364,1548633380,4646948716,
%U 13942943300,41833024204,125507461220,376539160876,1129651037060,3389020220044,10167194877860
%N a(0)=0 and a(n+1) = 3*a(n) + 2^(n+2).
%C Suggested by a discussion on the Sequence Fans Mailing List; the formula is due to Andrew V. Sutherland.
%C First differences of A255459. - _Klaus Purath_, Apr 25 2020
%H Vincenzo Librandi, <a href="/A145563/b145563.txt">Table of n, a(n) for n = 0..170</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (5,-6).
%F From _R. J. Mathar_, Mar 18 2009: (Start)
%F a(n) = 4*(3^n - 2^n) = 4*A001047(n).
%F G.f.: 4*x/((1-2*x)*(1-3*x)). (End)
%F a(n) = A056182(n)*2. - _Omar E. Pol_, Mar 18 2009
%F a(n) = A217764(n,7). - _Ross La Haye_, Mar 27 2013
%F From _Klaus Purath_, Apr 25 2020: (Start)
%F a(n) = 5*a(n-1) - 6*a(n-2).
%F a(n) = 2*A210448(n) - A056182(n).
%F a(n) = (A056182(n) + A245804(n+1))/2. (End)
%t CoefficientList[Series[4x/((1-2x)(1-3x)),{x,0,40}],x] (* or *) RecurrenceTable[{a[0]==0, a[n]==(3a[n-1]+2^(n+1))},a,{n,40}] (* _Harvey P. Dale_, Apr 24 2011 *)
%o (Magma) [ 4*(3^n - 2^n): n in [0..50]]; // _Vincenzo Librandi_, Apr 24 2011
%o (PARI) a(n) = 4*(3^n - 2^n) \\ _Felix Fröhlich_, Sep 01 2018
%Y Cf. A001047, A056182, A210448, A217764, A245804.
%K nonn
%O 0,2
%A _N. J. A. Sloane_, Mar 18 2009
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