%I #20 May 14 2024 10:23:53
%S 1,0,2,4,4,16,24,48,112,192,416,832,1600,3328,6528,13056,26368,52224,
%T 104960,209920,418816,839680,1677312,3354624,6713344,13418496,
%U 26845184,53690368,107364352,214761472,429490176,858980352,1718026240,3435921408,6871973888
%N Expansion of 1/(1-2*x^2-4*x^3). (2,4)-Padovan sequence.
%C See A000931 (Padovan), and the W. Lang link given there for a combinatorial interpretation and an explicit form.
%C a(n)/2^n equals the probability that n will occur as a partial sum in a randomly-generated infinite sequence of 2's and 3's. The limiting ratio is 2/5. - _Bob Selcoe_, Jul 12 2013
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (0,2,4).
%F O.g.f.: 1/((1+2*x+2*x^2)*(1-2*x)) = ((3+2*x)/(1+2*x+2*x^2) + 2/(1-2*x))/5.
%F a(n) = (3*b(n) + 2*b(n-1) + 2^(n+1))/5, with b(n):=A108520(n), and b(-1)=0.
%F a(n) = 2*a(n-2) + 4*a(n-3). - _Bob Selcoe_, Aug 26 2014
%F a(n) = 2^(n+1)/5 + Re((3-i)*(-1-i)^n)/5. - _Robert Israel_, Aug 26 2014
%F 5*a(n) = 2^(n+1) -A078069(n+1). - _R. J. Mathar_, May 14 2024
%e Combinatorics for (A,B)=(2,4) Padovan sequence with weighted (3,2)-Morse type code (see the W. Lang link under A000931): n=5, - -- and -- -, with weights 2^1*4^1 and 4^1*2^1, respectively, adding to 2*2*4=16=a(5).
%p seq(2^(n+1)/5 + Re((3-I)*(-1-I)^n)/5, n=0..100); # _Robert Israel_, Aug 26 2014
%t CoefficientList[Series[1/(1 - 2*x^2 - 4*x^3), {x, 0, 30}], x] (* _Wesley Ivan Hurt_, Aug 26 2014 *)
%Y Cf. A000931, A108520.
%K nonn,easy
%O 0,3
%A _Wolfdieter Lang_, Jul 14 2010