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A301483 a(n) = floor(a(n-1)/(2^(1/3)-1)) with a(1)=1.

%I A301483 #40 Dec 10 2023 18:11:42
%S A301483 1,3,11,42,161,619,2381,9160,35241,135583,521631,2006882,7721121,
%T A301483 29705639,114287161,439699520,1691665681,6508382763,25039844851,
%U A301483 96336348522,370636962881,1425959779059,5486126574341,21106896023080,81205027571321,312421897357543
%N A301483 a(n) = floor(a(n-1)/(2^(1/3)-1)) with a(1)=1.
%C A301483 a(n+1)/a(n) approaches 1/(2^(1/3)-1).
%F A301483 Conjectures from _Colin Barker_, Apr 01 2018: (Start)
%F A301483 G.f.: x*(1 - x - x^2) / ((1 - x)*(1 - 3*x - 3*x^2 - x^3)).
%F A301483 a(n) = 4*a(n-1) - 2*a(n-3) - a(n-4) for n>4.
%F A301483 (End)
%F A301483 a(n) = A195350(n) + A303647(n-2) - A195339(n-4) (conjectured).
%p A301483 a:=proc(n) option remember;
%p A301483    if n<1 then 0  else if n=1 then 1 else floor(a(n-1)/(2^(1/3)-1))
%p A301483 end if end if end proc:
%p A301483 seq(a(n), n=1..25);
%t A301483 RecurrenceTable[{a[1]==1, a[n]==Floor[a[n-1]/(2^(1/3)-1)]}, a, {n, 30}] (* _Vincenzo Librandi_, Apr 04 2018 *)
%o A301483 (PARI) a=vector(50); a[1]=1; for(n=2, #a, a[n]=a[n-1]\(2^(1/3)-1)); a \\ _Altug Alkan_, Mar 22 2018
%o A301483 (Magma) [n le 1 select 1 else Floor(Self(n-1)/(2^(1/3)-1)): n in [1..30]]; // _Vincenzo Librandi_, Apr 04 2018
%Y A301483 Cf. A024537, A195350 (also has 1/(2^(1/3)-1) ratio), A303647.
%K A301483 nonn
%O A301483 1,2
%A A301483 _Gregory Gerard Wojnar_, Mar 22 2018