A301483 a(n) = floor(a(n-1)/(2^(1/3)-1)) with a(1)=1.
1, 3, 11, 42, 161, 619, 2381, 9160, 35241, 135583, 521631, 2006882, 7721121, 29705639, 114287161, 439699520, 1691665681, 6508382763, 25039844851, 96336348522, 370636962881, 1425959779059, 5486126574341, 21106896023080, 81205027571321, 312421897357543
Offset: 1
Keywords
Programs
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Magma
[n le 1 select 1 else Floor(Self(n-1)/(2^(1/3)-1)): n in [1..30]]; // Vincenzo Librandi, Apr 04 2018
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Maple
a:=proc(n) option remember; if n<1 then 0 else if n=1 then 1 else floor(a(n-1)/(2^(1/3)-1)) end if end if end proc: seq(a(n), n=1..25);
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Mathematica
RecurrenceTable[{a[1]==1, a[n]==Floor[a[n-1]/(2^(1/3)-1)]}, a, {n, 30}] (* Vincenzo Librandi, Apr 04 2018 *)
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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
Formula
Conjectures from Colin Barker, Apr 01 2018: (Start)
G.f.: x*(1 - x - x^2) / ((1 - x)*(1 - 3*x - 3*x^2 - x^3)).
a(n) = 4*a(n-1) - 2*a(n-3) - a(n-4) for n>4.
(End)
Comments