OFFSET
1,2
COMMENTS
a(n+1)/a(n) approaches 1/(2^(1/3)-1).
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)
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);
MATHEMATICA
RecurrenceTable[{a[1]==1, a[n]==Floor[a[n-1]/(2^(1/3)-1)]}, a, {n, 30}] (* Vincenzo Librandi, Apr 04 2018 *)
PROG
(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
(Magma) [n le 1 select 1 else Floor(Self(n-1)/(2^(1/3)-1)): n in [1..30]]; // Vincenzo Librandi, Apr 04 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Gregory Gerard Wojnar, Mar 22 2018
STATUS
approved